Consider the electrochemical cell: $Pt \ | \ O_{2(g)} \ (1 \ bar) \ | \ HCl \ (aq) \ || \ M^{2+} \ (aq, 1.0 \ M) \ | \ M_{(s)}$. The pH above which, oxygen gas would start to evolve at the anode is . . . . . . (nearest integer). $\left[ \text{Given :} \ E^{\circ}_{M^{2+}/M} = 0.994 \ V, \ E^{\circ}_{O_{2}/H_{2}O} = 1.23 \ V, \ \frac{RT}{F}(2.303) = 0.059 \ V \ \text{at the given condition} \right]$

For a given half-cell,$Al^{3+} + 3e^{-} \rightarrow Al$,on increasing the concentration of aluminium ions,the electrode potential will

For the electrochemical cell shown below:
$Pt \mid H_{2}(p=1 \, atm) \mid H^{+}(aq., x \, M) \mid\mid Cu^{2+}(aq., 1.0 \, M) \mid Cu_{(s)}$
The potential is $0.49 \, V$ at $298 \, K$. The $pH$ of the solution is closest to:
[Given: Standard reduction potential,$E^{\circ}$ for $Cu^{2+}/Cu$ is $0.34 \, V$; Gas constant,$R = 8.31 \, J \, K^{-1} \, mol^{-1}$; Faraday constant,$F = 9.65 \times 10^{4} \, J \, V^{-1} \, mol^{-1}$]

At $298 \, K$,the standard reduction potential for $Cu^{2+}/Cu$ electrode is $0.34 \, V$.
Given: $K_{sp} \text{ of } Cu(OH)_2 = 1 \times 10^{-20}$
Take $\frac{2.303 RT}{F} = 0.059 \, V$
The reduction potential at $pH = 14$ for the above couple is $(-)x \times 10^{-2} \, V$. The value of $x$ is $........$.

If $E^{\circ}_{\text{cell}}$ is $1.049 \ V$ and the reaction involves the transfer of $2$ electrons,calculate the equilibrium constant $(K)$ of the cell.

Calculate cell potential at $298 \ K$ for the following cell.
$Ag_{(s)} | Ag^{+}(0.25 \ M) || Ag^{+}(0.75 \ M) | Ag_{(s)}$ $\left[ E_{Ag^{+} \mid Ag}^{o} = 0.80 \ V \right]$