During the discharging of a lead storage cell | vedclass

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(A) The reaction at the anode is: $Pb(s)   SO_4^{2-}(aq) \rightarrow PbSO_4(s)   2e^-$. The reaction at the cathode is: $PbO_2(s)   SO_4^{2-}(aq)   4H

Vedclass · Accepted Answer

$6$

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(A) The reaction at the anode is: $Pb(s)   SO_4^{2-}(aq) ightarrow PbSO_4(s)   2e^-$. The reaction at the cathode is: $PbO_2(s)   SO_4^{2-}(aq)   4H^ (aq)   2e^- ightarrow PbSO_4(s)   2H_2O(l)$. The net cell reaction is: $Pb(s)   PbO_2(s)   2H_2SO_4(aq) ightarrow 2PbSO_4(s)   2H_2O(l)$. From the stoichiometry,$2 \ mol$ of $H_2SO_4$ are consumed to produce $2 \ mol$ of electrons ($2 \ F$ of charge). Total mass of solution $= 	ext{Volume} 	imes 	ext{Density} = 4900 \ mL 	imes 1.2 \ g/mL = 5880 \ g$. Initial mass of $H_2SO_4 = 40\% 	ext{ of } 5880 \ g = 0.40 	imes 5880 = 2352 \ g$. Final mass of $H_2SO_4 = 30\% 	ext{ of } 5880 \ g = 0.30 	imes 5880 = 1764 \ g$. Mass of $H_2SO_4$ consumed $= 2352 \ g - 1764 \ g = 588 \ g$. Moles of $H_2SO_4$ consumed $= \frac{588 \ g}{98 \ g/mol} = 6 \ mol$. Since $2 \ mol$ of $H_2SO_4$ produce $2 \ F$ of charge,$6 \ mol$ of $H_2SO_4$ will produce $6 \ F$ of charge. Therefore,the total charge produced is $6 \ faraday$.

Column-$I$	Column-$II$
$A$. Leclanche cell (Dry cell)	$P$. Converts energy of combustion into electrical energy
$B$. Lead storage battery	$Q$. Reaction at cathode: $O_{2(g)} + 4H^+_{(aq)} + 4e^- \rightarrow 2H_2O_{(\ell)}$
$C$. Fuel cell	$R$. Reaction at cathode: $MnO_2 + NH_4^+ + e^- \rightarrow MnO(OH) + NH_3$
$D$. Rusting	$S$. Reaction at anode: $Pb_{\text{(s)}} + SO_4^{2-}{_{\text{(aq)}}} \rightarrow PbSO_{4\text{(s)}} + 2e^{-}$

During the discharging of a lead storage cell,the concentration of $H_2SO_4$ reduces from $40\% \ w/w$ to $30\% \ w/w$. Find the total charge produced in $faraday$. Given the volume of the solution $= 4.9 \ L$ and density $= 1.2 \ g/mL$. (Assume volume and density remain constant)

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