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CHM 112
Section 1

Homework Assignment #30
Due April 26, 2006 by 8:05 a.m.

Practice (not to be turned in): 18.53, 18.54

Part 1

For the electrochemical cells described below, complete and balance the reaction and find the thermodynamic equilibrium constant and Go (in units of kJ/mole) at 25 °C corresponding to the last digit of your Student ID Number.

0) Pt(s) | MnO2(s) | MnO4(aq) | pH=0 || Co2+(aq) | Co3+(aq) | Pt(s)

1) Pt(s) | Mn2+(aq) | MnO4(aq) | pH = 0 || Au3+(aq) | Au(s)

2) Pt(s) | Au3+(aq) | Au+(aq) || ClO3(aq) | Cl(aq) | pH = 0 | Pt(s)

3) Pt(s) | NO2(g) | HNO2(aq) | pH = 0 || MnO2(s) | Mn2+(aq) | pH = 0 | Pt(s)

4) Hg(l) | Hg22+(aq) || NO(g) | NO3(aq) | pH = 0 | Pt(s)

5) Pt(s) | MnO4(aq) | MnO42–(aq) || O2(g) | H2O2(aq) | pH = 0 | Pt(s)

6) Cu(s) | Cu2+(aq) || H2SO3(aq) | S(s) | pH = 0 | Pt(s)

7) Hg(l) | HgCl2(s) | Cl(aq) || VO2+(aq) | V3+(aq) | pH = 0 | Pt(s)

8) Fe(s) | Fe3+(aq) || H2S(g) | S(s) | pH = 0 | Pt(s)

9) Pb(s) | PbSO4(s) | SO42–(aq) || H3PO4(aq) | H3PO3(aq) | pH = 0 | Pt(s)


Part 2

Complete and balance the reaction corresponding to the eighth digit of your Student ID Number.

0) Cd(ClO4)2(aq) + H2S(aq)

1) Co(ClO4)2(aq) + H2S(aq)

2) Cu(ClO4)2(aq) + H2S(aq)

3) Fe(ClO4)2(aq) + H2S(aq)

4) Hg(ClO4)2(aq) + H2S(aq)

5) Mn(ClO4)2(aq) + H2S(aq)

6) Ni(ClO4)2(aq) + H2S(aq)

7) Pb(ClO4)2(aq) + H2S(aq)

8) Sn(ClO4)2(aq) + H2S(aq)

9) Zn(ClO4)2(aq) + H2S(aq)



Answers
Part 1

All of these reactions are done in acid:

0) Pt(s) | MnO2(s) | MnO4(aq) | pH=0 || Co2+(aq) | Co3+(aq) | Pt(s)

Anode: MnO2(s) + 2 H2O(l) MnO4(aq) + 4 H+(aq) + 3 e        Eoox = –1.70 V

Cathode: [ Co3+(aq) + e Co2+(aq) ] × 3        Eored = 1.82 V

Net: 3 Co3+(aq) + MnO2(s) + 2 H2O(l) 3 Co2+(aq) + MnO4(aq) + 4 H+(aq)

Eo = Eoox + Eored = –1.70 + 1.82 = +0.12 V

Go = –nFEo = –(3)(96500)(0.12) = –35000 J/mole = –35 kJ/mole

Keq = exp{nFEo/RT} = exp{(3)(96500)(0.12)/(8.314)(298)} = 1.2×106

1) Pt(s) | Mn2+(aq) | MnO4(aq) | pH = 0 || Au3+(aq) | Au(s)

Anode: [ Mn2+(aq) + 4 H2O(l) MnO4(aq) + 8 H+(aq) + 5 e ] × 3        Eoox = –1.51 V

Cathode: [ Au3+(aq) + 3 e Au(s) ] × 5        Eored = 1.52 V

Net: 3 Mn2+(aq) + 5 Au3+(aq) + 12 H2O(l) 5 Au(s) + 3 MnO4(aq) + 24 H+(aq)

Eo = Eoox + Eored = –1.51 + 1.52 = +0.01 V

Go = –nFEo = –(15)(96500)(0.01) = –14000 J/mole = –14 kJ/mole (stretching the significant figures...)

Keq = exp{nFEo/RT} = exp{(15)(96500)(0.01)/(8.314)(298)} = 3.4×102

2) Pt(s) | Au3+(aq) | Au+(aq) || ClO3(aq) | Cl(aq) | pH = 0 | Pt(s)

Anode: [ Au+(aq) Au3+(aq) + 2 e ] × 3        Eoox = –1.36 V

Cathode: ClO3(aq) + 6 H+(aq) + 6 e Cl(aq) + 3 H2O(l)        Eored = 1.450 V

Net: 3 Au+(aq) + ClO3(aq) + 6 H+(aq) 3 Au3+(aq) + Cl(aq) + 3 H2O(l)

Eo = Eoox + Eored = –1.36 + 1.450 = +0.09 V

Go = –nFEo = –(6)(96500)(0.09) = –52000 J/mole = –52 kJ/mole (stretching the significant figures...)

Keq = exp{nFEo/RT} = exp{(6)(96500)(0.09)/(8.314)(298)} = 1.4×109

3) Pt(s) | NO2(g) | HNO2(aq) | pH = 0 || MnO2(s) | Mn2+(aq) | pH = 0 | Pt(s)

Anode: [ HNO2(aq) NO2(g) + H+(aq) + e ] × 2        Eoox = –1.07 V

Cathode: MnO2(s) + 4 H+(aq) + 2 e Mn2+(aq) + 2 H2O(l)        Eored = 1.23 V

Net: 2 HNO2(aq) + MnO2(s) + 2 H+(aq) 2 NO2(g) + Mn2+(aq) + 2 H2O(l)

Eo = Eoox + Eored = –1.07 + 1.23 = +0.16 V

Go = –nFEo = –(2)(96500)(0.16) = –31000 J/mole = –31 kJ/mole

Keq = exp{nFEo/RT} = exp{(2)(96500)(0.16)/(8.314)(298)} = 2.6×105

4) Hg(l) | Hg22+(aq) || NO(g) | NO3(aq) | pH = 0 | Pt(s)

Anode: [ 2 Hg(l) Hg22+(aq) + 2 e ] × 3        Eoox = –0.80 V

Cathode: [ NO3(aq) + 4 H+(aq) + 3 e NO(g) + 2 H2O(l) ] × 2        Eored = 0.956 V

Net: 6 Hg(l) + 2 NO3(aq) + 8 H+(aq) 3 Hg22+(aq) + 2 NO(g) + 4 H2O(l)

Eo = Eoox + Eored = –0.80 + 0.956 = +0.16 V

Go = –nFEo = –(6)(96500)(0.16) = –93000 J/mole = –93 kJ/mole

Keq = exp{nFEo/RT} = exp{(6)(96500)(0.16)/(8.314)(298)} = 1.7×1016

5) Pt(s) | MnO4(aq) | MnO42–(aq) || O2(g) | H2O2(aq) | pH = 0 | Pt(s)

Anode: [ MnO42–(aq) MnO4(aq) + e ] × 2        Eoox = –0.56 V

Cathode: O2(g) + 2 H+(aq) + 2 e H2O2(aq)        Eored = 0.695 V

Net: 2 MnO42–(aq) + O2(g) + 2 H+(aq) 2 MnO4(aq) + H2O2(aq)

Eo = Eoox + Eored = –0.56 + 0.695 = +0.14 V

Go = –nFEo = –(2)(96500)(0.14) = –27000 J/mole = –27 kJ/mole

Keq = exp{nFEo/RT} = exp{(2)(96500)(0.14)/(8.314)(298)} = 5.4×104

6) Cu(s) | Cu2+(aq) || H2SO3(aq) | S(s) | pH = 0 | Pt(s)

Anode: [ Cu(s) Cu2+(aq) + 2 e ] × 2        Eoox = –0.340 V

Cathode: H2SO3(aq) + 4 H+(aq) + 4 e S(s) + 3 H2O(l)        Eored = 0.449 V

Net: 2 Cu(s) + H2SO3(aq) + 4 H+(aq) 2 Cu2+(aq) + S(s) + 3 H2O(l)

Eo = Eoox + Eored = –0.340 + 0.449 = +0.109 V

Go = –nFEo = –(4)(96500)(0.109) = –42100 J/mole = –42.1 kJ/mole

Keq = exp{nFEo/RT} = exp{(4)(96500)(0.109)/(8.314)(298)} = 2.4×107

7) Hg(l) | HgCl2(s) | Cl(aq) || VO2+(aq) | V3+(aq) | pH = 0 | Pt(s)

Anode: Hg(l) + 2 Cl(aq) HgCl2(s) + 2 e        Eoox = –0.2676 V

Cathode: [ VO2+(aq) + 2 H+(aq) + e V3+(aq) + H2O(l) ] × 2        Eored = 0.337 V

Net: Hg(l) + 2 Cl(aq) + 2 VO2+(aq) + 4 H+(aq) HgCl2(s) + 2 V3+(aq) + 2 H2O(l)

Eo = Eoox + Eored = –0.2676 + 0.337 = +0.069 V

Go = –nFEo = –(2)(96500)(0.069) = –13000 J/mole = –13 kJ/mole

Keq = exp{nFEo/RT} = exp{(2)(96500)(0.069)/(8.314)(298)} = 220

8) Fe(s) | Fe3+(aq) || H2S(g) | S(s) | pH = 0 | Pt(s)

Anode: [ Fe(s) Fe3+(aq) + 3 e ] × 2        Eoox = 0.04 V

Cathode: [ S(s) + 2 H+(aq) + 2 e H2S(g) ] × 3        Eored = 0.14 V

Net: 2 Fe(s) + 3 S(s) + 6 H+(aq) 2 Fe3+(aq) + 3 H2S(g)

Eo = Eoox + Eored = +0.04 + 0.14 = +0.18 V

Go = –nFEo = –(6)(96500)(0.18) = –100000 J/mole = –100 kJ/mole

Keq = exp{nFEo/RT} = exp{(6)(96500)(0.18)/(8.314)(298)} = 1.9×1018

9) Pb(s) | PbSO4(s) | SO42–(aq) || H3PO4(aq) | H3PO3(aq) | pH = 0 | Pt(s)

Anode: Pb(s) + SO42–(aq) PbSO4(s) + 2 e        Eoox = 0.356 V

Cathode: H3PO4(aq) + 2 H+(aq) + 2 e H3PO3(aq) + H2O(l)        Eored = –0.276 V

Net: Pb(s) + SO42–(aq) + H3PO4(aq) + 2 H+(aq) PbSO4(s) + H3PO3(aq) + H2O(l)

Eo = Eoox + Eored = +0.356 + –0.276 = +0.080 V

Go = –nFEo = –(2)(96500)(0.080) = –15000 J/mole = –15 kJ/mole

Keq = exp{nFEo/RT} = exp{(2)(96500)(0.080)/(8.314)(298)} = 5.1×102


Part 2

0) Cd(ClO4)2(aq) + H2S(aq)

Cd(ClO4)2(aq) + H2S(aq) + 2 H2O(l) CdS(s) + 2 H3O+(aq) + 2 ClO4(aq)

1) Co(ClO4)2(aq) + H2S(aq)

Co(ClO4)2(aq) + H2S(aq) + 2 H2O(l) CoS(s) + 2 H3O+(aq) + 2 ClO4(aq)

2) Cu(ClO4)2(aq) + H2S(aq)

Cu(ClO4)2(aq) + H2S(aq) + 2 H2O(l) CuS(s) + 2 H3O+(aq) + 2 ClO4(aq)

3) Fe(ClO4)2(aq) + H2S(aq)

Fe(ClO4)2(aq) + H2S(aq) + 2 H2O(l) FeS(s) + 2 H3O+(aq) + 2 ClO4(aq)

4) Hg(ClO4)2(aq) + H2S(aq)

Hg(ClO4)2(aq) + H2S(aq) + 2 H2O(l) HgS(s) + 2 H3O+(aq) + 2 ClO4(aq)

5) Mn(ClO4)2(aq) + H2S(aq)

Mn(ClO4)2(aq) + H2S(aq) + 2 H2O(l) MnS(s) + 2 H3O+(aq) + 2 ClO4(aq)

6) Ni(ClO4)2(aq) + H2S(aq)

Ni(ClO4)2(aq) + H2S(aq) + 2 H2O(l) NiS(s) + 2 H3O+(aq) + 2 ClO4(aq)

7) Pb(ClO4)2(aq) + H2S(aq)

Pb(ClO4)2(aq) + H2S(aq) + 2 H2O(l) PbS(s) + 2 H3O+(aq) + 2 ClO4(aq)

8) Sn(ClO4)2(aq) + H2S(aq)

Sn(ClO4)2(aq) + H2S(aq) + 2 H2O(l) SnS(s) + 2 H3O+(aq) + 2 ClO4(aq)

9) Zn(ClO4)2(aq) + H2S(aq)

Zn(ClO4)2(aq) + H2S(aq) + 2 H2O(l) ZnS(s) + 2 H3O+(aq) + 2 ClO4(aq)