Answer:
The value of the equilibrium constant at 25°C is 2.177
Explanation:
Based on the half-reactions:
Sn(s) → Sn²⁺ + 2e⁻ E° = -(Reduction potential) = 0.14V
Pb²⁺ + 2e⁻ → Pb(s) E° = -0.13V
Sn(s) + Pb²⁺ → Pb(s) + Sn²⁺ E° = 0.14V - 0.13V = 0.01V
Using:
Eºcell = 0.0592/n (log K) At 25°C
Where E° cell = 0.01V
n are moles of electrons = 2
K is equilibrium constant
Replaing:
0.01V = 0.0592/2 * (logK)
0.3378 = log K
2.177 = K
