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: $\frac{2.303RT}{F} = 0.059$)

The $EMF$ of the cell $M | M^{n+} (0.02 \, M) || H^{+} (1 \, M) | H_{2(g)} (1 \, atm), Pt$ at $25 \, ^\circ C$ is $0.81 \, V$. Calculate the valency of the metal $(n)$ if the standard oxidation potential of the metal is $0.76 \, V$. (Use $\frac{2.303 \, RT}{F} = 0.06, \log \, 2 = 0.3$)

For a cell reaction involving a two-electron change,the standard emf of the cell is found to be $0.295 \ V$ at $25 \ ^oC$. The equilibrium constant of the reaction at $25 \ ^oC$ will be

For the cell reaction,$3 Sn^{4+} + 2 Cr \longrightarrow 3 Sn^{2+} + 2 Cr^{3+}$,$E^{\circ}_{cell}$ is $0.89 \ V$. Then $\Delta G^{\circ}$ for the reaction is

If $Zn/Zn^{+2}$ electrode is diluted $100$ times,then the change in electromotive force will be:

In the cell $Pt_{(s)} | H_2(g, 1 \, bar) | HCl_{(aq)} | AgCl_{(s)} | Ag_{(s)} | Pt_{(s)}$,the cell potential is $0.92 \, V$ when a $10^{-6} \, m$ $HCl$ solution is used. The standard electrode potential of the $(AgCl/Ag, Cl^-)$ electrode is ............. $V$ $\{ \text{Given, } \frac{2.303RT}{F} = 0.06 \, V \text{ at } 298 \, K \}$