One half cell is a cathode and other is an anode.
The electrodes in both half cells are same as Sn.
Anodic reaction;
Sn(s) → Sn²⁺(aq) + 2e E⁰oxi = +0.137 V
Cathodic reaction;
Sn²⁺(aq) + 2e → Sn(s) E⁰red= -0.137 V
E⁰cell = E⁰red - E⁰oxi
E⁰cell = 0 V
TheNernst equation;
Ecell = E⁰cell - (RT/nF) ln[anode] / [cathode]
Ecell = 0.18 V
E⁰cell = 0 V
R = 8.314 J mol⁻¹ K⁻¹
T = 298 K
n = 2 mol (electrone moles that participated in)
F = 96485 C mol⁻¹ (Fara¹day constan)
By applying Nernst equation;
0.18 = 0 - ((8.314 x 298) / (2 x 96485)) ln [anode]/[cathode]
0.18 = 0 - 2.303 x 0.0128 log₁₀ [anode]/[cathode]
0.18 = 0 - 0.0295 log₁₀ [anode]/[cathode]
0.2095 = log₁₀ [anode]/[cathode]
[anode]/[cathode] = antilog 0.2095
= 1.62.
Hence Sn²⁺ concentration ration in two half cells = 1.62
