For Question 42, answer is B.
Passage itself states Pb (s) is at anode and gets oxidized to PbSO4
Electrons from that oxidation reduce the Pb4+ in PbO2 to Pb2+.
Even if B were true, how come C is false?
I'm not sure why B is correct, but C is false because the question is asking about recharging. PbO2 goes to Pb2+ and SO42- in the half reaction which is favored, and that would causes a current and discharge the battery. You have to regenerate the PbO2 that is lost, but I don't know why that would occur at the anode.
EDIT: Apparently the anode is defined as the place of oxidation (loss of electrons) and cathode is defined as the place of reduction (gain of electrons), so they switch when you are recharging the battery.
PbSO4 is a solid and sticks to the plates of both the (-) and (+) so on the PbO2 side (cathode during discharging) you reverse the current and pull 2e- out which flips the reaction and the PbSO4 combines with the H2O to form PbO2 (also 3H+ and HSO4- {passage numbers are 4 H+ and SO42-}) and 2 e-. Since electrons are insoluble in water (it would short out), the electrons go towards the negative terminal (anode during discharge) and the PbSO4 stuck there combines with the 2e- to make Pb(s) and SO42-(aq) (Pb(s) sticks to the bar and SO42- goes into the solution).
Since you're losing electrons from the PbO2 end (+) it is oxidation and that side is called the anode.
