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While charging the lead storage battery:A. `PbSO_(4)` anode is reduced to PbB. `PbSO_(4)` cathode is reduced to PbC. `PbSO_(4)` cathods is oxidised to PbD. `PbSO_(4)` anode is oxidised to `PbO_(2)` |
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Answer» Correct Answer - A While charging the lead storage battery the reaction occurring on cell is reversed and `PbSO_(4)` (s) on anode and cathode is converted into Pb and `PbO_(2)` respectively Hence, option (a) is the correct choice The electrode reactions are as follows At cathode `PbSO_(4)(s)+2e^(-)rarrPb(s)+SO_(4)^(2-)(aq)` (Reduction) At anode `PbSO_(4)(s)+2H_(2)Orarr PbO_(2)(s)+SO_(4)^(2-)+4H^(+)+2e^(-)` (Oxidation) Overall reaction `2PbSO_(4)(s)+2H_(2)O rarr Pb(s)+PbO_(2)(s)+4H^(+)(aq.)+2SO_(4)^(2-)(aq.)` |
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