At $298 \ K$ the equilibrium constant for the reaction $M_{(s)} + 2 Ag^{+}_{(aq)} \rightarrow M^{2+}_{(aq)} + 2 Ag_{(s)}$ is $10^{15}$. What is the $E_{cell}^{\ominus}$ (in $V$) for this reaction? $\left(\frac{2.303 RT}{F}\right) = 0.06 \ V$

At $298 \ K$, the $emf$ of the cell is ............ $V$.
$Pt | H_{2(2 \ atm)} | H_{(0.02 \ M)}^{+} || H_{(0.1 \ M)}^{+} | H_{2(1 \ atm)} | Pt$

Considering the reaction $Cl_{2(g)} + 2Br^{-}_{(aq)} \rightarrow 2Cl^{-}_{(aq)} + Br_{2(l)}$,calculate the cell $emf$ in $V$ when $[Cl^{-}] = [Br_2] = [Br^{-}] = 0.01 \ M$ and $Cl_2$ gas is at $1 \ atm$ pressure. (Given $E^o = 0.29 \ V$ for the reaction)

In a cell reaction,a $2$ electron change occurs. The standard $e.m.f.$ of the cell at $25^\circ C$ is $0.295 \ V$. What is the equilibrium constant for the reaction at $25^\circ C$?

The logarithm of the equilibrium constant for the reaction $Pd^{2+}{(aq)} + 4Cl^{-}{(aq)} \rightleftharpoons PdCl_4^{2-}{(aq)}$ is (Nearest integer).
Given: $\frac{2.303 RT}{F} = 0.06 \ V$
$Pd^{2+}{(aq)} + 2e^{-} \rightleftharpoons Pd_{(s)} \quad E^{\circ} = 0.83 \ V$
$PdCl_4^{2-}{(aq)} + 2e^{-} \rightleftharpoons Pd_{(s)} + 4Cl^{-}{(aq)} \quad E^{\circ} = 0.65 \ V$

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 $E_{H^{+}|H_2}^{\circ} = 0.0 \ V$.