181. Write the Nernst equation and emf of the following cells at 298 k.
Mg(s) | Mg²⁺ (0.001 M) || Cu²⁺ (0.0001 M) | Cu(s)

For an electrochemical cell reaction
aA + bB → cC + dD
The Nernst’s equation for cell reaction is
${\mathrm{E}}_{\mathrm{cell}} = {{\mathrm{E}}^0}_{\mathrm{cell}}-\frac{0.059}{\mathrm{n}} \log \frac{{\left[\mathrm{C}\right]}^{\mathrm{c}} {\left[\mathrm{D}\right]}^{\mathrm{d}}}{{\left[\mathrm{A}\right]}^{\mathrm{a}} {\left[\mathrm{B}\right]}^{\mathrm{b}}}$
The values of a, b, c, d and n can be obtained from the balanced cell reaction. (i) Anode reaction:
$\mathrm{Mg}\left(\mathrm{s}\right) \to {\mathrm{Mg}}^{2+}\left(\mathrm{aq}\right) + 2{\mathrm{e}}^{-}$
Cathode reaction:
${\mathrm{Cu}}^{2+}\left(\mathrm{aq}\right) + 2{\mathrm{e}}^{-} \to \mathrm{Cu}\left(\mathrm{s}\right)$
Overall cell reaction:
          $\mathrm{Mg}\left(\mathrm{s}\right) + {\mathrm{Cu}}^{2+}\left(\mathrm{aq}\right) \to {\mathrm{Mg}}^{2+}\left(\mathrm{aq}\right) + \mathrm{Cu}\left(\mathrm{s}\right)$
Here,  n= 2,  ${{\mathrm{E}}^0}_{\mathrm{cell}} = {{\mathrm{E}}^0}_{\mathrm{cathode}} - {{\mathrm{E}}^0}_{\mathrm{anode}} = 0.34 \mathrm{V} - (-2.37) \mathrm{V} = 2.71 \mathrm{V}$
The Nernst equation for E_cell at 298 can be written as
$$\begin{gathered} {\mathrm{E}}_{\mathrm{cell}} ={{\mathrm{E}}^0}_{\mathrm{cell}} \\ = -\frac{0.059}{\mathrm{n}}\log \frac{{10}^{-3}}{{10}^{-4}} \\ = 2.71- 0.0295 = 2.68 \mathrm{V} \end{gathered}$$