# Redox Help

1. May 21, 2009

### vipertongn

1. The problem statement, all variables and given/known data

Balance the equation (Note: the coefficients provided may be wrong):
3 Co(s) + 8 H+(aq) + 8 NO3-(aq) → 3 Co2+(aq) + 6 NO3-(aq) + 2 NO(g) + 4 H2O(l)

3. The attempt at a solution
I started out by doing a half reaction

Co → Co 2+ +2e- This one I just added electrons to the appropriate side

I get stuck here, I tried just balancing the elements...
4e- +4H+ + 2NO3 - → NO3 - + NO + 2H2O

No matter how i look at it I don't know why i have this answer incorrect

the real answer is 3 Co(s) + 8 H+(aq) + 8 NO3-(aq) → 3 Co2+(aq) + 6 NO3-(aq) + 2 NO(g) + 4 H2O(l)

but i don't know how it came to that

2. May 22, 2009

### Staff: Mentor

You have NO3- on both sides. Try to leave it only on the left.

3. May 22, 2009

### Querious

yeah the "real answer" you have looks rather strange. It has NO3- on both sides.

Something looks wrong with this though

"4e- +4H+ + 2NO3 - → NO3 - + NO + 2H2O "

How did you arrive at 4 electrons?