CHAPTER SEVENTEEN ELECTROCHEMISTRY SEVENTEEN ELECTROCHEMISTRY Review of Oxidation ... CHAPTER 17...

473

CHAPTER SEVENTEEN

ELECTROCHEMISTRY

Review of Oxidation - Reduction Reactions

13. Oxidation: increase in oxidation number; loss of electrons

Reduction: decrease in oxidation number; gain of electrons

14. See Table 4.2 in Chapter 4 of the text for rules for assigning oxidation numbers.

a. H (+1), O (-2), N (+5) b. Cl (-1), Cu (+2)

c. O (0) d. H (+1), O (-1)

e. Mg (+2), O (-2), S (+6) f. Ag (0)

g. Pb (+2), O (-2), S (+6) h. O (-2), Pb (+4)

i. Na (+1), O (-2), C (+3) j. O (-2), C (+4)

4 2 4 3 4 4k. (NH ) Ce(SO ) contains NH ions and SO ions. Thus, cerium exists as the Ce ion. + 2- 4+

H (+1), N (-3), Ce (+4), S (+6), O (-2)

l. O (-2), Cr (+3)

15. The species oxidized shows an increase in oxidation numbers and is called the reducing agent. Thespecies reduced shows a decrease in oxidation numbers and is called the oxidizing agent. The pertinentoxidation numbers are listed by the substance oxidized and the substance reduced.

Substance SubstanceRedox? Ox. Agent Red. Agent Oxidized Reduced

2 4 4 2a. Yes H O CH CH (C, -4 ÷ +2) H O (H, +1 ÷ 0)

3 3b. Yes AgNO Cu Cu (0 ÷ +2) AgNO (Ag, +1 ÷ 0)

c. Yes HCl Zn Zn (0 ÷ +2) HCl (H, +1 ÷ 0)

d. No; There is no change in any of the oxidation numbers.

CHAPTER 17 ELECTROCHEMISTRY474

16. See Chapter 4.10 of the text for rules on balancing oxidation-reduction reactions.

3a. Cr ÷ Cr + 3 e NO ÷ NO3+ - -

3 2 4 H + NO ÷ NO + 2 H O+ -

3 23 e + 4 H + NO ÷ NO + 2 H O- + -

Cr ÷ Cr + 3 e3+ -

3 2 3 e + 4 H + NO ÷ NO + 2 H O- + -

3 24 H (aq) + NO (aq) + Cr(s) ÷ Cr (aq) + NO(g) + 2 H O(l)+ - 3+

4b. (Al ÷ Al + 3 e ) × 5 MnO ÷ Mn3+ - - 2+

4 2 8 H + MnO ÷ Mn + 4 H O+ - 2+

4 2(5 e + 8 H + MnO ÷ Mn + 4 H O) × 3- + - 2+

5 Al ÷ 5 Al + 15 e3+ -

4 2 15 e + 24 H + 3 MnO ÷ 3 Mn + 12 H O- + - 2+

4 224 H (aq) + 3 MnO (aq) + 5 Al(s) ÷ 5 Al (aq) + 3 Mn (aq) + 12 H O(l)+ - 3+ 2+

3 2c. (Ce + e ÷ Ce ) × 6 CH OH ÷ CO4+ - 3+

2 3 2H O + CH OH ÷ CO + 6 H+

2 3 2H O + CH OH ÷ CO + 6 H + 6 e+ -

6 Ce + 6 e ÷ 6 Ce4+ - 3+

2 3 2 H O + CH OH ÷ CO + 6 H + 6 e+ -

2 3 2H O(l) + CH OH(aq) + 6 Ce (aq) ÷ 6 Ce (aq) + CO (g) + 6 H (aq)4+ 3+ +

3 4d. PO ÷ PO3- 3-

2 3 4(H O + PO ÷ PO + 2 H + 2 e ) × 33- 3- + -

4 2 MnO ÷ MnO-

4 2 2(3 e + 4 H + MnO ÷ MnO + 2 H O) × 2- + -

2 3 4 3 H O + 3 PO ÷ 3 PO + 6 H + 6 e3- 3- + -

4 2 26 e + 8 H + 2 MnO ÷ 2 MnO + 4 H O- + -

4 3 4 2 2 2 H + 2 MnO + 3 PO ÷ 3 PO + 2 MnO + H O+ - 3- 3-

2Now convert to a basic solution by adding 2 OH to both sides. 2 H + 2 OH ÷ 2 H O on the- + -

2reactant side. After converting H to OH , simplify the overall equation by crossing off 1 H O on+ -

each side of the reaction. The overall balanced equation is:

2 4 3 4 2H O(l) + 2 MnO (aq) + 3 PO (aq) ÷ 3 PO (aq) + 2 MnO (s) + 2 OH (aq)- 3- 3- -

2e. Mg ÷ Mg(OH) OCl ÷ Cl- -

CHAPTER 17 ELECTROCHEMISTRY 475

2 2 22 H O + Mg ÷ Mg(OH) + 2 H + 2 e 2 e + 2 H + OCl ÷ Cl + H O+ - - + - -

2 2 2 H O + Mg ÷ Mg(OH) + 2 H + 2 e+ -

2 2 e + 2 H + OCl ÷ Cl + H O- + - -

2 2OCl (aq) + H O(l) + Mg(s) ÷ Mg(OH) (s) + Cl (aq) - -

The final overall reaction does not contain H , so we are done.+

2 3 3 2 3f. H CO ÷ HCO Ag(NH ) ÷ Ag + 2 NH- +

2 2 3 3 2 32 H O + H CO ÷ HCO + 5 H + 4 e (e + Ag(NH ) ÷ Ag + 2 NH ) × 4- + - - +

2 2 3 2 H O + H CO ÷ HCO + 5 H + 4 e- + -

3 2 3 4 e + 4 Ag(NH ) ÷ 4 Ag + 8 NH- +

3 2 2 2 3 3 4 Ag(NH ) + 2 H O + H CO ÷ HCO + 5 H + 4 Ag + 8 NH+ - +

2Convert to a basic solution by adding 5 OH to both sides (5 H + 5 OH ÷ 5 H O). Then, cross- + -

2off 2 H O on both sides, which gives the overall balanced equation:

3 2 2 3 2 3 5 OH (aq) + 4 Ag(NH ) (aq) + H CO(aq) ÷ HCO (aq) + 3 H O(l) + 4 Ag(s) + 8 NH (aq)- + -

Questions

17. In a galvanic cell, a spontaneous redox reaction occurs which produces an electric current. In anelectrolytic cell, electricity is used to force a redox reaction that is not spontaneous to occur.

18. The salt bridge allows counter ions to flow into the two cell compartments to maintain electricalneutrality. Without a salt bridge, no sustained electron flow can occur.

19. a. Cathode: The electrode at which reduction occurs.

b. Anode: The electrode at which oxidation occurs.

c. Oxidation half-reaction: The half-reaction in which electrons are products. In a galvanic cell,the oxidation half-reaction always occurs at the anode.

d. Reduction half-reaction: The half-reaction in which electrons are reactants. In a galvanic cell,the reduction half-reaction always occurs at the cathode.

20. a. Purification by electrolysis is called electrorefining. See the text for a discussion of theelectrorefining of copper. Electrorefining is possible because of the selectivity of electrodereactions. The anode is made up of the impure metal. A potential is applied so just the metal ofinterest and all more easily oxidized metals are oxidized at the anode. The metal of interest is theonly metal plated at the cathode due to the careful control of the potential applied. The metal ionsthat could plate out at the cathode in preference to the metal we are purifying will not be insolution, because these metals were not oxidized at the anode.

b. A more easily oxidized metal is placed in electrical contact with the metal we are trying to protect.It is oxidized in preference to the protected metal. The protected metal becomes the cathode


electrode, thus, cathodic protection.

cell21. As a battery discharges, E decreases, eventually reaching zero. A charged battery is not at

cellequilibrium. At equilibrium, E = 0 and )G = 0. We get no work out of an equilibrium system. Abattery is useful to us because it can do work as it approaches equilibrium.

22. Standard reduction potentials are an intensive property, i.e., they do not depend on how many timesthe reaction occurs. As long as the concentrations of ions and gases are 1 M or 1 atm, standardreduction potentials and standard oxidation potentials are a constant and not dependent on thecoefficients in the balanced equation.

23. Both fuel cells and batteries are galvanic cells. However, fuel cells, unlike batteries, have the reactantscontinuously supplied and can produce a current indefinitely.

24. Moisture must be present to act as a medium for ion flow between the anodic and cathodic regions.Salt provides ions necessary to complete the electrical circuit in the corrosion process. Together, saltand water make up the salt bridge in this spontaneous electrochemical process.

Three methods discussed in the text to prevent corrosion are galvanizing, alloying and cathodicprotection. Galvanizing coats the metal of interest (usually iron) with zinc, which is an easily oxidizedmetal. Alloying mixes in metals that form durable, effective oxide coatings over the metal of interest.Cathodic protection connects, via a wire, a more easily oxidized metal to the metal we are trying toprotect. The more active metal is preferentially oxidized, thus protecting our metal object fromcorrosion.

Exercises

Galvanic Cells, Cell Potentials, Standard Reduction Potentials, and Free Energy

25. A typical galvanic cell diagram is:

The diagram for all cells will look like this. The contents of each half-cell compartment will beidentified for each reaction, with all solute concentrations at 1.0 M and all gases at 1.0 atm. ForExercises 17.25 and 17.26, the flow of ions through the salt bridge was not asked for in the questions.

If asked, however, cations always flow into the cathode compartment, and anions always flow into theanode compartment. This is required to keep each compartment electrically neutral.


a. Table 17.1 of the text lists balanced reduction half-reactions for many substances. For this overall

2 2 7reaction, we need the Cl to Cl reduction half-reaction and the Cr to Cr O oxidation half-- 3+ 2-

reaction. Manipulating these two half-reactions gives the overall balanced equation.

2 (Cl + 2 e ÷ 2 Cl ) × 3- -

2 2 7 7 H O + 2 Cr ÷ Cr O + 14 H + 6 e3+ 2- + -

2 2 2 77 H O(l) + 2 Cr (aq) + 3 Cl (g) ÷ Cr O (aq) + 6 Cl (aq) + 14 H (aq)3+ 2- - +

The contents of each compartment are:

2Cathode: Pt electrode; Cl bubbled into solution, Cl in solution-

2 7Anode: Pt electrode; Cr , H , and Cr O in solution3+ + 2-

We need a nonreactive metal to use as the electrode in each case, since all the reactants andproducts are in solution. Pt is a common choice. Another possibility is graphite.

b. Cu + 2 e ÷ Cu2+ -

Mg ÷ Mg + 2e 2+ -

Cu (aq) + Mg(s) ÷ Cu(s) + Mg (aq)2+ 2+

Cathode: Cu electrode; Cu in solution; Anode: Mg electrode; Mg in solution2+ 2+

26. Reference Exercise 17.25 for a typical galvanic cell diagram. The contents of each half-cellcompartment are identified below with all solute concentrations at 1.0 M and all gases at 1.0 atm.

a. Reference Table 17.1 for the balanced half-reactions.

3 2 2 5 e + 6 H + IO ÷ 1/2 I + 3 H O- + -

(Fe ÷ Fe + e ) × 52+ 3+ -

3 2 2 6 H + IO + 5 Fe ÷ 5 Fe + 1/2 I + 3 H O+ - 2+ 3+

3 2 2or 12 H (aq) + 2 IO (aq) + 10 Fe (aq) ÷ 10 Fe (aq) + I (aq) + 6 H O(l)+ - 2+ 3+

3 2 2 4Cathode: Pt electrode; IO , I and H SO (H source) in solution.- +

2Note: I (s) would make a poor electrode since it sublimes.

Anode: Pt electrode; Fe and Fe in solution2+ 3+

b. (Ag + e ÷ Ag) × 2+ -

Zn ÷ Zn + 2 e2+ -

Zn(s) + 2 Ag (aq) ÷ 2 Ag(s) + Zn (aq)+ 2+


Cathode: Ag electrode; Ag in solution; Anode: Zn electrode; Zn in solution+ 2+

27. To determine E° for the overall cell reaction, we must add the standard reduction potential to the

standard oxidation potential ( ). Reference Table 17.1 for values of standard reduction

potentials. Remember that and that standard potentials are not multiplied by the integer

used to obtain the overall balanced equation.

25a. = + = 1.36 V + (-1.33 V) = 0.03 V

25b. = + = 0.34 V + 2.37 V = 2.71 V

28. 26a. = + = 1.20 V + (-0.77 V) = 0.43 V

26b. = + = 0.80 V + 0.76 V = 1.56 V

29. Reference Exercise 17.25 for a typical galvanic cell design. The contents of each half-cellcompartment are identified below with all solute concentrations at 1.0 M and all gases at 1.0 atm. Foreach pair of half-reactions, the half-reaction with the largest standard reduction potential will be thecathode reaction, and the half-reaction with the smallest reduction potential will be reversed to becomethe anode reaction. Only this combination gives a spontaneous overall reaction, i.e., a reaction witha positive overall standard cell potential. Note that in a galvanic cell as illustrated in Exercise 17.25,the cations in the salt bridge migrate to the cathode, and the anions migrate to the anode.

2a. Cl + 2 e ÷ 2 Cl E° = 1.36 V- -

2 2 Br ÷ Br + 2 e -E° = -1.09 V- -

2 2Cl (g) + 2 Br (aq) ÷ Br (aq) + 2 Cl (aq) = 0.27 V- -

The contents of each compartment are:

2Cathode: Pt electrode; Cl (g) bubbled in, Cl in solution-

2Anode: Pt electrode; Br and Br in solution-

4 3 2b. (2 e + 2 H + IO ÷ IO + H O) × 5 E° = 1.60 V- + - -

2 4 (4 H O + Mn ÷ MnO + 8 H + 5 e ) × 2 -E° = -1.51 V2+ - + -

4 2 3 2 410 H + 5 IO + 8 H O + 2 Mn ÷ 5 IO + 5 H O + 2 MnO + 16 H = 0.09 V+ - 2+ - - +

This simplifies to:

2 4 3 43 H O(l) + 5 IO (aq) + 2 Mn (aq) ÷ 5 IO (aq) + 2 MnO (aq) + 6 H (aq) = 0.09 V- 2+ - - +

4 3 2 4Cathode: Pt electrode; IO , IO , and H SO (as a source of H ) in solution- - +

4 2 4Anode: Pt electrode; Mn , MnO and H SO in solution2+ -


30. Reference Exercise 17.25 for a typical galvanic cell design. The contents of each half-cellcompartment are identified below, with all solute concentrations at 1.0 M and all gases at 1.0 atm.

2 2 2a. H O + 2 H + 2 e ÷ 2 H O E° = 1.78 V+ -

2 2 2 H O ÷ O + 2 H + 2 e -E° = -0.68 V+ -

2 2 2 2 2 H O (aq) ÷ 2 H O(l) + O (g) = 1.10 V

2 2Cathode: Pt electrode; H O and H in solution+

2 2 2Anode: Pt electrode; O (g) bubbled in, H O and H in solution+

b. (Fe + 3 e ÷ Fe) × 2 E° = -0.036 V3+ -

(Mn ÷ Mn + 2 e ) × 3 -E° = 1.18 V2+ -

2 Fe (aq) + 3 Mn(s) ÷ 2 Fe(s) + 3 Mn (aq) = 1.14 V3+ 2+

Cathode: Fe electrode; Fe in solution; Anode: Mn electrode; Mn in solution3+ 2+

31. In standard line notation, the anode is listed first and the cathode is listed last. A double line separatesthe two compartments. By convention, the electrodes are on the ends with all solutes and gasestowards the middle. A single line is used to indicate a phase change. We also include allconcentrations.

2 7 225a. Pt * Cr (1.0 M), Cr O (1.0 M), H (1.0 M) ** Cl (1.0 atm)* Cl (1.0 M) * Pt3+ 2- + -

25b. Mg * Mg (1.0 M) ** Cu (1.0 M) * Cu2+ 2+

2 229a. Pt * Br (1.0 M), Br (1.0 M) ** Cl (1.0 atm) * Cl (1.0 M) * Pt- -

4 4 329b. Pt * Mn (1.0 M), MnO (1.0 M), H (1.0 M) ** IO (1.0 M), H (1.0 M), IO (1.0 M) * Pt2+ - + - + -

3 232. 26a. Pt * Fe (1.0 M), Fe (1.0 M) ** IO (1.0 M), H (1.0 M), I (1.0 M)* Pt2+ 3+ - +

26b. Zn * Zn (1.0 M) ** Ag (1.0 M) * Ag2+ +

2 2 2 2 230a. Pt * H O (1.0 M), H (1.0 M) | O (1.0 atm)** H O (1.0 M), H (1.0 M)*Pt+ +

30b. Mn * Mn (1.0 M) ** Fe (1.0 M) * Fe2+ 3+

33. Locate the pertinent half-reactions in Table 17.1, and then figure which combination will give apositive standard cell potential. In all cases, the anode compartment contains the species with thesmallest standard reduction potential. For part a, the copper compartment is the anode, and in partb, the cadmium compartment is the anode.


a. Au + 3 e ÷ Au E° = 1.50 V3+ -

(Cu ÷ Cu + e ) × 3 - E° = -0.16 V+ 2+ -

Au (aq) + 3 Cu (aq) ÷ Au(s) + 3 Cu (aq) = 1.34 V3+ + 2+

2 2b. (VO + 2 H + e ÷ VO + H O) × 2 E° = 1.00 V+ + - 2+

Cd ÷ Cd + 2e -E° = 0.40 V2+ -

2 22 VO (aq) + 4 H (aq) + Cd(s) ÷ 2 VO (aq) + 2 H O(l) + Cd (aq) = 1.40 V+ + 2+ 2+

2 2 234. a. (H O + 2 H + 2 e ÷ 2 H O) × 3 E° = 1.78 V+ -

2 2 7 2 Cr + 7 H O ÷ Cr O + 14 H + 6 e -E° = -1.33 V3+ 2- + -

2 2 2 2 73 H O (aq) + 2 Cr (aq) + H O(l) ÷ Cr O (aq) + 8 H (aq) = 0.45 V3+ 2- +

2b. (2 H + 2 e ÷ H ) × 3 E° = 0.00 V+ -

(Al ÷ Al + 3 e ) × 2 -E° = 1.66 V3+ -

26 H (aq) + 2 Al(s) ÷ 3 H (g) + 2 Al (aq) = 1.66 V+ 3+

35. a. 2 Ag + 2 e ÷ 2 Ag E° = 0.80 V+ -

Cu ÷ Cu + 2 e -E° = -0.34 V2+ -

2 Ag + Cu ÷ Cu + 2 Ag = 0.46 V Spontaneous at standard conditions + 2+

( > 0).

b. Zn + 2 e ÷ Zn E° = -0.76 V2+ -

Ni ÷ Ni + 2 e -E° = 0.23 V2+ -

Zn + Ni ÷ Zn + Ni = -0.53 V Not spontaneous at standard conditions2+ 2+

( < 0).

4 236. a. (5 e + 8 H + MnO ÷ Mn + 4 H O) × 2 E° = 1.51 V- + - 2+

2 (2 I ÷ I + 2 e ) × 5 -E° = -0.54 V- -

4 2 216 H + 2 MnO + 10 I ÷ 5 I + 2 Mn + 8 H O = 0.97 V Spontaneous+ - - 2+

4 2b. (5 e + 8 H + MnO ÷ Mn + 4 H O) × 2 E° = 1.51 V- + - 2+

2 (2 F ÷ F + 2 e ) × 5 -E° = -2.87 V- -

4 2 216 H + 2 MnO + 10 F ÷ 5 F + 2 Mn + 8 H O = -1.36 V Not spontaneous+ - - 2+

237. Cl + 2 e ÷ 2 Cl E° = 1.36 V- -

2 2 (ClO ÷ ClO + e ) × 2 -E° = -0.954 V- -

2 2 22 ClO (aq) + Cl (g) ÷ 2 ClO (aq) + 2 Cl (aq) = 0.41 V = 0.41 J/C- -

)G° = = -(2 mol e )(96,485 C/mol e )(0.41 J/C) = -7.9 × 10 J = -79 kJ- - 4


3 238. a. (4 H + NO + 3 e ÷ NO + 2 H O) × 2 E° = 0.96 V+ - -

(Mn ÷ Mn + 2 e ) × 3 -E° = 1.18 V2+ -

3 23 Mn(s) + 8 H (aq) + 2 NO (aq) ÷ 2 NO(g) + 4 H O(l) + 3 Mn (aq) = 2.14 V+ - 2+

4 3 2 (2 e + 2 H + IO ÷ IO + H O) × 5 E° = 1.60 V- + - -

2 4 (Mn + 4 H O ÷ MnO + 8 H + 5 e ) × 2 -E° = -1.51 V2+ - + -

4 2 3 45 IO (aq) + 2 Mn (aq) + 3 H O(l) ÷ 5 IO (aq) + 2 MnO (aq) + 6 H (aq) = 0.09 V- 2+ - - +

b. Nitric acid oxidation (see above for ):


Periodate oxidation (see above for ):

)G° = -(10 mol e )(96,485 C/mol e )(0.09 J/C)(1 kJ/1000 J) = -90 kJ- -

max39. Since the cells are at standard conditions, w = )G = )G° = . See Exercise 17.33 for thebalanced overall equations and for .

max33a. w = -(3 mol e )(96,485 C/mol e )(1.34 J/C) = -3.88 × 10 J = -388 kJ- - 5

max33b. w = -(2 mol e )(96,485 C/mol e )(1.40 J/C) = -2.70 × 10 J = -270. kJ- - 5

max40. Since the cells are at standard conditions, w = )G = )G° = . See Exercise 17.34 for thebalanced overall equations and for .

max34a. w = -(6 mol e )(96,485 C/mol e )(0.45 J/C) = -2.6 × 10 J = -260 kJ- - 5

max34b. w = -(6 mol e )(96,485 C/mol e )(1.66 J/C) = -9.61 × 10 J = -961 kJ- - 5

3 2 2 241. CH OH(l) + 3/2 O (g) ÷ CO (g) + 2 H O(l) )G° = 2(-237) + (-394) - [-166] = -702 kJ

The balanced half-reactions are:

2 3 2 2 2H O + CH OH ÷ CO + 6 H + 6 e and O + 4 H + 4 e ÷ 2 H O+ - + -

2For 3/2 mol O , 6 moles of electrons will be transferred (n = 6).

)G° = -nFE°, E° = = 1.21 J/C = 1.21 V

42. Fe + 2 e ÷ Fe E° = -0.44 V = -0.44 J/C2+ -

)G° = -nFE° = -(2 mol e )(96,485 C/mol e )(-0.44 J/C)(1 kJ/1000 J) = 85 kJ- -

85 kJ = 0 - , = -85 kJ


We can get two ways. Consider: Fe + e ÷ Fe E° = 0.77 V3+ - 2+

)G° = -(1 mol e)(96,485 C/mol e )(0.77 J/C) = -74,300 J = -74 kJ-

Fe ÷ Fe + e )G° = 74 kJ2+ 3+ -

Fe ÷ Fe + 2 e )G° = -85 kJ2+ -

Fe ÷ Fe + 3 e )G° = -11 kJ, = -11 kJ/mol3+ -

or consider: Fe + 3 e ÷ Fe E° = -0.036 V3+ -

)G° = -(3 mol e )(96,485 C/mol e )(-0.036 J/C) = 10,400 J . 10. kJ- -

10. kJ = 0 - , = -10. kJ/mol; Round-off error explains the 1 kJ discrepancy.

43. Good oxidizing agents are easily reduced. Oxidizing agents are on the left side of the reduction half-reactions listed in Table 17.1. We look for the largest, most positive standard reduction potentials tocorrespond to the best oxidizing agents. The ordering from worst to best oxidizing agents is:

2 2 4 3 K < H O < Cd < I < AuCl < IO+ 2+ - -

E°(V) -2.92 -0.83 -0.40 0.54 0.99 1.20

44. Good reducing agents are easily oxidized. The reducing agents are on the right side of the reductionhalf-reactions listed in Table 17.1. The best reducing agents have the most negative standard reductionpotentials (E°) or the most positive standard oxidation potentials, (= -E°).

2 F < Cr < Fe < H < Zn < Li- 3+ 2+

-E°(V) -2.87 -1.33 -0.77 0.00 0.76 3.05

245. a. 2 H + 2 e ÷ H E° = 0.00 V; Cu ÷ Cu + 2 e -E° = -0.34 V+ - 2+ -

= -0.34 V; No, H cannot oxidize Cu to Cu at standard conditions ( < 0).+ 2+

2b. Fe + e ÷ Fe E° = 0.77 V; 2 I ÷ I + 2 e -E° = -0.54 V3+ - 2+ - -

2 = 0.77 - 0.54 = 0.23 V; Yes, Fe can oxidize I to I .3+ -

2c. H ÷ 2 H + 2 e -E° = 0.00 V; Ag + e ÷ Ag E° = 0.80 V + - + -

2 = 0.80 V; Yes, H can reduce Ag to Ag at standard conditions ( > 0).+

d. Fe ÷ Fe + e -E° = -0.77 V; Cr + e ÷ Cr E° = -0.50 V2+ 3+ - 3+ - 2+

= -0.50 - 0.77 = -1.27 V; No, Fe cannot reduce Cr to Cr at standard conditions.2+ 3+ 2+

246. Cl + 2 e ÷ 2 Cl E° = 1.36 V Ag + e ÷ Ag E° = 0.80 V- - + -

Pb + 2 e ÷ Pb E° = -0.13 V Zn + 2 e ÷ Zn E° = -0.76 V2+ - 2+ -


Na + e ÷ Na E° = -2.71 V+ -

a. Oxidizing agents (species reduced) are on the left side of the above reduction half-reactions. Ofthe species available, Ag would be the best oxidizing agent since it has the largest E° value. Note+

2that Cl is a better oxidizing agent than Ag , but it is not one of the choices listed.+

b. Reducing agents (species oxidized) are on the right side of the reduction half-reactions. Of thespecies available, Zn would be the best reducing agent since it has the largest -E° value.

4 2 3 2 4c. SO + 4 H + 2 e ÷ H SO + H O E° = 0.20 V; SO can oxidize Pb and Zn at standard2- + - 2-

4conditions. When SO is coupled with these reagents, is positive.2-

d. Al ÷ Al + 3 e -E° = 1.66 V; Al can oxidize Ag and Zn at standard conditions since 3+ - + 2+

> 0.

2 247. a. 2 Br ÷ Br + 2 e -E° = -1.09 V; 2 Cl ÷ Cl + 2 e -E° = -1.36 V; E° > 1.09 V to oxidize- - - -

2 7 2 2 3Br ; E° < 1.36 V to not oxidize Cl ; Cr O , O , MnO , and IO are all possible since when all- - 2- -

of these oxidizing agents are coupled with Br , > 0, and when coupled with Cl , < 0- -

(assuming standard conditions).

b. Mn ÷ Mn + 2 e -E° = 1.18; Ni ÷ Ni + 2 e -E° = 0.23 V; Any oxidizing agent with2+ - 2+ -

4 2-0.23 V > E° > -1.18 V will work. PbSO , Cd , Fe , Cr , Zn and H O will be able to oxidize2+ 2+ 3+ 2+

Mn but not Ni (assuming standard conditions).

48. a. Cu + 2 e ÷ Cu E° = 0.34 V; Cu + e ÷ Cu E° = 0.16 V; To reduce Cu to Cu but2+ - 2+ - + 2+

not reduce Cu to Cu , the reducing agent must have a standard oxidation potential ( =2+ +

-E°) between -0.34 V and -0.16 V (so is positive only for the Cu to Cu reduction). The2+

reducing agents (species oxidized) are on the right side of the half-reactions in Table 17.1. Thereagents at standard conditions which have ( = -E°) between -0.34 V and -0.16 V are Ag (in

2 3.1.0 M Cl ) and H SO-

2 2b. Br + 2 e ÷ 2 Br E° = 1.09 V; I + 2 e ÷ 2 I E° = 0.54 V; From Table 17.1, VO , Au (in- - - - 2+

2 2 2 2 41.0 M Cl ), NO, ClO , Hg , Ag, Hg, Fe , H O and MnO are all capable at standard conditions- - 2+ 2+ -

2 2 2of reducing Br to Br but not reducing I to I . When these reagents are coupled with Br , >- -

20, and when coupled with I , < 0.

249. ClO + H O + 2 e ÷ 2 OH + Cl E° = 0.90 V- - - -

3 2 4 2 2 NH + 2 OH ÷ N H + 2 H O + 2 e -E° = 0.10 V- -

3 2 4 2 ClO (aq) + 2 NH (aq) ÷ Cl (aq) + N H (aq) + H O(l) = 1.00 V- -

Since is positive for this reaction, then at standard conditions ClO can spontaneously oxidize-

3 2 4NH to the somewhat toxic N H .

50. Tl + 2 e ÷ Tl E° = 1.25 V3+ - +

3 3 I ÷ I + 2 e -E° = -0.55 V- - -

3Tl + 3 I ÷ Tl + I = 0.70 V3+ - + -


3 3In solution, Tl can oxidize I to I . Thus, we expect TlI to be thallium(I) triiodide.3+ - -

The Nernst Equation

2 2 251. H O + 2 H + 2 e ÷ 2 H O E° = 1.78 V+ -

(Ag ÷ Ag + e ) × 2 -E° = -0.80 V+ -

2 2 2H O (aq) + 2 H (aq) + 2 Ag(s) ÷ 2 H O(l) + 2 Ag (aq) = 0.98 V+ +

a. A galvanic cell is based on spontaneous chemical reactions. At standard conditions, this reactionproduces a voltage of 0.98 V. Any change in concentration that increases the tendency of theforward reaction to occur will increase the cell potential. Conversely, any change in concentrationthat decreases the tendency of the forward reaction to occur (increases the tendency of the reversereaction to occur) will decrease the cell potential. Using Le Chatelier’s principle, increasing the

2 2reactant concentrations of H O and H from 1.0 M to 2.0 M will drive the forward reaction+

further to right (will further increase the tendency of the forward reaction to occur). Therefore,

cellE will be greater than .

b. Here, we decreased the reactant concentration of H and increased the product concentration of+

Ag from the standard conditions. This decreases the tendency of the forward reaction to occur+

cell cellwhich will decrease E as compared to (E < ).

52. The concentrations of Fe in the two compartments are now 0.01 M and 1 × 10 M. The driving force2+ -7

for this reaction is to equalize the Fe concentrations in the two compartments. This occurs if the2+

compartment with 1 × 10 M Fe becomes the anode (Fe will be oxidized to Fe ) and the-7 2+ 2+

compartment with the 0.01 M Fe becomes the cathode (Fe will be reduced to Fe). Electron flow,2+ 2+

as always for galvanic cells, goes from the anode to the cathode, so electron flow will go from the rightcompartment ([Fe ] = 1 × 10 M) to the left compartment ([Fe ] = 0.01 M).2+ -7 2+

53. For concentration cells, the driving force for the reaction is the difference in ion concentrations betweenthe anode and cathode. In order to equalize the ion concentrations, the anode always has the smallerion concentration. The general setup for this concentration cell is:

Cathode: Ag (x M) + e ÷ Ag E° = 0.80 V+ -

Anode: Ag ÷ Ag (y M) + e -E° = -0.80 V+ -

Ag (cathode, x M) ÷ Ag (anode, y M) = 0.00 V+ +

log Q = log

For each concentration cell, we will calculate the cell potential using the above equation. Rememberthat the anode always has the smaller ion concentration.

a. Since both compartments are at standard conditions ([Ag ] = 1.0 M) then = 0 V. No+

voltage is produced since no reaction occurs. Concentration cells only produce a voltage when theion concentrations are not equal.

b. Cathode = 2.0 M Ag ; Anode = 1.0 M Ag ; Electron flow is always from the anode to the+ +


cathode, so electrons flow to the right in the diagram.

log = log = 0.018 V

c. Cathode = 1.0 M Ag ; Anode = 0.10 M Ag ; Electrons flow to the left in the diagram.+ +

log = log = 0.059 V

d. Cathode = 1.0 M Ag ; Anode = 4.0 × 10 M Ag ; Electrons flow to the left in the diagram.+ -5 +

log = 0.26 V

anode cathode celle. Since the ion concentrations are the same, then log ([Ag ] /[Ag ] ) = log (1.0) = 0 and E+ +

= 0. No electron flow occurs.

54. As is the case for all concentration cells, = 0, and the smaller ion concentration is always in theanode compartment. The general Nernst equation for the Ni * Ni (x M) ** Ni (y M) * Ni2+ 2+

concentration cell is:

log Q = log

a. Since both compartments are at standard conditions ([Ni ] = 1.0 M), then = 0 V. No2+

electron flow occurs.

b. Cathode = 2.0 M Ni ; Anode = 1.0 M Ni ; Electron flow is always from the anode to the2+ 2+

cathode, so electrons flow to the right in the diagram.

log = log = 8.9 × 10 V-3

c. Cathode = 1.0 M Ni ; Anode = 0.10 M Ni ; Electrons flow to the left in the diagram.2+ 2+

log = 0.030 V

d. Cathode = 1.0 M Ni ; Anode = 4.0 × 10 M Ni ; Electrons flow to the left in the diagram.2+ -5 2+

log = 0.13 V

celle. Since both concentrations are equal, log (2.5/2.5) = log 1.0 = 0 and E = 0. No electron flowoccurs.

4 255. 5 e + 8 H + MnO ÷ Mn + 4 H O E° = 1.51 V- + - 2+

(Fe ÷ Fe + e ) × 5 -E° = -0.77 V2+ 3+ -

4 28 H (aq) + MnO (aq) + 5 Fe (aq) ÷ 5 Fe (aq) + Mn (aq) + 4 H O(l) = 0.74 V+ - 2+ 3+ 2+


cellE = - log Q = 0.74 V - ; pH = 4.0 so H =+

1 × 10 M-4

cellE = 0.74 -

cellE = 0.74 - log (1 × 10 ) = 0.74 V - 0.15 V = 0.59 V = 0.6 V (1 sig. fig. due to13

concentrations)

cellYes, E > 0 so the reaction will occur as written.

456. n = 2 for this reaction (lead goes from Pb ÷ Pb in PbSO ).2+

E = E° - log = 2.04 V - log

E = 2.04 V + 0.077 V = 2.12 V

57. Cu + 2 e ÷ Cu E° = 0.34 V2+ -

Zn ÷ Zn + 2 e -E° = 0.76 V2+ -

Cu (aq) + Zn(s) ÷ Zn (aq) + Cu(s) = 1.10 V2+ 2+

Since Zn is a product in the reaction, the Zn concentration increases from 1.00 M to 1.20 M. This2+ 2+

means that the reactant concentration of Cu must decrease from 1.00 M to 0.80 M (from the 1:1 mol2+

ratio in the balanced reaction).

log Q = 1.10 V - log

cellE = 1.10 V - log = 1.10 V - 0.0052 V = 1.09 V

58. (Pb + 2 e ÷ Pb) × 3 E° = -0.13 V2+ -

(Al ÷ Al + 3 e ) × 2 -E° = 1.66 V3+ -

3 Pb (aq) + 2 Al(s) ÷ 3 Pb(s) + 2 Al (aq) = 1.53 V2+ 3+

From the balanced reaction, when the Al has increased by 0.60 mol/L (Al is a product in the3+ 3+

spontaneous reaction), then the Pb concentration has decreased by 3/2 (0.60 mol/L) = 0.90 M.2+

cellE = 1.53 V - = 1.53 -

cellE = 1.53 V - 0.034 V = 1.50 V

259. Cu (aq) + H (g) ÷ 2 H (aq) + Cu(s) = 0.34 V - 0.00V = 0.34 V; n = 2 mol electrons2+ +

Since = 1.0 atm and [H ] = 1.0 M: - log +


a. = 0.34 V - log = 0.34 V - 0.11V = 0.23 V

b. 0.195 V = 0.34 V - log , log = 4.91, [Cu ] = 10 = 1.2 × 10 M2+ -4.91 -5

Note: When determining exponents, we will carry extra significant figures.

60. 3 Ni (aq) + 2 Al(s) ÷ 2 Al (aq) + 3 Ni(s) = -0.23 + 1.66 = 1.43 V; n = 6 mol electrons2+ 3+

cella. E = 1.43 V - = 1.43 -

cell E = 1.43 V - (-0.042 V) = 1.47 V

b. 1.62 V = 1.43 V - , log [Al ] = -19.293+ 2

[Al ] = 10 , [Al ] = 2.3 × 10 M3+ 2 -19.29 3+ -10

261. Cu (aq) + H (g) ÷ 2 H (aq) + Cu(s) = 0.34 V - 0.00 V = 0.34 V; n = 22+ +

Since = 1.0 atm and [H ] = 1.0 M: - log +

spUse the K expression to calculate the Cu concentration in the cell.2+

2 spCu(OH) (s) º Cu (aq) + 2 OH (aq) K = 1.6 × 10 = [Cu ] [OH ]2+ - -19 2+ - 2

From problem, [OH ] = 0.10 M, so: [Cu ] = = 1.6 × 10 M- 2+ -17

- log = 0.34 V - log = 0.34 - 0.50 = -0.16 V

cellSince E < 0, the forward reaction is not spontaneous, but the reverse reaction is spontaneous. The

cellCu electrode becomes the anode and E = 0.16 V for the reverse reaction. The cell reaction is: 2

2H (aq) + Cu(s) ÷ Cu (aq) + H (g).+ 2+

62. 3 Ni (aq) + 2 Al(s) ÷ 2 Al (aq) + 3 Ni(s) = -0.23 V + 1.66 V = 1.43 V; n = 62+ 3+

cellE = - , 1.82 V = 1.43 V -

log [Al ] = -39.59, [Al ] = 10 , [Al ] = 1.6 × 10 M3+ 2 3+ 2 -39.59 3+ -20

3 spAl(OH) (s) Al (aq) + 3 OH (aq) K = [Al ] [OH ] ; From the problem, [OH ]3+ - 3+ - 3 -

= 1.0 × 10 M.-4


spK = (1.6 × 10 ) (1.0 × 10 ) = 1.6 × 10-20 -4 3 -32

63. See Exercises 17.25, 17.27 and 17.29 for balanced reactions and standard cell potentials. Balancedreactions are necessary to determine n, the moles of electrons transferred.

2 2 2 725a. 7 H O + 2 Cr + 3 Cl ÷ Cr O + 6 Cl + 14 H = 0.03 V = 0.03 J/C3+ 2- - +


cell cellE = - log Q: At equilibrium, E = 0 and Q = K, so = log K

log K = = 3.05, K = 10 = 1 × 103.05 3

Note: When determining exponents, we will round off to the correct number of significantfigures after the calculation is complete in order to help eliminate excessive round-off error.

25b. )G°= -(2 mol e )(96,485 C/mol e )(2.71 J/C) = -5.23 × 10 J = -523 kJ- - 5

log K = = 91.709, K = 5.12 × 1091

29a. )G° = -(2 mol e )(96,485 C/mol e )(0.27 J/C) = -5.21 × 10 J = -52 kJ- - 4

log K = = 9.14, K = 1.4 × 109

29b. )G° = -(10 mol e )(96,485 C/mol e )(0.09 J/C) = -8.7 × 10 J = -90 kJ- - 4

log K = = 15.23, K = 2 × 1015

64. )G° = ; log K, log K =

26a. )G° = -(10 mol e )(96,485 C/mol e )(0.43 J/C) = -4.1 × 10 J = -410 kJ - - 5

log K = = 72.76, K = 10 = 5.8 × 1072.76 72

26b. )G° = -(2 mol e )(96,485 C/mol e )(1.56 J/C) = -3.01 × 10 J = -301 kJ- - 5

log K = = 52.792, K = 6.19 × 1052

30a. )G° = -(2 mol e )(96,485 C/mol e )(1.10 J/C) = -2.12 × 10 J = -212 kJ- - 5

log K = = 37.225, K = 1.68 × 1037

30b. )G° = -(6 mol e )(96,485 C/mol e )(1.14 J/C) = -6.60 × 10 J = -660. kJ- - 5

log K = =115.736, K = 5.45 × 10115


2 265. a. Possible reaction: I (s) + 2 Cl (aq) ÷ 2 I (aq) + Cl (g) = 0.54 V - 1.36 V = -0.82 V- -

This reaction is not spontaneous at standard conditions since < 0. No reaction occurs.

2 2b. Possible reaction: Cl (g) + 2 I (aq) ÷ I (s) + 2 Cl (aq) = 0.82 V; This reaction is- -

spontaneous at standard conditions since > 0. The reaction will occur.

2 2Cl (g) + 2 I (aq) ÷ I (s) + 2 Cl (aq) = 0.82 V = 0.82 J/C- -


E° = log K, log K = = = 27.75, K = 10 = 5.6 × 1027.75 27

c. Possible reaction: 2 Ag(s) + Cu (aq) ÷ Cu(s) + 2 Ag (aq) = -0.46 V; No reaction2+ +

occurs.

4 2 2d. Fe can be oxidized or reduced. The other species present are H , SO , H O, and O from air.2+ + 2-

2Only O in the presence of H has a large enough standard reduction potential to oxidize Fe to+ 2+

Fe (resulting in > 0). All other combinations, including the possible reduction of Fe , give3+ 2+

negative cell potentials. The spontaneous reaction is:

2 24 Fe (aq) + 4 H (aq) + O (g) ÷ 4 Fe (aq) + 2 H O(l) = 1.23 - 0.77 = 0.46 V2+ + 3+

)G° = = -(4 mol e )(96,485 C/mol e )(0.46 J/C)(1 kJ/1000 J) = -180 kJ- -

log K = = 31.13, K = 1.3 × 1031

66. a. Cu + e ÷ Cu E° = 0.52 V+ -

Cu ÷ Cu + e -E° = -0.16 V+ 2+ -

2 Cu (aq) ÷ Cu (aq) + Cu(s) = 0.36 V; Spontaneous+ 2+

)G° = = -(1 mol e )(96,485 C/mol e)(0.36 J/C) = -34,700 J = -35 kJ-

= log K, log K = = 6.09, K = 10 = 1.2 × 106.09 6

b. Fe + 2 e ÷ Fe E° = -0.44 V2+ -

(Fe ÷ Fe + e ) × 2 -E° = -0.77 V2+ 3+ -

3 Fe (aq) ÷ 2 Fe (aq) + Fe(s) = -1.21 V; Not spontaneous2+ 3+

2 2c. HClO + 2 H + 2 e ÷ HClO + H O E° = 1.65 V+ -

2 2 3 HClO + H O ÷ ClO + 3 H + 2 e -E° = -1.21 V- + -

2 3 2 HClO (aq) ÷ ClO (aq) + H (aq) + HClO(aq) = 0.44 V; Spontaneous- +


)G° = = -(2 mol e )(96,485 C/mol e )(0.44 J/C) = -84,900 J = -85 kJ- -

log K = = 14.89, K = 7.8 × 1014

67. a. Au + 3 e ÷ Au E° = 1.50 V3+ -

(Tl ÷ Tl + e ) × 3 -E° = 0.34 V+ -

Au (aq) + 3 Tl(s) ÷ Au(s) + 3 Tl (aq) = 1.84 V3+ +

b. )G° = = -(3 mol e )(96,485 C/mol e )(1.84 J/C) = -5.33 × 10 J = -533 kJ- - 5

log K = = 93.401, K = 10 = 2.52 × 1093.401 93

c. = 1.84 V - log = 1.84 - log

= 1.84 - (-0.20) = 2.04 V

68. (Cr ÷ Cr + e ) × 22+ 3+ -

Co + 2 e ÷ Co2+ -

2 Cr + Co ÷ 2 Cr + Co2+ 2+ 3+

= log K = log (2.79 × 10 ) = 0.220 V7

E = E° - log = 0.220 V - log = 0.151 V

)G = -nFE = -(2 mol e )(96,485 C/mol e )(0.151 J/C) = -2.91 × 10 J = -29.1 kJ - - 4

69. CdS + 2 e ÷ Cd + S E° = -1.21 V- 2-

Cd ÷ Cd + 2 e -E° = 0.402 V2+ -

sp CdS(s) ÷ Cd (aq) + S (aq) = -0.81 V K = ?2+ 2-

sp splog K = = -27.41, K = 10 = 3.9 × 10-27.41 -28

70. Al + 3 e ÷ Al E° = -1.66 V3+ -

6 Al + 6 F ÷ AlF + 3 e -E° = 2.07 V- 3- -


6Al (aq) + 6 F (aq) ÷ AlF (aq) = 0.41 V K = ?3+ - 3-

log K = = 20.81, K = 10 = 6.5 × 1020.81 20

71. Ag + e ÷ Ag E° = 0.80 V+ -

2 3 2 3 2 Ag + 2 S O ÷ Ag(S O ) + e -E° = -0.017 V2- 3- -

2 3 2 3 2 Ag (aq) + 2 S O (aq) ÷ Ag(S O ) (aq) = 0.78 V K = ?+ 2- 3-

For this overall reaction, log K

log K = = 13.20, K = 10 = 1.6 × 1013.20 13

72. CuI + e ÷ Cu + I = ?- -

Cu ÷ Cu + e -E°= -0.52 V+ -

CuI(s) ÷ Cu (aq) + I (aq) = - 0.52 V + -

spFor this overall reaction, K = K = 1.1 × 10 :-12

sp = log K = log (1.1 × 10 ) = -0.71 V-12

= -0.71 V = - 0.52, = -0.19 V

Electrolysis

73. a. Al + 3 e ÷ Al; 3 mol e are needed to produce 1 mol Al from Al .3+ - - 3+

1.0 × 10 g Al × = 1.07 × 10 s = 30. hours 3 5

b. 1.0 g Ni × = 33 s

c. 5.0 mol Ag × = 4.8 × 10 s = 1.3 hours3

74. The oxidation state of bismuth in BiO is +3 because oxygen has a -2 oxidation state in this ion. +

Therefore, 3 moles of electrons are required to reduce the bismuth in BiO to Bi(s).+

10.0 g Bi × = 554 s = 9.23 min

75. 15 A = = 5.4 × 10 C of charge passed in 1 hour4


a. 5.4 × 10 C × = 16 g Co4

b. 5.4 × 10 C × = 25 g Hf4

2 2c. 2 I ÷ I + 2 e ; 5.4 × 10 C × = 71 g I- - 4

3 3d. CrO (l) ÷ Cr + 3 O ; 6 mol e are needed to produce 1 mol Cr from molten CrO .6+ 2- -

5.4 × 10 C × = 4.9 g Cr4

2 376. Al is in the +3 oxidation in Al O , so 3 mol e are needed to convert Al into Al(s).- 3+

2.00 h × × = 6.71 × 10 g5

77. 1397 s × = 3.14 × 10 mol M where M = unknown metal-2

Molar mass = ; The element is scandium. Sc forms 3+ ions.

78. Alkaline earth metals form +2 ions, so 2 mol of e are transferred to form the metal, M.-

mol M = 748 s × = 1.94 × 10 mol M-2

2molar mass of M = = 24.3 g/mol; MgCl was electrolyzed.

2 279. F is produced at the anode: 2 F ÷ F + 2 e- -

2.00 h × = 0.746 mol e-

20.746 mol e × = 0.373 mol F ; PV = nRT, -

2 = 9.12 L F

K is produced at the cathode: K + e ÷ K+ -


0.746 mol e × = 29.2 g K-

80. The half-reactions for the electrolysis of water are:

2 2 (2 e + 2 H O ÷ H + 2 OH ) × 2- -

2 2 2 H O ÷ 4 H + O + 4 e+ -

2 2 2 22 H O(l) ÷ 2 H (g) + O (g) Note: 4 H + 4 OH ÷ 4 H O and n = 4+ -

for this reaction as it is written.

215.0 min × = 1.17 × 10 mol H-2

At STP, 1 mole of an ideal gas occupies a volume of 22.42 L (see Chapter 5 of the text).

2 21.17 × 10 mol H × = 0.262 L = 262 mL H-2

2 21.17 × 10 mol H × = 0.131 L = 131 mL O-2

81.

= 7.44 × 10 C/s or a current of 7.44 × 10 A4 4

82. Al + 3 e ÷ Al; 3 mol e are needed to produce Al from Al3+ - - 3+

2000 lb Al × = 1 × 10 C of electricity needed10

= 1 × 10 C/s = 1 × 10 A5 5

83. 2.30 min × = 138 s; 138 s × = 2.86 × 10 mol Ag-3

[Ag ] = 2.86 × 10 mol Ag /0.250 L = 1.14 × 10 M+ -3 + -2

84. 0.50 L × 0.010 mol Pt /L = 5.0 × 10 mol Pt4+ -3 4+

To plate out 99% of the Pt , we will produce 0.99 × 5.0 × 10 mol Pt.4+ -3

0.99 × 5.0 × 10 mol Pt × = 480 s-3


85. Au + 3 e ÷ Au E° = 1.50 V Ni + 2 e ÷ Ni E° = -0.23 V3+ - 2+ -

Ag + e ÷ Ag E° = 0.80 V Cd + 2 e ÷ Cd E° = -0.40 V+ - 2+ -

2 22 H O + 2e ÷ H + 2 OH E° = -0.83 V- -

Au(s) will plate out first since it has the most positive reduction potential, followed by Ag(s), whichis followed by Ni(s), and finally Cd(s) will plate out last since it has the most negative reductionpotential of the metals listed.

4 286. Species present: Fe , SO , H and H O. The possible cathode reactions are:2+ 2- +

4 2 3 2 SO + 4 H + 2 e ÷ H SO + H O E° = 0.20 V2- + -

2 2 H + 2 e ÷ H E° = 0.00 V+ -

Fe + 2 e ÷ Fe E° = -0.44 V2+ -

2 2 2 H O + 2 e ÷ H + 2 OH E° = -0.83 V- -

4Reduction of SO will occur at the cathode since is most positive. The possible anode2-

reactions are:

Fe ÷ Fe + e -E° = -0.77 V2+ 3+ -

2 22 H O ÷ O + 4 H + 4 e -E° = -1.23 V+ -

Oxidation of Fe will occur at the anode since is most positive.2+

87. Reduction occurs at the cathode, and oxidation occurs at the anode. First, determine all the speciespresent, then look up pertinent reduction and/or oxidation potentials in Table 17.1 for all these species.The cathode reaction will be the reaction with the most positive reduction potential, and the anodereaction will be the reaction with the most positive oxidation potential.

2a. Species present: Ni and Br ; Ni can be reduced to Ni, and Br can be oxidized to Br (from2+ - 2+ -

Table 17.1). The reactions are:

Cathode: Ni + 2e ÷ Ni E° = -0.23 V2+ -

2Anode: 2 Br ÷ Br + 2 e -E° = -1.09 V- -

b. Species present: Al and F ; Al can be reduced, and F can be oxidized. The reactions are:3+ - 3+ -

Cathode: Al + 3 e ÷ Al E° = -1.66 V3+ -

2Anode: 2 F ÷ F + 2 e -E° = -2.87 V- -

c. Species present: Mn and I ; Mn can be reduced, and I can be oxidized. The reactions are:2+ - 2+ -

Cathode: Mn + 2 e ÷ Mn E° = -1.18 V2+ -

2Anode: 2 I ÷ I + 2 e -E° = -0.54 V- -

288. These are all in aqueous solutions, so we must also consider the reduction and oxidation of H O inaddition to the potential redox reactions of the ions present. For the cathode reaction, the species with


the most positive reduction potential will be reduced, and for the anode reaction, the species with themost positive oxidation potential will be oxidized.

2a. Species present: Ni , Br and H O. Possible cathode reactions are:2+ -

Ni + 2e ÷ Ni E° = -0.23 V2+ -

2 22 H O + 2 e ÷ H + 2 OH E° = -0.83 V- -

2Since it is easier to reduce Ni than H O (assuming standard conditions), Ni will be reduced by2+ 2+

the above cathode reaction.

Possible anode reactions are:

2 2 Br ÷ Br + 2 e -E° = -1.09 V- -

2 2 2 H O ÷ O + 4 H + 4 e -E° = -1.23 V+ -

2Since Br is easier to oxidize than H O (assuming standard conditions), then Br will be oxidized- -

by the above anode reaction.

2 2b. Species present: Al , F and H O; Al and H O can be reduced. The reduction potentials are3+ - 3+

2 2E° = -1.66 V for Al and E° = -0.83 V for H O (assuming standard conditions). H O will be3+

2 2reduced at the cathode (2 H O + 2 e ÷ H + 2 OH ).- -

2F and H O can be oxidized. The oxidation potentials are -E° = -2.87 V for F and -E° = -1.23 V for- -

2 2H O (assuming standard conditions). From the potentials, we would predict H O to be oxidized at the

2 2anode (2 H O ÷ O + 4 H + 4 e ).+ -

2 2c. Species present: Mn , I and H O; Mn and H O can be reduced. The possible cathode reactions2+ - 2+

are:

Mn + 2 e ÷ Mn E° = -1.18 V2+ -

2 2 2 H O + 2 e ÷ H + 2 OH E° = -0.83 V- -

2Reduction of H O will occur at the cathode since is most positive.

2I and H O can be oxidized. The possible anode reactions are:-

2 2 I ÷ I + 2 e -E° = -0.54 V- -

-E° = -1.23 V2 22 H O ÷ O + 4 H + 4 e+ -

Oxidation of I will occur at the anode since is most positive. -

Additional Exercises

89. The half-reaction for the SCE is:

2 2Hg Cl + 2 e ÷ 2 Hg + 2 Cl = 0.242 V- -


cellFor a spontaneous reaction to occur, E must be positive. Using the standard reduction potentialsin Table 17.1 and the given SCE potential, deduce which combination will produce a positive overallcell potential.

a. Cu + 2 e ÷ Cu E° = 0.34 V2+ -

cellE = 0.34 - 0.242 = 0.10 V; SCE is the anode.

b. Fe + e ÷ Fe E° = 0.77 V3+ - 2+

cellE = 0.77 - 0.242 = 0.53 V; SCE is the anode.

c. AgCl + e ÷ Ag + Cl E° = 0.22 V- -

cell E = 0.242 - 0.22 = 0.02 V; SCE is the cathode.

d. Al + 3 e ÷ Al E° = -1.66 V3+ -

cell E = 0.242 + 1.66 = 1.90 V; SCE is the cathode.

e. Ni + 2 e ÷ Ni E° = -0.23 V2+ -

cellE = 0.242 + 0.23 = 0.47 V; SCE is the cathode.

390. The potential oxidizing agents are NO and H . Hydrogen ion cannot oxidize Pt under either- +

condition. Nitrate cannot oxidize Pt unless there is Cl in the solution. Aqua regia has both Cl and- -

3NO . The overall reaction is:-

3 2 (NO + 4 H + 3 e ÷ NO + 2 H O) × 2 E° = 0.96 V- + -

4 (4 Cl + Pt ÷ PtCl + 2 e ) × 3 -E° = -0.755 V- 2- -

3 4 212 Cl (aq) + 3 Pt(s) + 2 NO (aq) + 8 H (aq) ÷ 3 PtCl (aq) + 2 NO(g) + 4 H O(l) = 0.21 V- - + 2-

91. 2 Ag (aq) + Cu(s) ÷ Cu (aq) + 2 Ag(s) = 0.80 - 0.34 V = 0.46 V; A galvanic cell produces+ 2+

a voltage as the forward reaction occurs. Any stress that increases the tendency of the forwardreaction to occur will increase the cell potential, while a stress that decreases the tendency of theforward reaction to occur will decrease the cell potential.

a. Added Cu (a product ion) will decrease the tendency of the forward reaction to occur,2+

which will decrease the cell potential.

3 3 4b. Added NH removes Cu in the form of Cu(NH ) . Removal of a product ion will increase2+ 2+

the tendency of the forward reaction to occur, which will increase the cell potential.

c. Added Cl removes Ag in the form of AgCl(s). Removal of a reactant ion will decrease the- +

tendency of the forward reaction to occur, which will decrease the cell potential.


1d. Q = ; As the volume of solution is doubled, each concentration is halved.

l = = 2 Q

The reaction quotient is doubled as the concentrations are halved. Since reactions are spontaneous when Q < K and since Q increases when the solution volume doubles, the

reaction is closer to equilibrium, which will decrease the cell potential.

e. Since Ag(s) is not a reactant in this spontaneous reaction, and since solids do not appear in thereaction quotient expressions, replacing the silver electrode with a platinum electrode will have noeffect on the cell potential.

92. (Al + 3 e ÷ Al) × 2 E° = -1.66 V3+ -

(M ÷ M + 2 e ) × 3 -E° = ?2+ -

3 M(s) + 2 Al (aq) ÷ 2 Al(s) + 3 M (aq) = -E° - 1.66 V3+ 2+

)G° = -nF , -411 × 10 J = -(6 mol e )(96,485 C/mol e ) , = 0.71 V3 - -

= -E° - 1.66 V = 0.71 V, -E° = 2.37 or E° = -2.37

From table 17.1, the reduction potential for Mg + 2 e ÷ Mg is -2.37 V, which fits the data. 2+ -

Hence, the metal is magnesium.

93. a. )G° = = 2(-480.) + 3(86) - [3(-40.)] = -582 kJ

From oxidation numbers, n = 6. )G° = -nFE°, E° = = 1.01 V

log K = = 102.538, K = 10 = 3.45 × 10102.538 102

2b. 2 e + Ag S ÷ 2 Ag + S ) × 3 = ?- 2-

(Al ÷ Al + 3 e ) × 2 -E° = 1.66 V3+ -

2 3 Ag S(s) + 2 Al(s) ÷ 6 Ag(s) + 3 S (aq) + 2 Al (aq) = 1.01 V = + 1.66V2- 3+

= 1.01 V - 1.66 V = -0.65 V

94. Zn ÷ Zn + 2 e -E° = 0.76 V; Fe ÷ Fe + 2 e -E° = 0.44 V2+ - 2+ -

It is easier to oxidize Zn than Fe, so the Zn would be oxidized, protecting the iron of the Monitor'shull.

2 2 2 795. From Exercise 17.27a: 3 Cl (g) + 2 Cr (aq) + 7 H O(l) º 14 H (aq) + Cr O (aq) + 6 Cl (aq)3+ + 2- -

= 0.03 V


-

2 2 7 2 4When K Cr O and Cl are added to concentrated H SO , Q becomes a large number due to [H ]- + 14

cellterm. The log of a large number is positive. E becomes negative, which means the reverse reaction

2becomes spontaneous. The pungent fumes were Cl (g).

96. Aluminum has the ability to form a durable oxide coating over its surface. Once the HCl dissolves thisoxide coating, Al is exposed to H and is easily oxidized to Al , i.e., the Al foil disappears after the+ 3+

oxide coating is dissolved.

97. Consider the strongest oxidizing agent combined with the strongest reducing agent from Table 17.1:

2 F + 2 e ÷ 2 F E° = 2.87 V- -

(Li ÷ Li + e ) × 2 -E° = 3.05 V+ -

2F (g) + 2 Li(s) ÷ 2 Li (aq) + 2 F (aq) = 5.92 V+ -

The claim is impossible. The strongest oxidizing agent and reducing agent when combined only givean value of about 6 V.

2 2 298. 2 H (g) + O (g) ÷2 H O(l); Oxygen goes from the zero oxidation state to the -2 oxidation state in

2H O. Since two mol O appear in the balanced reaction, then n = 4 mol electrons transferred.

a. = log K = log (1.28 × 10 ), = 1.23 V83

)G° = -nF = -(4 mol e )(96,485 C/mol e )(1.23 J/C) = -4.75 × 10 J = -475 kJ- - 5

b. Since mol of gas decrease as reactants are converted into products, then )S° will be negative (unfavorable). Since the value of )G° is negative, then )H° must be negative ()G° =)H° - T)S°).

maxc. )G = w = )H - T)S. Since )S is negative, then as T increases, )G becomes more

maxpositive (closer to zero). Therefore, w will decrease as T increases.

2 299. a. O + 2 H O + 4 e ÷ 4 OH E° = 0.40 V- -

2 2 (H + 2 OH ÷ 2 H O + 2 e ) × 2 -E° = 0.83 V- -

2 2 2 2 H (g) + O (g) ÷ 2 H O(l) = 1.23 V = 1.23 J/C

max 2Since standard conditions are assumed, then w = )G° for 2 mol H O produced.

)G° = = -(4 mol e )(96,485 C/mol e )(1.23 J/C) = -475,000 J = -475 kJ- -

2 maxFor 1.00 × 10 g H O produced, w is:3

2 max1.00 × 10 g H O × = -13,200 kJ = w3


The work done can be no larger than the free energy change. The best that could happen is that all ofthe free energy released would go into doing work, but this does not occur in any real process sincethere is always waste energy in a real process. Fuel cells are more efficient in converting chemicalenergy into electrical energy; they are also less massive. The major disadvantage is that they are

2 2expensive. In addition, H (g) and O (g) are an explosive mixture if ignited; much more so than fossilfuels.

2100. Cadmium goes from the zero oxidation state to the +2 oxidation state in Cd(OH) . Since one mol ofCd appears in the balanced reaction, then n = 2 mol electrons transferred. At standard conditions:

max max w = )G° = -nFE°, w = -(2 mol e )(96,485 C/mol e )(1.10 J/C) = -2.12 × 10 J = -212 kJ- - 5

2101. (CO + O ÷ CO + 2 e ) × 22- -

2 O + 4 e ÷ 2 O - 2-

2 22 CO + O ÷ 2 CO

)G = -nFE, E = = = 0.98 V

2102. In the electrolysis of aqueous sodium chloride, H O is reduced in preference to Na , and Cl is oxidized+ -

2 2 2in preference to H O. The anode reaction is 2 Cl ÷ Cl + 2 e , and the cathode reaction is 2 H O +- -

22 e ÷ H + 2 OH . The overall reaction is:- -

2 2 2 2 H O(l) + 2 Cl (aq) ÷ Cl (g) + H (g) + 2 OH (aq).- -

2 2 2From the 1:1 mol ratio between Cl and H in the overall balanced reaction, if 257 L of Cl (g) are

2produced, then 257 L of H (g) will also be produced since moles and volume of gas are directlyproportional at constant T and P (see Chapter 5 of text).

103. mol e = 50.0 min × = 7.77 × 10 mol e- -2 -

mol Ru = 2.618 g Ru × = 2.590 × 10 mol Ru-2

= 3.00; The charge on the ruthenium ions is +3 (Ru + 3 e ÷ Ru).3+ -

104. 15 kWh = = 5.4 × 10 J or 5.4 × 10 kJ (Hall process)7 4

To melt 1.0 kg Al requires: 1.0 × 10 g Al × = 4.0 × 10 kJ3 2

It is feasible to recycle Al by melting the metal because, in theory, it takes less than 1% of the energyrequired to produce the same amount of Al by the Hall process.


Challenge Problems

105. )G° = -nFE° = )H° - T)S°, E° =

If we graph E° vs. T we should get a straight line (y = mx + b). The slope of the line is equal to)S°/nF, and the y-intercept is equal to -)H°/nF. From the equation above, E° will have a smalltemperature dependence when )S° is close to zero.

106. a. We can calculate )G° from )G° = )H° - T)S° and then E° from )G° = -nFE°; or we can usethe equation derived in Exercise 17.105. For this reaction, n = 2 (from oxidation states).

= 1.98 J/C = 1.98 V

b. ln Q = 1.98 V - ln

-20E = 1.98 V - ln = 1.98 V + 0.066 V = 2.05 V

c. From Exercise 17.56, E = 2.12 V at 25°C. As the temperature decreases, the cell potentialdecreases. Also, oil becomes more viscous at lower temperatures, which adds to the difficulty ofstarting an engine on a cold day. The combination of these two factors results in batteries failingmore often on cold days than on warm days.

107. (Ag + e ÷ Ag) × 2 E° = 0.80 V+ -

Pb ÷ Pb + 2 e -E° = -(-0.13)2+ -

2 Ag + Pb ÷ 2 Ag + Pb = 0.93 V+ 2+

E = E° - log , 0.83 V = 0.93 V - log

log = = 3.4, = 10 , [Ag ] = 0.027 M3.4 +

2 4 4 sp 4 Ag SO (s) 2 Ag (aq) + SO (aq) K = [Ag ] [SO ]+ 2- + 2 2-

Initial s = solubility (mol/L) 0 0Equil. 2s s

From problem: 2s = 0.027 M, s = 0.027/2

spK = (2s) (s) = (0.027) (0.027/2) = 9.8 × 102 2 -6

cell108. a. Zn(s) + Cu (aq) ÷ Zn (aq) + Cu(s) = 1.10 V; E = 1.10 V - 2+ 2+


cellE = 1.10 V - = 1.10 V + 0.041 V = 1.14 V

b. 10.0 h × = 1.87 mol Cu produced

The Cu concentration will decrease by 1.87 mol/L, and the Zn concentration will increase by2+ 2+

1.87 mol/L.

[Cu ] = 2.50 - 1.87 = 0.63 M; [Zn ] = 0.10 + 1.87 = 1.97 M2+ 2+

cellE = 1.10 V - = 1.10 V - 0.015 V = 1.09 V

c. 1.87 mol Zn consumed × = 122 g Zn; Mass of electrode = 200. - 122 = 78 g Zn

1.87 mol Cu formed × = 119 g Cu; Mass of electrode = 200. + 119 = 319 g Cu

d. Three things could possibly cause this battery to go dead:

1. All of the Zn is consumed.2. All of the Cu is consumed.2+

cell3. Equilibrium is reached (E = 0).

We began with 2.50 mol Cu and 200. g Zn × 1 mol Zn/65.38 g Zn = 3.06 mol Zn. Cu is the2+ 2+

limiting reagent and will run out first. To react all the Cu requires:2+

2.50 mol Cu × = 13.4 h2+

For equilibrium to be reached: E = 0 = 1.10 V -

= K = 10 = 1.68 × 102(1.10)/0.0591 37

This is such a large equilibrium constant that virtually all of the Cu must react to reach2+

equilibrium. So, the battery will go dead in 13.4 hours.

2109. 2 H + 2 e ÷ H E° = 0.000 V+ -

Fe ÷ Fe + 2 e -E° = -(-0.440V)2+ -

22 H (aq) + Fe(s) ÷ H (g) + Fe (aq) = 0.440 V + 2+

- log Q, where n = 2 and Q =

aTo determine K for the weak acid, first use the electrochemical data to determine the H concentration+

in the half-cell containing the weak acid.

0.333 V = 0.440 V -


= 10 = 4.18 × 10 , [H ] = 4.89 × 10 M3.621 3 + -4

aNow we can solve for the K value of the weak acid HA through the normal setup for a weak acidproblem.

a HA H + A K = + -

Initial 1.00 M ~0 0Equil. 1.00 - x x x

a aK = where x = [H ] = 4.89 × 10 M, K = = 2.39 × 10+ -4 -7

110. a. Nonreactive anions are present in each half-cell to balance the cation charges.

b. Au (aq) + 3 Fe (aq) ÷ 3 Fe (aq) + Au(s) = 1.50 - 0.77 = 0.73 V3+ 2+ 3+

cellE = log Q = 0.73 V -

Since [Fe ] = [Fe ] = 1.0 M: 0.31 V = 0.73 V - 3+ 2+

, log [Au ] = -21.32, [Au ] = 10 = 4.8 × 10 M3+ 3+ -21.32 -22

4 4Au + 4 Cl º AuCl ; Since the equilibrium Au concentration is so small, assume [AuCl ] .3+ - - 3+ -

o[Au ] . 1.0 M, i.e., assume K is large, so the reaction essentially goes to completion.3+

K = = 2.1 × 10 ; Assumption good (K is large).25


cell ref111. a. E = E + 0.05916 pH, 0.480 V = 0.250 V + 0.05916 pH

pH = = 3.888; Uncertainty = ± 1 mV = ± 0.001 V

max minpH = = 3.905; pH = = 3.871

So, if the uncertainty in potential is ± 0.001 V, the uncertainty in pH is ± 0.017 or about ± 0.02pH units. For this measurement, [H ] = 10 = 1.29 × 10 M. For an error of +1 mV, [H ] =+ -3.888 -4 +

10 = 1.24 × 10 M. For an error of -1 mV, [H ] = 10 = 1.35 × 10 M. So, the-3.905 -4 + -3.871 -4

uncertainty in [H ] is ± 0.06 × 10 M = ± 6 × 10 M.+ -4 -6

b. From part a, we will be within ± 0.02 pH units if we measure the potential to the nearest ± 0.001V (1 mV).

2 2112. a. From Table 17.1: 2 H O + 2 e ÷ H + 2 OH E° = -0.83 V- -

= -0.83 V + 2.36 V = 1.53 V

2 2Yes, the reduction of H O to H by Zr is spontaneous at standard conditions since > 0.

2 2b. (2 H O + 2 e ÷ H + 2 OH ) × 2- -

2 2 2 Zr + 4 OH ÷ ZrO CH O + H O + 4 e- -

2 2 2 23 H O(l) + Zr(s) ÷ 2 H (g) + ZrO CH O(s)

c. )G° = -nFE° = -(4 mol e )(96,485 C/mol e )(1.53 J/C) = -5.90 × 10 J = -590. kJ- - 5

E = E° - log Q; At equilibrium, E = 0 and Q = K.

E° = log K, log K = = 104, K . 10104

2d. 1.00 × 10 kg Zr × = 2.19 × 10 mol H3 4

2 22.19 × 10 mol H × = 4.42 × 10 g H4 4

2V = = 2.3 × 10 L H6

2e. Probably yes; Less radioactivity overall was released by venting the H than what would have

2been released if the H had exploded inside the reactor (as happened at Chernobyl). Neither


alternative is pleasant, but venting the radioactive hydrogen is the less unpleasant of the twoalternatives.

113. a. (Ag + e ÷ Ag) × 2 E° = 0.80 V+ -

Cu ÷ Cu + 2 e !E° = !0.34 V2+ -

2 Ag (aq) + Cu(s) ÷ 2 Ag(s) + Cu (aq) = 0.46 V+ 2+

- log Q where n = 2 and Q =

cell spTo calculate E , we need to use the K data to determine [Ag ].+

spAgCl(s) Ag (aq) + Cl (aq) K = 1.6 × 10 = [Ag ] [Cl ]+ - -10 + -

Initial s = solubility (mol/L) 0 0Equil. s s

spK = 1.6 × 10 = s , s = [Ag ] = 1.3 × 10 mol/L-10 2 + -5

cellE = 0.46 V - log = 0.46 V - 0.30 = 0.16 V

3 4 4b. Cu (aq) + 4 NH (aq) Cu(NH ) (aq) K = 1.0 × 10 = 2+ 2+ 13

3 4Since K is very large for the formation of Cu(NH ) ,the forward reaction is dominant. At2+

3 4equilibrium, essentially all of the 2.0 M Cu will react to form 2.0 M Cu(NH ) . This reaction2+ 2+

3requires 8.0 M NH to react with all of the Cu in the balanced equation. Therefore, the mol of2+

3 3NH added to 1.0 L solution will be larger than 8.0 mol since some NH must be present at

3equilibrium. In order to calculate how much NH is present at equilibrium, we need to use theelectrochemical data to determine the Cu concentration.2+

- log Q, 0.52 V = 0.46 V - log

log = -2.03, = 10 = 9.3 × 10-2.03 -3

[Cu ] = 1.6 × 10 = 2 × 10 M (We carried extra significant figures in the calculation.)2+ -12 -12

Note: Our assumption that the 2.0 M Cu essentially reacts to completion is excellent as only 22+

3× 10 M Cu remains after this reaction. Now we can solve for the equilibrium [NH ].-12 2+


3K = 1.0 × 10 = , [NH ] = 0.6 M13

3 3Since 1.0 L of solution is present, then 0.6 mol NH remains at equilibrium. The total mol of NH

3 3 4added is 0.6 mol plus the 8.0 mol NH necessary to form 2.0 M Cu(NH ) . Therefore, 8.0 + 0.62+

3= 8.6 mol NH were added.

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CHAPTER SEVENTEEN ELECTROCHEMISTRY SEVENTEEN ELECTROCHEMISTRY Review of Oxidation ... CHAPTER 17...

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