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.
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...
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