oxidation - reduction doubt

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MnO₂ + Pb₃O₄ + HNO₃ ---> HMnO₄ + Pb(NO₃)₂ + H₂O

Balance this equation

I an having a doubt due to fractional charge on Pb on LHS

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2 Answers

govind ·2010-03-18 08:59:09

Do it in this way...Pb₃O₄ is a mixed oxide of 2PbO and PbO₂....so Pb is changing it's oxidation state from +4 to +2 in PbO₂ and in PbO it's oxidation state remains same....

MnO₂ + PbO₂ +PbO + HNO₃ â†’ Pb(NO₃)₂ + HMnO₄ + H₂O

so Pb is changing it's oxidation state from +4 to +2 and Mn from +4 to +7..so write the equation as .
so mutiply MnO₂ by 2 and PbO₂ by 3..
Now to form a mixed oxide the number or moles of PbO₂ = 2 * the number of moles of PbO..so multiply PbO by 6...

So u will get the balanced equation as
2MnO₂ + 3Pb₃O₄ + 18HNO₃ â†’ 9Pb(NO₃)₂ + 2HMnO₄ + 8H₂O