While charging the lead storage battery:A. `PbSO_(4)` anode is reduced to PbB. `PbSO_(4)` cathode is reduced to PbC. `PbSO_(4)` cathode is oxidised to PbD. `PbSO_(4)` anode is oxidised to `PbO_(2)`

While charging the lead storage battery:A. `PbSO_(4)` anode is reduced

1.	While charging the lead storage battery:A. `PbSO_(4)` anode is reduced to PbB. `PbSO_(4)` cathode is reduced to PbC. `PbSO_(4)` cathode is oxidised to PbD. `PbSO_(4)` anode is oxidised to `PbO_(2)`
Answer» While charging the lead storage battery the reaction occuring on cell is reversed and `PbSO_(4)(s)` on anode and cathode is converted into Pb and `PbO_(2)` respectively Hence, opiton (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)OrarrPbO _(2)(s)+SO_(4)^(2-)+4H^(+)+2e^(-)` (Oxidation) Overall reaction `2PbSO_(4)(s)+2H_(2)OrarrPb(s)+PbO_(2)(s)+4H^(+)(aq)+2SO_(4)^(2-)(aq.)`

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What are elementary reactions ?

While charging the lead storage battery:A. `PbSO_(4)` anode is reduced to PbB. `PbSO_(4)` cathode is reduced to PbC. `PbSO_(4)` cathode is oxidised to PbD. `PbSO_(4)` anode is oxidised to `PbO_(2)`

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