Q.

Which of the following statements is/are true for the above figure?

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Cathode reaction Pb(s)+SO42−(aq)⟶PbSO4(s)+2e−

Anode reaction PbO2(s)+SO42−(aq)+4H+(aq)+2e−⟶PbSO4(s) + 2 H2O(l)

On charging the battery PbSO4 give Pb at anode and PbO2. at cathode.

All of the above

answer is C.

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Detailed Solution

Anode Pb(s)+SO42−(aq)⟶PbSO4(s)+2e− Cathode PbO2(s)+SO42−(aq)+4H++2e−⟶PbSO4(s) +2 H2O(l)Overall cell reaction Pb(s)+PbO2(s)+2H2SO4(aq)⟶2PbSO4(s)+2H2O(l)At the time of charging, a reverse reaction takes place. In which PbSO4 (s) at anode and cathode is converted into Pb and PbO2 respectively.

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