Explanation

(b)
Electrode potential of cell must be positive for spontaneous reaction.
${Zn}^{2 +}\ \longrightarrow Zn\ ;\ E^{{^\circ}} = \ - 0.76\ V$
${Cu}^{2 +}\ \longrightarrow Cu\ ;\ E^{{^\circ}} = \ - 0.34\ V$
Redox reaction is
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ Zn\ \rightarrow \ \ {Zn}^{2 +} + \ {2e}^{-}$ (oxidation)
$\ {Cu}^{2 +} + \ {2e}^{-}\ \rightarrow Cu$ (reduction)
[table] $Zn + \ {Cu}^{2 +}\ \rightarrow \ {Zn}^{2 +} + \ Cu$ [/table]
$E_{cell} = \ E_{cathode}^{{^\circ}}\ - \ E_{anode}^{{^\circ}}$
= - 0.34 – (- 0.76)
= + 0.42 V
$E_{cell}$ is positive , so above reaction is feasible.