Chapter 17, Question 60 A \"dead\" 12 V lead storage battery has 7.71 g of PbSO4

Question

Chapter 17, Question 60 A "dead" 12 V lead storage battery has 7.71 g of PbSO4 deposited on each of its anodes. It is connected to a "trickle charger" that supplies 0.120 A of current at a voltage of 13 V. The reaction for the discharge of the lead storage battery follows. Pb(s) + 2HSO4"(aq) + PbO2(s) + 2H3O+(aq) 2PbSO4(s) + 4H20(1) To which electrode, Pb or Pbo2, should the anode from the charger be connectedi? 2 Edit LINK TO TEXT Compute the time in hours needed to convert all 7.71 g of PbSO4 back to HSO4 ons in solution. The Faraday constant is 96485 C/mol the tolerance is +/-396

Explanation / Answer

As w.During discharging the Pb(S) was converted into PbSO4, this means electrons were lost by Pb and thus acting as anode and gained by PbO2(*because oxidation state of Pb changed from +4 to +2) and thus acting as cathode..Now during recharge of battery the reaction will be reverse.So, anode will act as cathode and cathode will act as anode.So, anode from the charger will be connected to Pb that will act as cathode.

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