Question

This is what I have. Could you please correct my work? Thank you.


Compute the equilibrium constant for Ni^2+ (aq) and Cd (s).

reduction
Ni²⁺(aq) + 2 e⁻ ⇌ Ni(s) ; E°_red = -0.25V
-oxidation
Cd(s) ⇌ Cd²⁺(aq) + 2 e⁻ ; E°_ox = 0.40V
(negative value of the listed values, cause we consider reaction in opposite direction)

Overall
Ni²⁺(aq) + Cd(s) ⇌ Ni(s) + Cd²⁺(aq) + 2 e⁻ ; E = 0.40V
E° = E_ox + E°_red = 0.15V

∆G° = -n∙F∙E°
(n number of electrons exchanged, F faraday's constant)


∆G° = -R∙T∙ln(K)


n∙F∙E° = R∙T∙ln(K)
=>
K = e^{ n∙F∙E°/ (R∙T) }

standard temperature T = 298K
n = 2

K = e^{ 2 ∙ 96485J/Vmol ∙ 0.15V / (8.3145J/mol ∙ 298K) }
= 1.18×10⁵

Answer 1

1.18X10^10

Answer 2

I ran through the work and obtained 1.18 x 10^5, your first answer before you re-posted with 1.18 x 10^10

Answer 3

You don't list any concns in the reaction. I have assumed they must be 1 M and E for the half cell is Eo for the half cell.