Aconcentration cell consists of two sn/sn2 half-cells. the cell has a potential of 0.18 v at 25 ? c. part a what is the ratio of the sn2 concentrations in the two half-cells?

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

Because it was a dramatically big guy