When the lead storage battery is in use (during discharge),the reaction that occurs at the anode is:

  • A
    $PbSO_{4\text{(s)}} + 2H_2O_{\text{(l)}} \rightarrow PbO_{2\text{(s)}} + SO_4^{2-}{_{\text{(aq)}}} + 4H^{+}{_{\text{(aq)}}} + 2e^{-}$
  • B
    $Pb_{(s)} + PbO_{2(s)} + 2H_2SO_{4(aq)} \longrightarrow 2PbSO_{4(s)} + 2H_2O(\ell)$
  • C
    $Pb_{\text{(s)}} + SO_4^{2-}{_{\text{(aq)}}} \rightarrow PbSO_{4\text{(s)}} + 2e^{-}$
  • D
    $PbO_{2\text{(s)}} + SO_4^{2-}{_{\text{(aq)}}} + 4H^{+}{_{\text{(aq)}}} + 2e^{-} \rightarrow PbSO_{4\text{(s)}} + 2H_2O_{\text{(l)}}$

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For the decomposition of $Al_2O_3$ at $500\,^{\circ}C$,the Gibbs energy change is given by $\frac{2}{3} Al_2O_3 \rightarrow \frac{4}{3} Al + O_2$,$\Delta G = 966 \, kJ \, mol^{-1}$. What is the potential difference (in $V$) required for the electrolytic reduction of $Al_2O_3$ at $500\,^{\circ}C$?

Given below are two statements: one is labelled as Assertion $A$ and the other is labelled as Reason $R$.
Assertion $A$: In equation $\Delta_r G = -nFE_{\text{cell}}$,the value of $\Delta_r G$ depends on $n$.
Reason $R$: $E_{\text{cell}}$ is an intensive property and $\Delta_r G$ is an extensive property.
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