Electrochemistry — Chemistry STD 12 Science — Question

1. PbSO₄ anode is reduced to Pb.

Explanation:

While charging the lead storage battery the reaction occurring on cell is reversed and PbSO₄(s) on anode and cathode is converted into Pb and Pb0₂ 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 $2\text{PbSO}_4(\text{s})+2\text{H}_2\text{O}\xrightarrow{\ \ \ \ \ \ \ \ \ \ \ \ \ \ }\text{Pb}(\text{s})+\text{PbO}_2(\text{s})+4\text{H}^+(\text{aq.})+2\text{SO}^{2-}_4(\text{aq.})$