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 $E^o_{Cu^{2+}/Cu} = 0.34 \ V$.

The relationship between standard reduction potential of a cell and the equilibrium constant is shown by:

What is the change in potential of the following cell $Zn_{(s)}|Zn^{2+} (1 \ M)||Pb^{2+} (1 \ M)|Pb_{(s)}$ if the concentration of ions at the anode is increased $10$ times?

Calculate the cell potential at $298 \ K$ for the following cell:
$Zn_{(s)} | Zn^{2+} (0.6 \ M) || Cu^{2+} (0.3 \ M) | Cu_{(s)} \quad [E_{cell}^{o} = 1.1 \ V]$

The potential for the given half cell at $298 \ K$ is $(-) \ldots \ldots \ldots \times 10^{-2} \ V.$
$2 H^{+}_{(aq)} + 2 e^- \rightarrow H_{2(g)}$
$[H^{+}] = 1 \ M, P_{H_2} = 2 \ atm$
(Given: $2.303 RT / F = 0.06 \ V, \log 2 = 0.3$)

The electrode potential of the following half cell at $298 \ K$ is given by the cell reaction:
$X | X^{2+}(0.001 \ M) || Y^{2+}(0.01 \ M) | Y$
The cell potential is $....... \times 10^{-2} \ V$ (Nearest integer).
Given: $E^0_{X^{2+} | X} = -2.36 \ V$,$E^0_{Y^{2+} | Y} = +0.36 \ V$,$\frac{2.303 \ RT}{F} = 0.06 \ V$.