Write the Nernst equation and calculate the $emf$ of the following cells at $298 \, K$:
$(i) \; Mg_{(s)} | Mg^{2+}(0.001 \, M) || Cu^{2+}(0.0001 \, M) | Cu_{(s)}$
$(ii) \; Fe_{(s)} | Fe^{2+}(0.001 \, M) || H^{+}(1 \, M) | H_{2(g)}(1 \, bar) | Pt_{(s)}$
$(iii) \; Sn_{(s)} | Sn^{2+}(0.050 \, M) || H^{+}(0.020 \, M) | H_{2(g)}(1 \, bar) | Pt_{(s)}$
$(iv) \; Pt_{(s)} | Br_{2(l)} | Br^{-}(0.010 \, M), H^{+}(0.030 \, M) || H_{2(g)}(1 \, bar) | Pt_{(s)}$

(N/A) $(i)$ For the reaction $Mg + Cu^{2+} \rightarrow Mg^{2+} + Cu$,$n=2$ and $E_{cell}^{\Theta} = 0.34 - (-2.36) = 2.70 \, V$.
$E_{cell} = 2.70 - \frac{0.0591}{2} \log \frac{0.001}{0.0001} = 2.70 - 0.02955 \log 10 = 2.67 \, V$.
$(ii)$ For the reaction $Fe + 2H^{+} \rightarrow Fe^{2+} + H_2$,$n=2$ and $E_{cell}^{\Theta} = 0 - (-0.44) = 0.44 \, V$.
$E_{cell} = 0.44 - \frac{0.0591}{2} \log \frac{0.001}{(1)^2} = 0.44 - 0.02955(-3) = 0.5286 \, V \approx 0.53 \, V$.
$(iii)$ For the reaction $Sn + 2H^{+} \rightarrow Sn^{2+} + H_2$,$n=2$ and $E_{cell}^{\Theta} = 0 - (-0.14) = 0.14 \, V$.
$E_{cell} = 0.14 - \frac{0.0591}{2} \log \frac{0.050}{(0.020)^2} = 0.14 - 0.02955 \log(125) = 0.14 - 0.062 = 0.078 \, V$.
$(iv)$ For the reaction $Br_2 + H_2 \rightarrow 2Br^{-} + 2H^{+}$,$n=2$ and $E_{cell}^{\Theta} = 1.09 - 0 = 1.09 \, V$. Since the cell is written as $Pt | Br_2 | Br^{-}, H^{+} || H_2 | Pt$,the reaction is $2Br^{-} + 2H^{+} \rightarrow Br_2 + H_2$,so $E_{cell}^{\Theta} = 0 - 1.09 = -1.09 \, V$.
$E_{cell} = -1.09 - \frac{0.0591}{2} \log \frac{1}{(0.010)^2 (0.030)^2} = -1.09 - 0.02955 \log(1.11 \times 10^7) = -1.298 \, V$.