If a cell has a standard electrode potential of $0.295 \ V$ and $n = 2$,calculate its equilibrium constant at $298 \ K$.

At $298 \, K$,the standard reduction potential for $Cu^{2+}/Cu$ electrode is $0.34 \, V$.
Given: $K_{sp} \text{ of } Cu(OH)_2 = 1 \times 10^{-20}$
Take $\frac{2.303 RT}{F} = 0.059 \, V$
The reduction potential at $pH = 14$ for the above couple is $(-)x \times 10^{-2} \, V$. The value of $x$ is $........$.

Calculate the equilibrium constant of the following reaction:
$\text{Cu(s)} + 2\text{Ag}^+_{\text{(aq)}} \rightarrow \text{Cu}^{2+}_{\text{(aq)}} + 2\text{Ag(s)}$, given $E^\circ_{\text{cell}} = 0.46 \text{ V}$.

$Zn_{(s)} + Cl_2(1 \ atm) \to Zn^{2+} + 2Cl^-$. The $E^0_{cell}$ of the cell is $2.12 \ V$. To increase $E_{cell}$:

The magnitude of the change in oxidising power of the $MnO_4^- / Mn^{2+}$ couple is $x \times 10^{-4} \, V$,if the $H^{+}$ concentration is decreased from $1 \, M$ to $10^{-4} \, M$ at $25^{\circ} C$. (Assume concentration of $MnO_4^-$ and $Mn^{2+}$ to be same on change in $H^{+}$ concentration). The value of $x$ is ....... .
(Rounded off to the nearest integer)
$[\text{Given} : \frac{2.303 RT}{F} = 0.059]$

At $298 \ K$,the following reaction takes place for a hydrogen electrode:
$H^{+}_{(aq)} + e^{-} \longrightarrow \frac{1}{2} H_2(1 \ bar)$
The solution $pH$ is $10.0$. What is the hydrogen electrode potential in volts?
$\left(\frac{2.303 RT}{F} = 0.06 \ V\right)$