- Joined
- Jul 29, 2010
- Messages
- 839
- Reaction score
- 9
PbO2 → PbSO4 : lead goes from +4 → +2 therefore PbO2 is reduced and is the cathode (choice A is correct)
Pb → PbSO4 : lead goes from zero → +2 therefore Pb(s) is oxidized and is the anode
For the first reaction I understand how lead is +4, but I do not understand how it is +2 in product. I know oxygen will be -8, but for some reason I am stumped on the Pb and S in PbSO4. What am I missing?
Pb → PbSO4 : lead goes from zero → +2 therefore Pb(s) is oxidized and is the anode
For the first reaction I understand how lead is +4, but I do not understand how it is +2 in product. I know oxygen will be -8, but for some reason I am stumped on the Pb and S in PbSO4. What am I missing?