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Electrochemistry Chapter 20.

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1 Electrochemistry Chapter 20

2 Electrochemistry = the study of relationships between electricity and chemical reactions
Topics Include: Batteries Spontaneity of reacation Corrosion of metals Electroplating

3 Redox Reactions (Chapter 4.4)
REDOX reactions always involve a transfer of electrons Oxidation = loss of electrons Reduction = gain of electrons 2 ways to remember: LEO says GER Loss Electrons Oxidation; Gain Electrons Reduction OIL RIG Oxidation Involves Loss; Reduction Involves Gain

4 Redox Terminology Oxidation Number  a positive or negative value assigned to each atom in a compound or ion for the purpose of keeping track of electrons Oxidizing Agent  the substance that makes it possible for another substance to be oxidized Oxidizing agents remove electrons. They pick them up and become reduced (become more negative)

5 Reducing Agent  the substance that gives up electrons
Reducing agents lose electrons, become oxidized, and become less negative/more positive

6 Rules for Assigning Ox. #’s
Summary 1. Ox. # of the atom of a free element is Zero Element = 0 2. Ox # of a monatomic ion equals its charge Calcium = +2 Sulfide = -2 3. In compounds, Oxygen is always (-2), except in peroxides (-1) H2O; O=-2 H2O2; O = -1 4. In compounds, Hydrogen is always +1 Hydrogen = +1 5. In compounds, Fluorine is ALWAYS -1 Fluorine = -1 6. Sum of the oxidation states for an electrically neutral compound must be zero

7 Helpful Number Line +4 +3 +2 +1 oxidation 0 reduction 1 2 3 4
Reducing Agent Oxidizing Agent

8 Example The nickel-cadmium battery, a rechargeable “dry cell” used in battery-operated devices, uses the following redox reaction to generate electricity: Cd(s) + NiO2(s) + H2O(l) Cd(OH)2(s) + Ni(OH)2(s) Assign all the oxidation numbers Identify the substances that are oxidized and reduced. Identify which are oxidizing agents and which are reducing agents.

9 Redox Every redox reaction has both oxidation and reduction occurring.
What is being oxidized? reduced ? Cd(s) + NiO2(s) + 2H2O(l) --> Cd(OH)2(s) + Ni(OH)2(s)    2+1 Identify the oxidation numbers Identify the substances which change in oxidation no. Cd ---> Cd e Ni e --> Ni Cd is oxidized Nickel is reduced Ni is the oxidizing agent Cd is the reducing agent

10 Practice Label the oxidation numbers for all elements on both sides of the reaction. Determine the Oxidizing Agent and the Reducing Agent in each reaction. Zn(s) + 2H+(aq)  Zn+2(aq) + H2(g) 2H2 (g) + O2 (g)  2H2O(l)

11 Balancing Redox Redox reactions are sometimes difficult to balance.
Must balance both atoms of each element AND charge. We only show the substances involved in the reaction, and add H, OH and/or H2O Requires following a series of steps and practice!

12 Half Reactions Although REDOX takes place simultaneously, it is easier to look at it as two separate processes. For Example: Sn+2 + 2Fe+3  Sn+4 + 2Fe+2 Oxidation: Sn+2  Sn e- Reduction: 2Fe e-  2Fe+2 Notice: In oxidation, electrons are products, while in reduction electrons are reactants

13 Balancing using Half Reactions
MnO4- + C2O4-2  Mn+2 + CO2 Lets look at the half reactions: MnO4-  Mn+2 C2O4-2  CO2 First add coefficients to balance out atoms (this may need to be done in an acidic or basic solution) Note: if the process is done in acidic solution, H+ and H2O can be added; in basic solution, OH - and H2O can be added.

14 MnO4- + C2O4-2  Mn+2 + CO2 MnO4-  Mn+2 + H2O (added water)
H+ + MnO4-  Mn+2 + H2O (added H) Now we can balance: 8H+ + MnO4-  Mn+2 + 4H2O Now equal types of atoms on both sides, but charges aren’t equal, so need to add electrons: 5e- + 8H+ + MnO4-  Mn+2 + 4H2O Now to balance the oxalate ½ reaction…

15 C2O4-2  CO2 (add coefficients to balance)
C2O4-2  2CO2(add electrons to balance charge) C2O4-2  2CO2 + 2e- Now we have to put both balanced half reactions together to get the overall balanced equation: 5e- + 8H+ + MnO4-  Mn+2 + 4H2O We need the same number of electrons to appear on both sides of the reaction so they will cancel out.

16 10e- + 16H+ + 2MnO4-  2Mn+2 + 8H2O 5C2O4-2  10CO2 + 10e- 16H+ + 2MnO4- + 5C2O4-2  2Mn+2 + 8H2O + 10CO2 The best way to do this is through practice and we will do various examples in class and for HW

17 AP Type question For the following: 1) write a balanced equation and 2) answer the question about the reaction. 1)Calcium metal is heated strongly in nitrogen gas. 2) What is the change in oxidation number of the nitrogen in this RXN? 1)Acidified potassium dichromate solution is mixed with potassium iodide solution. 2) What is the reducing agent in the above reaction? *Hint: Water is one of the products for the above reaction.

18 Redox Reactions - the reality: electricity
Moving electrons is electric current. A voltaic cell (or galvanic cell) is a device which allows the transfer of e to take place. Voltaic cells produce electric current. A battery is a voltaic cell. Batteries contain chemicals which undergo redox reactions.

19 Features of a Voltaic Cell
Two different substances connected by a conductor in 2 separate compartments containing an electrolytic solution which is allowed to flow between the 2 compartments When assembled: a complete circuit forms which allows electrons to flow in a one-way direction. Direct Current (DC)

20 The one-way direction is the direction of the spontaneous reaction.
Each substance is called an electrode. Anode - electrode where oxidation occurs Cathode - electrode where reduction occurs Electrons always flow from anode to the cathode.

21 Both metals in same mixture?
Metal ions of one metal is deposited on another Reaction happens without doing useful work Zn Cu Cu+2 H+ Zn+2 SO4-2 but if separate, the electricity is available to do work

22  Connected this way the reaction starts, but
Stops immediately because charge builds up. For a cell to work, the solutions in each half cell must remain electrically neutral. 2e  2e  2e  Zn (s) --> Zn+2 + 2e Cu e --> Cu(s) Zn Cu Cu+2 NO3- Zn+2 NO3- A half cell

23 A bridge is needed Salt Bridge allows current to flow Zn+2 Cu+2 NO3-

24 Now, Electricity travels in a complete circuit
Zn (s) --> Zn+2 + 2e e  Cu e --> Cu(s) Zn Cu Zn+2 NO3- Cu+2 NO3-

25 Instead of a salt bridge
Porous Disk or Barrier works too Zn+2 NO3- Cu+2 NO3-

26 e- e- e- e- e- e- Reducing Agent Zn Oxidizing Agent Cu+2  +
Reduction occurs at the cathode, Oxidation at the anode e- e- e- e- e are available to do work Zn (s) --> Zn+2 + 2e Cu e --> Cu(s) Anode (Zn) Is oxidized Cathode (Cu) Is reduced Porous barrier e- e- + Reducing Agent Zn Oxidizing Agent Cu+2

27 The Complete Picture Zn (s) + Cu+2 --> Cu(s) + Zn+2
Red. at cathode: Cu e --> Cu(s) Oxid. at anode: Zn (s) --> Zn+2 + 2e 2e 2e 2e salt bridge NaNO3 Zn Cu Anode Cathode NO3 Cu+2 SO4-2 Na+ NO3 Zn+2 SO Na+ CuSO4 Solution (electrolyte) ZnSO4 Solution All + ions flow this way All  ions flow this way

28 While Operating . . . The Zn electrode is oxidized and loses mass, while [Zn+2] increases in solution. The Cu electrode is reduced, gains mass, while the [Cu+2] decreases in solution. Salt bridge allows ions to flow in both directions. ions move toward the anode +ions move toward the cathode Salt bridge contains an electrolyte that doesn’t react with anything in the cell. Salt bridge connects the 2 cells keeping + and  ions neutral. The actual charge on an electrode is zero.

29 Cr2O7-2 + 14H+ + 6I- ---> 2Cr+3 +3I2 + 7H2O
For the above reaction, a simple voltaic cell is assembled. Which is the anode? Which is the cathode? I is the anode ; Cr2O7-2 is the cathode What reaction occurs at the anode? 6 I ---> 3I2 + 6e What reaction occurs at the cathode? Cr2O H e ---> 2Cr H2O What is the direction of electron flow? Electrons flow from the anode, I , to the cathode, Cr2O7-2 What is the sign at each electrode? I Anode is () and Cr2O7-2, cathode is ()

30 Cell Potential Ecell Electrons flow from anode to cathode because there is a potential energy difference between them. Oxidizing agent pulls the electron. Reducing agent pushes the electron. The push or pull (“driving force”) is called the cell potential, Ecell Also called the electromotive force (EMF) Unit is the volt(V) = 1 joule of work/coulomb of charge Measured with a voltmeter

31 The magnitude of Ecell depends upon:
Specific reactions at each electrode Concentrations of reactants / products Temperature The value of Ecell Is what the voltmeter reads. Is the difference between the 2 half cell potentials (which we look up in tables) Eºcell = Eºcathode  Eºanode

32 Standard Hydrogen Electrode
This is the reference all other oxidations are compared to Eº = 0 º indicates standard states of 25ºC, 1 atm, 1 M solutions. Platinum wire H2 in H+ Cl- 1 M HCl Platinum electrode

33 H2 in Cathode Anode H+ Cl- Zn+2 SO4-2 1 M ZnSO4 1 M HCl 0.76
For example Zn ---> Zn+2 + 2e- 0.76 H2 in Cathode Anode H+ Cl- Zn+2 SO4-2 1 M ZnSO4 1 M HCl

34 Cell Potential The total cell potential is the sum of the potential at each electrode. Eºcell = Eºcathode + Eºanode We can look up reduction potentials in the table. Tables are for reduction potential. The oxidation reaction must be reversed, so we change its sign. Since we always change the sign of the oxidation reaction, the formula is expressed as: Eºcell = Eºreduction  Eºoxidation OR Eºcell = Eºcathode  Eºanode

35 Calculating Ecell Co(s) --> Co+2 (aq) + 2e Ered = 0.277 V
AgCl(s) + e --> Ag(s) + Cl Ered = V What reaction occurs at the anode? What reaction occurs at the cathode? What is the standard cell potential, Ecell? Oxidation Co(s) --> Co+2 (aq) + 2e Reduction AgCl(s) + e --> Ag(s) + Cl Eºcell = Eºcathode  Eºanode Eºcell =  (0.277) = V Note: multiplying a reaction by a coefficient doesn’t effect the Value of Eo.

36 Cell Potential Determine the cell potential for a voltaic cell based on the redox reaction: Cu(s) + Fe+3(aq) ® Cu+2(aq) + Fe+2(aq) Fe+3(aq) + e-® Fe+2(aq) Eºred = V Cu+2(aq)+2e- ® Cu(s) Eºred = V Eºcell = Eºcathode  Eºanode Eºcell = 0.77  (+0.34) = V Remember, to operate a cell must have a (+) cell potential.

37 In Summary A voltaic cell operates on a driving force called the cell potential, Ecell or EMF. An operating voltaic cell must have a + Ecell value. In any voltaic cell, the reaction at the cathode (reduction occurs) has a more + E°red value than does the reaction at the anode (oxidation occurs). The strongest oxidizing agents (they gain electrons) have the most + E°red values. Standard Reduction Potential tables can be used to mix and match the strongest voltaic cells possible.

38 Spontaneity of Redox Reactions
A complete circuit always flows in a one way direction. A Direct Current (DC) results. The one-way direction is the direction of the spontaneous reaction. To be spontaneous, Ecell must be (+) If Ecell is (-), it will not work!! The stronger oxidizing agent has the greater “driving force” and forces the reducing agent to run in reverse as an oxidation reaction

39 Activity Series and Eºred
“The ions of the metal below will oxidize the solid metal above.” The activity series is experimentally determined with the strongest reducing agents listed at the top. The activity series is really a list of oxidation potentials - the flipped up-side-down list of our standard reduction potentials. The Ered table has the strongest oxidizing agents listed at the top. We now know why certain metal ions react with other metals: potential energy exists as a driving force, + Ecell And, why certain metals don’t react,  Ecell

40 Working Voltaic Cells-Why ?
Recall that ∆G, Gibbs Free Energy determines spontaneity. ∆G = spontaneous reaction + Ecell = spontaneous redox reaction Michael Faraday determined the relationship between Free Energy and cell potential: ∆G = nFE where n = no. of electrons transferred (always (+) ) F = Faraday’s constant: 96,500 J/V*mol e- E = Ecell or emf

41 DG makes the cell operational
DGº = -nFEº if E º = +, then DGº  spontaneous if E º = , then DGº  nonspontaneous, the reverse is spontaneous.

42 Practice Problem DGº = -nFEº DGº = -(2)(96,500J/V•mol)(0.78V)
Calculate DGº for the following reaction: Cu+2(aq)+ Fe(s) ® Cu(s)+ Fe+2(aq) Fe+2(aq) + 2e-® Fe(s) Eº = 0.44 V Cu+2(aq)+2e- ® Cu(s) Eº = 0.34 V Eºcell = Eºcathode  Eºanode Eºcell =  (0.44) Eºcell = V DGº = -nFEº DGº = -(2)(96,500J/V•mol)(0.78V) DGº = 150,540J/mol = 150.4 kJ/mol

43 Practice Problem DGº = -nFEº DGº = -(6)(96,500J/V•mol)(-0.12V)
3Ni Cr(OH)3 (s) + 10 OH ----> 3 Ni (s) + 2CrO42 + 8 H2O Calculate ∆G for this reaction. 3Ni e ----> 3 Ni (s) E = 0.25V CrO42 + 4H2O + 3e ---> Cr(OH)3 + 5 OH E = 0.13V Eºcell = 0.25  (0.13) = -0.12V) DGº = -nFEº DGº = -(6)(96,500J/V•mol)(-0.12V) DGº = 69,480 J/mol  69 kJ/mol

44 Why don’t batteries last forever?
As a voltaic operates, reactants are consumed and products are generated. As the conc. of reactants decreases, so does the Ecell. At some point the battery goes “dead” and the Ecell = 0 Walther Nernst determined the relationship between the conc. and EMF

45 The Nernst Equation E = Eº  RT lnQ nF
DG = DGº + R T lnQ (standard free energy change) -nFE = -nFEº + RTlnQ (substituting DGº = -nFEº) E = Eº  RT lnQ nF At standard conditions, RT/F = V So, E = Eº  V log Q n Q is the reaction quotient, [prod]/[react] Remember, we leave solids out of the equilibrium expression

46 The Nernst Equation Remember, always determine n by balancing the redox reaction. What it means: Equation relates concentrations to the emf (voltage). If concentration can give us voltage, then from voltage we can tell concentration.

47 Equilibrium and Cell emf
When E =0, the cell reaction has reached equilibrium and no net reaction is occurring in the voltaic cell. Recall that at equilibrium, ∆G = 0 and Q = Keq Substituting, simplifying and rearranging the Nernst Equation gives us: nE log Keq = (at 25C) 0.0592 Thus, the equilibrium constant for a redox reaction can be determined from the E value.

48 Equilibrium and Cell emf
As a voltaic operates, reactants are consumed and products are generated. When “dead”, the cell at E =0 has reached equilibrium, ∆G =0. Where’s the equilibrium? Qualitatively - we can predict direction of change in E from LeChâtelier. E = Eº  V log Q n

49 E = Eº  0.0592 V log Q E = Eº  0.0592 V log [Al+3]2 Practice Problem
n 2Al(s) + 3Mn+2(aq) ® 2Al+3(aq) + 3Mn(s) Predict if Ecell will be greater or less than Eºcell if [Al+3] = 1.5 M and [Mn+2] = 1.0 M E = Eº  V log [Al+3]2 n [Mn+2]3 if [Al+3] = 1.0 M and [Mn+2] = 1.5M if [Al+3] = 1.5 M and [Mn+2] = 1.5 M Less than Ecell Greater greater

50 Practice Problem Calculate the emf at 298K generated by a cell for the equation below when [Cr2O7-2]=2.OM, [H+]=1.0M, and [Cr+3]=1.0x10-5M. Cr2O7-2(aq) + 14H+(aq)+ 6I-(aq)  Cr+3 (aq) + 3I2(s) + 7H2O(l) Solve for Q Q = 5.0 x 10-11 Figure out n n = 6 Plug into Nernst Equation and get 0.89 V

51 Batteries are Voltaic Cells
Car batteries are lead storage batteries. Pb +PbO2 +H2SO4 ®PbSO4(s) +H2O Dry Cell Zn + NH4+ +MnO2 ® Zn+2 + NH3 + H2O+ MnO(OH) (s) Alkaline Zn +MnO2 ® ZnO+ Mn2O3 (in base) NiCad NiO(OH) s + Cd + 2H2O ® Cd(OH)2 +Ni(OH)2

52 Corrosion Rusting - spontaneous oxidation.
Most structural metals have reduction potentials that are less positive than O2 . Fe+2 +2e- ® Fe Eº= 0.44 V O2 + 2H2O + 4e- ® 4OH- Eº= 0.40 V Fe+2 + O2 + H2O ® Fe2 O3 + H+ Reaction happens in two places.

53 Salt speeds up process by increasing conductivity
Water O2 Fe+2 Rust Fe2O3 Fe+2 Fe+2 e- Iron Dissolves in Water: Fe ® Fe+2

54 Preventing Corrosion Coating to keep out air and water.
Galvanizing - Putting on a zinc coat Has a lower reduction potential, so it is more easily oxidized. Alloying with metals that form oxide coats. Cathodic Protection - Attaching large pieces of an active metal like magnesium that get oxidized instead.

55 Electrolysis Running a voltaic cell backwards.
Put a voltage bigger than the potential and reverse the direction of the redox reaction. Used for electroplating.

56 1.10 e- e- Zn Cu 1.0 M Zn+2 1.0 M Cu+2 Anode Cathode

57 A battery >1.10V e- e- Zn Cu 1.0 M Zn+2 1.0 M Cu+2 Cathode Anode

58 Calculations for plating
Have to count charge. Measure current I (in amperes) 1 amp = 1 coulomb of charge per second q = I x t q/nF = moles of metal Mass of plated metal How long must 5.00 amp current be applied to produce 15.5 g of Ag from Ag+?

59 q = I x t and q t = I = 5.00 amp I q = moles x Fn
How long must 5.00 amp current be applied to produce 15.5 g of Ag from Ag+? q = I x t and q t = I = 5.00 amp I q = moles x Fn q = mol(96,500) (1) q = 13,862 C 13,862 C t = = 2772 sec = 46.2 min. 5.00 amp


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