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Redox Equation Balance Help

  • Thread starter vipertongn
  • Start date
  • #1
98
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Homework Statement



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)

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
 

Answers and Replies

  • #2
Borek
Mentor
28,326
2,708
You have NO3- on both sides. Try to leave it only on the left.
 
  • #3
5
0
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?
 

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