VO{2+} (aq) + Zn (s) --->  V{3+} (aq) + Zn{2+} (aq) [in acidic solution]

so V is reduced (+6 --> +3) and Zn is oxidized (0 --> +2). After I seperated the half reactions and balanced the number of electrons out, I ended up with:

2VO{2+} (aq) + 3Zn (s) ---> 2V{3+} (aq) + 3Zn{2+} (aq)

I can't figure out how to balance the charges/the # of H and O atoms after this step.

okay that's good
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