Question

Consider the following half reactions at 298 K

Fe²⁺ + 2 e^- → Fe    E^o = -0.441 V
Cd²⁺ + 2 e^- → Cd    E^o = -0.403 V

A galvanice cell based on these half reactions is set up under standard conditions where each solution is 1.00 L and each electrode weighs exactly 100.0 g. How much will the Cd electrode weigh when the nonstandard potential of the cell is 0.02804 V?

Answer 1

First, find initial concentrations:

mol of Fe = mass/MW = 100/55.8450 = 1.7906

mol of Cd = mass/MW = 100/112.4110 = 0.88959

initially

[Fe+2] = 1.7906

[Cd+2] = 0.88959

after reaction

[Fe+2] = 1.7906 + x

[Cd+2] = 0.88959 - x

Solve for

E° = Ered - Eox

Clearly, the most positive is Cd, so it will reduce

E° = (-0.403--0.441) = 0.038

Apply Nernst equation

E = E° - 0.0592/n * log(Q)

E = 0.02804, n = 2 electrons

Q = [Fe+2]/[Cd+2]

so

E = 0.038 - 0.0592/2 * log([Fe+2]/[Cd+2])

0.02804 = 0.038  - 0.0592/2 * log((1.7906 + x)/(0.88959 - x))

(0.02804 -0.038 ) *2 /(-0.0592) =  log((1.7906 + x)/(0.88959 - x))

10^0.33648 = (1.7906 + x)/(0.88959 - x)

2.1701*(0.88959 - x) = 1.7906+x

2.1701*(0.88959) - 2.1701x = 1.7906+x

2.1701*(0.88959) - 1.7906 = 3.1701x

x = (2.1701*(0.88959) - 1.7906) /3.1701

x= 0.04413

substitute:

[Fe+2] = 1.7906 + 0.04413 = 1.83473 mol /L

[Cd+2] = 0.88959 - 0.04413 = 0.84546 mol /L

For Cd:

V = 1 L, so

mol Cd = 0.84546 mol

mass = mol*Mw = 0.84546* 112.4110 = 95.039 g of Cd will be produced