Consider the cell, $\mathrm{Ag}\overline{)\mathrm{AgCl}\left(\mathrm{s}\right)}\left|\mathrm{KCl} \left(1\mathrm{M}\right)\right|{\mathrm{Hg}}_2{\mathrm{Cl}}_2\left(\mathrm{s}\right)$$\overline{)\mathrm{Hg} \left(\mathrm{l}\right)}$$|\mathrm{Pt}.$ The cell potential

In the following electrochemical cell: $\mathrm{Pt} \overline{) {\mathrm{H}}_2} \left(\mathrm{x} \mathrm{atm}\right) \overline{) {\mathrm{H}}^{+} }\left(\mathrm{pH}=\mathrm{y}\right) \Vert {\mathrm{Zn}}^{2+} \left(\mathrm{z} \mathrm{M}\right) \overline{) \mathrm{Zn}}$

${\mathrm{E}}_{\mathrm{cell}}^{\mathrm{o}}={\mathrm{E}}_{\text{cell }}.$ This will be possible when

The cell potential for the unbalanced chemical reaction :

${\mathrm{Hg}}_2^{2+}+{\mathrm{NO}}_3^{-}+{\mathrm{H}}^{+}⟶{\mathrm{Hg}}^{2+}+{\mathrm{HNO}}_2+{\mathrm{H}}_2\mathrm{O};{\mathrm{E}}_{\mathrm{cell} }^{\mathrm{o}}=0.02 \mathrm{V}$

$\left(\frac{2.303\mathrm{RT}}{F}=0.06\right)$

Then choose the correct statement.