Write the Nernst equation and calculate the emf of the following cell at 298 K.

Cu (s) / Cu⁺⁺ ( 0.130 M) // Ag⁺ ( 1.0× 10^-4 M) / Ag (s) given : E Cu++ / Cu = 0.34V and E Ag+/Ag = 0.80V

Cell reaction:

$\mathrm{Cu}\left(\mathrm{s}\right) + 2 {\mathrm{Ag}}^{+}\left(\mathrm{aq}\right) \to {\mathrm{Cu}}^{2+} \left(\mathrm{aq}\right) + 2 \mathrm{Ag} \left(\mathrm{s}\right)$

Half cell reactions:

$$\begin{gathered} \mathrm{Cathode} \left(\mathrm{reduction}\right): 2{\mathrm{Ag}}^{+}\left(\mathrm{aq}\right) + 2{\mathrm{e}}^{-} \to 2\mathrm{Ag} \left(\mathrm{s}\right) \\ \mathrm{Anode} \left(\mathrm{oxidation}\right): \mathrm{Cu}\left(\mathrm{s}\right) \to {\mathrm{Cu}}^{2+}\left(\mathrm{aq}\right) + 2 {\mathrm{e}}^{-} \end{gathered}$$

Silver electrode acts as a cathode and copper electrode as anode. The cell can be represented as:

Cu(s) | Cu²⁺ (0.130 M) || Ag⁺ ( 1.0× 10^-4 M) | Ag(s)

E^o_cell = E_right - Eleft
         = E_Ag⁺|Ag - E_Cu²⁺|Cu
​         = 0.80 - (0.34) V
        = 0.46 V

Emf at 298 K can be calculated by the Nernst equation:

$E_{cell} = {E^o}_{cell} - \frac{RT \times 2.303}{2F} \log \frac{\left[C{u}^{2+}\right]}{[A{g}^{+}{]}^2}$

where,
E^o = cell potential at standard conditions
R = gas constant
T = temperature
n = number of electrons exchanged (2 in this case)
F = Faraday's constant

Now substituting the values in the Nernst equation, we get

$$\begin{gathered} {\mathrm{E}}_{\mathrm{cell}} = 0.46 - \frac{8.314\times 298\times 2.303}{2\times 96500} \log \frac{(0.130)}{(1\times {10}^{-4}{)}^2} \\ = 0.46 - 0.029 \times 7.11 \\ =0.2538 \mathrm{V} \end{gathered}$$