Posted by David Scott on Thursday, January 26, 2012 at 11:51pm.
Cu changes from zero on the left to +2 on the right.
N changes from +5 on the left to + 2 on the right (for NO--don't concern yourself yet with the nitrate on the right since it doesn't change.
So Cu to Cu^2+ is a loss of 2e and N is a gain of 3e. Multiply the Cu half by 3 and the HNO3/NO half by 2.
That gives us
3Cu + 2HNO3 ==> 3Cu^2+ + 2NO + 4H2O
That balances everything but H and the charge. Since we have a Cu^2+ here it will use nitrate to balance it and form Cu(NO3)2 so we add six more HNO3 on the left to form 3Cu(NO3)2 which makes 6 more H^+ added on the left at the same time.
3Cu + 8HNO3 ==> 3Cu(NO3)2 + 2NO + 4H2O
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