Question

im having trouble understanding this hw question,

okay here's the question:
A 0.326 8-g unknown containing Pb(CH3CHOHCO2)2 (lead lactate,
FM 385.3) plus inert material was electrolyzed to produce
0.111 1 g of PbO2 (FM 239.2). Was the PbO2 deposited at the anode or
at the cathode?

The answers states that PbO2 is formed at the anode and is an oxidation. But when i write out the eq here:

Pb(lactate)2 --> PbO2
Pb2+ + 2e- --> Pb(s)
This is a reduction reaction...

Answer 1

You are both right and wrong. The second equation is wrong. The first one is right.
Pb^2+ ==> Pb02 + 2e.
The lead ion from lead lactate is +2 and it is going to PbO2 where Pb is +4. That is a loss of electrons which is oxidiation. And the PbO2 will deposit at the anode because oxidation occurs at the anode.