Question

Use the data in Appendix L to determine the equilibrium constant for the following reactions. Assume 298.15 K if no temperature is given. (a) AgCl(s) ⇌ Ag+(aq) + Cl−(aq) (b) CdS(s) ⇌ Cd2+(aq) + S2−(aq) at 377 K (c) Hg2+(aq) + 4Br−(aq) ⇌ [HgBr4 ] 2−(aq) (d) H2 O(l) ⇌ H+(aq) + OH−(aq) at 25 °C

Answer 1

a. AgCl(s)⇌Ag+(aq)+Cl−(aq)

oxidation half reaction: AgCl (s) + e- ⇌ Ag (s) + Cl- E˚ = 0.2223

reduction half reaction: Ag+(aq) + e- ⇌ Ag (s)    E˚ = 0.7996

E˚cell = E˚reduction - E˚oxidation = 0.2223 - 0.7996 = -0.5773

E˚cell = [(RT)/nF] ln(K) = (.0591v/n) log(K)

-0.5773 v = (0.0591v/1) log(K)

k = 1.7 x 10-10

b. CdS(s)⇌Cd2+(aq)+S2−(aq) at 377 K

Cd(s) ⇌ Cd2+(aq) + 2e-    E˚ = -0.4030

CdS(s) + 2e- ⇌ Cd(s) + S2-(aq) E˚ = -1.17

E˚cell = E˚reduction - E˚oxidation = -1.17 - (-0.4030) = -0.767

E˚cell = [(RT)/nF] ln(K) = [(8.3145 J/mol K)(377K)/(2)(96485)] ln(K)

K = 2.6 x 10-21

c. Hg2+(aq)+4Br−(aq)⇌[HgBr4]2−(aq)

this is not a redox reaction

Hg: +2 -> +2

Br: -1 -> -1

d. H2O(l)⇌H+(aq)+OH−(aq) at 25°C

H2(g) ⇌ 2H+(aq) + 2e-    E˚ = 0.0v

H2O(l) + e- ⇌ 1/2*H2(g) + OH-(aq) E˚ = -0.8277v

The K value corresponds to the Kwater, which has a value of 1 x 10-14