At $25\,^{\circ}C$,calculate the equilibrium constant for the cell reaction,
$X_{(s)} + Y_{(aq)}^{2+} \rightleftharpoons Y_{(s)} + X_{(aq)}^{2+}$
Given:
$E_{X^{2+}/X}^{o} = -1.36\,V$;
$E_{Y^{2+}/Y}^{o} = -0.76\,V$; $\frac{2.303\,RT}{F} = 0.06$

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

What is the potential of the cell consisting of two hydrogen electrodes as shown below (in $V$)?
$Pt | H_2(g) | H^+_{(aq)} (10^{-8} \ M) || H^+_{(aq)} (0.001 \ M) | H_2(g) | Pt$

If $E^{\circ}(Mg^{+2}_{(aq)} \mid Mg_{(s)}) = -2.37 \ V$. What is the potential for $Mg_{(s)} \rightarrow Mg^{+2}_{(0.01 \ M)} + 2 \overline{e}$ at $298 \ K$?

Consider the cell at $25^{\circ} C$:
$Zn | Zn^{2+}_{(aq)} (1 \ M) || Fe^{3+}_{(aq)}, Fe^{2+}_{(aq)} | Pt_{(s)}$
The fraction of total iron present as $Fe^{3+}$ ion at the cell potential of $1.500 \ V$ is $X \times 10^{-2}$. The value of $X$ is $.....$ (Nearest integer).
(Given $E^{0}_{Fe^{3+} / Fe^{2+}} = 0.77 \ V, E^{0}_{Zn^{2+} / Zn} = -0.76 \ V$)

What is the value of $E_{cell}$ at $298 \ K$ for the reaction,$Zn_{(s)} + Cu^{+2}(0.1 \ M) \rightarrow Zn^{+2}(0.1 \ M) + Cu_{(s)}$ if $E^{\circ}_{cell} = 1.1 \ V$ (in $V$)?