For the cell reaction $Zn(s) + Cu^{2+}(aq) (1.0 \, M) \to Cu(s) + Zn^{2+}(aq) (0.1 \, M)$,the measured $e.m.f.$ at $25 \, ^oC$ is $1.3 \, V$. Calculate the $E^o$ value for the cell reaction. (in $, V$)

$Cu_{(s)} + Sn^{2+}(0.001 \ M) \rightarrow Cu^{2+}(0.01 \ M) + Sn_{(s)}$
The Gibbs free energy change for the above reaction at $298 \ K$ is $x \times 10^{-1} \ kJ \ mol^{-1}$;
The value of $x$ is ..... [nearest integer] $\left[\text{Given}: E^{\ominus}_{Cu^{2+}/Cu} = 0.34 \ V; E^{\ominus}_{Sn^{2+}/Sn} = -0.14 \ V; F = 96500 \ C \ mol^{-1}\right]$

Calculate the equilibrium constant $(K_C)$ for the cell obtained by connecting two electrodes with standard electrode potentials $E^o_{(Sn^{2+}|Sn)} = -0.14 \ V$ and $E^o_{(Ni^{2+}|Ni)} = -0.23 \ V$ at $298 \ K$.

The standard electrode potential for $Cu^{+2}/Cu$ is $0.34 \ V$. Calculate the reduction potential at $pH = 14$ for the above couple $V$ $[K_{sp}[Cu(OH)_2] = 1 \times 10^{-19}]$

Calculate the emf of the half-cell given below: $Pt(s) | H_2(g, 2 \text{ atm}) | HCl(aq, 0.02 \text{ M})$, $E^\circ_{H^+/H_2} = 0 \text{ V}$. (Given: $\frac{2.303RT}{F} = 0.059$, $\log 2 = 0.3010$)

By how much will the potential of the half-cell $Cu^{2+}/Cu$ change if the solution is diluted to $100$ times at $298 \ K$?