Electrochemistry.

Electrochemistry

Electrochemistry Study of the interchange of chemical and electrical energy …

Electrochemistry Study of the interchange of chemical and electrical energy … Galvanic Cells ...

Electrochemistry Study of the interchange of chemical and electrical energy … Galvanic Cells ... Devices in which chemical energy is converted into electrical energy ...

Electrochemistry Study of the interchange of chemical and electrical energy … Galvanic Cells ... Devices in which chemical energy is converted into electrical energy ... Cathode ...

Electrochemistry Study of the interchange of chemical and electrical energy … Galvanic Cells ... Devices in which chemical energy is converted into electrical energy ... Cathode ... Reduction occurs here ...

Electrochemistry Study of the interchange of chemical and electrical energy … Galvanic Cells ... Devices in which chemical energy is converted into electrical energy ... Cathode ... Reduction occurs here ... Species undergoing reduction ...

Electrochemistry Study of the interchange of chemical and electrical energy … Galvanic Cells ... Devices in which chemical energy is converted into electrical energy ... Cathode ... Reduction occurs here ... Species undergoing reduction ... a.k.a. oxidizing agent …

Electrochemistry Study of the interchange of chemical and electrical energy … Galvanic Cells ... Devices in which chemical energy is converted into electrical energy ... Cathode ... Reduction occurs here ... Species undergoing reduction ... a.k.a. oxidizing agent … receives electrons from the cathode ...

Electrochemistry Study of the interchange of chemical and electrical energy … Galvanic Cells ... Devices in which chemical energy is converted into electrical energy ... Cathode ... Reduction occurs here ... Species undergoing reduction ... a.k.a. oxidizing agent … receives electrons from the cathode ... Anode ...

Electrochemistry Study of the interchange of chemical and electrical energy … Galvanic Cells ... Devices in which chemical energy is converted into electrical energy ... Cathode ... Reduction occurs here ... Species undergoing reduction ... a.k.a. oxidizing agent … receives electrons from the cathode ... Anode ... Oxidation occurs here ...

Electrochemistry Study of the interchange of chemical and electrical energy … Galvanic Cells ... Devices in which chemical energy is converted into electrical energy ... Cathode ... Reduction occurs here ... Species undergoing reduction ... a.k.a. oxidizing agent … receives electrons from the cathode ... Anode ... Oxidation occurs here ... aka reducing agent …

Electrochemistry Study of the interchange of chemical and electrical energy … Galvanic Cells ... Devices in which chemical energy is converted into electrical energy ... Cathode ... Reduction occurs here ... Species undergoing reduction ... a.k.a. oxidizing agent … receives electrons from the cathode ... Anode ... Oxidation occurs here ... aka reducing agent … gives up electrons to cathode here ...

Electrochemistry Study of the interchange of chemical and electrical energy … Galvanic Cells ... Devices in which chemical energy is converted into electrical energy ... Cathode ... Reduction occurs here ... Species undergoing reduction ... a.k.a. oxidizing agent … receives electrons from the cathode ... Anode ... Oxidation occurs here ... aka reducing agent … gives up electrons to cathode here ... Salt bridge

Electrochemistry Study of the interchange of chemical and electrical energy … Galvanic Cells ... Devices in which chemical energy is converted into electrical energy ... Cathode ... Reduction occurs here ... Species undergoing reduction ... a.k.a. oxidizing agent … receives electrons from the cathode ... Anode ... Oxidation occurs here ... aka reducing agent … gives up electrons to cathode here ... Salt bridge (or porous disk) …

Electrochemistry Study of the interchange of chemical and electrical energy … Galvanic Cells ... Devices in which chemical energy is converted into electrical energy ... Cathode ... Reduction occurs here ... Species undergoing reduction ... a.k.a. oxidizing agent … receives electrons from the cathode ... Anode ... Oxidation occurs here ... aka reducing agent … gives up electrons to cathode here ... Salt bridge (or porous disk) … allows flow of electrons to keep electrical neutrality ...

Electrochemistry Study of the interchange of chemical and electrical energy … Galvanic Cells ... Devices in which chemical energy is converted into electrical energy ... Cathode ... Reduction occurs here ... Species undergoing reduction ... a.k.a. oxidizing agent … receives electrons from the cathode ... Anode ... Oxidation occurs here ... aka reducing agent … gives up electrons to cathode here ... Salt bridge (or porous disk) … allows flow of electrons to keep electrical neutrality ... electroactve solutions remain separated …

Electrochemistry Study of the interchange of chemical and electrical energy … Galvanic Cells ... Devices in which chemical energy is converted into electrical energy ... Cathode ... Reduction occurs here ... Species undergoing reduction ... a.k.a. oxidizing agent … receives electrons from the cathode ... Anode ... Oxidation occurs here ... aka reducing agent … gives up electrons to cathode here ... Salt bridge (or porous disk) … allows flow of electrons to keep electrical neutrality ... electroactve solutions remain separated … Electromotive force (emf) ...

Electrochemistry Study of the interchange of chemical and electrical energy … Galvanic Cells ... Devices in which chemical energy is converted into electrical energy ... Cathode ... Reduction occurs here ... Species undergoing reduction ... a.k.a. oxidizing agent … receives electrons from the cathode ... Anode ... Oxidation occurs here ... aka reducing agent … gives up electrons to cathode here ... Salt bridge (or porous disk) … allows flow of electrons to keep electrical neutrality ... electroactve solutions remain separated … Electromotive force (emf) ... the driving force with which electrons are pulled through a wire ...

Electrochemistry Study of the interchange of chemical and electrical energy … Galvanic Cells ... Devices in which chemical energy is converted into electrical energy ... Cathode ... Reduction occurs here ... Species undergoing reduction ... a.k.a. oxidizing agent … receives electrons from the cathode ... Anode ... Oxidation occurs here ... aka reducing agent … gives up electrons to cathode here ... Salt bridge (or porous disk) … allows flow of electrons to keep electrical neutrality ... electroactve solutions remain separated … Electromotive force (emf) ... the driving force with which electrons are pulled through a wire ... Volt (V) ...

Electrochemistry Study of the interchange of chemical and electrical energy … Galvanic Cells ... Devices in which chemical energy is converted into electrical energy ... Cathode ... Reduction occurs here ... Species undergoing reduction ... a.k.a. oxidizing agent … receives electrons from the cathode ... Anode ... Oxidation occurs here ... aka reducing agent … gives up electrons to cathode here ... Salt bridge (or porous disk) … allows flow of electrons to keep electrical neutrality ... electroactve solutions remain separated … Electromotive force (emf) ... the driving force with which electrons are pulled through a wire ... Volt (V) ... the unit of electrical potential …

Electrochemistry Study of the interchange of chemical and electrical energy … Galvanic Cells ... Devices in which chemical energy is converted into electrical energy ... Cathode ... Reduction occurs here ... Species undergoing reduction ... a.k.a. oxidizing agent … receives electrons from the cathode ... Anode ... Oxidation occurs here ... aka reducing agent … gives up electrons to cathode here ... Salt bridge (or porous disk) … allows flow of electrons to keep electrical neutrality ... electroactve solutions remain separated … Electromotive force (emf) ... the driving force with which electrons are pulled through a wire ... Volt (V) ... the unit of electrical potential … it equals 1 Joule/coulomb ...

Electrochemistry Direction of electron flow in Galvanic Cells …

Electrochemistry Direction of electron flow in Galvanic Cells … From anaode to cathode ...

Electrochemistry Direction of electron flow in Galvanic Cells … From anaode to cathode ... FATCAT ...

Electrochemistry Direction of electron flow in Galvanic Cells … From anode to cathode ... FATCAT ... Species undergoing reduction receive electrons from the cathode ...

Electrochemistry Direction of electron flow in Galvanic Cells … From anode to cathode ... FATCAT ... Species undergoing reduction receive electrons from the cathode ... Species undergoing oxidation donate electrons to the anode ...

Electrochemistry Direction of electron flow in Galvanic Cells … From anode to cathode ... FATCAT ... Species undergoing reduction receive electrons from the cathode ... Species undergoing oxidation donate electrons to the anode ... Known: a. Mg(s) + 2H+(aq) ----> Mg+2(aq) + H2(g)

Electrochemistry Direction of electron flow in Galvanic Cells … From anode to cathode ... FATCAT ... Species undergoing reduction receive electrons from the cathode ... Species undergoing oxidation donate electrons to the anode ... Known: a. Mg(s) + 2H+(aq) ----> Mg+2(aq) + H2(g) b. MnO4-1(aq) + 5Fe+2(aq) + 8H+(aq) ---> Mn+2(aq) + 5Fe+3(aq) + 4H2O(l)

Electrochemistry Direction of electron flow in Galvanic Cells … From anode to cathode ... FATCAT ... Species undergoing reduction receive electrons from the cathode ... Species undergoing oxidation donate electrons to the anode ... Known: a. Mg(s) + 2H+(aq) ----> Mg+2(aq) + H2(g) b. MnO4-1(aq) + 5Fe+2(aq) + 8H+(aq) ---> Mn+2(aq) + 5Fe+3(aq) + 4H2O(l) Unknown:

Electrochemistry Direction of electron flow in Galvanic Cells … From anode to cathode ... FATCAT ... Species undergoing reduction receive electrons from the cathode ... Species undergoing oxidation donate electrons to the anode ... Known: a. Mg(s) + 2H+(aq) ----> Mg+2(aq) + H2(g) b. MnO4-1(aq) + 5Fe+2(aq) + 8H+(aq) ---> Mn+2(aq) + 5Fe+3(aq) + 4H2O(l) Unknown: Identify the species that would receive electrons from the cathode and which would lose electrons at the anode in each of the following galvanic cells ...

Electrochemistry Direction of electron flow in Galvanic Cells … From anode to cathode ... FATCAT ... Species undergoing reduction receive electrons from the cathode ... Species undergoing oxidation donate electrons to the anode ... Known: a. Mg(s) + 2H+(aq) ----> Mg+2(aq) + H2(g) b. MnO4-1(aq) + 5Fe+2(aq) + 8H+(aq) ---> Mn+2(aq) + 5Fe+3(aq) + 4H2O(l) Unknown: Identify the species that would receive electrons from the cathode and which would lose electrons at the anode in each of the following galvanic cells ... Solution:

Electrochemistry Direction of electron flow in Galvanic Cells … From anode to cathode ... FATCAT ... Species undergoing reduction receive electrons from the cathode ... Species undergoing oxidation donate electrons to the anode ... Known: a. Mg(s) + 2H+(aq) ----> Mg+2(aq) + H2(g) b. MnO4-1(aq) + 5Fe+2(aq) + 8H+(aq) ---> Mn+2(aq) + 5Fe+3(aq) + 4H2O(l) Unknown: Identify the species that would receive electrons from the cathode and which would lose electrons at the anode in each of the following galvanic cells ... Solution: reduction ½ rxs receive e-1 at cathode ...

Electrochemistry Direction of electron flow in Galvanic Cells … From anode to cathode ... FATCAT ... Species undergoing reduction receive electrons from the cathode ... Species undergoing oxidation donate electrons to the anode ... Known: a. Mg(s) + 2H+(aq) ----> Mg+2(aq) + H2(g) b. MnO4-1(aq) + 5Fe+2(aq) + 8H+(aq) ---> Mn+2(aq) + 5Fe+3(aq) + 4H2O(l) Unknown: Identify the species that would receive electrons from the cathode and which would lose electrons at the anode in each of the following galvanic cells ... Solution: reduction ½ rxs receive cathode ... oxidation ½ rxs. lose anode ...

Electrochemistry Known: a. Mg(s) + 2H+(aq) ----> Mg+2(aq) + H2(g)
b. MnO4-1(aq) + 5Fe+2(aq) + 8H+(aq) ---> Mn+2(aq) + 5Fe+3(aq) + 4H2O(l) Unknown: Identify the species that would receive electrons from the cathode and which would lose electrons at the anode in each of the following galvanic cells ... Solution: reduction ½ rxs receive cathode ... oxidation ½ rxs. lose anode ... a. Oxidation (anode):

b. MnO4-1(aq) + 5Fe+2(aq) + 8H+(aq) ---> Mn+2(aq) + 5Fe+3(aq) + 4H2O(l) Unknown: Identify the species that would receive electrons from the cathode and which would lose electrons at the anode in each of the following galvanic cells ... Solution: reduction ½ rxs receive cathode ... oxidation ½ rxs. lose anode ... a. Oxidation (anode): Mg(s) ----> Mg+2(aq) + 2e-1

b. MnO4-1(aq) + 5Fe+2(aq) + 8H+(aq) ---> Mn+2(aq) + 5Fe+3(aq) + 4H2O(l) Unknown: Identify the species that would receive electrons from the cathode and which would lose electrons at the anode in each of the following galvanic cells ... Solution: reduction ½ rxs receive cathode ... oxidation ½ rxs. lose anode ... a. Oxidation (anode): Mg(s) ----> Mg+2(aq) + 2e-1 Reduction (cathode):

b. MnO4-1(aq) + 5Fe+2(aq) + 8H+(aq) ---> Mn+2(aq) + 5Fe+3(aq) + 4H2O(l) Unknown: Identify the species that would receive electrons from the cathode and which would lose electrons at the anode in each of the following galvanic cells ... Solution: reduction ½ rxs receive cathode ... oxidation ½ rxs. lose anode ... a. Oxidation (anode): Mg(s) ----> Mg+2(aq) + 2e-1 Reduction (cathode): 2H+(aq) + 2e > H2(g)

b. MnO4-1(aq) + 5Fe+2(aq) + 8H+(aq) ---> Mn+2(aq) + 5Fe+3(aq) + 4H2O(l) Unknown: Identify the species that would receive electrons from the cathode and which would lose electrons at the anode in each of the following galvanic cells ... Solution: reduction ½ rxs receive cathode ... oxidation ½ rxs. lose anode ... a. Oxidation (anode): Mg(s) ----> Mg+2(aq) + 2e-1 Reduction (cathode): 2H+(aq) + 2e > H2(g) b. Oxidation (anode):

b. MnO4-1(aq) + 5Fe+2(aq) + 8H+(aq) ---> Mn+2(aq) + 5Fe+3(aq) + 4H2O(l) Unknown: Identify the species that would receive electrons from the cathode and which would lose electrons at the anode in each of the following galvanic cells ... Solution: reduction ½ rxs receive cathode ... oxidation ½ rxs. lose anode ... a. Oxidation (anode): Mg(s) ----> Mg+2(aq) + 2e-1 Reduction (cathode): 2H+(aq) + 2e > H2(g) b. Oxidation (anode): 5Fe+2(aq) ----> 5Fe+3(aq) + 5e-1

b. MnO4-1(aq) + 5Fe+2(aq) + 8H+(aq) ---> Mn+2(aq) + 5Fe+3(aq) + 4H2O(l) Unknown: Identify the species that would receive electrons from the cathode and which would lose electrons at the anode in each of the following galvanic cells ... Solution: reduction ½ rxs receive cathode ... oxidation ½ rxs. lose anode ... a. Oxidation (anode): Mg(s) ----> Mg+2(aq) + 2e-1 Reduction (cathode): 2H+(aq) + 2e > H2(g) b. Oxidation (anode): 5Fe+2(aq) ----> 5Fe+3(aq) + 5e-1 Reduction (cathode):

b. MnO4-1(aq) + 5Fe+2(aq) + 8H+(aq) ---> Mn+2(aq) + 5Fe+3(aq) + 4H2O(l) Unknown: Identify the species that would receive electrons from the cathode and which would lose electrons at the anode in each of the following galvanic cells ... Solution: reduction ½ rxs receive cathode ... oxidation ½ rxs. lose anode ... a. Oxidation (anode): Mg(s) ----> Mg+2(aq) + 2e-1 Reduction (cathode): 2H+(aq) + 2e > H2(g) b. Oxidation (anode): 5Fe+2(aq) ----> 5Fe+3(aq) + 5e-1 Reduction (cathode): MnO4-1(aq) + 8H+(aq) + 5e > Mn+2(aq) + 4H2O(l)

Electrochemistry Standard Reduction Potentials …

Electrochemistry Standard Reduction Potentials …
Electromotive force (emf) of a galvanic cell is a combination of the potentials of the two half-reactions ...

Electromotive force (emf) of a galvanic cell is a combination of the potentials of the two half-reactions ... Cathodic potentials are described relative to anodic reactions ...

Electromotive force (emf) of a galvanic cell is a combination of the potentials of the two half-reactions ... Cathodic potentials are described relative to anodic reactions ... An absolute standard against which all other half-reactions can be compared is required ...

Electromotive force (emf) of a galvanic cell is a combination of the potentials of the two half-reactions ... Cathodic potentials are described relative to anodic reactions ... An absolute standard against which all other half-reactions can be compared is required ... The standard is the standard hydrogen electrode ...

2H+(aq) + 2e-1 ---> H2(g) E’ = 0.00 V
Electrochemistry Standard Reduction Potentials … Electromotive force (emf) of a galvanic cell is a combination of the potentials of the two half-reactions ... Cathodic potentials are described relative to anodic reactions ... An absolute standard against which all other half-reactions can be compared is required ... The standard is the standard hydrogen electrode ... 2H+(aq) + 2e-1 ---> H2(g) E’ = 0.00 V

2H+(aq) + 2e-1 ---> H2(g) E’ = 0.00 V (exactly, by definition)
Electrochemistry Standard Reduction Potentials … Electromotive force (emf) of a galvanic cell is a combination of the potentials of the two half-reactions ... Cathodic potentials are described relative to anodic reactions ... An absolute standard against which all other half-reactions can be compared is required ... The standard is the standard hydrogen electrode ... 2H+(aq) + 2e-1 ---> H2(g) E’ = 0.00 V (exactly, by definition)

Electrochemistry Standard Reduction Potentials … Electromotive force (emf) of a galvanic cell is a combination of the potentials of the two half-reactions ... Cathodic potentials are described relative to anodic reactions ... An absolute standard against which all other half-reactions can be compared is required ... The standard is the standard hydrogen electrode K & 1 atm ... 2H+(aq) + 2e-1 ---> H2(g) E’ = 0.00 V (exactly, by definition)

Electrochemistry Standard Reduction Potentials … Electromotive force (emf) of a galvanic cell is a combination of the potentials of the two half-reactions ... Cathodic potentials are described relative to anodic reactions ... An absolute standard against which all other half-reactions can be compared is required ... The standard is the standard hydrogen electrode K & 1 atm ... 2H+(aq) + 2e-1 ---> H2(g) E’ = 0.00 V (exactly, by definition) Galvanic cells require E’cell > 0 ...

Electrochemistry Standard Reduction Potentials … Electromotive force (emf) of a galvanic cell is a combination of the potentials of the two half-reactions ... Cathodic potentials are described relative to anodic reactions ... An absolute standard against which all other half-reactions can be compared is required ... The standard is the standard hydrogen electrode K & 1 atm ... 2H+(aq) + 2e-1 ---> H2(g) E’ = 0.00 V (exactly, by definition) Galvanic cells require E’cell > 0 ... One of the tabulated cell potentials have to be reversed ...

Electrochemistry Standard Reduction Potentials … Electromotive force (emf) of a galvanic cell is a combination of the potentials of the two half-reactions ... Cathodic potentials are described relative to anodic reactions ... An absolute standard against which all other half-reactions can be compared is required ... The standard is the standard hydrogen electrode K & 1 atm ... 2H+(aq) + 2e-1 ---> H2(g) E’ = 0.00 V (exactly, by definition) Galvanic cells require E’cell > 0 ... One of the tabulated cell potentials have to be reversed ... to form an oxidation half-reaction in EVERY E’ calculation …

Electrochemistry The standard is the standard hydrogen electrode K & 1 atm ... 2H+(aq) + 2e-1 ---> H2(g) E’ = 0.00 V (exactly, by definition) Galvanic cells require E’cell > 0 ... One of the tabulated cell potentials have to be reversed ... to form an oxidation half-reaction in EVERY E’ calculation …

Electrochemistry The standard is the standard hydrogen electrode K & 1 atm ... 2H+(aq) + 2e-1 ---> H2(g) E’ = 0.00 V (exactly, by definition) Galvanic cells require E’cell > 0 ... One of the tabulated cell potentials have to be reversed ... to form an oxidation half-reaction in EVERY E’ calculation … To determine which reaction is to be reversed ...

Electrochemistry The standard is the standard hydrogen electrode K & 1 atm ... 2H+(aq) + 2e-1 ---> H2(g) E’ = 0.00 V (exactly, by definition) Galvanic cells require E’cell > 0 ... One of the tabulated cell potentials have to be reversed ... to form an oxidation half-reaction in EVERY E’ calculation … To determine which reaction is to be reversed ... the sum of the oxidation and reduction half-reactions must be > 0 in a galvanic cell …

Electrochemistry The standard is the standard hydrogen electrode K & 1 atm ... 2H+(aq) + 2e-1 ---> H2(g) E’ = 0.00 V (exactly, by definition) Galvanic cells require E’cell > 0 ... One of the tabulated cell potentials have to be reversed ... to form an oxidation half-reaction in EVERY E’ calculation … To determine which reaction is to be reversed ... the sum of the oxidation and reduction half-reactions must be > 0 in a galvanic cell … When you reverse a reaction ...

Electrochemistry The standard is the standard hydrogen electrode K & 1 atm ... 2H+(aq) + 2e-1 ---> H2(g) E’ = 0.00 V (exactly, by definition) Galvanic cells require E’cell > 0 ... One of the tabulated cell potentials have to be reversed ... to form an oxidation half-reaction in EVERY E’ calculation … To determine which reaction is to be reversed ... the sum of the oxidation and reduction half-reactions must be > 0 in a galvanic cell … When you reverse a reaction ... E’ gets the opposite sign ...

Electrochemistry The standard is the standard hydrogen electrode K & 1 atm ... 2H+(aq) + 2e-1 ---> H2(g) E’ = 0.00 V (exactly, by definition) Galvanic cells require E’cell > 0 ... One of the tabulated cell potentials have to be reversed ... to form an oxidation half-reaction in EVERY E’ calculation … To determine which reaction is to be reversed ... the sum of the oxidation and reduction half-reactions must be > 0 in a galvanic cell … When you reverse a reaction ... E’ gets the opposite sign ... When you multiply a reaction by a coefficient (for purposes of balancing) ...

Electrochemistry The standard is the standard hydrogen electrode K & 1 atm ... 2H+(aq) + 2e-1 ---> H2(g) E’ = 0.00 V (exactly, by definition) Galvanic cells require E’cell > 0 ... One of the tabulated cell potentials have to be reversed ... to form an oxidation half-reaction in EVERY E’ calculation … To determine which reaction is to be reversed ... the sum of the oxidation and reduction half-reactions must be > 0 in a galvanic cell … When you reverse a reaction ... E’ gets the opposite sign ... When you multiply a reaction by a coefficient (for purposes of balancing) ... E’ is not changed ...

Electrochemistry Cell Emf …

Electrochemistry Cell Emf … Known:
a. Mg(s) + 2H+(aq) ----> Mg+2(aq) + H2(g)

Mg(s) + 2H+(aq) ----> Mg+2(aq) + H2(g) Cu+2(aq) + 2Ag(s) ----> Cu(s) + 2Ag+1(aq)

a. Mg(s) + 2H+(aq) ----> Mg+2(aq) + H2(g) b. Cu+2(aq) + 2Ag(s) ----> Cu(s) + 2Ag+1(aq) c. 2Zn+2(aq) + 4OH-1(aq) ----> 2Zn(aq) + O2(g) + 2H2O(l)

Electrochemistry Cell Emf … Known: Unknown:
a. Mg(s) + 2H+(aq) ----> Mg+2(aq) + H2(g) E’cell = ? V b. Cu+2(aq) + 2Ag(s) ----> Cu(s) + 2Ag+1(aq) c. 2Zn+2(aq) + 4OH-1(aq) ----> 2Zn(aq) + O2(g) + 2H2O(l)

a. Mg(s) + 2H+(aq) ----> Mg+2(aq) + H2(g) E’cell = ? V b. Cu+2(aq) + 2Ag(s) ----> Cu(s) + 2Ag+1(aq) Galvanic? c. 2Zn+2(aq) + 4OH-1(aq) ----> 2Zn(aq) + O2(g) + 2H2O(l)

a. Mg(s) + 2H+(aq) ----> Mg+2(aq) + H2(g) E’cell = ? V b. Cu+2(aq) + 2Ag(s) ----> Cu(s) + 2Ag+1(aq) Galvanic? c. 2Zn+2(aq) + 4OH-1(aq) ----> 2Zn(aq) + O2(g) + 2H2O(l) Solution:

a. Mg(s) + 2H+(aq) ----> Mg+2(aq) + H2(g) E’cell = ? V b. Cu+2(aq) + 2Ag(s) ----> Cu(s) + 2Ag+1(aq) Galvanic? c. 2Zn+2(aq) + 4OH-1(aq) ----> 2Zn(aq) + O2(g) + 2H2O(l) Solution: Split each rx into two ½ rxs …

a. Mg(s) + 2H+(aq) ----> Mg+2(aq) + H2(g) E’cell = ? V b. Cu+2(aq) + 2Ag(s) ----> Cu(s) + 2Ag+1(aq) Galvanic? c. 2Zn+2(aq) + 4OH-1(aq) ----> 2Zn(aq) + O2(g) + 2H2O(l) Solution: Split each rx into two ½ rxs … look up cell potentials ...

a. Mg(s) + 2H+(aq) ----> Mg+2(aq) + H2(g) E’cell = ? V b. Cu+2(aq) + 2Ag(s) ----> Cu(s) + 2Ag+1(aq) Galvanic? c. 2Zn+2(aq) + 4OH-1(aq) ----> 2Zn(aq) + O2(g) + 2H2O(l) Solution: Split each rx into two ½ rxs … look up cell potentials ... a. reduction:

a. Mg(s) + 2H+(aq) ----> Mg+2(aq) + H2(g) E’cell = ? V b. Cu+2(aq) + 2Ag(s) ----> Cu(s) + 2Ag+1(aq) Galvanic? c. 2Zn+2(aq) + 4OH-1(aq) ----> 2Zn(aq) + O2(g) + 2H2O(l) Solution: Split each rx into two ½ rxs … look up cell potentials ... a. reduction: 2H+(aq) + 2e > H2(g)

a. Mg(s) + 2H+(aq) ----> Mg+2(aq) + H2(g) E’cell = ? V b. Cu+2(aq) + 2Ag(s) ----> Cu(s) + 2Ag+1(aq) Galvanic? c. 2Zn+2(aq) + 4OH-1(aq) ----> 2Zn(aq) + O2(g) + 2H2O(l) Solution: Split each rx into two ½ rxs … look up cell potentials ... a. reduction: 2H+(aq) + 2e > H2(g) E’ = 0.00 V

a. Mg(s) + 2H+(aq) ----> Mg+2(aq) + H2(g) E’cell = ? V b. Cu+2(aq) + 2Ag(s) ----> Cu(s) + 2Ag+1(aq) Galvanic? c. 2Zn+2(aq) + 4OH-1(aq) ----> 2Zn(aq) + O2(g) + 2H2O(l) Solution: Split each rx into two ½ rxs … look up cell potentials ... a. reduction: 2H+(aq) + 2e > H2(g) E’ = 0.00 V oxidation: Mg(s) ----> Mg+2(aq) + 2e-1

a. Mg(s) + 2H+(aq) ----> Mg+2(aq) + H2(g) E’cell = ? V b. Cu+2(aq) + 2Ag(s) ----> Cu(s) + 2Ag+1(aq) Galvanic? c. 2Zn+2(aq) + 4OH-1(aq) ----> 2Zn(aq) + O2(g) + 2H2O(l) Solution: Split each rx into two ½ rxs … look up cell potentials ... a. reduction: 2H+(aq) + 2e > H2(g) E’ = 0.00 V oxidation: Mg(s) ----> Mg+2(aq) + 2e-1 E’ = V

a. Mg(s) + 2H+(aq) ----> Mg+2(aq) + H2(g) E’cell = ? V b. Cu+2(aq) + 2Ag(s) ----> Cu(s) + 2Ag+1(aq) Galvanic? c. 2Zn+2(aq) + 4OH-1(aq) ----> 2Zn(aq) + O2(g) + 2H2O(l) Solution: Split each rx into two ½ rxs … look up cell potentials ... a. reduction: 2H+(aq) + 2e > H2(g) E’ = V oxidation: Mg(s) ----> Mg+2(aq) + 2e-1 E’ = V E’cell = V

a. Mg(s) + 2H+(aq) ----> Mg+2(aq) + H2(g) E’cell = ? V b. Cu+2(aq) + 2Ag(s) ----> Cu(s) + 2Ag+1(aq) Galvanic? c. 2Zn+2(aq) + 4OH-1(aq) ----> 2Zn(aq) + O2(g) + 2H2O(l) Solution: Split each rx into two ½ rxs … look up cell potentials ... a. reduction: 2H+(aq) + 2e > H2(g) E’ = V oxidation: Mg(s) ----> Mg+2(aq) + 2e-1 E’ = V E’cell = V (Galvanic)

a. Mg(s) + 2H+(aq) ----> Mg+2(aq) + H2(g) E’cell = ? V b. Cu+2(aq) + 2Ag(s) ----> Cu(s) + 2Ag+1(aq) Galvanic? c. 2Zn+2(aq) + 4OH-1(aq) ----> 2Zn(aq) + O2(g) + 2H2O(l) Solution: Split each rx into two ½ rxs … look up cell potentials ... a. reduction: 2H+(aq) + 2e > H2(g) E’ = V oxidation: Mg(s) ----> Mg+2(aq) + 2e-1 E’ = V E’cell = V (Galvanic) b. reduction: Cu+2(aq) + 2e > Cu(s)

a. Mg(s) + 2H+(aq) ----> Mg+2(aq) + H2(g) E’cell = ? V b. Cu+2(aq) + 2Ag(s) ----> Cu(s) + 2Ag+1(aq) Galvanic? c. 2Zn+2(aq) + 4OH-1(aq) ----> 2Zn(aq) + O2(g) + 2H2O(l) Solution: Split each rx into two ½ rxs … look up cell potentials ... a. reduction: 2H+(aq) + 2e > H2(g) E’ = V oxidation: Mg(s) ----> Mg+2(aq) + 2e-1 E’ = V E’cell = V (Galvanic) b. reduction: Cu+2(aq) + 2e > Cu(s) E’ = V

a. Mg(s) + 2H+(aq) ----> Mg+2(aq) + H2(g) E’cell = ? V b. Cu+2(aq) + 2Ag(s) ----> Cu(s) + 2Ag+1(aq) Galvanic? c. 2Zn+2(aq) + 4OH-1(aq) ----> 2Zn(aq) + O2(g) + 2H2O(l) Solution: Split each rx into two ½ rxs … look up cell potentials ... a. reduction: 2H+(aq) + 2e > H2(g) E’ = V oxidation: Mg(s) ----> Mg+2(aq) + 2e-1 E’ = V E’cell = V (Galvanic) b. reduction: Cu+2(aq) + 2e > Cu(s) E’ = V oxidation: 2Ag(s) ----> 2Ag+1(aq) + 2e-1

a. Mg(s) + 2H+(aq) ----> Mg+2(aq) + H2(g) E’cell = ? V b. Cu+2(aq) + 2Ag(s) ----> Cu(s) + 2Ag+1(aq) Galvanic? c. 2Zn+2(aq) + 4OH-1(aq) ----> 2Zn(aq) + O2(g) + 2H2O(l) Solution: Split each rx into two ½ rxs … look up cell potentials ... a. reduction: 2H+(aq) + 2e > H2(g) E’ = V oxidation: Mg(s) ----> Mg+2(aq) + 2e-1 E’ = V E’cell = V (Galvanic) b. reduction: Cu+2(aq) + 2e > Cu(s) E’ = V oxidation: 2Ag(s) ----> 2Ag+1(aq) + 2e-1 E’ = V

a. Mg(s) + 2H+(aq) ----> Mg+2(aq) + H2(g) E’cell = ? V b. Cu+2(aq) + 2Ag(s) ----> Cu(s) + 2Ag+1(aq) Galvanic? c. 2Zn+2(aq) + 4OH-1(aq) ----> 2Zn(aq) + O2(g) + 2H2O(l) Solution: Split each rx into two ½ rxs … look up cell potentials ... a. reduction: 2H+(aq) + 2e > H2(g) E’ = V oxidation: Mg(s) ----> Mg+2(aq) + 2e-1 E’ = V E’cell = V (Galvanic) b. reduction: Cu+2(aq) + 2e > Cu(s) E’ = V oxidation: 2Ag(s) ----> 2Ag+1(aq) + 2e-1 E’ = V E’cell = V

a. Mg(s) + 2H+(aq) ----> Mg+2(aq) + H2(g) E’cell = ? V b. Cu+2(aq) + 2Ag(s) ----> Cu(s) + 2Ag+1(aq) Galvanic? c. 2Zn+2(aq) + 4OH-1(aq) ----> 2Zn(aq) + O2(g) + 2H2O(l) Solution: Split each rx into two ½ rxs … look up cell potentials ... a. reduction: 2H+(aq) + 2e > H2(g) E’ = V oxidation: Mg(s) ----> Mg+2(aq) + 2e-1 E’ = V E’cell = V (Galvanic) b. reduction: Cu+2(aq) + 2e > Cu(s) E’ = V oxidation: 2Ag(s) ----> 2Ag+1(aq) + 2e-1 E’ = V E’cell = V (Not Galvanic)

a. Mg(s) + 2H+(aq) ----> Mg+2(aq) + H2(g) E’cell = ? V b. Cu+2(aq) + 2Ag(s) ----> Cu(s) + 2Ag+1(aq) Galvanic? c. 2Zn+2(aq) + 4OH-1(aq) ----> 2Zn(aq) + O2(g) + 2H2O(l) Solution: Split each rx into two ½ rxs … look up cell potentials ... a. reduction: 2H+(aq) + 2e > H2(g) E’ = V oxidation: Mg(s) ----> Mg+2(aq) + 2e-1 E’ = V E’cell = V (Galvanic) b. reduction: Cu+2(aq) + 2e > Cu(s) E’ = V oxidation: 2Ag(s) ----> 2Ag+1(aq) + 2e-1 E’ = V E’cell = V (Not Galvanic) Note: will not run in this direction ...

Electrochemistry Known: Cell Emf … Unknown:
a. Mg(s) + 2H+(aq) ----> Mg+2(aq) + H2(g) E’cell = ? V b. Cu+2(aq) + 2Ag(s) ----> Cu(s) + 2Ag+1(aq) Galvanic? c. 2Zn+2(aq) + 4OH-1(aq) ----> 2Zn(aq) + O2(g) + 2H2O(l) Solution: Split each rx into two ½ rxs … look up cell potentials ... a. reduction: 2H+(aq) + 2e > H2(g) E’ = V oxidation: Mg(s) ----> Mg+2(aq) + 2e-1 E’ = V E’cell = V (Galvanic) b. reduction: Cu+2(aq) + 2e > Cu(s) E’ = V oxidation: 2Ag(s) ----> 2Ag+1(aq) + 2e-1 E’ = V E’cell = V (Not Galvanic) Note: will not run in this direction ... c. Reduction: 2Zn+2(aq) + 4e > 2Zn(s)

a. Mg(s) + 2H+(aq) ----> Mg+2(aq) + H2(g) E’cell = ? V b. Cu+2(aq) + 2Ag(s) ----> Cu(s) + 2Ag+1(aq) Galvanic? c. 2Zn+2(aq) + 4OH-1(aq) ----> 2Zn(aq) + O2(g) + 2H2O(l) Solution: Split each rx into two ½ rxs … look up cell potentials ... a. reduction: 2H+(aq) + 2e > H2(g) E’ = V oxidation: Mg(s) ----> Mg+2(aq) + 2e-1 E’ = V E’cell = V (Galvanic) b. reduction: Cu+2(aq) + 2e > Cu(s) E’ = V oxidation: 2Ag(s) ----> 2Ag+1(aq) + 2e-1 E’ = V E’cell = V (Not Galvanic) Note: will not run in this direction ... c. Reduction: 2Zn+2(aq) + 4e > 2Zn(s) E’ = V

a. Mg(s) + 2H+(aq) ----> Mg+2(aq) + H2(g) E’cell = ? V b. Cu+2(aq) + 2Ag(s) ----> Cu(s) + 2Ag+1(aq) Galvanic? c. 2Zn+2(aq) + 4OH-1(aq) ----> 2Zn(aq) + O2(g) + 2H2O(l) Solution: Split each rx into two ½ rxs … look up cell potentials ... a. reduction: 2H+(aq) + 2e > H2(g) E’ = V oxidation: Mg(s) ----> Mg+2(aq) + 2e-1 E’ = V E’cell = V (Galvanic) b. reduction: Cu+2(aq) + 2e > Cu(s) E’ = V oxidation: 2Ag(s) ----> 2Ag+1(aq) + 2e-1 E’ = V E’cell = V (Not Galvanic) Note: will not run in this direction ... c. Reduction: 2Zn+2(aq) + 4e > 2Zn(s) E’ = V Oxidation: 4OH-1(aq) ----> O2(g) + 2H2O(l) + 4e-1

a. Mg(s) + 2H+(aq) ----> Mg+2(aq) + H2(g) E’cell = ? V b. Cu+2(aq) + 2Ag(s) ----> Cu(s) + 2Ag+1(aq) Galvanic? c. 2Zn+2(aq) + 4OH-1(aq) ----> 2Zn(aq) + O2(g) + 2H2O(l) Solution: Split each rx into two ½ rxs … look up cell potentials ... a. reduction: 2H+(aq) + 2e > H2(g) E’ = V oxidation: Mg(s) ----> Mg+2(aq) + 2e-1 E’ = V E’cell = V (Galvanic) b. reduction: Cu+2(aq) + 2e > Cu(s) E’ = V oxidation: 2Ag(s) ----> 2Ag+1(aq) + 2e-1 E’ = V E’cell = V (Not Galvanic) Note: will not run in this direction ... c. Reduction: 2Zn+2(aq) + 4e > 2Zn(s) E’ = V Oxidation: 4OH-1(aq) ----> O2(g) + 2H2O(l) + 4e-1 E’ = V

a. Mg(s) + 2H+(aq) ----> Mg+2(aq) + H2(g) E’cell = ? V b. Cu+2(aq) + 2Ag(s) ----> Cu(s) + 2Ag+1(aq) Galvanic? c. 2Zn+2(aq) + 4OH-1(aq) ----> 2Zn(aq) + O2(g) + 2H2O(l) Solution: Split each rx into two ½ rxs … look up cell potentials ... a. reduction: 2H+(aq) + 2e > H2(g) E’ = V oxidation: Mg(s) ----> Mg+2(aq) + 2e-1 E’ = V E’cell = V (Galvanic) b. reduction: Cu+2(aq) + 2e > Cu(s) E’ = V oxidation: 2Ag(s) ----> 2Ag+1(aq) + 2e-1 E’ = V E’cell = V (Not Galvanic) Note: will not run in this direction ... c. Reduction: 2Zn+2(aq) + 4e > 2Zn(s) E’ = V Oxidation: 4OH-1(aq) ----> O2(g) + 2H2O(l) + 4e-1 E’ = V E’cell = V

a. Mg(s) + 2H+(aq) ----> Mg+2(aq) + H2(g) E’cell = ? V b. Cu+2(aq) + 2Ag(s) ----> Cu(s) + 2Ag+1(aq) Galvanic? c. 2Zn+2(aq) + 4OH-1(aq) ----> 2Zn(aq) + O2(g) + 2H2O(l) Solution: Split each rx into two ½ rxs … look up cell potentials ... a. reduction: 2H+(aq) + 2e > H2(g) E’ = V oxidation: Mg(s) ----> Mg+2(aq) + 2e-1 E’ = V E’cell = V (Galvanic) b. reduction: Cu+2(aq) + 2e > Cu(s) E’ = V oxidation: 2Ag(s) ----> 2Ag+1(aq) + 2e-1 E’ = V E’cell = V (Not Galvanic) Note: will not run in this direction ... c. Reduction: 2Zn+2(aq) + 4e > 2Zn(s) E’ = V Oxidation: 4OH-1(aq) ----> O2(g) + 2H2O(l) + 4e-1 E’ = V E’cell = V (Not Gal)

Electrochemistry Key Point …

Electrochemistry Key Point … The more positive the E’ Value ...

Electrochemistry Key Point …
The more positive the E’ Value ... the more likely the species is to be reduced …

The more positive the E’ Value ... the more likely the species is to be reduced … it is a stronger oxidizing agent ...

The more positive the E’ Value ... the more likely the species is to be reduced … it is a stronger oxidizing agent ... The more negative the E’ Value ...

The more positive the E’ Value ... the more likely the species is to be reduced … it is a stronger oxidizing agent ... The more negative the E’ Value ... the more likely the species is to be oxidized …

The more positive the E’ Value ... the more likely the species is to be reduced … it is a stronger oxidizing agent ... The more negative the E’ Value ... the more likely the species is to be oxidized … it is a stronger reducing agent ...

The more positive the E’ Value ... the more likely the species is to be reduced … it is a stronger oxidizing agent ... The more negative the E’ Value ... the more likely the species is to be oxidized … it is a stronger reducing agent ... Ex: Reduction: Br2(l)+ 2e > 2Br-1(aq) E’ = V

The more positive the E’ Value ... the more likely the species is to be reduced … it is a stronger oxidizing agent ... The more negative the E’ Value ... the more likely the species is to be oxidized … it is a stronger reducing agent ... Ex: Reduction: Br2(l)+ 2e > 2Br-1(aq) E’ = V Oxidation: 2I > I2(l) + 2e-1 E’ = V

The more positive the E’ Value ... the more likely the species is to be reduced … it is a stronger oxidizing agent ... The more negative the E’ Value ... the more likely the species is to be oxidized … it is a stronger reducing agent ... Ex: Reduction: Br2(l)+ 2e > 2Br-1(aq) E’ = V Oxidation: 2I > I2(l) + 2e-1 E’ = V E’cell = +.55 V

Electrochemistry Known: Reduction Strength

Electrochemistry Known: Reduction Strength
Fe+2(aq) + 2e > Fe(s) E’ = V

Fe+2(aq) + 2e > Fe(s) E’ = V ClO2(aq) + e > ClO2-1(aq) E’ = V

Fe+2(aq) + 2e > Fe(s) E’ = V ClO2(aq) + e > ClO2-1(aq) E’ = V F2(l) + 2e > 2F-1(aq) E’ = V

Fe+2(aq) + 2e > Fe(s) E’ = V ClO2(aq) + e > ClO2-1(aq) E’ = V F2(l) + 2e > 2F-1(aq) E’ = V AgCl(s) + e > Ag(s) + Cl-1(aq) E’ = V

Electrochemistry Known: Reduction Strength Unknown:
Fe+2(aq) + 2e > Fe(s) E’ = V Rank in order of strength as ClO2(aq) + e > ClO2-1(aq) E’ = V oxidizing agent … F2(l) + 2e > 2F-1(aq) E’ = V AgCl(s) + e > Ag(s) + Cl-1(aq) E’ = V

Fe+2(aq) + 2e > Fe(s) E’ = V Rank in order of strength as ClO2(aq) + e > ClO2-1(aq) E’ = V oxidizing agent … F2(l) + 2e > 2F-1(aq) E’ = V AgCl(s) + e > Ag(s) + Cl-1(aq) E’ = V Solution: good oxidizing agents are more likely to be in reduced state as they readily strip electrons from other substances …

Fe+2(aq) + 2e > Fe(s) E’ = V Rank in order of strength as ClO2(aq) + e > ClO2-1(aq) E’ = V oxidizing agent … F2(l) + 2e > 2F-1(aq) E’ = V AgCl(s) + e > Ag(s) + Cl-1(aq) E’ = V Solution: good oxidizing agents are more likely to be in reduced state as they readily strip electrons from other substances … the more positive the E’ value ...

Fe+2(aq) + 2e > Fe(s) E’ = V Rank in order of increasing ClO2(aq) + e > ClO2-1(aq) E’ = V strength as oxidizing agent … F2(l) + 2e > 2F-1(aq) E’ = V AgCl(s) + e > Ag(s) + Cl-1(aq) E’ = V Solution: good oxidizing agents are more likely to be in reduced state as they readily strip electrons from other substances … the more positive the E’ value ... the more likely a substance will be in reduced form …

Fe+2(aq) + 2e > Fe(s) E’ = V Rank in order of increasing ClO2(aq) + e > ClO2-1(aq) E’ = V strength as oxidizing agent … F2(l) + 2e > 2F-1(aq) E’ = V AgCl(s) + e > Ag(s) + Cl-1(aq) E’ = V Solution: good oxidizing agents are more likely to be in reduced state as they readily strip electrons from other substances … the more positive the E’ value ... the more likely a substance will be in reduced form … Fe+2 <

Fe+2(aq) + 2e > Fe(s) E’ = V Rank in order of increasing ClO2(aq) + e > ClO2-1(aq) E’ = V strength as oxidizing agent … F2(l) + 2e > 2F-1(aq) E’ = V AgCl(s) + e > Ag(s) + Cl-1(aq) E’ = V Solution: good oxidizing agents are more likely to be in reduced state as they readily strip electrons from other substances … the more positive the E’ value ... the more likely a substance will be in reduced form … Fe+2 < AgCl <

Fe+2 < AgCl < ClO2 <
Electrochemistry Known: Reduction Strength Unknown: Fe+2(aq) + 2e > Fe(s) E’ = V Rank in order of increasing ClO2(aq) + e > ClO2-1(aq) E’ = V strength as oxidizing agent … F2(l) + 2e > 2F-1(aq) E’ = V AgCl(s) + e > Ag(s) + Cl-1(aq) E’ = V Solution: good oxidizing agents are more likely to be in reduced state as they readily strip electrons from other substances … the more positive the E’ value ... the more likely a substance will be in reduced form … Fe+2 < AgCl < ClO2 <

Fe+2 < AgCl < ClO2 < F2
Electrochemistry Known: Reduction Strength Unknown: Fe+2(aq) + 2e > Fe(s) E’ = V Rank in order of increasing ClO2(aq) + e > ClO2-1(aq) E’ = V strength as oxidizing agent … F2(l) + 2e > 2F-1(aq) E’ = V AgCl(s) + e > Ag(s) + Cl-1(aq) E’ = V Solution: good oxidizing agents are more likely to be in reduced state as they readily strip electrons from other substances … the more positive the E’ value ... the more likely a substance will be in reduced form … Fe+2 < AgCl < ClO2 < F2

Electrochemistry Composing Galvanic Cells …

Electrochemistry Composing Galvanic Cells … Known:
a. Ni+2 + 2e > Ni E’ = V

a. Ni+2 + 2e > Ni E’ = V O2 + 4H+1 + 4e >2H2O E’ = V

a. Ni+2 + 2e > Ni E’ = V O2 + 4H+1 + 4e >2H2O E’ = V b. Ce+4 + e > Ce+3 E’ = V

a. Ni+2 + 2e > Ni E’ = V O2 + 4H+1 + 4e >2H2O E’ = V b. Ce+4 + e > Ce+3 E’ = V Sn+2 + 2e > Sn E’ = V

Electrochemistry Composing Galvanic Cells … Known: Unknown:
a. Ni+2 + 2e > Ni E’ = V Anode? O2 + 4H+1 + 4e >2H2O E’ = V b. Ce+4 + e > Ce+3 E’ = V Sn+2 + 2e > Sn E’ = V

a. Ni+2 + 2e > Ni E’ = V Anode? O2 + 4H+1 + 4e >2H2O E’ = V Cathode? b. Ce+4 + e > Ce+3 E’ = V Sn+2 + 2e > Sn E’ = V

a. Ni+2 + 2e > Ni E’ = V Anode? O2 + 4H+1 + 4e >2H2O E’ = V Cathode? Balance ... b. Ce+4 + e > Ce+3 E’ = V Sn+2 + 2e > Sn E’ = V

a. Ni+2 + 2e > Ni E’ = V Anode? O2 + 4H+1 + 4e >2H2O E’ = V Cathode? Balance ... b. Ce+4 + e > Ce+3 E’ = V Write Overall Cell Rx ... Sn+2 + 2e > Sn E’ = V

a. Ni+2 + 2e > Ni E’ = V Anode? O2 + 4H+1 + 4e >2H2O E’ = V Cathode? Balance ... b. Ce+4 + e > Ce+3 E’ = V Write Overall Cell Rx ... Sn+2 + 2e > Sn E’ = V Calculate E’cell Solution:

a. Ni+2 + 2e > Ni E’ = V Anode? O2 + 4H+1 + 4e >2H2O E’ = V Cathode? Balance ... b. Ce+4 + e > Ce+3 E’ = V Write Overall Cell Rx ... Sn+2 + 2e > Sn E’ = V Calculate E’cell Solution: Goal ... Which arrangement will make E’cell > 0 ... a. Reversing Ni+2 will create a galvanic cell …

a. Ni+2 + 2e > Ni E’ = V Anode? O2 + 4H+1 + 4e >2H2O E’ = V Cathode? Balance ... b. Ce+4 + e > Ce+3 E’ = V Write Overall Cell Rx ... Sn+2 + 2e > Sn E’ = V Calculate E’cell Solution: Goal ... Which arrangement will make E’cell > 0 ... a. Reversing Ni+2 will create a galvanic cell … Cathode: O2 + 4H+1 + 4e >2H2O E’ = V

a. Ni+2 + 2e > Ni E’ = V Anode? O2 + 4H+1 + 4e >2H2O E’ = V Cathode? Balance ... b. Ce+4 + e > Ce+3 E’ = V Write Overall Cell Rx ... Sn+2 + 2e > Sn E’ = V Calculate E’cell Solution: Goal ... Which arrangement will make E’cell > 0 ... a. Reversing Ni+2 will create a galvanic cell … multiple Ni+2 by 2 ... Cathode: O2 + 4H+1 + 4e >2H2O E’ = V

a. Ni+2 + 2e > Ni E’ = V Anode? O2 + 4H+1 + 4e >2H2O E’ = V Cathode? Balance ... b. Ce+4 + e > Ce+3 E’ = V Write Overall Cell Rx ... Sn+2 + 2e > Sn E’ = V Calculate E’cell Solution: Goal ... Which arrangement will make E’cell > 0 ... a. Reversing Ni+2 will create a galvanic cell … multiple Ni+2 by 2 ... cathode: O2 + 4H+1 + 4e >2H2O E’ = V anode: 2Ni ----> 2Ni+2 + 4e-1 E’ = V

a. Ni+2 + 2e > Ni E’ = V Anode? O2 + 4H+1 + 4e >2H2O E’ = V Cathode? Balance ... b. Ce+4 + e > Ce+3 E’ = V Write Overall Cell Rx ... Sn+2 + 2e > Sn E’ = V Calculate E’cell Solution: Goal ... Which arrangement will make E’cell > 0 ... a. Reversing Ni+2 will create a galvanic cell … multiple Ni+2 by 2 ... cathode: O2 + 4H+1 + 4e >2H2O E’ = V anode: 2Ni ----> 2Ni+2 + 4e-1 E’ = V overall: O2 + 4H+1 + 2Ni ----> 2Ni+2 + 2H2O

a. Ni+2 + 2e > Ni E’ = V Anode? O2 + 4H+1 + 4e >2H2O E’ = V Cathode? Balance ... b. Ce+4 + e > Ce+3 E’ = V Write Overall Cell Rx ... Sn+2 + 2e > Sn E’ = V Calculate E’cell Solution: Goal ... Which arrangement will make E’cell > 0 ... a. Reversing Ni+2 will create a galvanic cell … multiple Ni+2 by 2 ... cathode: O2 + 4H+1 + 4e >2H2O E’ = V anode: 2Ni ----> 2Ni+2 + 4e-1 E’ = V overall: O2 + 4H+1 + 2Ni ----> 2Ni+2 + 2H2O E’cell = V

a. Ni+2 + 2e > Ni E’ = V Anode? O2 + 4H+1 + 4e >2H2O E’ = V Cathode? Balance ... b. Ce+4 + e > Ce+3 E’ = V Write Overall Cell Rx ... Sn+2 + 2e > Sn E’ = V Calculate E’cell Solution: Goal ... Which arrangement will make E’cell > 0 ... b. Reversing Sn+2 reaction will create a galvanic cell …

a. Ni+2 + 2e > Ni E’ = V Anode? O2 + 4H+1 + 4e >2H2O E’ = V Cathode? Balance ... b. Ce+4 + e > Ce+3 E’ = V Write Overall Cell Rx ... Sn+2 + 2e > Sn E’ = V Calculate E’cell Solution: Goal ... Which arrangement will make E’cell > 0 ... b. Reversing Sn+2 reaction will create a galvanic cell … multiply Ce+4 by 2 ... cathode: 2Ce+4 + 2e > 2Ce+3 E’ = V

a. Ni+2 + 2e > Ni E’ = V Anode? O2 + 4H+1 + 4e >2H2O E’ = V Cathode? Balance ... b. Ce+4 + e > Ce+3 E’ = V Write Overall Cell Rx ... Sn+2 + 2e > Sn E’ = V Calculate E’cell Solution: Goal ... Which arrangement will make E’cell > 0 ... b. Reversing Sn+2 reaction will create a galvanic cell … multiply Ce+4 by 2 ... cathode: 2Ce+4 + 2e > 2Ce+3 E’ = V anode: Sn ----> Sn+2 + 2e-1 E’ = V

a. Ni+2 + 2e > Ni E’ = V Anode? O2 + 4H+1 + 4e >2H2O E’ = V Cathode? Balance ... b. Ce+4 + e > Ce+3 E’ = V Write Overall Cell Rx ... Sn+2 + 2e > Sn E’ = V Calculate E’cell Solution: Goal ... Which arrangement will make E’cell > 0 ... b. Reversing Sn+2 reaction will create a galvanic cell … multiply Ce+4 by 2 ... cathode: 2Ce+4 + 2e > 2Ce+3 E’ = V anode: Sn ----> Sn+2 + 2e-1 E’ = V overall rx: 2Ce+4 + Sn ----> Sn+2 + 2Ce+3

a. Ni+2 + 2e > Ni E’ = V Anode? O2 + 4H+1 + 4e >2H2O E’ = V Cathode? Balance ... b. Ce+4 + e > Ce+3 E’ = V Write Overall Cell Rx ... Sn+2 + 2e > Sn E’ = V Calculate E’cell Solution: Goal ... Which arrangement will make E’cell > 0 ... b. Reversing Sn+2 reaction will create a galvanic cell … multiply Ce+4 by 2 ... cathode: 2Ce+4 + 2e > 2Ce+3 E’ = V anode: Sn ----> Sn+2 + 2e-1 E’ = V overall rx: 2Ce+4 + Sn ----> Sn+2 + 2Ce+3 E’cell = V

Electrochemistry Line Notation ...

Electrochemistry Line Notation ... A method of representing electrochemical cells ...

Electrochemistry Line Notation ... A method of representing electrochemical cells ... shorthand notation used to describe galvanic cells ...

Electrochemistry Line Notation ... A method of representing electrochemical cells ... shorthand notation used to describe galvanic cells ... The general line notation formation is ...

anode side || cathode side
Electrochemistry Line Notation ... A method of representing electrochemical cells ... shorthand notation used to describe galvanic cells ... The general line notation formation is ... anode side || cathode side Porous disk or Salt bridge

Electrochemistry Line Notation ... A method of representing electrochemical cells ... shorthand notation used to describe galvanic cells ... The general line notation formation is ... anode side || cathode side Porous disk or Salt bridge Each side will have a phase boundary separating, e.g., a solid electrode from ions in solution …

Electrochemistry Line Notation ... A method of representing electrochemical cells ... shorthand notation used to describe galvanic cells ... The general line notation formation is ... anode side || cathode side Porous disk or Salt bridge Each side will have a phase boundary separating, e.g., a solid electrode from ions in solution … such as Cu(s) and Cu+2(aq) ...

Electrochemistry Line Notation ... A method of representing electrochemical cells ... shorthand notation used to describe galvanic cells ... The general line notation formation is ... anode side || cathode side Porous disk or Salt bridge Each side will have a phase boundary separating, e.g., a solid electrode from ions in solution … such as Cu(s) and Cu+2(aq) ... denoted by a single vertical line (“|”) with the solid phase on the outside …

Electrochemistry Consider the reaction between copper and zinc ...

Electrochemistry Consider the reaction between copper and zinc ...
anode: (oxidation) Zn ----> Zn+2 + 2e-1

anode: (oxidation) Zn ----> Zn+2 + 2e-1 cathode: (reduction) Cu+2 + 2e > Cu

anode: (oxidation) Zn ----> Zn+2 + 2e-1 cathode: (reduction) Cu+2 + 2e > Cu The line notation is ...

Zn(s) | Zn+2(aq) || Cu+2(aq) | Cu(s)
Electrochemistry Consider the reaction between copper and zinc ... anode: (oxidation) Zn ----> Zn+2 + 2e-1 cathode: (reduction) Cu+2 + 2e > Cu The line notation is ... Zn(s) | Zn+2(aq) || Cu+2(aq) | Cu(s)

Zn(s) | Zn+2(aq) || Cu+2(aq) | Cu(s)
Electrochemistry Consider the reaction between copper and zinc ... anode: (oxidation) Zn ----> Zn+2 + 2e-1 cathode: (reduction) Cu+2 + 2e > Cu The line notation is ... Zn(s) | Zn+2(aq) || Cu+2(aq) | Cu(s) Note: On occasion ...

Zn(s) | Zn+2(aq) || Cu+2(aq) | Cu(s)
Electrochemistry Consider the reaction between copper and zinc ... anode: (oxidation) Zn ----> Zn+2 + 2e-1 cathode: (reduction) Cu+2 + 2e > Cu The line notation is ... Zn(s) | Zn+2(aq) || Cu+2(aq) | Cu(s) Note: On occasion ... only ions (rather than solids) ...

Zn(s) | Zn+2(aq) || Cu+2(aq) | Cu(s)
Electrochemistry Consider the reaction between copper and zinc ... anode: (oxidation) Zn ----> Zn+2 + 2e-1 cathode: (reduction) Cu+2 + 2e > Cu The line notation is ... Zn(s) | Zn+2(aq) || Cu+2(aq) | Cu(s) Note: On occasion ... only ions (rather than solids) ... Will be involved in the redox process ...

Zn(s) | Zn+2(aq) || Cu+2(aq) | Cu(s)
Electrochemistry Consider the reaction between copper and zinc ... anode: (oxidation) Zn ----> Zn+2 + 2e-1 cathode: (reduction) Cu+2 + 2e > Cu The line notation is ... Zn(s) | Zn+2(aq) || Cu+2(aq) | Cu(s) Note: On occasion ... only ions (rather than solids) ... Will be involved in the redox process ... in those cases “inert’ ecltrodes are used ...

Electrochemistry More on line notations …

Electrochemistry More on line notations … write line notations for the following galvanic cells ...

Electrochemistry More on line notations … write line notations for the following galvanic cells ... Hg+2 + Cd ----> Hg + Cd+2

Electrochemistry More on line notations … write line notations for the following galvanic cells ... Hg+2 + Cd ----> Hg + Cd+2 anode: Cd ----> Cd+2 + 2e-1

Electrochemistry More on line notations … write line notations for the following galvanic cells ... Hg+2 + Cd ----> Hg + Cd+2 anode: Cd ----> Cd+2 + 2e-1 cathode: Hg+2 + 2e > Hg

Electrochemistry More on line notations … write line notations for the following galvanic cells ... Hg+2 + Cd ----> Hg + Cd+2 anode: Cd ----> Cd+2 + 2e-1 cathode: Hg+2 + 2e > Hg line notation: Cd(s)| Cd+2(aq) || Hg+2(aq) | Hg(s)

Electrochemistry More on line notations … write line notations for the following galvanic cells ... Hg+2 + Cd ----> Hg + Cd+2 anode: Cd ----> Cd+2 + 2e-1 cathode: Hg+2 + 2e > Hg line notation: Cd(s)| Cd+2(aq) || Hg+2(aq) | Hg(s) Pb + 2Cr > Pb+2 + 2Cr+2

Electrochemistry More on line notations … write line notations for the following galvanic cells ... Hg+2 + Cd ----> Hg + Cd+2 anode: Cd ----> Cd+2 + 2e-1 cathode: Hg+2 + 2e > Hg line notation: Cd(s)| Cd+2(aq) || Hg+2(aq) | Hg(s) Pb + 2Cr > Pb+2 + 2Cr+2 anode: Pb ----> Pb+2 + 2e-1

Electrochemistry More on line notations … write line notations for the following galvanic cells ... Hg+2 + Cd ----> Hg + Cd+2 anode: Cd ----> Cd+2 + 2e-1 cathode: Hg+2 + 2e > Hg line notation: Cd(s)| Cd+2(aq) || Hg+2(aq) | Hg(s) Pb + 2Cr > Pb+2 + 2Cr+2 anode: Pb ----> Pb+2 + 2e-1 cathode: 2Cr+3 + 2e > 2 Cr+2

Electrochemistry More on line notations … write line notations for the following galvanic cells ... Hg+2 + Cd ----> Hg + Cd+2 anode: Cd ----> Cd+2 + 2e-1 cathode: Hg+2 + 2e > Hg line notation: Cd(s)| Cd+2(aq) || Hg+2(aq) | Hg(s) Pb + 2Cr > Pb+2 + 2Cr+2 anode: Pb ----> Pb+2 + 2e-1 cathode: 2Cr+3 + 2e > 2 Cr+2 line notation: Pb(s)| Pb+2(aq) || Cr+3(aq) , Cr+2(aq) | Pt(s)

Electrochemistry More on line notations … write line notations for the following galvanic cells ... Hg+2 + Cd ----> Hg + Cd+2 anode: Cd ----> Cd+2 + 2e-1 cathode: Hg+2 + 2e > Hg line notation: Cd(s)| Cd+2(aq) || Hg+2(aq) | Hg(s) Pb + 2Cr > Pb+2 + 2Cr+2 anode: Pb ----> Pb+2 + 2e-1 cathode: 2Cr+3 + 2e > 2 Cr+2 line notation: Pb(s)| Pb+2(aq) || Cr+3(aq) , Cr+2(aq) | Pt(s) Cu+2 + 2Pu+4 + 4H2O -----> Cu + 2PuO H+1

Electrochemistry More on line notations … write line notations for the following galvanic cells ... Hg+2 + Cd ----> Hg + Cd+2 anode: Cd ----> Cd+2 + 2e-1 cathode: Hg+2 + 2e > Hg line notation: Cd(s)| Cd+2(aq) || Hg+2(aq) | Hg(s) Pb + 2Cr > Pb+2 + 2Cr+2 anode: Pb ----> Pb+2 + 2e-1 cathode: 2Cr+3 + 2e > 2 Cr+2 line notation: Pb(s)| Pb+2(aq) || Cr+3(aq) , Cr+2(aq) | Pt(s) Cu+2 + 2Pu+4 + 4H2O -----> Cu + 2PuO H+1 anode: 2Pu+4 + 4H2O ----> 2PuO H+1 + 2e-1

Electrochemistry More on line notations … write line notations for the following galvanic cells ... Hg+2 + Cd ----> Hg + Cd+2 anode: Cd ----> Cd+2 + 2e-1 cathode: Hg+2 + 2e > Hg line notation: Cd(s)| Cd+2(aq) || Hg+2(aq) | Hg(s) Pb + 2Cr > Pb+2 + 2Cr+2 anode: Pb ----> Pb+2 + 2e-1 cathode: 2Cr+3 + 2e > 2 Cr+2 line notation: Pb(s)| Pb+2(aq) || Cr+3(aq) , Cr+2(aq) | Pt(s) Cu+2 + 2Pu+4 + 4H2O -----> Cu + 2PuO H+1 anode: 2Pu+4 + 4H2O ----> 2PuO H+1 + 2e-1 cathode: Cu+2 + 2e > Cu

Electrochemistry More on line notations … write line notations for the following galvanic cells ... Hg+2 + Cd ----> Hg + Cd+2 anode: Cd ----> Cd+2 + 2e-1 cathode: Hg+2 + 2e > Hg line notation: Cd(s)| Cd+2(aq) || Hg+2(aq) | Hg(s) Pb + 2Cr > Pb+2 + 2Cr+2 anode: Pb ----> Pb+2 + 2e-1 cathode: 2Cr+3 + 2e > 2 Cr+2 line notation: Pb(s)| Pb+2(aq) || Cr+3(aq) , Cr+2(aq) | Pt(s) Cu+2 + 2Pu+4 + 4H2O -----> Cu + 2PuO H+1 anode: 2Pu+4 + 4H2O ----> 2PuO H+1 + 2e-1 cathode: Cu+2 + 2e > Cu line notation: Pt|Pu+4, PuO2+1, H+1 || Cu+2 | Cu

Electrochemistry Describing Galvanic Cells …

Electrochemistry Describing Galvanic Cells …
Cu+2 + 2e > Cu E’ = V

Cu+2 + 2e > Cu E’ = V Cr2O H+1 + 6e > 2Cr+3 + 7H2O E’ = V 1st: E’cell > 0 ...

Cu+2 + 2e > Cu E’ = V Cr2O H+1 + 6e > 2Cr+3 + 7H2O E’ = V 1st: E’cell > 0 ... anode: 3Cu ----> 3Cu+2 + 6e-1 E’ = V

Cu+2 + 2e > Cu E’ = V Cr2O H+1 + 6e > 2Cr+3 + 7H2O E’ = V 1st: E’cell > 0 ... anode: 3Cu ----> 3Cu+2 + 6e-1 E’ = V cathode: Cr2O H+1 + 6e > 2Cr+3 + 7H2O E’ = V

Cu+2 + 2e > Cu E’ = V Cr2O H+1 + 6e > 2Cr+3 + 7H2O E’ = V 1st: E’cell > 0 ... anode: 3Cu ----> 3Cu+2 + 6e E’ = V cathode: Cr2O H+1 + 6e > 2Cr+3 + 7H2O E’ = V overall rx: 3Cu(s) + Cr2O7-2(aq) + 14H+1(aq) ----> 3Cu+2(aq) + 2Cr+3(aq) + 7H2O(l) E’ = V

Cu+2 + 2e > Cu E’ = V Cr2O H+1 + 6e > 2Cr+3 + 7H2O E’ = V 1st: E’cell > 0 ... anode: 3Cu ----> 3Cu+2 + 6e E’ = V cathode: Cr2O H+1 + 6e > 2Cr+3 + 7H2O E’ = V overall rx: 3Cu(s) + Cr2O7-2(aq) + 14H+1(aq) ----> 3Cu+2(aq) + 2Cr+3(aq) + 7H2O(l) E’cell = V Line notation: Cu(s) | Cu+2(aq) || Cr2O7-2(aq) , Cr+3(aq) , H+1(aq) | Pt(s)

Cu+2 + 2e > Cu E’ = V Cr2O H+1 + 6e > 2Cr+3 + 7H2O E’ = V 1st: E’cell > 0 ... anode: 3Cu ----> 3Cu+2 + 6e E’ = V cathode: Cr2O H+1 + 6e > 2Cr+3 + 7H2O E’ = V overall rx: 3Cu(s) + Cr2O7-2(aq) + 14H+1(aq) ----> 3Cu+2(aq) + 2Cr+3(aq) + 7H2O(l) E’cell = V Line notation: Cu(s) | Cu+2(aq) || Cr2O7-2(aq) , Cr+3(aq) , H+1(aq) | Pt(s) Note: Electron flow is from Anode to Cathode, FATCAT, from Copper to Platinum ...

Electrochemistry Cell Potential, Electrical Work & Free Energy …

Electrochemistry Cell Potential, Electrical Work & Free Energy …
In general, the actual work that can be achieved is always less than the theoretical work available ...

In general, the actual work that can be achieved is always less than the theoretical work available ... The relationship between free energy (maximum is assumed though not attainable) .... And cell potential ... ∆G = -nFE

In general, the actual work that can be achieved is always less than the theoretical work available ... The relationship between free energy (maximum is assumed though not atainable) .... And cell potential ... ∆G = -nFE @ standard conditions ...

In general, the actual work that can be achieved is always less than the theoretical work available ... The relationship between free energy (maximum is assumed though not atainable) .... And cell potential ... ∆G = -nFE @ standard conditions ... ∆G’ = -nFE’

In general, the actual work that can be achieved is always less than the theoretical work available ... The relationship between free energy (maximum is assumed though not atainable) .... And cell potential ... ∆G = -nFE @ standard conditions ... ∆G’ = -nFE’ Where ...

In general, the actual work that can be achieved is always less than the theoretical work available ... The relationship between free energy (maximum is assumed though not atainable) .... And cell potential ... ∆G = -nFE @ standard conditions ... ∆G’ = -nFE’ Where ... ∆G = free energy (J)

In general, the actual work that can be achieved is always less than the theoretical work available ... The relationship between free energy (maximum is assumed though not atainable) .... And cell potential ... ∆G = -nFE @ standard conditions ... ∆G’ = -nFE’ Where ... ∆G = free energy (J) n = moles of electrons exchanged in the redox reaction …

In general, the actual work that can be achieved is always less than the theoretical work available ... The relationship between free energy (maximum is assumed though not atainable) .... And cell potential ... ∆G = -nFE @ standard conditions ... ∆G’ = -nFE’ Where ... ∆G = free energy (J) n = moles of electrons exchanged in the redox reaction … F = Faraday, a constant (96,486 Coulombs per mole of electrons)

In general, the actual work that can be achieved is always less than the theoretical work available ... The relationship between free energy (maximum is assumed though not atainable) .... And cell potential ... ∆G = -nFE @ standard conditions ... ∆G’ = -nFE’ Where ... ∆G = free energy (J) n = moles of electrons exchanged in the redox reaction … F = Faraday, a constant (96,486 Coulombs per mole of electrons) E = cell voltage (J/C)

In general, the actual work that can be achieved is always less than the theoretical work available ... The relationship between free energy (maximum is assumed though not atainable) .... And cell potential ... ∆G = -nFE @ standard conditions ... ∆G’ = -nFE’ Where ... ∆G = free energy (J) n = moles of electrons exchanged in the redox reaction … F = Faraday, a constant (96,486 Coulombs per mole of electrons) E = cell voltage (J/C) Note: ∆G and E’ have opposite signs... for a spontaneous process ... ∆G is “-” & E’ = “+”

Known: Zn+2(aq) + Cu(s) ----> Zn(s) + Cu+2(aq)

Known: Unknown: Zn+2(aq) + Cu(s) ----> Zn(s) + Cu+2(aq) ∆G’ = ? KJ

Known: Unknown: Zn+2(aq) + Cu(s) ----> Zn(s) + Cu+2(aq) ∆G’ = ? KJ Will Zn ions plate on Cu strip?

Known: Unknown: Zn+2(aq) + Cu(s) ----> Zn(s) + Cu+2(aq) ∆G’ = ? KJ Will Zn ions plate on Cu strip? Solution:

Known: Unknown: Zn+2(aq) + Cu(s) ----> Zn(s) + Cu+2(aq) ∆G’ = ? KJ Will Zn ions plate on Cu strip? Solution: 1st: Write the half-reactions ...

Known: Unknown: Zn+2(aq) + Cu(s) ----> Zn(s) + Cu+2(aq) ∆G’ = ? KJ Will Zn ions plate on Cu strip? Solution: 1st: Write the half-reactions ... oxidation rx: Cu(S) ----> Cu+2(aq) + 2e-1 E’ = V reduction rx: Zn+2(aq) + 2e > Zn(s) E’ = V

Known: Unknown: Zn+2(aq) + Cu(s) ----> Zn(s) + Cu+2(aq) ∆G’ = ? KJ Will Zn ions plate on Cu strip? Solution: 1st: Write the half-reactions ... oxidation rx: Cu(S) ----> Cu+2(aq) + 2e-1 E’ = V reduction rx: Zn+2(aq) + 2e > Zn(s) E’ = V E’cell = V

Known: Unknown: Zn+2(aq) + Cu(s) ----> Zn(s) + Cu+2(aq) ∆G’ = ? KJ Will Zn ions plate on Cu strip? Solution: 1st: Write the half-reactions ... oxidation rx: Cu(S) ----> Cu+2(aq) + 2e-1 E’ = V reduction rx: Zn+2(aq) + 2e > Zn(s) E’ = V E’cell = V 2nd: Calculate ∆G’ ....

Known: Unknown: Zn+2(aq) + Cu(s) ----> Zn(s) + Cu+2(aq) ∆G’ = ? KJ Will Zn ions plate on Cu strip? Solution: 1st: Write the half-reactions ... oxidation rx: Cu(S) ----> Cu+2(aq) + 2e-1 E’ = V reduction rx: Zn+2(aq) + 2e > Zn(s) E’ = V E’cell = V 2nd: Calculate ∆G’ .... ∆G’ = ? kJ||

Known: Unknown: Zn+2(aq) + Cu(s) ----> Zn(s) + Cu+2(aq) ∆G’ = ? KJ Will Zn ions plate on Cu strip? Solution: 1st: Write the half-reactions ... oxidation rx: Cu(S) ----> Cu+2(aq) + 2e-1 E’ = V reduction rx: Zn+2(aq) + 2e > Zn(s) E’ = V E’cell = V 2nd: Calculate ∆G’ .... ∆G’ = ? kJ|| -nFE’ =

Known: Unknown: Zn+2(aq) + Cu(s) ----> Zn(s) + Cu+2(aq) ∆G’ = ? KJ Will Zn ions plate on Cu strip? Solution: 1st: Write the half-reactions ... oxidation rx: Cu(S) ----> Cu+2(aq) + 2e-1 E’ = V reduction rx: Zn+2(aq) + 2e > Zn(s) E’ = V E’cell = V 2nd: Calculate ∆G’ .... ∆G’ = ? kJ|| -nFE’ = -(2 mole-)(96,486 C/mole-)(-1.10 J/C) =

Known: Unknown: Zn+2(aq) + Cu(s) ----> Zn(s) + Cu+2(aq) ∆G’ = ? KJ Will Zn ions plate on Cu strip? Solution: 1st: Write the half-reactions ... oxidation rx: Cu(S) ----> Cu+2(aq) + 2e-1 E’ = V reduction rx: Zn+2(aq) + 2e > Zn(s) E’ = V E’cell = V 2nd: Calculate ∆G’ .... ∆G’ = ? kJ|| -nFE’ = -(2 mole-)(96,486 C/mole-)(-1.10 J/C) = 2.12 x 105 J

Known: Unknown: Zn+2(aq) + Cu(s) ----> Zn(s) + Cu+2(aq) ∆G’ = ? KJ Will Zn ions plate on Cu strip? Solution: 1st: Write the half-reactions ... oxidation rx: Cu(S) ----> Cu+2(aq) + 2e-1 E’ = V reduction rx: Zn+2(aq) + 2e > Zn(s) E’ = V E’cell = V 2nd: Calculate ∆G’ .... ∆G’ = ? kJ|| -nFE’ = -(2 mole-)(96,486 C/mole-)(-1.10 J/C) = 2.12 x 105 J = 212 kJ

Known: Unknown: Zn+2(aq) + Cu(s) ----> Zn(s) + Cu+2(aq) ∆G’ = ? KJ Will Zn ions plate on Cu strip? Solution: 1st: Write the half-reactions ... oxidation rx: Cu(S) ----> Cu+2(aq) + 2e-1 E’ = V reduction rx: Zn+2(aq) + 2e > Zn(s) E’ = V E’cell = V 2nd: Calculate ∆G’ .... ∆G’ = ? kJ|| -nFE’ = -(2 mole-)(96,486 C/mole-)(-1.10 J/C) = 2.12 x 105 J = 212 kJ Note: Zn+2 ions will not plate out on a Cu(s) at Standard Conditions ...

Known: Unknown: Zn+2(aq) + Cu(s) ----> Zn(s) + Cu+2(aq) ∆G’ = ? KJ Will Zn ions plate on Cu strip? Solution: 1st: Write the half-reactions ... oxidation rx: Cu(S) ----> Cu+2(aq) + 2e-1 E’ = V reduction rx: Zn+2(aq) + 2e > Zn(s) E’ = V E’cell = V 2nd: Calculate ∆G’ .... ∆G’ = ? kJ|| -nFE’ = -(2 mole-)(96,486 C/mole-)(-1.10 J/C) = 2.12 x 105 J = 212 kJ Note: Zn+2 ions will not plate out on a Cu(s) at Standard Conditions ... E’cell < 0

Electrochemistry Dependence of Cell Potential on Concentration …

Electrochemistry Dependence of Cell Potential on Concentration …
Concentration Cells ...

Concentration Cells ... A cell in which current flows due only to a difference in concentration of an ion in two different compartments of a cell ...

Concentration Cells ... A cell in which current flows due only to a difference in concentration of an ion in two different compartments of a cell ... Le Chatelier’s Principle is applicable here ...

Concentration Cells ... A cell in which current flows due only to a difference in concentration of an ion in two different compartments of a cell ... Le Chatelier’s Principle is applicable here ... In a cell where there is an equal concentration of metal ions on both sides ...

Concentration Cells ... A cell in which current flows due only to a difference in concentration of an ion in two different compartments of a cell ... Le Chatelier’s Principle is applicable here ... In a cell where there is an equal concentration of metal ions on both sides ... E’cell = 0 ...

Electrochemistry Concentration Cells …

Electrochemistry Concentration Cells … Known:
A cell has on its left side a 0.20 M Cu+2 solution ...

A cell has on its left side a 0.20 M Cu+2 solution ... The right side has a M Cu+2 solution ...

A cell has on its left side a 0.20 M Cu+2 solution ... The right side has a M Cu+2 solution ... the compartments are connected by Cu electrodes and a salt bridge ...

A cell has on its left side a 0.20 M Cu+2 solution ... The right side has a M Cu+2 solution ... the compartments are connected by Cu electrodes and a salt bridge ... Unknown:

A cell has on its left side a 0.20 M Cu+2 solution ... The right side has a M Cu+2 solution ... the compartments are connected by Cu electrodes and a salt bridge ... Unknown: Designate the cathode, anode, and direction of current ...

A cell has on its left side a 0.20 M Cu+2 solution ... The right side has a M Cu+2 solution ... the compartments are connected by Cu electrodes and a salt bridge ... Unknown: Designate the cathode, anode, and direction of current ... Solution:

A cell has on its left side a 0.20 M Cu+2 solution ... The right side has a M Cu+2 solution ... the compartments are connected by Cu electrodes and a salt bridge ... Unknown: Designate the cathode, anode, and direction of current ... Solution: Current will flow until the [Cu+2] is equal in both compartments ...

A cell has on its left side a 0.20 M Cu+2 solution ... The right side has a M Cu+2 solution ... the compartments are connected by Cu electrodes and a salt bridge ... Unknown: Designate the cathode, anode, and direction of current ... Solution: Current will flow until the [Cu+2] is equal in both compartments ... this means that the concentration of Cu+2 in the left-hand side (0.20 M) must be reduced by …

A cell has on its left side a 0.20 M Cu+2 solution ... The right side has a M Cu+2 solution ... the compartments are connected by Cu electrodes and a salt bridge ... Unknown: Designate the cathode, anode, and direction of current ... Solution: Current will flow until the [Cu+2] is equal in both compartments ... this means that the concentration of Cu+2 in the left-hand side (0.02 M) must be reduced by … Cu+2 + 2e > Cu

A cell has on its left side a 0.20 M Cu+2 solution ... The right side has a M Cu+2 solution ... the compartments are connected by Cu electrodes and a salt bridge ... Unknown: Designate the cathode, anode, and direction of current ... Solution: Current will flow until the [Cu+2] is equal in both compartments ... this means that the concentration of Cu+2 in the left-hand side (0.02 M) must be reduced by … Cu+2 + 2e > Cu The left-hand side will be the cathode ...

A cell has on its left side a 0.20 M Cu+2 solution ... The right side has a M Cu+2 solution ... the compartments are connected by Cu electrodes and a salt bridge ... Unknown: Designate the cathode, anode, and direction of current ... Solution: Current will flow until the [Cu+2] is equal in both compartments ... this means that the concentration of Cu+2 in the left-hand side (0.02 M) must be reduced by … Cu+2 + 2e > Cu The left-hand side will be the cathode ... the right-hand side (0.050 M Cu+2) will be the anode …

A cell has on its left side a 0.20 M Cu+2 solution ... The right side has a M Cu+2 solution ... the compartments are connected by Cu electrodes and a salt bridge ... Unknown: Designate the cathode, anode, and direction of current ... Solution: Current will flow until the [Cu+2] is equal in both compartments ... this means that the concentration of Cu+2 in the left-hand side (0.02 M) must be reduced by … Cu+2 + 2e > Cu The left-hand side will be the cathode ... the right-hand side (0.050 M Cu+2) will be the anode … current will flow from right to left ...

A cell has on its left side a 0.20 M Cu+2 solution ... The right side has a M Cu+2 solution ... the compartments are connected by Cu electrodes and a salt bridge ... Unknown: Designate the cathode, anode, and direction of current ... Solution: Current will flow until the [Cu+2] is equal in both compartments ... this means that the concentration of Cu+2 in the left-hand side (0.02 M) must be reduced by … Cu+2 + 2e > Cu The left-hand side will be the cathode ... the right-hand side (0.050 M Cu+2) will be the anode … current will flow from right to left ... anode to cathode …

Electrochemistry Concentration Cells …

Electrochemistry Concentration Cells … produce a very small voltage ...

Electrochemistry Concentration Cells … produce a very small voltage ... nonstandard concentrations produce a cell voltage that is different from that at standard concentrations ...

Ecell(nonstandard) =/ E’cell
Electrochemistry Concentration Cells … produce a very small voltage ... nonstandard concentrations produce a cell voltage that is different from that at standard concentrations ... Ecell(nonstandard) =/ E’cell

Ecell(nonstandard) =/ E’cell
Electrochemistry Concentration Cells … produce a very small voltage ... nonstandard concentrations produce a cell voltage that is different from that at standard concentrations ... Ecell(nonstandard) =/ E’cell Nernst Equation ...

Electrochemistry Concentration Cells … produce a very small voltage ... nonstandard concentrations produce a cell voltage that is different from that at standard concentrations ... Ecell(nonstandard) =/ E’cell Nernst Equation ... Ecell = E’cell – (0.0592/n) log Q T = 25 ‘C)

Electrochemistry Concentration Cells … produce a very small voltage ... nonstandard concentrations produce a cell voltage that is different from that at standard concentrations ... Ecell(nonstandard) =/ E’cell Nernst Equation ... Ecell = E’cell – (0.0592/n) log Q T = 25 ‘C) Where ... Q = [products]o / [reactants]o

Electrochemistry Concentration Cells … produce a very small voltage ... nonstandard concentrations produce a cell voltage that is different from that at standard concentrations ... Ecell(nonstandard) =/ E’cell Nernst Equation ... Ecell = E’cell – (0.0592/n) log Q T = 25 ‘C) Where ... Q = [products]o / [reactants]o If [reactants]o > [products]o ....

Electrochemistry Concentration Cells … produce a very small voltage ... nonstandard concentrations produce a cell voltage that is different from that at standard concentrations ... Ecell(nonstandard) =/ E’cell Nernst Equation ... Ecell = E’cell – (0.0592/n) log Q T = 25 ‘C) Where ... Q = [products]o / [reactants]o If [reactants]o > [products]o .... log Q < 0 …

Electrochemistry Concentration Cells … produce a very small voltage ... nonstandard concentrations produce a cell voltage that is different from that at standard concentrations ... Ecell(nonstandard) =/ E’cell Nernst Equation ... Ecell = E’cell – (0.0592/n) log Q T = 25 ‘C) Where ... Q = [products]o / [reactants]o If [reactants]o > [products]o .... log Q < 0 … Ecell will be > E’cell ...

Electrochemistry Concentration Cells … produce a very small voltage ... nonstandard concentrations produce a cell voltage that is different from that at standard concentrations ... Ecell(nonstandard) =/ E’cell Nernst Equation ... Ecell = E’cell – (0.0592/n) log Q T = 25 ‘C) Where ... Q = [products]o / [reactants]o If [reactants]o > [products]o .... log Q < 0 … Ecell will be > E’cell ... consistent with Le Chatelier’s Principle and concentration cells ….

Electrochemistry Concentration Cells … produce a very small voltage ... nonstandard concentrations produce a cell voltage that is different from that at standard concentrations ... Ecell(nonstandard) =/ E’cell Nernst Equation ... Ecell = E’cell – (0.0592/n) log Q T = 25 ‘C) Where ... Q = [products]o / [reactants]o If [reactants]o > [products]o .... log Q < 0 … Ecell will be > E’cell ... consistent with Le Chatelier’s Principle and concentration cells …. When a battery is fully discharged ...

Electrochemistry Concentration Cells … produce a very small voltage ... nonstandard concentrations produce a cell voltage that is different from that at standard concentrations ... Ecell(nonstandard) =/ E’cell Nernst Equation ... Ecell = E’cell – (0.0592/n) log Q T = 25 ‘C) Where ... Q = [products]o / [reactants]o If [reactants]o > [products]o .... log Q < 0 … Ecell will be > E’cell ... consistent with Le Chatelier’s Principle and concentration cells …. When a battery is fully discharged ... it is at equilibrium ...

Electrochemistry Concentration Cells … produce a very small voltage ... nonstandard concentrations produce a cell voltage that is different from that at standard concentrations ... Ecell(nonstandard) =/ E’cell Nernst Equation ... Ecell = E’cell – (0.0592/n) log Q T = 25 ‘C) Where ... Q = [products]o / [reactants]o If [reactants]o > [products]o .... log Q < 0 … Ecell will be > E’cell ... consistent with Le Chatelier’s Principle and concentration cells …. When a battery is fully discharged ... it is at equilibrium ... Ecell =

Electrochemistry Nernst Equation ….

Electrochemistry Nernst Equation …. T = 25 ‘C

Electrochemistry Nernst Equation …. Known: @ T = 25 ‘C
Cd+2 + 2e > Cd E’ = V

Cd+2 + 2e > Cd E’ = V Pb+2 + 2e > Pb E’ = V

Cd+2 + 2e > Cd E’ = V Pb+2 + 2e > Pb E’ = V [Cd+2] = M

Cd+2 + 2e > Cd E’ = V Pb+2 + 2e > Pb E’ = V [Cd+2] = M [Pb+2] = M

Electrochemistry Nernst Equation …. Known: @ T = 25 ‘C Unknown:
Cd+2 + 2e > Cd E’ = V Ecell = ? V Pb+2 + 2e > Pb E’ = V [Cd+2] = M [Pb+2] = M

Cd+2 + 2e > Cd E’ = V Ecell = ? V Pb+2 + 2e > Pb E’ = V [Cd+2] = M [Pb+2] = M Solution: 1st: Ecell > 0 V to be Galvanic ...

Cd+2 + 2e > Cd E’ = V Ecell = ? V Pb+2 + 2e > Pb E’ = V [Cd+2] = M [Pb+2] = M Solution: 1st: Ecell > 0 V to be Galvanic ... Cd ----> Cd+2 + 2e-1 E’ = V

Cd+2 + 2e > Cd E’ = V Ecell = ? V Pb+2 + 2e > Pb E’ = V [Cd+2] = M [Pb+2] = M Solution: 1st: Ecell > 0 V to be Galvanic ... Cd ----> Cd+2 + 2e-1 E’ = V Pb+2 + 2e > Pb E’ = V

Cd+2 + 2e > Cd E’ = V Ecell = ? V Pb+2 + 2e > Pb E’ = V [Cd+2] = M [Pb+2] = M Solution: 1st: Ecell > 0 V to be Galvanic ... Cd ----> Cd+2 + 2e-1 E’ = V Pb+2 + 2e > Pb E’ = V Cd(s) + Pb+2(aq) ----> Cd+2(aq) + Pb(s)

Cd+2 + 2e > Cd E’ = V Ecell = ? V Pb+2 + 2e > Pb E’ = V [Cd+2] = M [Pb+2] = M Solution: 1st: Ecell > 0 V to be Galvanic ... Cd ----> Cd+2 + 2e-1 E’ = V Pb+2 + 2e > Pb E’ = V Cd(s) + Pb+2(aq) ----> Cd+2(aq) + Pb(s) E’cell = V

Cd+2 + 2e > Cd E’ = V Ecell = ? V Pb+2 + 2e > Pb E’ = V [Cd+2] = M [Pb+2] = M Solution: 1st: Ecell > 0 V to be Galvanic ... Cd ----> Cd+2 + 2e-1 E’ = V Pb+2 + 2e > Pb E’ = V Cd(s) + Pb+2(aq) ----> Cd+2(aq) + Pb(s) E’cell = V 2nd: Calculate nonstandard conditions ...

Cd+2 + 2e > Cd E’ = V Ecell = ? V Pb+2 + 2e > Pb E’ = V [Cd+2] = M [Pb+2] = M Solution: 1st: Ecell > 0 V to be Galvanic ... Cd ----> Cd+2 + 2e-1 E’ = V Pb+2 + 2e > Pb E’ = V Cd(s) + Pb+2(aq) ----> Cd+2(aq) + Pb(s) E’cell = V 2nd: Calculate nonstandard conditions ... Ecell = ? V||

Cd+2 + 2e > Cd E’ = V Ecell = ? V Pb+2 + 2e > Pb E’ = V [Cd+2] = M [Pb+2] = M Solution: 1st: Ecell > 0 V to be Galvanic ... Cd ----> Cd+2 + 2e-1 E’ = V Pb+2 + 2e > Pb E’ = V Cd(s) + Pb+2(aq) ----> Cd+2(aq) + Pb(s) E’cell = V 2nd: Calculate nonstandard conditions ... Ecell = ? V|| E’cell – (0.0592/n) log Q

Cd+2 + 2e > Cd E’ = V Ecell = ? V Pb+2 + 2e > Pb E’ = V [Cd+2] = M [Pb+2] = M Solution: 1st: Ecell > 0 V to be Galvanic ... Cd ----> Cd+2 + 2e-1 E’ = V Pb+2 + 2e > Pb E’ = V Cd(s) + Pb+2(aq) ----> Cd+2(aq) + Pb(s) E’cell = V 2nd: Calculate nonstandard conditions ... Ecell = ? V|| E’cell – (0.0592/n) log Q = | E’cell – (0.0592/n) log [[Cd+1]/[Pb+2]] =

Cd+2 + 2e > Cd E’ = V Ecell = ? V Pb+2 + 2e > Pb E’ = V [Cd+2] = M [Pb+2] = M Solution: 1st: Ecell > 0 V to be Galvanic ... Cd ----> Cd+2 + 2e-1 E’ = V Pb+2 + 2e > Pb E’ = V Cd(s) + Pb+2(aq) ----> Cd+2(aq) + Pb(s) E’cell = V 2nd: Calculate nonstandard conditions ... Ecell = ? V|| E’cell – (0.0592/n) log Q = | E’cell – (0.0592/n) log [[Cd+1]/[Pb+2]] = = V – (0.0592/2)log [[0.010][0.100]]

Cd+2 + 2e > Cd E’ = V Ecell = ? V Pb+2 + 2e > Pb E’ = V [Cd+2] = M [Pb+2] = M Solution: 1st: Ecell > 0 V to be Galvanic ... Cd ----> Cd+2 + 2e-1 E’ = V Pb+2 + 2e > Pb E’ = V Cd(s) + Pb+2(aq) ----> Cd+2(aq) + Pb(s) E’cell = V 2nd: Calculate nonstandard conditions ... Ecell = ? V|| E’cell – (0.0592/n) log Q = | E’cell – (0.0592/n) log [[Cd+1]/[Pb+2]] = = V – (0.0592/2)log [[0.010][0.100]] = – (0.0296) log (0.10) =

Cd+2 + 2e > Cd E’ = V Ecell = ? V Pb+2 + 2e > Pb E’ = V [Cd+2] = M [Pb+2] = M Solution: 1st: Ecell > 0 V to be Galvanic ... Cd ----> Cd+2 + 2e-1 E’ = V Pb+2 + 2e > Pb E’ = V Cd(s) + Pb+2(aq) ----> Cd+2(aq) + Pb(s) E’cell = V 2nd: Calculate nonstandard conditions ... Ecell = ? V|| E’cell – (0.0592/n) log Q = | E’cell – (0.0592/n) log [[Cd+1]/[Pb+2]] = = V – (0.0592/2)log [[0.010][0.100]] = – (0.0296) log (0.10) = +0.30V

Cd+2 + 2e > Cd E’ = V Ecell = ? V Pb+2 + 2e > Pb E’ = V [Cd+2] = M [Pb+2] = M Solution: 1st: Ecell > 0 V to be Galvanic ... Cd ----> Cd+2 + 2e-1 E’ = V Pb+2 + 2e > Pb E’ = V Cd(s) + Pb+2(aq) ----> Cd+2(aq) + Pb(s) E’cell = V 2nd: Calculate nonstandard conditions ... Ecell = ? V|| E’cell – (0.0592/n) log Q = | E’cell – (0.0592/n) log [[Cd+1]/[Pb+2]] = = V – (0.0592/2)log [[0.010][0.100]] = – (0.0296) log (0.10)

Electrochemistry Nernst Equation …. T = 25 ‘C

EQ1: FeO H+1 + 3e > Fe+3 + 4H2O E’ = V

EQ1: FeO H+1 + 3e > Fe+3 + 4H2O E’ = V EQ2: O2+ 4H+1 + 4e > 2 H2O E’ = V

EQ1: FeO H+1 + 3e > Fe+3 + 4H2O E’ = V EQ2: O2+ 4H+1 + 4e > 2 H2O E’ = V [FeO4-2] = 2.0 x 10-3 M

EQ1: FeO H+1 + 3e > Fe+3 + 4H2O E’ = V EQ2: O2+ 4H+1 + 4e > 2 H2O E’ = V [FeO4-2] = 2.0 x 10-3 M [Fe+3] = 1.0 x 10-3 M

EQ1: FeO H+1 + 3e > Fe+3 + 4H2O E’ = V EQ2: O2+ 4H+1 + 4e > 2 H2O E’ = V [FeO4-2] = 2.0 x 10-3 M [O2] = 1.0 x 10-5 atm [Fe+3] = 1.0 x 10-3 M

EQ1: FeO H+1 + 3e > Fe+3 + 4H2O E’ = V EQ2: O2+ 4H+1 + 4e > 2 H2O E’ = V [FeO4-2] = 2.0 x 10-3 M [O2] = 1.0 x 10-5 atm [Fe+3] = 1.0 x 10-3 M pH = 5.2

EQ1: FeO H+1 + 3e > Fe+3 + 4H2O E’ = V E’cell = ? V EQ2: O2+ 4H+1 + 4e > 2 H2O E’ = V [FeO4-2] = 2.0 x 10-3 M [O2] = 1.0 x 10-5 atm [Fe+3] = 1.0 x 10-3 M pH = 5.2

EQ1: FeO H+1 + 3e > Fe+3 + 4H2O E’ = V E’cell = ? V EQ2: O2+ 4H+1 + 4e > 2 H2O E’ = V [FeO4-2] = 2.0 x 10-3 M [O2] = 1.0 x 10-5 atm [Fe+3] = 1.0 x 10-3 M pH = 5.2 Solution:

EQ1: FeO H+1 + 3e > Fe+3 + 4H2O E’ = V E’cell = ? V EQ2: O2+ 4H+1 + 4e > 2 H2O E’ = V [FeO4-2] = 2.0 x 10-3 M [O2] = 1.0 x 10-5 atm [Fe+3] = 1.0 x 10-3 M pH = 5.2 Solution: EQ1 has higher reduction potential ...

EQ1: FeO H+1 + 3e > Fe+3 + 4H2O E’ = V E’cell = ? V EQ2: O2+ 4H+1 + 4e > 2 H2O E’ = V [FeO4-2] = 2.0 x 10-3 M [O2] = 1.0 x 10-5 atm [Fe+3] = 1.0 x 10-3 M pH = 5.2 Solution: EQ1 has higher reduction potential ... It wil be reduced ...

EQ1: FeO H+1 + 3e > Fe+3 + 4H2O E’ = V E’cell = ? V EQ2: O2+ 4H+1 + 4e > 2 H2O E’ = V [FeO4-2] = 2.0 x 10-3 M [O2] = 1.0 x 10-5 atm [Fe+3] = 1.0 x 10-3 M pH = 5.2 Solution: EQ1 has higher reduction potential ... It wil be reduced ... EQ2 will be oxidized ...

EQ1: FeO H+1 + 3e > Fe+3 + 4H2O E’ = V E’cell = ? V EQ2: O2+ 4H+1 + 4e > 2 H2O E’ = V [FeO4-2] = 2.0 x 10-3 M [O2] = 1.0 x 10-5 atm [Fe+3] = 1.0 x 10-3 M pH = 5.2 Solution: EQ1 has higher reduction potential ... It wil be reduced ... EQ2 will be oxidized ... EQ1 is multiplied by 4 ...

EQ1: FeO H+1 + 3e > Fe+3 + 4H2O E’ = V E’cell = ? V EQ2: O2+ 4H+1 + 4e > 2 H2O E’ = V [FeO4-2] = 2.0 x 10-3 M [O2] = 1.0 x 10-5 atm [Fe+3] = 1.0 x 10-3 M pH = 5.2 Solution: EQ1 has higher reduction potential ... It wil be reduced ... EQ2 will be oxidized ... EQ1 is multiplied by EQ2 is multiplied by

EQ1: FeO H+1 + 3e > Fe+3 + 4H2O E’ = V E’cell = ? V EQ2: O2+ 4H+1 + 4e > 2 H2O E’ = V [FeO4-2] = 2.0 x 10-3 M [O2] = 1.0 x 10-5 atm [Fe+3] = 1.0 x 10-3 M pH = 5.2 Solution: EQ1 has higher reduction potential ... It wil be reduced ... EQ1 will be oxidized ... EQ1 is multiplied by EQ2 is multiplied by 1st: Calculate E’cell .... 4FeO H e > 4Fe H2O E’ = V

EQ1: FeO H+1 + 3e > Fe+3 + 4H2O E’ = V E’cell = ? V EQ2: O2+ 4H+1 + 4e > 2 H2O E’ = V [FeO4-2] = 2.0 x 10-3 M [O2] = 1.0 x 10-5 atm [Fe+3] = 1.0 x 10-3 M pH = 5.2 Solution: EQ1 has higher reduction potential ... It wil be reduced ... EQ1 will be oxidized ... EQ1 is multiplied by EQ2 is multiplied by 1st: Calculate E’cell .... 4FeO H e > 4Fe H2O E’ = V 6 H2O ----> 3O2+ 12H e-1 E’ = V

EQ1: FeO H+1 + 3e > Fe+3 + 4H2O E’ = V E’cell = ? V EQ2: O2+ 4H+1 + 4e > 2 H2O E’ = V [FeO4-2] = 2.0 x 10-3 M [O2] = 1.0 x 10-5 atm [Fe+3] = 1.0 x 10-3 M pH = 5.2 Solution: EQ1 has higher reduction potential ... It wil be reduced ... EQ1 will be oxidized ... EQ1 is multiplied by EQ2 is multiplied by 1st: Calculate E’cell .... 4FeO H e > 4Fe H2O E’ = V 6 H2O ----> 3O2+ 12H e-1 E’ = V 4FeO4-2(aq) + 20H+1(aq) ----> 4Fe+3 + 3O2(g) + 10H2O(l)

EQ1: FeO H+1 + 3e > Fe+3 + 4H2O E’ = V E’cell = ? V EQ2: O2+ 4H+1 + 4e > 2 H2O E’ = V [FeO4-2] = 2.0 x 10-3 M [O2] = 1.0 x 10-5 atm [Fe+3] = 1.0 x 10-3 M pH = 5.2 Solution: EQ1 has higher reduction potential ... It wil be reduced ... EQ1 will be oxidized ... EQ1 is multiplied by EQ2 is multiplied by 1st: Calculate E’cell .... 4FeO H e > 4Fe H2O E’ = V 6 H2O ----> 3O2+ 12H e-1 E’ = V 4FeO4-2(aq) + 20H+1(aq) ----> 4Fe+3 + 3O2(g) + 10H2O(l) E’cell = V

EQ1: FeO H+1 + 3e > Fe+3 + 4H2O E’ = V E’cell = ? V EQ2: O2+ 4H+1 + 4e > 2 H2O E’ = V [FeO4-2] = 2.0 x 10-3 M [O2] = 1.0 x 10-5 atm [Fe+3] = 1.0 x 10-3 M pH = 5.2 Solution: EQ1 has higher reduction potential ... It wil be reduced ... EQ1 will be oxidized ... EQ1 is multiplied by EQ2 is multiplied by 1st: Calculate E’cell .... 4FeO H e > 4Fe H2O E’ = V 6 H2O ----> 3O2+ 12H e-1 E’ = V 4FeO4-2(aq) + 20H+1(aq) ----> 4Fe+3 + 3O2(g) + 10H2O(l) E’cell = V 2nd: Calculate the E’cell ...

EQ1: FeO H+1 + 3e > Fe+3 + 4H2O E’ = V E’cell = ? V EQ2: O2+ 4H+1 + 4e > 2 H2O E’ = V [FeO4-2] = 2.0 x 10-3 M [O2] = 1.0 x 10-5 atm [Fe+3] = 1.0 x 10-3 M pH = 5.2 Solution: EQ1 has higher reduction potential ... It wil be reduced ... EQ1 will be oxidized ... EQ1 is multiplied by EQ2 is multiplied by 1st: Calculate E’cell .... 4FeO H e > 4Fe H2O E’ = V 6 H2O ----> 3O2+ 12H e-1 E’ = V 4FeO4-2(aq) + 20H+1(aq) ----> 4Fe+3 + 3O2(g) + 10H2O(l) E’cell = V 2nd: Calculate the E’cell ... E = ? V ||

EQ1: FeO H+1 + 3e > Fe+3 + 4H2O E’ = V E’cell = ? V EQ2: O2+ 4H+1 + 4e > 2 H2O E’ = V [FeO4-2] = 2.0 x 10-3 M [O2] = 1.0 x 10-5 atm [Fe+3] = 1.0 x 10-3 M pH = 5.2 Solution: EQ1 has higher reduction potential ... It wil be reduced ... EQ1 will be oxidized ... EQ1 is multiplied by EQ2 is multiplied by 1st: Calculate E’cell .... 4FeO H e > 4Fe H2O E’ = V 6 H2O ----> 3O2+ 12H e-1 E’ = V 4FeO4-2(aq) + 20H+1(aq) ----> 4Fe+3 + 3O2(g) + 10H2O(l) E’cell = V 2nd: Calculate the E’cell ... E = ? V || E’cell – (0.059/n) log Q =

EQ1: FeO H+1 + 3e > Fe+3 + 4H2O E’ = V E’cell = ? V EQ2: O2+ 4H+1 + 4e > 2 H2O E’ = V [FeO4-2] = 2.0 x 10-3 M [O2] = 1.0 x 10-5 atm [Fe+3] = 1.0 x 10-3 M pH = 5.2 Solution: EQ1 has higher reduction potential ... It wil be reduced ... EQ1 will be oxidized ... EQ1 is multiplied by EQ2 is multiplied by 1st: Calculate E’cell .... 4FeO H e > 4Fe H2O E’ = V 6 H2O ----> 3O2+ 12H e-1 E’ = V 4FeO4-2(aq) + 20H+1(aq) ----> 4Fe+3 + 3O2(g) + 10H2O(l) E’cell = V 2nd: Calculate the E’cell ... E = ? V || E’cell – (0.059/n) log Q = E’cell – (0.059/n) log [[Fe+3]4[O2]3/[FeO4-2]4[H+]20]

EQ1: FeO H+1 + 3e > Fe+3 + 4H2O E’ = V E’cell = ? V EQ2: O2+ 4H+1 + 4e > 2 H2O E’ = V [FeO4-2] = 2.0 x 10-3 M [O2] = 1.0 x 10-5 atm [Fe+3] = 1.0 x 10-3 M pH = 5.2 Solution: EQ1 has higher reduction potential ... It wil be reduced ... EQ1 will be oxidized ... EQ1 is multiplied by EQ2 is multiplied by 1st: Calculate E’cell .... 4FeO H e > 4Fe H2O E’ = V 6 H2O ----> 3O2+ 12H e-1 E’ = V 4FeO4-2(aq) + 20H+1(aq) ----> 4Fe+3 + 3O2(g) + 10H2O(l) E’cell = V 2nd: Calculate the E’cell ... E = ? V || E’cell – (0.059/n) log Q = E’cell – (0.059/n) log [[Fe+3]4[O2]3/[FeO4-2][H+]20] = V – (0.059/12) log [[1.0x10-3]4[1.0x10-5]3 /[2.0x10-3]4[6.31x10-6]20

EQ1: FeO H+1 + 3e > Fe+3 + 4H2O E’ = V E’cell = ? V EQ2: O2+ 4H+1 + 4e > 2 H2O E’ = V [FeO4-2] = 2.0 x 10-3 M [O2] = 1.0 x 10-5 atm [Fe+3] = 1.0 x 10-3 M pH = 5.2 Solution: EQ1 has higher reduction potential ... It wil be reduced ... EQ1 will be oxidized ... EQ1 is multiplied by EQ2 is multiplied by 1st: Calculate E’cell .... 4FeO H e > 4Fe H2O E’ = V 6 H2O ----> 3O2+ 12H e-1 E’ = V 4FeO4-2(aq) + 20H+1(aq) ----> 4Fe+3 + 3O2(g) + 10H2O(l) E’cell = V 2nd: Calculate the E’cell ... E = ? V || E’cell – (0.059/n) log Q = E’cell – (0.059/n) log [[Fe+3]4[O2]3/[FeO4-2][H+]20] = V – (0.059/12) log [[1.0x10-3]4[1.0x10-5]3 /[2.0x10-3]4[6.31x10-6]20 = V – (4.9 x 10-3) log (6.3 x 1087)

EQ1: FeO H+1 + 3e > Fe+3 + 4H2O E’ = V E’cell = ? V EQ2: O2+ 4H+1 + 4e > 2 H2O E’ = V [FeO4-2] = 2.0 x 10-3 M [O2] = 1.0 x 10-5 atm [Fe+3] = 1.0 x 10-3 M pH = 5.2 Solution: EQ1 has higher reduction potential ... It wil be reduced ... EQ1 will be oxidized ... EQ1 is multiplied by EQ2 is multiplied by 1st: Calculate E’cell .... 4FeO H e > 4Fe H2O E’ = V 6 H2O ----> 3O2+ 12H e-1 E’ = V 4FeO4-2(aq) + 20H+1(aq) ----> 4Fe+3 + 3O2(g) + 10H2O(l) E’cell = V 2nd: Calculate the E’cell ... E = ? V || E’cell – (0.059/n) log Q = E’cell – (0.059/n) log [[Fe+3]4[O2]3/[FeO4-2][H+]20] = V – (0.059/12) log [[1.0x10-3]4[1.0x10-5]3 /[2.0x10-3]4[6.31x10-6]20 = V – (4.9 x 10-3) log (6.3 x 1087) = V – 0.43 V =

EQ1: FeO H+1 + 3e > Fe+3 + 4H2O E’ = V E’cell = ? V EQ2: O2+ 4H+1 + 4e > 2 H2O E’ = V [FeO4-2] = 2.0 x 10-3 M [O2] = 1.0 x 10-5 atm [Fe+3] = 1.0 x 10-3 M pH = 5.2 Solution: EQ1 has higher reduction potential ... It wil be reduced ... EQ1 will be oxidized ... EQ1 is multiplied by EQ2 is multiplied by 1st: Calculate E’cell .... 4FeO H e > 4Fe H2O E’ = V 6 H2O ----> 3O2+ 12H e-1 E’ = V 4FeO4-2(aq) + 20H+1(aq) ----> 4Fe+3 + 3O2(g) + 10H2O(l) E’cell = V 2nd: Calculate the E’cell ... E = ? V || E’cell – (0.059/n) log Q = E’cell – (0.059/n) log [[Fe+3]4[O2]3/[FeO4-2][H+]20] = V – (0.059/12) log [[1.0x10-3]4[1.0x10-5]3 /[2.0x10-3]4[6.31x10-6]20 = V – (4.9 x 10-3) log (6.3 x 1087) = V – 0.43 V = 0.54 V

Electrochemistry Nernst Equation ….

Electrochemistry Nernst Equation Equilibrium ...

Electrochemistry Nernst Equation Equilibrium ... E & ∆G = Thus ...

Electrochemistry Nernst Equation Equilibrium ... E & ∆G = Thus ... log K = nE’ / @ 25 ‘C

Electrochemistry Equilibrium Constants and Cell Potential …

Electrochemistry.

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