Electrochemistry Ch 13 pg 225 Princeton Review

Oxidation half-reaction (lose e-) Electrochemical processes are oxidation-reduction reactions in which: the energy released by a spontaneous reaction is converted to electricity or electrical energy is used to cause a nonspontaneous reaction to occur 2+ 2- 2Mg (s) + O2 (g) 2MgO (s) 2Mg 2Mg2+ + 4e- Oxidation half-reaction (lose e-) O2 + 4e- 2O2- Reduction half-reaction (gain e-) 19.1

Copyright © Cengage Learning. All rights reserved Cell Potential A galvanic cell consists of an oxidizing agent in one compartment that pulls electrons through a wire from a reducing agent in the other compartment. The “pull”, or driving force, on the electrons is called the cell potential ( ), or the electromotive force (emf) of the cell. Unit of electrical potential is the volt (V). 1 joule of work per coulomb of charge transferred. Copyright © Cengage Learning. All rights reserved

Galvanic Cell Flow of e- Flow of current Voltmeter e– e– Zinc anode Copper cathode Cl– K+ Salt bridge Cotton plugs SO 4 2– Cu2+ Zn2+ SO 4 2– ZnSO4 solution CuSO4 solution 2e– 2e– Cu2+ Zn Zn2+ Cu Zn is oxidized to Zn2+ at anode. Cu2+ is reduced to Cu at cathode. Zn(s) Zn2+(aq) + 2e– 2e– + Cu2+(aq) Cu(s) Net reaction Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s)

Anode Reaction Cathode Reaction

SO 4 2– Cu2+ Zn2+ SO 4 2– ZnSO4 solution CuSO4 solution

Voltmeter e– e– Zinc anode Copper cathode Cl– K+ Salt bridge Cotton plugs SO 4 2– Cu2+ Zn2+ SO 4 2– ZnSO4 solution CuSO4 solution Salt bridge provides electrical neutrality by providing negative anions to equal the positive cations being created at the Zn anode during oxidation. And cations ions (K+) to replace Cu 2+ being used up at reduction.

Voltmeter e– e– Zinc anode Copper cathode Cl– K+ Salt bridge Cotton plugs SO 4 2– Cu2+ Zn2+ SO 4 2– ZnSO4 solution CuSO4 solution Zn

Voltmeter e– e– Zinc anode Copper cathode Cl– K+ Salt bridge Cotton plugs SO 4 2– Cu2+ Zn2+ SO 4 2– ZnSO4 solution CuSO4 solution 2e– Zn Zn2+ Zn is oxidized to Zn2+ at anode. Zn(s) Zn2+(aq) + 2e–

Voltmeter e– e– Zinc anode Copper cathode Cl– K+ Salt bridge Cotton plugs SO 4 2– Cu2+ Zn2+ SO 4 2– ZnSO4 solution CuSO4 solution 2e– Zn Zn2+ Cu Zn is oxidized to Zn2+ at anode. Zn(s) Zn2+(aq) + 2e–

Voltmeter e– e– Zinc anode Copper cathode Cl– K+ Salt bridge Cotton plugs SO 4 2– Cu2+ Zn2+ SO 4 2– ZnSO4 solution CuSO4 solution 2e– 2e– Cu2+ Zn Zn2+ Cu Zn is oxidized to Zn2+ at anode. Zn(s) Zn2+(aq) + 2e–

Voltmeter e– e– Zinc anode Copper cathode Cl– K+ Salt bridge Cotton plugs SO 4 2– Cu2+ Zn2+ SO 4 2– ZnSO4 solution CuSO4 solution 2e– 2e– Cu2+ Zn Zn2+ Cu Zn is oxidized to Zn2+ at anode. Cu2+ is reduced to Cu at cathode. Zn(s) Zn2+(aq) + 2e– 2e– + Cu2+(aq) Cu(s)

Voltmeter e– e– Zinc anode Copper cathode Cl– K+ Salt bridge Cotton plugs SO 4 2– Cu2+ Zn2+ SO 4 2– ZnSO4 solution CuSO4 solution 2e– 2e– Cu2+ Zn Zn2+ Cu Zn is oxidized to Zn2+ at anode. Cu2+ is reduced to Cu at cathode. Zn(s) Zn2+(aq) + 2e– 2e– + Cu2+(aq) Cu(s) Net reaction Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s)

Galvanic Cells anode oxidation cathode reduction spontaneous redox reaction 19.2

Current The flow of positive charge. Current is always in the opposite directions from electron flow. Flow of e- Flow of current

Copyright © Cengage Learning. All rights reserved CONCEPT CHECK! Sketch a cell using the following solutions and electrodes. Include: The potential of the cell The direction of electron flow Labels on the anode and the cathode Zn electrode in 1.0 M Zn2+(aq) and Cu electrode in 1.0 M Cu2+(aq) Zinc is the anode, copper is the cathode (electrons flow from zinc to copper). The cell potential is 1.10 V. Copyright © Cengage Learning. All rights reserved

Copyright © Cengage Learning. All rights reserved CONCEPT CHECK! Consider the cell from part b. What would happen to the potential if you increase the [Cu2+]? Explain. The cell potential should increase. Since the copper(II) ion is the reactant in the overall equation of the cell, the cell potential should increase (LeChâtelier's principle applies here). Copyright © Cengage Learning. All rights reserved

Copyright © Cengage Learning. All rights reserved Line Notation Used to describe electrochemical cells. Anode components are listed on the left. Cathode components are listed on the right. Separated by double vertical lines which indicated salt bridge or porous disk. The concentration of aqueous solutions should be specified in the notation when known. Example: Mg(s)|Mg2+(aq)||Al3+(aq)|Al(s) Mg → Mg2+ + 2e– (anode) Al3+ + 3e– → Al (cathode) Copyright © Cengage Learning. All rights reserved

Electrolytic Cells The difference in electrical potential between the anode and cathode is called: cell voltage electromotive force (emf) cell potential Used in electro plating Cell Diagram Zn (s) + Cu2+ (aq) Cu (s) + Zn2+ (aq) [Cu2+] = 1 M & [Zn2+] = 1 M Zn (s) | Zn2+ (1 M) || Cu2+ (1 M) | Cu (s) Anode OX Cathode RED 19.2

Reduction Potentials reduction potential (E0): is the voltage associated with a reduction reaction The more positive E0 the greater the tendency for the substance to be reduced On the AP test you will be given a chart of reduction potentials. You can reverse them and change the sign on the voltage to get oxidation potentials. 19.3

The half-cell reactions are reversible The sign of E0 changes when the reaction is reversed Changing the stoichiometric coefficients of a half-cell reaction does not change the value of E0 19.3

Using the Table TOP: F2 + 2e-  2F- = E° = +2.87 Large reduction potential = more likely to be reduced and thus a strong oxidizing agent. Bottom: Li+ + e-  Li = E° = -3.05 Li Li+ + e- = E° = +3.05 Large oxidation potential = more likely to lose an electron and become oxidized thus it’s a good reducing agent.

Spontaneity of Redox Reactions Redox rxns will occur spontaneously if its cell potential has a positive value. Or if its free energy (G) is negative. DG0 =-nFEcell 19.4

Example Look at the spontaneous rxn: Zn + Ag+  Zn2+ + Ag E0 = + 1.56 Zn  Zn2+ + 2e- (LEO) E0 = -0.76 reverse = +0.76 Ag+ + e- Ag (GER) E0 = + 0.80 E = Eox + Ered = .76 + .80 = + 1.56 E is positive so G is negative and the rxn is spontaneous

Example with coefficients The number of electrons lost must equal the number of electrons gained. In order to insure this we must multiply the half reaction by integers to balance. This does not change the Cell potential E°.

Example Fe3+ + Cu  Cu2+ + Fe 2+ ½ rxns GER Fe3+ + e-  Fe 2+ E° = 0.77 V LEO Cu  Cu2+ + 2 e- E° = -0.34 V The reduction must occue 2x for every 1 OX rxn Note the voltage of the OX is reversed because our chart is of reduction potentials E° cell = - 0. 34 V + 0.77 E° = 0.43 V

∆G° = - nFE° G = Gibbs free energy kJ/mol n = moles of e- exchanged Under standard conditions the maximum cell potential (E°) is directly related to the free energy (G) difference between the reactants and the products. This equation allows us to test for ∆G° in the lab. ∆G° = - nFE° G = Gibbs free energy kJ/mol n = moles of e- exchanged F = Faraday’s constant 96,500 coulombs/mole (how much charge is produced for every 1 mole of e- ) E° = Standard reduction potential

Example Calculate ∆G° for the reaction Fe + Cu2+  Cu + Fe 2+ Write ½ rxns to ID red.ox atoms Fe +  Fe 2+ 0 +2 LEO (rev potential) Cu2+  Cu +2 0 GER Use chart to find reduction potentials Fe +  Fe 2+ E = -0.44 = +0.44 v Cu2+  Cu E = +0.34 v

E cell = 0.44+0.34 = 0.78 F = 96,500 n = 2 mole (2 e- per atom) 3. Plug in G = -2(96,500)(0.78) = - 1.5 x 105 J 4. Conclude G = negative so rxn is spontaneous E cell = positive spontaneous ** if E is negative rxn will not occur and it is not spontaneous

Copyright © Cengage Learning. All rights reserved Nernst Equation The relationship between cell potential and concentrations of cell components At 25°C: or (at equilibrium) K and Q = [P]/[R] n = # of electrons Copyright © Cengage Learning. All rights reserved

Copyright © Cengage Learning. All rights reserved CONCEPT CHECK! Explain the difference between E and E°. When is E equal to zero? ε is the cell potential at any condition When is E° equal to zero? ε is the cell potential under standard conditions (1.0M, or 1 atm and 25C) ε is the cell potential at any condition, and ε is the cell potential under standard conditions (1.0 M or 1 atm, and 25C). ε equals zero when the cell is in equilibrium ("dead" battery). ε is equal to zero for a concentration cell. Copyright © Cengage Learning. All rights reserved

Q = reaction quotient Q = concentration of products [ ] raised to its coefficient divided by [ ] reactants raised to its coefficient

Example Calculate E for the following reaction given [Al3+] = 1.5M [Mn2+] = 0.5 M Ecell = 0.48 2 Al + 3 Mn2+  2Al 3+ + 3 Mn Q = 1.52 = 18 0.53 n = 2 Al  2Al 3+ 3 Mn2+  3 Mn 0 +3 = 3(2) = 6 +2 0 = 2(3)= 6 n = 6

Cont. Plug it in: E = 0.48 – (0.0591/6) log 18 = 0.47 V

Electrolysis and Mass Changes charge (C) = current (A) x time (s) 1 mole e- = 96,500 C 19.8

How much Ca will be produced in an electrolytic cell of molten CaCl2 if a current of 0.452 A is passed through the cell for 1.5 hours? Anode: 2Cl- (l) Cl2 (g) + 2e- Cathode: 2 mole e- = 1 mole Ca Ca2+ (l) + 2e- Ca (s) Ca2+ (l) + 2Cl- (l) Ca (s) + Cl2 (g) mol Ca = 0.452 C s x 1.5 hr x 3600 s hr 96,500 C 1 mol e- x 2 mol e- 1 mol Ca x = 0.0126 mol Ca = 0.50 g Ca 19.8