The equilibrium constant for the following general reaction is $10^{30}$. Calculate $E^o$ for the cell at $298 \ K$.
$2 X_2(s)+3 Y^{2+}(a q) \rightarrow 2 X_2^{3+}(a q)+3 Y(s)$

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

For the following electrochemical cell at $298 \ K$,
$Pt_{(s)} \mid H_2(g, 1 \ bar) \mid H^{+}(aq, 1 \ M) \parallel M^{4+}_{(aq)}, M^{2+}_{(aq)} \mid Pt_{(s)}$
$E_{\text{cell}} = 0.092 \ V$ when $\frac{[M^{2+}_{(aq)}]}{[M^{4+}_{(aq)}]} = 10^x$
Given : $E^0_{M^{4+}/M^{2+}} = 0.151 \ V$; $2.303 \frac{RT}{F} = 0.059 \ V$
The value of $x$ is

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 $298 \ K$.

An oxidation-reduction reaction in which $3$ electrons are transferred has a $\Delta G^{\circ}$ of $17.37 \ kJ \ mol^{-1}$ at $25^{\circ} C$. The value of $E_{\text{cell}}^{\circ}$ (in $V$) is........ $\times 10^{-2} \ V$.
$(1 \ F = 96,500 \ C \ mol^{-1})$

Calculate the $pH$ of an $HCl$ solution at $298 \ K$ for the following cell:
$Pt_{(s)} \mid H_2 \ (1 \ bar) \mid HCl \ (xM) \parallel Ag^+ \ (0.01 \ M) \mid Ag_{(s)}$
Given that the standard cell potential $E^\circ_{cell} = 1.05 \ V$. (in $.73$)