Question

Part A:

Balance the following redox reactions occurring in acidic aqueous solution.

MnO4−(aq)+Al(s)→Mn2+(aq)+Al3+(aq)

Part B:

What mass of lead sulfate is formed in a lead-acid storage battery when 1.40 g of Pb undergoes oxidation?

Answer 1

Part A

Reduction half :-

MnO4- -------> Mn2+
balance oxidation no. ...Mn is +7 in left and +2 in right ...so add 5e- in left
MnO4- + 5e- ----------> Mn2+
now balance charge...... total of -6 in left and +2 in right ...so add 8H+ in left as we are in acidic condition
MnO4- + 5e- + 8H+ ----------> Mn2+
now add H2O to balance H and O
MnO4- + 5e- + 8H+ ----------> Mn2+ + 4H2O....(1)
This is balanced reduction half reaction
Oxidation half :-
Al -------> Al+3
balance oxidation no.
Al --------> Al+3 + 3e- ....(2)
this is balanced oxidation half reaction
multiply (1) by 3 and (2) by 5 and add them
3MnO4- + 15e- + 24H+ ----------> 3Mn2+ + 12H2O
5Al --------> 5Al+3 + 15e-
15e- will cancel out ...
3MnO4- + 24H+ + 5Al -----------> 3Mn2+ + 12H2O + 5Al3+
This is the balanced redox reaction.

Part B

There is no lead (??) sulfite formed in a lead acid battery. The anion is sulfate, SO4^2-.During the discharge of a lead acid battery, lead (IV) oxide gets reduced to PbSO4, and Pb metal gets oxidized to PbSO4.
In charging a lead acid battery, lead (II) disproportionates to form metallic lead and lead (IV) oxide, PbO2.
As to your question you MUST consider the entire redox reaction, not just the oxidation of lead.
Pb(s) + PbO2(s) + 2H2SO4(aq) --> 2PbSO4(s) + 2H2O(l)
1.40 g Pb x (1 mole Pb / 207.2g Pb) x (2 mol PbSO4 / 1 mol Pb) x (303.3 g PbSO4 / 1 mol PbSO4) is answer.