ManimalJax

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I'm trying to figure this question out as I was studying for the MCAT:

As2O3 (s) + NO3- (aq) ----> H3As)4 (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?
 

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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 As2O3 to H3AsO4, As has gone from +3 to +5, so that's a loss of 2 e- per As. Thus, the oxidation half-reaction is:

As2O3 —> 2 H3AsO4 + 4 e-

In going from NO3- 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:

NO3- + 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 As2O3 —> 6 H3AsO4 + 12 e-

and

4 NO3- + 12 e- —> 4 NO

Overall we have:

3 As2O3(s) + 4 NO3-(aq) —> 6 H3AsO4(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 As2O3(s) + 4 NO3-(aq) —> 6 H3AsO4(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 H2O to the left side.

The final balanced reaction is:

7 H2O(l) + 4 H+(aq) + 3 As2O3(s) + 4 NO3-(aq) —> 6 H3AsO4(aq) + 4 NO(g)
 
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