What is the value of the equilibrium constant at 25 oC for the reaction between the pair

Question

What is the value of the equilibrium constant at 25 ^oC for the reaction between the pair:

Cr(s) and Fe²⁺(aq) to give Fe(s) and Cr³⁺(aq)

Use the reduction potential values for Cr³⁺(aq) of -0.74 V and for Fe²⁺(aq) of -0.44 V

The correct answer is 3 *10³⁰

Answer 1

Fe²⁺ + 2e- ==> Fe(s) Eº = -0.44 V (reduction half reaction) x 3

Cr(s) ===> Cr3+ + 3e- Eº = -0.74 V (oxidation half reaction) x2

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3Fe2+ + 6e- ==> 3Fe(s)

2Cr(s) ===> 2Cr3+ + 6e-

2Cr(s) + 3Fe²⁺(aq) ==> 3Fe(s) + 2Cr³⁺(aq) ... overal reaction

Eºcell = cathode - anode = reduction - oxidation = -0.44 - (-0.74) = 0.30V

Eºcell = RT/nF lnK and at 25ºC and substituting the values for R and F, and log base 10, we obtain...

Eºcell = 0.0592/n (log K)

0.30 = 0.0592/6 log K

0.30 = 0.00987 log K

log K = 30.4

K = 2.5x10³⁰ and if we round to 1 significant figure we have 3x10³⁰ = K