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 $\frac{2.303 \, RT}{F} = 0.06, \log \, 2 = 0.3$)

If the standard electrode potential for a cell is $2 \ V$ at $300 \ K,$ the equilibrium constant $(K)$ for the reaction $Zn_{(s)} + Cu^{2+}_{(aq)} \rightleftharpoons Zn^{2+}_{(aq)} + Cu_{(s)}$ at $300 \ K$ is approximately $(R = 8 \ J \ K^{-1} \ mol^{-1}, F = 96000 \ C \ mol^{-1})$

Write a note on the relation between Gibbs free energy and cell potential for a cell reaction.

The concentration of potassium ions inside a biological cell is at least twenty times higher than the outside. The resulting potential difference across the cell is important in several processes such as transmission of nerve impulses and maintaining the ion balance. $A$ simple model for such a concentration cell involving a metal $M$ is:
$M_{(s)} \mid M^{+}(aq; 0.05 \ M) \parallel M^{+}(aq; 1 \ M) \mid M_{(s)}$
For the above electrolytic cell the magnitude of the cell potential $|E_{cell}|=70 \ mV$.
$1.$ For the above cell
$(A)$ $E_{cell} < 0 ; \Delta G > 0$ $(B)$ $E_{cell} > 0 ; \Delta G < 0$
$(C)$ $E_{cell} < 0 ; \Delta G^{\circ} > 0$ $(D)$ $E_{cell} > 0 ; \Delta G^{\circ} > 0$
$2.$ If the $0.05 \ M$ solution of $M^{+}$ is replaced by $0.0025 \ M$ $M^{+}$ solution,then the magnitude of the cell potential would be
$(A)$ $35 \ mV$ $(B)$ $70 \ mV$ $(C)$ $140 \ mV$ $(D)$ $700 \ mV$
Give the answer for questions $1$ and $2$.

The cell reaction of the given cell is spontaneous if:
$Pt | Cl_2 (P_1 \, atm) | Cl^{-} (1 \, M) || Cl^{-} (1 \, M) | Cl_2 (P_2 \, atm) | Pt$

For the cell $Cu_{(s)}|Cu^{2+}_{(aq)}(0.1 \ M) || Ag^{+}_{(aq)}(0.01 \ M)| Ag_{(s)}$,the cell potential $E_{1} = 0.3095 \ V$. For the cell $Cu_{(s)}|Cu^{2+}_{(aq)}(0.01 \ M) || Ag^{+}_{(aq)}(0.001 \ M)| Ag_{(s)}$,the cell potential $= ..... \times 10^{-2} \ V$. (Round off to the Nearest Integer). [Use: $\frac{2.303 \ RT}{F} = 0.059$]