Consider two reactions

 $\mathrm{X}⟶\mathrm{Y}, \mathrm{\Delta G}°=-386\frac{\mathrm{kJ}}{\mathrm{mol}}$

${\mathrm{M}}^{2+}⟶{\mathrm{M}}^{4+}+2{\mathrm{e}}^{-},\mathrm{E}°=-\mathrm{pV}$

It is given that all the energy released from the conversion of $\mathrm{X}$ to $\mathrm{Y}$ is used in the oxidation of ${\mathrm{M}}^{2+}$ to ${\mathrm{M}}^{4+}$. Under standard conditions, on conversion of 1 mole of $\mathrm{X}, 4$ moles of ${\mathrm{M}}^{2+}$ are oxidised. The value of $100\mathrm{p}$ is (use $\mathrm{F}=96500{\mathrm{Cmol}}^{-1}$ )

The reaction occurs in which of the given galvanic cell?

$\frac{1}{2}{\mathrm{H}}_2( \mathrm{g})+\mathrm{AgCl}\left(\mathrm{s}\right)\rightleftharpoons {\mathrm{H}}^{+}\left(\mathrm{aq}\right)+{\mathrm{Cl}}^{-}\left(\mathrm{aq}\right)+\mathrm{Ag}\left(\mathrm{s}\right)$

At $298 \mathrm{K}$, the $\mathrm{e}.\mathrm{m}.\mathrm{f}.$ of the galvanic cell mentioned below is $\left({\mathrm{E}}^0=1.1 \mathrm{V}\right)$

$\mathrm{Zn}\left(\mathrm{s}\right)\left|{\mathrm{Zn}}^{2+}\left({10}^{-3}\mathrm{M}\right)\Vert {\mathrm{Cu}}^{2+}\left(0.1\mathrm{M}\right)\right|\mathrm{Cu}\left(\mathrm{s}\right)$

For the cell $\mathrm{Zn} \left(\mathrm{s}\right)\left|{\mathrm{ZnSO}}_4 \left(\mathrm{aq}\right)\right|\left|{\mathrm{CuSO}}_4 \left(\mathrm{aq}\right)\right|\mathrm{Cu} \left(\mathrm{s}\right)$, when the concentration of ${\mathrm{Zn}}^{2+}$ is $10$ times the concentration of ${\mathrm{Cu}}^{2+}$, the expression for $\mathrm{\Delta G} \left(\mathrm{in}\mathrm{ }\mathrm{J}\mathrm{ }{\mathrm{mol}}^{-1}\right)$ is

$[\mathrm{F}=$Faraday's constant, $\mathrm{R}=$universal gas constant, $\mathrm{T}=$temperature, ${\mathrm{E}}^{\mathrm{o}}\mathrm{cell}=1.1\mathrm{V}]$

The following electrochemical cell has been set up.

$\begin{array}{l}\mathrm{Pt}\left(1\right)|{\mathrm{Fe}}^{3+},{\mathrm{Fe}}^{2+}(\mathrm{a}=1)||{\mathrm{Ce}}^{4+},{\mathrm{Ce}}^{3+}(\mathrm{a}=1\left)\right|\mathrm{Pt}\left(2\right)\\ \mathrm{E}°({\mathrm{Fe}}^{3+},{\mathrm{Fe}}^{2+})=0.77\mathrm{V}:\mathrm{E}°({\mathrm{Ce}}^{4+},{\mathrm{Ce}}^{3+})=1.61\mathrm{V}\end{array}$
If an ammeter is connected between the two platinum electrodes, predict the direction of flow of current.