Consider the figure and answer the following questions.
$(i)$ Cell $'A'$ has $E_{cell} = 2 \ V$ and cell $'B'$ has $E_{cell} = 1.1 \ V$. Which of the two cells,$'A'$ or $'B'$,will act as an electrolytic cell? Which electrode reactions will occur in this cell?
$(ii)$ If cell $'A'$ has $E_{cell} = 0.5 \ V$ and cell $'B'$ has $E_{cell} = 1.1 \ V$,then what will be the reaction at the anode and cathode?

(N/A) $(i)$ Cell $'B'$ will act as an electrolytic cell because it has a lower $emf$ $(1.1 \ V < 2 \ V)$.
$\therefore$ The electrode reactions in cell $'B'$ will be:
Reduction at cathode: $Zn^{2+} + 2e^{-} \rightarrow Zn$
Oxidation at anode: $Cu \rightarrow Cu^{2+} + 2e^{-}$
$(ii)$ Now,cell $'B'$ acts as a galvanic cell because it has a higher $emf$ $(1.1 \ V > 0.5 \ V)$ and will push electrons into cell $'A'$.
The electrode reactions will be:
At anode: $Zn \rightarrow Zn^{2+} + 2e^{-}$
At cathode: $Cu^{2+} + 2e^{-} \rightarrow Cu$