$3.482 \times 10^{-1} \mathrm{~g}$ of Fe gets deposited when an aqueous solution of Ferric sulphate is electrolysed for 20 minutes using a current of " $x$ " amperes. Find " $x$ ". (Atomic mass of $\mathrm{Fe}=56 \mathrm{amu}$ ).

For the cell reaction $4 \mathrm{Br}^{-}+\mathrm{O}_2+4 \mathrm{H}^{+} \rightarrow 2 \mathrm{Br}_2+2 \mathrm{H}_2 \mathrm{O}$ at 298 K , the $\mathrm{E}^0$ cell $=0.16 \mathrm{~V}$. What would be the $\mathrm{K}_{\mathrm{c}}$ (Equilibrium constant) value if the reverse reaction were to take place?

Two statements, one Assertion and the other Reason, are given. Choose the correct option.

Assertion : During the electrolysis of aqueous $\mathrm{NaCl}, \mathrm{Cl}_2$ is liberated at the anode in preference to $\mathrm{O}_2$ and water gets reduced to $\mathrm{H}_2$ at cathode.

Reason : The reaction at Anode with lower oxidation potential is not preferred due to over potential of Oxygen.

For a cell $2 \mathrm{M}_{(\mathrm{S})}+\mathrm{O}_2(\mathrm{~g})+4 \mathrm{H}^{+} \rightarrow 2 \mathrm{M}^{2+}(\mathrm{aq})+2 \mathrm{H}_2 \mathrm{O}_{(\mathrm{l})}$ the $\mathrm{E}^0$ cell $=+1.67 \mathrm{~V}$. When $\left[\mathrm{M}^{2+}\right]$ is $1.0 \times 10^{-3} \mathrm{M}$ and $\mathrm{p}\left(\mathrm{O}_2\right)$ is 0.1 atm , the EMF of the cell becomes +1.57 V . Calculate the pH of the electrochemical cell.