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    Balance the following reactions:

    Pb(s) + PbO2(s) + H2SO4 --> PbSO4(s)
    H3AsO4 + Zn(s) --> As + Zn^+2
    O3 + NO --> O2 + NO2

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    Please see the attached file for solutions.

    Balance-ACID-Pb(s) + PbO2(s) + H2SO4 arrow PbSO4(s)

    Must first divide into 2 half reactions.....
    This one is hard, unless you are familiar with disproportionation reactions.....This above question is like a reverse disproportionation.

    Pb  PbSO4 (1)
    Lead metal (Pb0) going to Pb2+

    PbO2  PbSO4 (2)
    Pb4+ going to Pb2+

    For a simplification, ignore the anion associated with the acid...

    Pb0  Pb2+ (1)
    Everything else other than O and H+ (bear with me) not needed
    Balance O, then balance H+ (you will se in just a bit) not needed
    Balance out charge by adding electrons to the side with the more positive charge...add 2 e-'s to right hand side (RHS)
    Pb0  Pb2+ + 2 e- (1)
    Done......

    PbO2  Pb2+ (2)
    Everything else other than O and H+ (bear with me) not needed
    Balance O ...

    Solution Summary

    This question is like a reverse disproportionation reaction. You must first divide it into 2 half reactions. For a simplification, ignore the anion associated with the acid. The next thing to do is to add the two reactions together in such a way as to eliminate electrons. Now we can in the acid anions to both sides to balance out.

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