Electrolysis confusing....

[sandi]

For each of the follwing electrolytic cells:
 i)write the appropriate half-equation that would occur at anode
 ii)write the appropriate half-equation that would occur at cathode

a)An aqueous solution of lead nitrate using lead electrodes
b)An aqueous solution containing a mixture of silver nitrate and zinc nitrate using unreactive electrodes.

for a) i thought it would be 2H₂O--->4e+4H⁺ +O₂ at the anode and Pb²⁺ +2e---->Pb at the cathode but this was wrong, instead it reacted with itself.

for b) i thought the anode reaction would be Ag--->Ag⁺ + e^- and the cathode reaction would be Zn²⁺ +2e---->Zn

Can someone explain this?

[stephanieteddy]

In electrolysis, always look for the highest species on the left hand side (the strongest oxidant) and the lowest species on the right hand side (strongest reductant)

Soooo, for a, all of the species we have are: Pb2+, Pb, H20 as a reductant and H20 as an oxidant. If you circle them on the electrochemical series, you will see that the Pb2+ as an oxidant is higher than water as the oxidant so therefore Pb2+ + 2e- > Pb will occur at the cathode.

Then if we look at the right hand side of the series, we try to find the lowest species, and that is Pb! Seeing as the lowest water species is at 1.23 V. So Pb is the strongest reductant and, Pb > Pb2+ + 2e- will occur at the anode!

Is this what you mean by reacting with itself?

[daniel034]

Hi for a) The species present are Pb2+, Pb and H2O. Pb2+ is a stronger oxidant than water, Pb is a stronger reductant than water as both these reactions (the same reaction in reverse) are both higher and lower than water on the electrochemical series. Therefore it will react ahead of water in both cases.

b) The species present are Ag+, Zn2- and H2O because you are using inert electrodes there is no Ag(s) like you have suggested below. As H2O is the only reductant, it will react at the anode. As Ag+ is higher than Zn2-, it will be reduced at the cathode.

Hope that helps!