Balancing redox equations in acidic solutions

General Forums Questions and Answers
1

I’m pretty tired right now so I’m not thinking very clearly so this is probably stupidly simple, but I’ve got a bit of a hangup on a question I’m doing.

I’m suppposed to balance this equation assuming it’s in an acid solution (in the presence of H+ ions)

2CuS(s) + HNO3 --> Cu(NO3)2 + H20 + NO (g) + S (s)

Ordinarily I’d balance the half reactions (excluding H and O), add H20 to compensate for O, H+ to compensate for H, and electrons to compensate for charge. But this equation already seems half done. The H2O has been accounted for, etc.

I tried just straight up balancing the equation but it’s sort of unruly. Trying to break it apart into the half reactions also gave me troubles. Any ideas?
Thanks.

EDIT: Forgot to put in the sulphur on the right side of the equation, it’s there now though.

2

Sulfur doesn’t appear on the left side so, unless you are the creator, this is impossible to balance.

3 Cu (s) + 8 HNO3(aq) 3 Cu(NO3)2(aq) + 4 H2O + 2 NO (g)

( I googled it without the sulfur)

3

BUMP

Thanks for the reply, I forgot to add a sulphur but it’s there now.

Anyone out there that can help explain this question to me? I’ve got a few similar to this.

4

Sorry, but I’m math and science retarded.