Assume a cell with the following reaction:
$Cu_{(s)} + 2 Ag^{+} (1 \times 10^{-3} \, M) \rightarrow Cu^{2+} (0.250 \, M) + 2 Ag_{(s)}$
$E_{Cell}^{\ominus} = 2.97 \, V$
$E_{cell}$ for the above reaction is $.... \, V.$ (Nearest integer)
[Given: $\log 2.5 = 0.3979, T = 298 \, K]$

Derive the Nernst equation for the following galvanic cell: $Ni_{(s)}|Ni_{(aq)}^{2+}\parallel Ag_{(aq)}^{+}|Ag_{(s)}$

Consider the cell at $25^{\circ} C$:
$Zn | Zn^{2+}_{(aq)} (1 \ M) || Fe^{3+}_{(aq)}, Fe^{2+}_{(aq)} | Pt_{(s)}$
The fraction of total iron present as $Fe^{3+}$ ion at the cell potential of $1.500 \ V$ is $X \times 10^{-2}$. The value of $X$ is $.....$ (Nearest integer).
(Given $E^{0}_{Fe^{3+} / Fe^{2+}} = 0.77 \ V, E^{0}_{Zn^{2+} / Zn} = -0.76 \ V$)

What will be the reduction potential of $Cu$ in an aqueous solution with $pH = 12$? Given that the $K_{sp}$ of $Cu(OH)_2$ is $1 \times 10^{-19}$ and $E^{\circ}_{Cu^{+2}/Cu} = 0.34 \ V$.

For the cell,$Zn_{(s)} | Zn^{2+} (1 \ M) || Ag^{+} (1 \ M) | Ag_{(s)}$. If the concentration of $Zn^{2+}$ decreases to $0.1 \ M$ at $298 \ K$,then the $EMF$ of the cell:

If the standard electrode potential for a cell is $2 \ V$ at $300 \ K,$ the equilibrium constant $(K)$ for the reaction $Zn_{(s)} + Cu^{2+}_{(aq)} \rightleftharpoons Zn^{2+}_{(aq)} + Cu_{(s)}$ at $300 \ K$ is approximately $(R = 8 \ J \ K^{-1} \ mol^{-1}, F = 96000 \ C \ mol^{-1})$