Question
While charging the lead storage battery $........$

Answer

While charging the lead storage battery the reaction occurring on cell is reversed and $\ce{PbSO_4(s)}$ on anode and cathode is converted into $Pb$ and $Pb0_2$ respectively.
Hence, option $(i)$ is the correct choice The electrode reactions are as follows$:$
At cathode $\text{PbSO}_4(\text{s})+2\text{e}^{-}\xrightarrow{\ \ \ \ \ \ \ \ \ \ \ \ \ }\text{Pb}(\text{s})+\text{SO}^{2-}_4(\text{aq})(\text{Reduction})$
At anode $\text{PbSO}_4(\text{s})+2\text{H}_2\text{O}\xrightarrow{\ \ \ \ \ \ \ \ \ \ \ \ \ \ }\text{PbO}_2(\text{s})+\text{SO}^{2-}_4+4\text{H}^{+}+2\text{e}^{-}(\text{Oxidation})$
Overall reaction $2\text{PbSO}_4(\text{s})+2\text{H}_2\text{O}\xrightarrow{\ \ \ \ \ \ \ \ \ \ \ \ \ \ }\text{Pb}(\text{s})+\text{PbO}_2(\text{s})+4\text{H}^+(\text{aq.})+2\text{SO}^{2-}_4(\text{aq.})$

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