Answer the following in brief.

Question

Why do the cell potential of lead accumulators decrease when it generates electricity ? How the cell potential can be increased ?

Accepted Answer

Working of a lead accumulator :
(1) Discharging : When the electric current is withdrawn from lead accumulator, the following reactions take place :Oxidation at the - ve electrode or anode :
$
\begin{array}{lr}
Pb ( s ) \rightarrow Pb _{( aq )}^{2+}+2 e ^{-} & \text {(oxidation) } \\
Pb _{( aq )}^{2+}+ SO _{4( aq )}^{2-} \rightarrow PbSO _{4( s )} & \text { (precipitation) } \\
\hline Pb ( s )+ SO _{4( aq )}^{2-} \rightarrow PbSO _{4( s )}+2 e ^{-} &
\$overall \ oxidation \ at \ anode)
\end{array}
$
Reduction at the + ve electrode or cathode:
$PbO _2^{( s )}+4 H _{( aq )}^{+}+2 e ^{-} \rightarrow Pb _{( aq )}^{2+}+2 H _2 O _{( I )}
\text { (reduction) }$
$\frac{ Pb _{( aq )}^{2+}+ SO _{4( aq )}^{2-} \rightarrow PbSO _4^{( s )} \quad \text { (precipitation) }}{ PbO _{2^{( s )}}+4 H _{( aq )}^{+}+ SO _{4( aq )}^{2-}+2 e ^{-} \rightarrow PbSO _4^{( s )}+2 H _2 O _{( l )}} \$overall \ reduction \ at \ cathode)$
(2) Net cell reaction :
(i) Thus, the overall cell reaction during discharging is
$\begin{aligned} Pb _{( s )}+ PbO _{2( s )}+4 H _{( aq )}^{+} & +2 SO _{4( aq )}^{2-} \longrightarrow 2 PbSO _{4( s )}+2 H _2 O _{( i )}\end{aligned}$
OR
$Pb_{(s)}+PbO_{2(s)}+2 H_2 SO_{4(aq)} \rightarrow 2 PbSO_{4(s)}+2 H_2 O_{(l)}$
The cell potential or emf of the cell depends upon the concentration of sulphuric acid. During the operation, the acid is consumed and its concentration decreases and specific gravity decreases from $1.28$ to $1.17$. As a result, the emf of the cell decreases. The emf of a fully charged cell is about $2.0 V$ .
(ii) Recharging of the cell : When the discharged battery is connected to external electric source and a higher external potential is applied the cell reaction gets reversed generating $H _2 SO _4$.
Reduction at the -ve electrode or cathode:
$PbSO _{( s )}+2 e ^{-} \rightarrow Pb _{( s )}+ SO _{4( aq )}^{2-}$
Oxidation at the + ve electrode or anode:
$PbSO _4^{( s )}+2 H _{2( l )} O \rightarrow PbO _2^{( s )}+4 H _{( aq )}^{+}+ SO _{4( aq )}^{2-}+2 e ^{-}$
The net reaction during charging is
$2 PbSO _{4( s )}+2 H _2 O _{( i )} \rightarrow Pb _{( s )}+ PbO _{2( s )}+4 H _{\text {(aq) }}^{+}+2 SO _{4( aq )}^{2-}$
$O R$
$2 PbSO _4^{(s)}+2 H _2 O \rightarrow Pb (s)+ PbO _{2^{(s)}}+2 H _2 SO _{4( m )}$
The emf of the accumulator depends only on the concentration of $H_2SO_4$.

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