Cell notation of a galvanic cell is represented as $\mathrm{Ni}\left(\mathrm{s}\right)/{\mathrm{Ni}}^{+2}\left({\mathrm{M}}_1\right)//{\mathrm{Cl}}^{-}\left({\mathrm{M}}_2\right)/{\mathrm{Cl}}_2\left(1\mathrm{atm}\right), \mathrm{Pt}\left(\mathrm{s}\right).$ Emf of the cell is maximum when the values of ${\mathrm{M}}_1$ and ${\mathrm{M}}_2$ are

Anode (LHE) $\mathrm{Ni}\left(\mathrm{s}\right)\stackrel{ }{\to }{\mathrm{Ni}}^{+2}\left({\mathrm{M}}_1\right)+2\overline{\mathrm{e}};$

Cathode (RHE) ${\mathrm{Cl}}_2\left(1\mathrm{atm}\right)+2\overline{\mathrm{e}}\stackrel{ }{\to }2{\mathrm{Cl}}^{-}\left({\mathrm{M}}_2\right)$

Net cell reaction : $\mathrm{Ni}\left(\mathrm{s}\right)+{\mathrm{Cl}}_2\left(1\mathrm{atm}\right)\stackrel{ }{\to }{\mathrm{Ni}}^{+2}\left({\mathrm{M}}_1\right)+2{\mathrm{Cl}}^{-}\left({\mathrm{M}}_2\right)$

$$\begin{gathered} {\mathrm{E}}_{\mathrm{cell}}={\mathrm{E}}_{\mathrm{cell}}^0-\frac{0.0591}{2}\log \left[{\mathrm{Ni}}^{+2}\right]{\left[{\mathrm{Cl}}^{-}\right]}^2 \\ \therefore {\mathrm{E}}_{\mathrm{cell}}\propto -\log \left[{\mathrm{Ni}}^{+2}\right]{\left[{\mathrm{Cl}}^{-}\right]}^2 \end{gathered}$$

Thus lesser the product of $\left[{\mathrm{Ni}}^{+2}\right]{\left[{\mathrm{Cl}}^{-}\right]}^2$ higher will be the ${\mathrm{E}}_{\mathrm{cell}}$;

The product of  $\left[{\mathrm{Ni}}^{+2}\right]{\left[{\mathrm{Cl}}^{-}\right]}^2$in each case is

(1) option ….${10}^{-6}$

(2) option ….1

(3) option ….${10}^{-5}$

(4) option ….${10}^{-4}$

maximum emf of the cell is when $\left[{\mathrm{Ni}}^{+2}\right]$=0.01 M and $\left[{\mathrm{Cl}}^{-}\right]$ =0.01 M