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Manganese II nitrate + lead dioxide + nitric acid ----> hydrogen permanganate + lead II nitrate + water


(Could you please tell me the oxidation #'s for the original, too? :) )

2007-11-14 14:14:28 · 1 answers · asked by Anonymous in Science & Mathematics Chemistry

1 answers

In terms of the main actors, Mn+2 is oxidized to Mn+7, and Pb+4, the oxidizing agent, is reduced to Pb+2. Nitric acid is needed to supply hydrogen for water and extra nitrate, because, as will be seen, the nitrate from Mg(NO3)2 is insufficient for the Pb(NO3)2 formed.

You need to do half-reactions:
the first is for the oxidant, PbO2:
PbO2 + 4H+ >> Pb+2 + 2H2O + 2e
(you dont worry about nitrate, since none is changed in the reactions)
for the Mn+2 being oxidized:
Mn+2 + 8H2O >> HMnO4 + 7H+ + 5e
Now, we multiply the two equations by 5 and 2 respectively and add them up. The electrons will drop out, and waters and H+ can be adjusted. You will get
5PbO2 + 6H+ + 2 Mn+2 >>
5Pb+2 + 2 HMnO4 + 2H2O
Now we address the nitrate balance. 6 will come with the H+ as nitric acid and 4 will come with the Mn(NO3)2. The 10 wind up as the nitrate in Pb(NO3)2.

2007-11-14 15:37:19 · answer #1 · answered by cattbarf 7 · 0 0

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