Electrochemistry

Oxidation – Reduction Reactions Consider the reaction of Copper wire and AgNO3(aq) AgNO3(aq) Cu(s) Ag(s)

Oxidation – Reduction Reactions If you leave the reaction a long time the solution goes blue! The blue is due to Cu2+(aq)

Oxidation-Reduction Reactions So when we mix Ag+(aq) with Cu(s) we get Ag(s) and Cu2+(aq) Ag+(aq) + 1e-  Ag(s) Cu(s)  Cu2+(aq) + 2e- The electrons gained by Ag+ must come from the Cu Can’t have reduction without oxidation (redox) Each Cu can reduce 2 Ag+ 2Ag+(aq) + 2e-  2Ag(s) 2Ag+(aq) + 2e- + Cu(s) 2Ag(s)+ Cu2+(aq) + 2e- gain electrons = reduction lose electrons = oxidation

Redox Cu/Ag E Cu electron flow Ag+ Ag+

Redox Cu/Ag E Cu2+ ΔE = e.Ecell Ag Ag e = charge on an electron Ecell = Voltage in a electrochemical cell If we could separate the two reactions we could use the energy gained by the e to do work Redox Cu/Ag E Cu2+ ΔE = e.Ecell Ag Ag

The maximum amount of energy available to do work is a definition of the free energy ΔG Redox Cu/Ag cell voltage Free energy for redox rxn E Cu2+ # electrons in redox rxn F = Faraday’s constant = 96485 C/mol Ag Ag

ΔEcell: the cell potential In a redox reaction electrons are transferred to a more stable state Most of the free energy of the reaction is due to these electrons Can we access this energy? Yes, by conducting the two half reactions in separate cells 2Ag+(aq) + 2e-  2Ag(s) Cu(s)  Cu2+(aq) + 2e- This is called the Voltaic electrochemical cell The cell potential can be measured in such a cell with a voltmeter

Where oxidation happens Voltaic Cell Cu/Ag Ecell= 0.460 V DGcell= -2F x 0.460 V cations to the cathode electrons to the cathode Anions to the anode Where oxidation happens Where reduction happens Cu(s)  Cu2+(aq) + 2e- 2Ag+(aq) + 2e-  2Ag(s)

Electric Current Flowing Directly Between Atoms

Electrochemistry and Cells electrochemistry is the study of redox reactions that produce or require an electric current the conversion between chemical energy and electrical energy is carried out in an electrochemical cell spontaneous redox reactions take place in a voltaic cell (galvanic cell) nonspontaneous redox reactions can be made to occur in an electrolytic cell by the addition of electrical energy

Where oxidation happens Voltaic Cell as a Battery: Electric Current Flowing Indirectly Between Atoms Electrons do work here electrons to the cathode Anions to the anode cations to the cathode Where oxidation happens Where reduction happens

Electrodes Anode (donates electrons to the cathode) electrode where oxidation occurs In a Galvanic cell it is the –ve terminal and in an electrolytic cell it is the +ve terminal anions attracted to it connected to positive end of battery in electrolytic cell loses weight in electrolytic cell Cathode (attracts electrons from the anode) electrode where reduction occurs In a Galvanic cell it is the +ve terminal and in an electrolytic cell it is the -ve terminal cations attracted to it connected to negative end of battery in electrolytic cell gains weight in electrolytic cell electrode where plating takes place in electroplating

Current and Voltage the number of electrons that flow through the system per second is the current unit = Ampere (A) 1 A of current = 1 Coulomb (C) of charge flowing by each second 1 A = 6.242 x 1018 electrons/second Electrode surface area dictates the number of electrons that can flow the difference in potential energy between the reactants and products per Coulomb of charge is called the Voltage unit = Volt (V) 1 V of force = 1 J of energy/Coulomb of charge the voltage is what drives electrons through the external circuit amount of force pushing the electrons through the wire is called the electromotive force, emf, (also units of V) the cell potential under standard conditions is called the standard emf, E°cell 25°C, 1 atm for gases, 1 M concentration of solution sum of the cell potentials for the half-reactions

Cell Potential the difference in potential energy between the anode the cathode in a voltaic cell is called the cell potential the cell potential depends on the relative ease with which the oxidizing agent is reduced at the cathode and the reducing agent is oxidized at the anode the cell potential under standard conditions is called the standard emf, E°cell 25°C, 1 atm for gases, 1 M concentration of solution sum of the cell potentials for the half-reactions

Cell Notation shorthand description of Voltaic cell electrode | electrolyte || electrolyte | electrode oxidation half-cell on left (anode), reduction half-cell on the right (cathode) single | = phase barrier if multiple electrolytes in same phase, a comma is used rather than | often use an inert electrode double line || = salt bridge

Fe(s) | Fe2+(aq) || MnO4-(aq), Mn2+(aq), H+(aq) | Pt(s)

Practice - Sketch and Label the Voltaic Cell Fe(s) | Fe2+(aq) || Pb2+(aq) | Pb(s) Write the Half-Reactions and Overall Reaction, and Determine the Cell Potential under Standard Conditions. Fe(s) | Fe2+(aq) || Pb2+(aq) | Pb(s) anode cathode

ox: Fe(s)  Fe2+(aq) + 2 e− Eox = +0.45 V red: Pb2+(aq) + 2 e−  Pb(s) Ered = −0.13 V tot: Pb2+(aq) + Fe(s)  Fe2+(aq) + Pb(s) Ecell = +0.32 V

Measuring the Tendency to Reduce: Standard Reduction Potential when two half-cells are connected, one side will oxidize and one side will reduce, but which will do what? clearly the electrons will flow so that the half-reaction with the stronger tendency to reduce will reduce, making the other half-cell reaction oxidize we cannot measure the absolute tendency of a half-reaction to reduce, we can only measure it relative to another half-reaction To measure the tendency to reduce we measure the E (the voltage) in a voltaic cell where one of the electrodes is a standard hydrogen electrode (see right) We assign a potential difference = 0.0 V to the following reaction 2H+(aq) +2e-  H2(g) Eored = 0.0V E°cell = Eoox + Eored = Eoox + 0.0 V = Eoox In so doing scientists can obtain standard reduction potentials as follows

The measured voltage can then be placed in a table like this one, where we write the voltage for the reduction reaction

Half-Cell Potentials ΔE°cell = E°oxidation + E°reduction SHE reduction potential is defined to be exactly 0 V half-reactions with a stronger tendency toward reduction than the SHE have a + value for E°red half-reactions with a stronger tendency toward oxidation than the SHE have a - value for E°red ΔE°cell = E°oxidation + E°reduction E°oxidation = -E°reduction when adding E° values for the half-cells, do not multiply the half-cell E° values (voltage is the energy per unit charge), even if you need to multiply the half-reactions to balance the equation ΔGocell=-nFΔE°cell Since ΔGocell < 0 to be spontaneous ΔE°cell > 0 for a redox reaction to be spontaneous if ΔE°cell < 0 the reaction will not be spontaneous

red: NO3−(aq) + 4 H+(aq) + 3 e−  NO(g) + 2 H2O(l) Calculate E°cell for the reaction at 25°C Al(s) + NO3−(aq) + 4 H+(aq)  Al3+(aq) + NO(g) + 2 H2O(l) Separate the reaction into the oxidation and reduction half-reactions ox: Al(s)  Al3+(aq) + 3 e− red: NO3−(aq) + 4 H+(aq) + 3 e−  NO(g) + 2 H2O(l) find the Eo for each half-reaction from the standard reduction table. If you need the oxidation potential of one Eoox=-Eored and sum to get Eocell Eoox = −Eored = +1.66 V Eored = +0.96 V Eocell = (+1.66 V) + (+0.96 V) = +2.62 V reverse

red: Mg2+(aq) + 2 e−  Mg(s) Predict if the following reaction is spontaneous under standard conditions Fe(s) + Mg2+(aq)  Fe2+(aq) + Mg(s) Separate the reaction into the oxidation and reduction half-reactions ox: Fe(s)  Fe2+(aq) + 2 e− red: Mg2+(aq) + 2 e−  Mg(s) look up the relative positions of the reduction half-reactions red: Mg2+(aq) + 2 e−  Mg(s) Ered = -2.37V ox: Fe(S)  Fe2+(aq) + 2 e− Eox =+0.45V Ecell = Eox + Ered = 0.45V – 2.37V = -1.92V Since Ecell < 0 the cell reaction is not spontaneous as written reverse

Keeping track of electrons Redox reactions involve the transfer of electrons from the donor to the acceptor The donor loses electrons and is oxidized The acceptor acquires electrons and is reduced To know if a reaction is a redox reaction we need a way to keep track of how many valence electrons each element has We define the oxidation state of an element to be +n (n integer) if it has n less electrons than it does as the free atom, and –p if it has p more electrons than it does in the free atom

Redox Reaction one or more elements change oxidation number all single displacement (A+BX--> AX+B), and combustion, some synthesis and decomposition always have both oxidation and reduction split reaction into oxidation half-reaction and a reduction half-reaction aka electron transfer reactions half-reactions include electrons oxidizing agent is reactant molecule that causes oxidation contains element reduced reducing agent is reactant molecule that causes reduction contains the element oxidized

Oxidation & Reduction oxidation is the process that occurs when oxidation number of an element increases element loses electrons compound adds oxygen compound loses hydrogen half-reaction has electrons as products reduction is the process that occurs when oxidation number of an element decreases element gains electrons compound loses oxygen compound gains hydrogen half-reactions have electrons as reactants

Assigning Oxidation Numbers The sum of the oxidation numbers Q of all the atoms in a compound/ion up to the charge on the compound/ion. This means …. free elements have an oxidation state = 0 QNa = 0 and QCl = 0 in 2 Na(s) + Cl2(g) monatomic ions have an oxidation state equal to their charge QNa = +1 and QCl = -1 in NaCl The sum of the oxidation numbers of all the atoms in a neutral compound is 0 the sum of the oxidation numbers of all the atoms in a polyatomic ion equals the charge on the ion (a) Group I metals have an oxidation state of +1 in all their compounds QNa = +1 in NaCl (b) Group II metals have an oxidation state of +2 in all their compounds QMg = +2 in MgCl2 F has an oxidation number QF = -1 H has an oxidation number QH = +1 O has an oxidation number QO = -2

Oxidation and Reduction oxidation occurs when an atom’s oxidation state increases during a reaction reduction occurs when an atom’s oxidation state decreases during a reaction rule 4 rule 1 rule 5 rule 4 CH4 + 2 O2 → CO2 + 2 H2O -4 +1 0 +4 –2 +1 -2 oxidation reduction

Oxidation–Reduction 2 Na(s) + Cl2(g) → 2 Na+Cl–(s) oxidation and reduction must occur simultaneously if an atom loses electrons another atom must take them the reactant that reduces an element in another reactant is called the reducing agent the reducing agent contains the element that is oxidized the reactant that oxidizes an element in another reactant is called the oxidizing agent the oxidizing agent contains the element that is reduced 2 Na(s) + Cl2(g) → 2 Na+Cl–(s) Na is oxidized, Cl is reduced Na is the reducing agent, Cl2 is the oxidizing agent

Identify the Oxidizing and Reducing Agents in Each of the Following 3 H2S + 2 NO3– + 2 H+  3 S + 2 NO + 4 H2O MnO2 + 4 HBr  MnBr2 + Br2 + 2 H2O

Identify the Oxidizing and Reducing Agents in Each of the Following red ag ox ag 3 H2S + 2 NO3– + 2 H+  3 S + 2 NO + 4 H2O MnO2 + 4 HBr  MnBr2 + Br2 + 2 H2O +1 -2 +5 -2 +1 0 +2 -2 +1 -2 reduction oxidation ox ag red ag +4 -2 +1 -1 +2 -1 0 +1 -2 oxidation reduction

Common Oxidizing Agents

Common Reducing Agents

Balancing Redox Reactions assign oxidation numbers determine element oxidized and element reduced write ox. & red. half-reactions, including electrons ox. electrons on right, red. electrons on left of arrow balance half-reactions by mass first balance elements other than H and O add H2O where need O add H+1 where need H neutralize H+ with OH- in base balance half-reactions by charge balance charge by adjusting electrons balance electrons between half-reactions add half-reactions check

Balance the equation: I-(aq) + MnO4-(aq)  I2(aq) + MnO2(s) in basic solution Assign Oxidation States Separate into half-reactions ox: I-(aq)  I2(aq) red: MnO4-(aq)  MnO2(s) Assign Oxidation States I-(aq) + MnO4-(aq)  I2(aq) + MnO2(s) Separate into half-reactions ox: red:

Ex 18.3 – Balance the equation: I-(aq) + MnO4-(aq)  I2(aq) + MnO2(s) in basic solution Balance half-reactions by mass in base, neutralize the H+ with OH- ox: 2 I-(aq)  I2(aq) red: 4 H+(aq) + MnO4-(aq)  MnO2(s) + 2 H2O(l) 4 H+(aq) + 4 OH-(aq) + MnO4-(aq)  MnO2(s) + 2 H2O(l) + 4 OH-(aq) 4 H2O(aq) + MnO4-(aq)  MnO2(s) + 2 H2O(l) + 4 OH-(aq) MnO4-(aq) + 2 H2O(l)  MnO2(s) + 4 OH-(aq) Balance half-reactions by mass then H by adding H+ ox: 2 I-(aq)  I2(aq) red: 4 H+(aq) + MnO4-(aq)  MnO2(s) + 2 H2O(l) Balance half-reactions by mass ox: 2 I-(aq)  I2(aq) red: MnO4-(aq)  MnO2(s) Balance half-reactions by mass ox: I-(aq)  I2(aq) red: MnO4-(aq)  MnO2(s) Balance half-reactions by mass then O by adding H2O ox: 2 I-(aq)  I2(aq) red: MnO4-(aq)  MnO2(s) + 2 H2O(l)

Ex 18.3 – Balance the equation: I-(aq) + MnO4-(aq)  I2(aq) + MnO2(s) in basic solution Balance Half-reactions by charge ox: 2 I-(aq)  I2(aq) + 2 e- red: MnO4-(aq) + 2 H2O(l) + 3 e-  MnO2(s) + 4 OH-(aq) Balance electrons between half-reactions ox: { 2 I-(aq)  I2(aq) + 2 e- } x 3 red: {MnO4-(aq) + 2 H2O(l) + 3 e-  MnO2(s) + 4 OH-(aq) } x 2 ox: 6 I-(aq)  3 I2(aq) + 6 e- red: 2 MnO4-(aq) + 4 H2O(l) + 6 e-  2 MnO2(s) + 8 OH-(aq)

Balance the equation: I-(aq) + MnO4-(aq)  I2(aq) + MnO2(s) in basic solution Add the Half-reactions ox: 6 I-(aq)  3 I2(aq) + 6 e- red: 2 MnO4-(aq) + 4 H2O(l) + 6 e-  2 MnO2(s) + 8 OH-(aq) tot: 6 I-(aq)+ 2 MnO4-(aq) + 4 H2O(l)  3 I2(aq)+ 2 MnO2(s) + 8 OH-(aq) Check Reactant Count Element Product 6 I 2 Mn 12 O 8 H 8- charge

Practice - Balance the Equation H2O2 + KI + H2SO4  K2SO4 + I2 + H2O

Practice - Balance the Equation H2O2 + KI + H2SO4  K2SO4 + I2 + H2O ox: 2 I-1  I2 + 2e-1 red: H2O2 + 2e-1 + 2 H+  2 H2O tot 2 I-1 + H2O2 + 2 H+  I2 + 2 H2O H2O2 + 2 KI + H2SO4  K2SO4 + I2 + 2 H2O

Practice - Balance the Equation ClO3-1 + Cl-1  Cl2 (in acid)

Practice - Balance the Equation ClO3-1 + Cl-1  Cl2 (in acid) +5 -2 -1 0 oxidation reduction ox: 2 Cl-1  Cl2 + 2 e-1 } x 5 red: 2 ClO3-1 + 10 e-1 + 12 H+  Cl2 + 6 H2O} x 1 tot 10 Cl-1 + 2 ClO3-1 + 12 H+  6 Cl2 + 6 H2O 1 ClO3-1 + 5 Cl-1 + 6 H+1  3 Cl2 + 3 H2O

Standard Reduction Potential a half-reaction with a strong tendency to occur has a large half-cell potential when two half-cells are connected, the electrons will flow so that the half-reaction with the stronger tendency will occur we cannot measure the absolute tendency of a half-reaction, we can only measure it relative to another half-reaction we select as a standard half-reaction, the reduction of H+ to H2 under standard conditions, which we assign a potential difference = 0 V standard hydrogen electrode, SHE

Half-Cell Potentials SHE reduction potential is defined to be exactly 0 V half-reactions with a stronger tendency toward reduction than the SHE have a + value for E°red half-reactions with a stronger tendency toward oxidation than the SHE have a - value for E°red E°cell = E°oxidation + E°reduction E°oxidation = -E°reduction when adding E° values for the half-cells, do not multiply the half-cell E° values, even if you need to multiply the half-reactions to balance the equation

red: NO3−(aq) + 4 H+(aq) + 3 e−  NO(g) + 2 H2O(l) Calculate E°cell for the reaction at 25°C Al(s) + NO3−(aq) + 4 H+(aq)  Al3+(aq) + NO(g) + 2 H2O(l) Separate the reaction into the oxidation and reduction half-reactions ox: Al(s)  Al3+(aq) + 3 e− red: NO3−(aq) + 4 H+(aq) + 3 e−  NO(g) + 2 H2O(l) find the Eo for each half-reaction and sum to get Eocell Eoox = −Eored = +1.66 v Eored = +0.96 v Eocell = (+1.66 v) + (+0.96 v) = +2.62 v

the reaction is spontaneous in the reverse direction tot: Mg(s) + Fe2+(aq)  Mg2+(aq) + Fe(s) ox: Mg(s)  Mg2+(aq) + 2 e− red: Fe2+(aq) + 2 e−  Fe(s) sketch the cell and label the parts – oxidation occurs at the anode; electrons flow from anode to cathode

Predicting Whether a Metal Will Dissolve in an Acid acids dissolve in metals if the reduction of the metal ion is easier than the reduction of H+(aq) metals whose ion reduction reaction lies below H+ reduction on the table will dissolve in acid

Electrochemical Cell Summary The cell consists of two half cells, one where oxidation occurs, and one where reduction occurs A device for either harnessing the electrical power of a redox reaction (battery), or a device for using electricity to induce non-spontaneous redox reactions (electrolytic cell) The anode and cathode are connect by a wire through which electrons can flow, and the electrolytes in each half cell are connected by a salt bridge through which cations and anions can flow The electrode which loses electrons we call the anode, and the electrode that gains electrons we call the cathode Each half cell consists of an electrode and an electrolyte e- anode Zn (s)--> Zn2+ (aq)+ 2e- salt bridge cathode Cu2+(aq)+ 2e- --> Cu(s)

Electrochemical Cell Summary The cell potential can be calculated knowing the standard reduction potentials. These can be used to find Eored for the reaction at the cathode, and Eoox (= - Eored). Then Eocell = Eoox+ Eored This manifests as a potential difference Ecell, across the electrodes. Where -qEcell is the change in potential energy when an amount of negative charge (-q) passes from the anode to the cathode The cell potential is related to the free energy of the reaction according to the relation Gcell = -nFEcell The differing stability of reactants, (Zn(s), Cu2+(aq)), and products (Zn2+, and Cu(s)), creates a potential energy gradient through which the charges migrate (from high energy to low). Zn(s) --> Zn2+(aq) + 2e- Eox= 0.76V Zn2+(aq) + 2e- --> Zn(s) -0.76V Cu2+(aq) + 2e- --> Cu(s) Ered=0.34V Ecell = 0.76V+0.34V = 1.1V Ecell=1.1 V e- anode Zn (s)--> Zn2+ (aq)+ 2e- salt bridge cathode Cu2+(aq)+ 2e- --> Cu(s)

The Relationship between ΔGo and Eocell Remember ΔGo is the maximum work the system can do Eocell is the cell potential energy (standard emf) per unit of charge Since the potential energy is the maximum amount of work that can be done on the surrounding we can write The total charge q = nF, n = number moles of electrons in the balanced equation and F is Faraday’s constant. Then we can write Δ 𝐺 𝑐𝑒𝑙𝑙 =Δ 𝐺 𝑐𝑒𝑙𝑙 𝑜 +𝑅𝑇𝑙𝑛𝑄 −𝑛𝐹Δ 𝐸 𝑐𝑒𝑙𝑙 =−𝑛𝐹Δ 𝐸 𝑐𝑒𝑙𝑙 𝑜 +𝑅𝑇𝑙𝑛𝑄 Δ 𝐸 𝑐𝑒𝑙𝑙 =Δ 𝐸 𝑐𝑒𝑙𝑙 𝑜 − 𝑅𝑇 𝑛𝐹 𝑙𝑛𝑄= Δ 𝐸 𝑐𝑒𝑙𝑙 𝑜 − 0.0592𝑉 𝑛 𝑙𝑜𝑔𝑄 Δ 𝐸 𝑐𝑒𝑙𝑙 𝑜 = 0.0592𝑉 𝑛 𝑙𝑜𝑔𝐾

E°cell, ΔG° and K for a spontaneous reaction one proceeds in the forward direction with the chemicals in their standard states ΔG° < 1 (negative) E° > 1 (positive) K > 1 ΔG° = −RTlnK = −nFE°cell n is the number of electrons F = Faraday’s Constant = 96,485 C/mol e−

I2(s) + 2 Br−(aq) → Br2(l) + 2 I−(aq) ΔG°, (J) Example 18.6- Calculate ΔG° for the reaction I2(s) + 2 Br−(aq) → Br2(l) + 2 I−(aq) Given: Find: I2(s) + 2 Br−(aq) → Br2(l) + 2 I−(aq) ΔG°, (J) Concept Plan: Relationships: E°ox, E°red E°cell ΔG° Solve: ox: 2 Br−(aq) → Br2(l) + 2 e− E° = −1.09 V red: I2(l) + 2 e− → 2 I−(aq) E° = +0.54 V tot: I2(l) + 2Br−(aq) → 2I−(aq) + Br2(l) E° = −0.55 V Answer: since ΔG° is +, the reaction is not spontaneous in the forward direction under standard conditions

Nonstandard Conditions - the Nernst Equation ΔG = ΔG° + RT ln Q E = E° - (0.0592/n) log Q at 25°C when Q = K, E = 0 use to calculate E when concentrations not 1 M

Example 18.7- Calculate K at 25°C for the reaction Cu(s) + 2 H+(aq) → H2(g) + Cu2+(aq) Given: Find: Cu(s) + 2 H+(aq) → H2(g) + Cu2+(aq) K Concept Plan: Relationships: E°ox, E°red E°cell K Solve: ox: Cu(s) → Cu2+(aq) + 2 e− E° = −0.34 V red: 2 H+(aq) + 2 e− → H2(aq) E° = +0.00 V tot: Cu(s) + 2H+(aq) → Cu2+(aq) + H2(g) E° = −0.34 V Answer: since K < 1, the position of equilibrium lies far to the left under standard conditions

E° at Nonstandard Conditions

Calculate Ecell at 25°C for the reaction 3 Cu(s) + 2 MnO4−(aq) + 8 H+(aq) → 2 MnO2(s) + 3Cu2+(aq) + 4 H2O(l) Given [Cu2+] = 0.010 M, [MnO4−] = 2.0 M, [H+] = 1.0 M 3 Cu(s) + 2 MnO4−(aq) + 8 H+(aq) → 2 MnO2(s) + 3Cu2+(aq) + 4 H2O(l) [Cu2+] = 0.010 M, [MnO4−] = 2.0 M, [H+] = 1.0 M Ecell Given: Find: Concept Plan: Relationships: E°ox, E°red E°cell Ecell Solve: ox: Cu(s) → Cu2+(aq) + 2 e− }x3 E° = −0.34 V red: MnO4−(aq) + 4 H+(aq) + 3 e− → MnO2(s) + 2 H2O(l) }x2 E° = +1.68 V tot: 3 Cu(s) + 2 MnO4−(aq) + 8 H+(aq) → 2 MnO2(s) + Cu2+(aq) + 4 H2O(l)) E° = +1.34 V Check: units are correct, Ecell > E°cell as expected because [MnO4−] > 1 M and [Cu2+] < 1 M

Concentration Cells it is possible to get a spontaneous reaction when the oxidation and reduction reactions are the same, as long as the electrolyte concentrations are different the difference in energy is due to the entropic difference in the solutions the more concentrated solution has lower entropy than the less concentrated electrons will flow from the electrode in the less concentrated solution to the electrode in the more concentrated solution till the concentrations are equal

Concentration Cell when the cell concentrations are equal there is no difference in energy between the half-cells and no electrons flow when the cell concentrations are different, electrons flow from the side with the less concentrated solution (anode) to the side with the more concentrated solution (cathode) Cu(s)| Cu2+(aq) (0.010 M) || Cu2+(aq) (2.0 M)| Cu(s)

LeClanche’ Acidic Dry Cell electrolyte in paste form ZnCl2 + NH4Cl or MgBr2 anode = Zn (or Mg) Zn(s)  Zn2+(aq) + 2 e- cathode = graphite rod MnO2 is reduced 2 MnO2(s) + 2 NH4+(aq) + 2 H2O(l) + 2 e-  2 NH4OH(aq) + 2 Mn(O)OH(s) cell voltage = 1.5 V expensive, nonrechargeable, heavy, easily corroded

Zn(s) + 2OH- Zn(OH)2(s) + 2 e- Alkaline Dry Cell same basic cell as acidic dry cell, except electrolyte is alkaline KOH paste anode = Zn (or Mg) Zn(s) + 2OH- Zn(OH)2(s) + 2 e- cathode = graphite MnO2 is reduced 2 MnO2(s) + 2 H2O(l) + 2 e- 2 Mn(O)OH(s) + 2 OH-(aq) Overall reaction Zn(s)+2MnO2(s)+2H2O(l) Zn(OH)2(s) + 2MnO(OH)(s) cell voltage = 1.54 V longer shelf life than acidic dry cells and rechargeable, little corrosion of zinc

Voltaic Cells as Batteries Use chemical reactions to provide a source of energized electrons that can power electronics Batteries in everything, phones, laptops, tablets, watches, cars Actively seeking batteries that are light, powerful, and retain their effectiveness after many recharge cycles Molten metal batteries developed for storing electricity in power stations (infinitely rechargable) Huge area of scientific and commercial interest Batteries are being made ever smaller for energy efficient processor applications Nanotechnology, even biotechnology using viruses! A benign bacteria virus serves as a template in water to grow manganese oxide nanowires that, when combined with palladium, increases the energy potential of a lithium battery

Voltaic Cells as Batteries Lead Storage Battery Anode: Pb Pb  Pb2+ + 2e- Pb(s) + SO42-(aq)  PbSO4(s) + 2 e- Cathode: Pb coated with PbO2 Pb4+ +2e- Pb2+ PbO2(s) + 4 H+(aq) + SO42-(aq) + 2 e- PbSO4(s) + 2 H2O(l) Electrolyte: 30% H2SO4 cell voltage = 2.09 V 6 in series makes 12V rechargeable, heavy

NiCad Battery electrolyte is concentrated KOH solution anode = Cd Cd(s) + 2 OH-1(aq) ® Cd(OH)2(s) + 2 e-1 E0 = 0.81 V cathode = Ni coated with NiO2 NiO2 is reduced NiO2(s) + 2 H2O(l) + 2 e-1 ® Ni(OH)2(s) + 2OH-1 E0 = 0.49 V cell voltage = 1.30 V rechargeable, long life, light – however recharging incorrectly can lead to battery breakdown

Ni-MH Battery Tro, Chemistry: A Molecular Approach electrolyte is concentrated KOH solution anode = metal alloy with dissolved hydrogen oxidation of H from H0 to H+1 M∙H(s) + OH-1(aq) ® M(s) + H2O(l) + e-1 E° = 0.89 V cathode = Ni coated with NiO2 NiO2 is reduced NiO2(s) + 2 H2O(l) + 2 e-1 ® Ni(OH)2(s) + 2OH-1 E0 = 0.49 V cell voltage = 1.30 V rechargeable, long life, light, more environmentally friendly than NiCad, greater energy density than NiCad Tro, Chemistry: A Molecular Approach

Voltaic Cells as Batteries Lithium Ion Battery anode = graphite impregnated with Li ions cathode = Li - transition metal oxide Electrolyte is an organic solvent like dimethylcarbonate, with a Li salt like LiPF6 rechargeable, long life, very light, more environmentally friendly, greater energy density High surface area electrodes Tiny ions that move fast between electrodes

Voltaic Cells as Batteries Why do we use Li-ion batteries in our computers and phones and not say Pb Acid Batteries? We want to make batteries as portable (ie small and light) as possible

H2 Fuel Cell like batteries in which reactants are constantly being added so it never runs down! Anode and Cathode both Pt coated metal (may change with cheaper technology) Electrolyte is OH– solution Anode Reaction: 2 H2 + 4 OH– → 4 H2O(l) + 4 e- Cathode Reaction: O2 + 4 H2O + 4 e- → 4 OH– Toyota Mirai available in CA $499/month 312 miles per tank

Electrolysis electrolysis is the process of using electricity to make non-spontaneous redox reactions happen electrolysis is done in an electrolytic cell electrolytic cells can be used to separate elements from their compounds generate H2 from water for fuel cells recover metals from their ores Make halogen gasses from halide salts

Electrolytic Cell uses electrical energy to overcome the energy barrier and cause a non-spontaneous reaction must be DC source the + terminal of the battery connects to the anode the - terminal of the battery connects to the cathode cations attracted to the cathode, anions to the anode Oxidation at the anode, reduction at the cathode some electrolysis reactions require more voltage than Etot, called the overvoltage

Applications of electrolysis: electroplating In electroplating, the work piece is the cathode. Cations are reduced at cathode and plate to the surface of the work piece. The anode is made of the plate metal. The anode oxidizes and replaces the metal cations in the solution Gold plated terminals in electronics (Au is the most conductive of all metals)

Electrochemical Cells: Voltaic vs Electrolytic in all electrochemical cells, oxidation occurs at the anode, reduction occurs at the cathode in voltaic cells, anode is the source of electrons and has a (−) charge cathode draws electrons and has a (+) charge in electrolytic cells electrons are drawn off the anode, so it must have a place to release the electrons, the + terminal of the battery electrons are forced toward the cathode, so it must have a source of electrons, the − terminal of the battery

Applications of Electrolysis: Making H2 With sufficient voltage across 2 Pt electrodes one can electrolyze H2O to get O2 and H2 This may become an important process if we move toward fuel cell driven cars For example use wind power to electrolyze water to make H2 for cars

Electrolysis of Pure Compounds must be in molten (liquid) state electrodes normally graphite cations are reduced at the cathode to metal element anions oxidized at anode to nonmetal element

Electrolysis of NaCl(l)

Mixtures of Ions when more than one cation is present, the cation that is easiest to reduce will be reduced first at the cathode least negative or most positive E°red when more than one anion is present, the anion that is easiest to oxidize will be oxidized first at the anode least negative or most positive E°ox

Electrolysis of Aqueous Solutions Complicated by more than one possible oxidation and reduction possible cathode reactions reduction of cation to metal reduction of water to H2 2 H2O + 2 e-1  H2 + 2 OH-1 E° = -0.83 V @ stand. cond. E° = -0.41 V @ pH 7 possible anode reactions oxidation of anion to element oxidation of H2O to O2 2 H2O  O2 + 4e-1 + 4H+1 E° = -1.23 V @ stand. cond. E° = -0.82 V @ pH 7 oxidation of electrode particularly Cu graphite doesn’t oxidize half-reactions that lead to least negative Etot will occur unless overvoltage changes the conditions

Electrolysis of NaI(aq): Inert Electrodes Write down ions/molecules/metals present and don’t forget water Here Na+, I-, H2O are what is present (the electrodes here are inert) Write down the reduction reactions involving these species I2(aq)+2e-  2I-(aq) Ered = 0.54V O2(g) + 4e-1 + 4H+1(aq) 2 H2O(l) Ered = 0.82V 2 H2O + 2 e-1  H2 + 2 OH-1 Ered = 0.41V Na+(aq) + e-  Na(s) Ered = -2.71V Rearrange the equations so the reactants are on the left (reverse the sign of E if you flip the equation – these will be the oxidation reactions possible oxidations (anode –ve terminal) 2 I-1  I2 + 2 e-1 E° = −0.54 v 2 H2O  O2 + 4e-1 + 4H+1 E° = −0.82 v possible oxidations 2 I-1  I2 + 2 e-1 E° = −0.54 v 2 H2O  O2 + 4e-1 + 4H+1 E° = −0.82 v possible reductions (cathode +ve terminal) Na+1 + 1e-1  Na0 E° = −2.71 v 2 H2O + 2 e-1  H2(g) + 2 OH-1E° = -0.41V possible reductions Na+1 + 1e-1  Na0 E° = −2.71 v 2 H2O + 2 e-1  H2 + 2 OH-1 E° = −0.41 v brown liquid overall reaction 2 I−(aq) + 2 H2O(l)  I2(aq) + H2(g) + 2 OH-1(aq) bubbles Ecell = -0.54V – 0.41V = -0.95V The battery needs to at least be 1V to make this happen

Electrolysis of NaI(aq) with Inert Electrodes possible oxidations 2 I-1  I2 + 2 e-1 E° = −0.54 v 2 H2O  O2 + 4e-1 + 4H+1 E° = −0.82 v possible oxidations 2 I-1  I2 + 2 e-1 E° = −0.54 v 2 H2O  O2 + 4e-1 + 4H+1 E° = −0.82 v possible reductions Na+1 + 1e-1  Na0 E° = −2.71 v 2 H2O + 2 e-1  H2 + 2 OH-1 E° = −0.41 v possible reductions Na+1 + 1e-1  Na0 E° = −2.71 v 2 H2O + 2 e-1  H2 + 2 OH-1 E° = −0.41 v overall reaction 2 I−(aq) + 2 H2O(l)  I2(aq) + H2(g) + 2 OH-1(aq)

Electrolysis in aqueous solutions: Overpotentials There are (potentially) competing processes in the electrolysis of an aqueous solution consider now NaCl(aq) with C electrodes Cathode Anode We choose those reactions that are thermodynamically favored But…chlorine is evolved at the anode!

Overpotentials Thermodynamics true at equilibrium but when the process has a current flowing kinetics plays a role Overpotential represents the additional voltage that must be applied to drive the process In the NaCl(aq) solution the overpotential for evolution of oxygen is greater than that for chlorine, and so chlorine is evolved preferentially The limiting process in electrolysis is usually diffusion of the ions in the electrolyte Overpotential will depend on the electrolyte and electrode Driving the cell at the least current will give rise to the smallest overpotential

Faraday’s Law the amount of metal deposited during electrolysis is directly proportional to the charge on the cation, the current (A), and the length of time (s) the cell runs charge = current x time Q = I x t

Calculate the mass of Au that can be plated in 25 min using 5 Calculate the mass of Au that can be plated in 25 min using 5.5 A for the half-reaction Au3+(aq) + 3 e− → Au(s) 3 mol e− : 1 mol Au, current = 5.5 amps, time = 25 min mass Au, g Given: Find: t(s), amp charge (C) mol e− mol Au g Au Concept Plan: Relationships: Solve: Check: units are correct, answer is reasonable since 10 A running for 1 hr ~ 1/3 mol e−

Corrosion corrosion is the spontaneous oxidation of a metal by chemicals in the environment since many materials we use are active metals (metals that like to make ions), corrosion can be a very big problem

Rusting rust is hydrated iron(III) oxide moisture must be present water is a reactant required for flow between cathode and anode electrolytes (sea-water) promote rusting enhances current flow acids (acid rain) promote rusting lower pH = lower E°red

O2(g) + 2H2O(l) + 4e-  4OH-(aq) Rusting Anode (inside the droplet): Fe(s)  Fe2+(aq) + 2e- Cathode (at the surface of the droplet: O2(g) + 2H2O(l) + 4e-  4OH-(aq) Within the droplet Fe2+(aq) + 2OH-(aq)  Fe(OH)2(s) Rust is then quickly produced by the oxidation of the precipitate at the edge of the droplet. 4Fe(OH)2(s) + O2(g)  2Fe2O3 •H2O(s) + 2H2O(l)

Preventing Corrosion Zn  Zn2+ + 2e- 0.76V Fe  Fe2+ + 2e- 0.45V one way to reduce or slow corrosion is to coat the metal surface to keep it from contacting corrosive chemicals in the environment paint some metals, like Al, form an oxide that strongly attaches to the metal surface, preventing the rest from corroding another method to protect one metal is to attach it to a more reactive metal that is cheap sacrificial electrode galvanized steel (Zinc coating) Zn  Zn2+ + 2e- 0.76V Fe  Fe2+ + 2e- 0.45V

Sacrificial Anode Zn  Zn2+ + 2e- 0.76V Fe  Fe2+ + 2e- 0.45V Galvanic anode on the hull of a ship

The Final Long Questions on the Following: Kinetics Method of Initial Rates to find rate law Arrhenius plot to find Activation energy Ea and the collision factor A Acid-Base Titration of a Weak Acid Calculate the initial pH of the acid given Ka Calculate the pH in the buffer region using Henderson-Hasselbalch Calculate the pH at the end point using Kb Galvanic and Electrolytic Cell An extra credit question on kinetics mechanisms (worth 15% of the final) Multiple Choice Redo for Electrochemistry A short (optional multiple choice on electrochemistry for those who want a second chance) The Final