183. Write the Nernst equation and emf of the following cells at 298 k.
Sn(s) | Sn²⁺ (0.050 M) || H⁺(0.020 M) H₂(g) (1bar) Pt (s)

For the given cell Anode reaction:
Sn(s) → Sn²⁺ (aq) + 2e^– 

Cathode reaction:
2H⁺(aq) + 2e^– → H₂(g) 

Overall cell reaction:
Sn(s) + 2H⁺ (aq) → Sn²⁺(aq) + H₂(g)

Here, n = 2, E°_cell = E°_cathode – E°_anode = 0 – (– 0.14 V) = + 0.14 V.
The Nernst equation for E_cell and 298 k can be written as:
$$\begin{gathered} {\mathrm{E}}_{\mathrm{cell}} = {{\mathrm{E}}^0}_{\mathrm{cell}} - \frac{0.059}{\mathrm{n}}\log \frac{\left[{\mathrm{Sn}}^{2+}\right]}{{\left[{\mathrm{H}}^{+}\right]}^2} \\ = 0.14 - \frac{0.059}{\mathrm{n}} \log \frac{\left[0.05\right]}{{\left[0.02\right]}^2} \\ = 0.14 \mathrm{V} - 0.0295 (\log 1.25 \times {10}^2) \\ = 0.14 \mathrm{V} -0.0295 \times 2.0969 \\ = 0.14 - 0.0619 \\ = 0.0781 \mathrm{V} \end{gathered}$$