The $\mathrm{e}.\mathrm{m}.\mathrm{f}.$ of the following cell

$\frac{\mathrm{Ni}\left(s\right)}{{\mathrm{NiSO}}_4\left(1.0\mathrm{M}\right)}\Vert \frac{{\mathrm{H}}^{+}\left(1.0\mathrm{M}\right)}{{\mathrm{H}}_2}\left(1\mathrm{atm}\right), \mathrm{Pt}$ at $25°\mathrm{C}$ is $0.236 \mathrm{V}.$ The electrical energy which can be produced is :-

${\mathrm{Hg}}_2{\mathrm{Cl}}_2$ is produced by the electrolytic reduction of ${\mathrm{Hg}}^{2+}$ ion in presence of ${\mathrm{Cl}}^{-}$ion is $2{\mathrm{Hg}}^{2+}+2{\mathrm{Cl}}^{-}+2{\mathrm{e}}^{\mathrm{\Theta }}\to {\mathrm{Hg}}_2{\mathrm{Cl}}_2.$ Calculate the current required to have a rate production of $44 \mathrm{g}$ per hour of ${\mathrm{Hg}}_2{\mathrm{Cl}}_2.$

[Atomic weight of $\mathrm{Hg}=200.6 ] :-$

Consider the following Galvanic cell as shown in the figure:

By what value will the cell voltage change when concentration of ions in anodic and cathodic compartments are both increased by a factor of $10$ at $298\mathrm{K}$.