How much electricity in terms of Faraday is required to carry out the reduction of one mole of PbO2 in lead storage battery
Answers
Answer:
The cell reaction that takes place in a lead storage battery are:
Pb + PbO2 + 4H+ + SO4-2 ---------> 2PbSO4 + 2H2O
In this reaction Pb shows an oxidation state of +4. It gets reduced to PbSO4 which shows an oxidation state of +2.
In this reaction, 2 electrons are necessary for the reduction of PbO2.
Pb4+ + 2e- -----------> Pb2+
Therefore, as 2 electrons were needed for the reduction, the Faraday’s law of electrolysis stated that:
The reduction of 1 mole of PbO2 required 2 F electricity.
Answer:
Pb contain +4 oxidation state of PbO2 and this electron by +2 in PbSO4, 2 Electron is need enough held this reaction
Pb4+ + 2e - - - - - - - > Pb2+
By Faraday’s law of electrolysis, 2 Electron is required to proceed this reaction.
1 mole of Pb need 2F eclecticity