Emf of the cell $\mathrm{Zn}/{\mathrm{Zn}}^{+2}//{\mathrm{Cu}}^{+2}/\mathrm{Cu}$will be highest when 

${\mathrm{E}}_{\mathrm{cell}}={\mathrm{E}}_{\mathrm{cell}}^{\mathrm{o}}-\frac{0.0591}{\mathrm{n}}\log \frac{\left[\mathrm{Products}\right]}{\left[\mathrm{Reactants}\right]}$

$\mathrm{Anode}-{\mathrm{Zn}}^{ }\to {\text{\hspace{0.17em}Zn}}^{+2}+2{\mathrm{e}}^{-}$

$\mathrm{Cathode} {\mathrm{Cu}}^{+2} +2{\mathrm{e}}^{-}\to \mathrm{Cu}$

                 $\mathrm{Zn} + {\mathrm{Cu}}^{+2} \to {\mathrm{Zn}}^{+2} +\mathrm{Cu}$

${{\mathrm{E}}^0}_{\mathrm{cell}} \mathrm{is} \mathrm{a} \mathrm{constant}$

${\mathrm{E}}_{\mathrm{cell}}\propto -\log \frac{\left[{\mathrm{Zn}}^{+2}\right]}{\left[{\mathrm{Cu}}^{+2}\right]}$

$\mathrm{Lesser}\mathrm{the}\frac{\left[{\mathrm{Zn}}^{+2}\right]}{\left[{\mathrm{Cu}}^{+2}\right]},\mathrm{higher}\mathrm{will}\mathrm{be}\mathrm{the}{\mathrm{E}}_{\mathrm{cell}}$

$\therefore \left(2\right) \mathrm{is} \mathrm{correct}$