Question

From the standard reduction potentials in Table 11.1 in the Appendix, calculate the standard cell potential and the equilibrium constant at 298.15 K for the following reactions:

(a) 4 NiOOH(s) + 2 H2O(l) ↔ 4 Ni(OH)2(s) + O2(g)

(b) 4 NO3 - (aq) + 4 H+ (aq) ↔ 4 NO(g) + 2 H2O (l) + 3 O2(g)

Answer 1

a)

NiOOH + H2O + e ↔ Ni(OH)2 + OH- 0.45

4 OH−(aq) ⇌O2(g) + 2 H2O + 4 e− E = -0.401

Reverse:

4NiOOH + 4H2O + 4e ↔ 4Ni(OH)2 + 4OH- 0.45

4OH−(aq) ⇌O2(g) + 2 H2O + 4 e− E = -0.401

add all

4OH−(aq) +4NiOOH + 4H2O + 4e ↔ 4Ni(OH)2 + 4OH- + O2(g) + 2 H2O + 4 e− E = 0.45-0.401

cancel common terms

4NiOOH + 2H2O↔ 4Ni(OH)2 + O2(g) E = 0.45-0.401 = 0.049 V

n = 4

Ecell = 0.049

dG = -nF*Ecell = -4*96500*0.049 = -18,914

dG = -RT*ln(K)

K = exp(-dG/(RT)

K = exp(18914 / (8.31*298)) =2,075.093

b)

4NO3−(aq) + 16H+ + 12e− ⇌ 4NO(g) + 8H2O(l) +0.958

6H2O ⇌ 3O2(g) + 12H+ + 12 e− E = -1.229

Add all

6H2O + 4NO3−(aq) + 16H+ + 12e− ⇌ 4NO(g) + 8H2O(l) + 3O2(g) + 12H+ + 12 e−

cancel common terms

4NO3−(aq) + 4H+ ⇌ 4NO(g) + 2H2O(l) + 3O2(g)

n = 12 electrons

Ecell = Ered + Eox = 0.958 - 1.229 = -0.271

dG = -nF*Ecell = -12*96500*(-0.271) = 313,818 J/mol = 313.818 kJ/mol

dG = -RT*ln(K)

k = exp(-dG/(RT))

K = exp(-313818/(8.314*298)) = 9.7888247*10^-56