I'm trying to figure this question out as I was studying for the MCAT:
As2O3 (s) + NO3- (aq) ----> H3AsO4 (aq) + NO (g)
How would I balance this redox reaction by using the half-reaction method? Would I be trying to form a net-ionic equation?
In going from As
2O
3 to H
3AsO
4, As has gone from +3 to +5, so that's a loss of 2 e- per As. Thus, the oxidation half-reaction is:
As
2O
3 > 2 H
3AsO
4 + 4 e-
In going from NO
3- to NO, N has gone from +5 to +2, so that's a gain of 3 e- per N. Thus, the reduction half-reaction is:
NO
3- + 3 e- > NO
From there, you need to cross multiply to balance the two half reactions (at 12 electrons each, because 12 is the smallest common mulitple of 3 and 4).
3 As
2O
3 > 6 H
3AsO
4 + 12 e-
and
4 NO
3- + 12 e- > 4 NO
Overall we have:
3 As
2O
3(s) + 4 NO
3-
(aq) > 6 H
3AsO
4(aq) + 4 NO
(g)
We need to balance the charges on each side, because right now there is a net ionic charge on the left side of 4- while the right side is uncharged. This is balanced by adding 4 H+ to the left side (which means that it really was 4 nitric acid molecules, HNO3).
4 H+
(aq) + 3 As
2O
3(s) + 4 NO
3-
(aq) > 6 H
3AsO
4(aq) + 4 NO
(g)
As far as atoms go, there are 6 As and 4 N on each side, but there are only 4 H and 21 O on the left side while there are 18 H and 28 O on the right side. The finally step in balancing would require adding 7 H
2O to the left side.
The final balanced reaction is:
7 H
2O
(l) + 4 H+
(aq) + 3 As
2O
3(s) + 4 NO
3-
(aq) > 6 H
3AsO
4(aq) + 4 NO
(g)
