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

What is the oxidation potential of $0.05 \, M \, H_2SO_4$ in volts?

The equilibrium constant of the reaction $Cu_{(s)} + 2Ag^{+}_{(aq)} \to Cu^{2+}_{(aq)} + 2Ag_{(s)}$ with $E^{\circ} = 0.46 \ V$ at $298 \ K$ is approximately:

What would be the electrode potential for the given half-cell reaction at $pH = 5$ (in $V$)?
$2 H_2O \rightarrow O_2 + 4 H^{\oplus} + 4 e^{-} ; E_{red}^{0} = 1.23 \ V$
$(R = 8.314 \ J \ mol^{-1} \ K^{-1} ; \text{Temperature} = 298 \ K ; \text{oxygen under std. atm. pressure of } 1 \ bar)$

The cell potential for the following cell notation is approximately
$M_{(s)} | M^{3+}(aq, 0.01 \ M) || N^{2+}(aq, 0.1 \ M) | N_{(s)}$
$E_{M^{3+} / M}^0 = 0.6 \ V$ and $E_{N^{2+} / N}^0 = 0.1 \ V$ (in $V$)

The oxidation potential of a hydrogen electrode at $pH = 10$ and $P_{H_2} = 1 \, atm$ is ........... $V$.