${\mathrm{Zn}}_{\left(\mathrm{s}\right)}+{\mathrm{Cl}}_2\left(1\mathrm{atm}\right)⟶{\mathrm{Zn}}^{2+}+2{\mathrm{Cl}}^{-}$ 
$\text{ The }{\mathrm{E}}^{\mathrm{o}}\text{ of the cell is }2.12\mathrm{V}\text{. To increase }\mathrm{E}$ 

The redox equation is given as 

$$\begin{gathered} {\mathrm{Zn}}_{\left(\mathrm{s}\right)}+\hspace{0.17em}{\mathrm{Cl}}_{2\left(1\mathrm{atm}\right)}\to {\mathrm{Zn}}^{2+}+\hspace{0.17em}2{\mathrm{Cl}}^{-} \\ \mathrm{Anode}:\hspace{0.17em}\mathrm{Zn}\to {\mathrm{Zn}}^{2+}+\hspace{0.17em}2{\mathrm{e}}^{-} \\ \mathrm{Cathode}:\hspace{0.17em}{\mathrm{Cl}}_2+\hspace{0.17em}2{\mathrm{e}}^{-}\to 2{\mathrm{Cl}}^{-} \\ {\mathrm{E}}_{\mathrm{cell}}^{ }= {\mathrm{E}}_{\mathrm{cell}}^{\mathrm{o}}- \frac{0.0591}{2}\log \left[{\mathrm{Zn}}^{2+}\right] {\left[{\mathrm{Cl}}^{-}\right]}^2 \end{gathered}$$

The E_cell will be increased when concentration of both Zn²⁺ and Cl^- will be decreased.