The equilibrium constant of the reaction $A_{(s)} + 2B^{2+}_{(aq)} \rightleftharpoons A^{2+}_{(aq)} + 2B_{(s)}$ with $E^{\circ}_{\text{cell}} = 0.0295 \ V$ is (Given: $\frac{2.303RT}{F} = 0.059$)

Cell potential for the following reaction at $298 \ K$ is $0.096 \ V$. Calculate the $pH$ of the $HCl$ solution for the cell: $Sn_{(s)} | Sn^{2+} (0.05 \ M) || H^{+} (x \ M) | H_{2(g)} (1 \ bar) | Pt$ given $E_{Sn^{2+}|Sn}^{\circ} = -0.14 \ V$ and $E_{H^{+}|H_2}^{\circ} = 0.0 \ V$.

What will be the electromotive force of the following cell at $298 \ K$ ............. $V$
$Pt \, | \, Br_{2(\ell), 0.1M} \, | \, Br^{-}_{(aq), 0.1M} \, || \, H^{+}_{(aq), 0.1M} \, | \, H_{2(g), 1atm} \, | \, Pt$
Given: $E^{0}_{Br^{-}/Br_{2}} = -1.06 \ V$

Calculate the equilibrium constant $(K_c)$ of the reaction: $Ni_{(s)} + 2Ag_{(aq)}^{+} \rightarrow Ni_{(aq)}^{2+} + 2Ag_{(s)}$; $E_{cell}^{\circ} = 1.05 \ V$. (Given: $\frac{2.303 \ RT}{F} = 0.06$)

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

If $E^{\circ}(Ag^{+}_{(aq)} \mid Ag_{(s)}) = +0.80 \ V$,what is the potential developed for $Ag_{(s)} \rightarrow Ag^{+}_{(aq)} (0.01 \ M) + e^{-}$ at $298 \ K$?