Here's the full molecular equation:

Zn + H₃AsO₄ + HNO₃ ---> AsH₃ + Zn(NO₃)₂ + H₂O

The change in zinc's oxidation state should be rather obvious. It makes the first half-reaction:

Zn ---> Zn²⁺ + 2e¯

Notice that I already balanced it. In addition, note that the nitrate has been removed. The nitrogen in the nitrate is neither oxidized nor reduced.

The arsenic should be a fairly obvious candidate for something and, in fact, it gets reduced. In the arsenate, the arsenic is a +5 and in arsine it is a +3. Here's the unbalanced half-reaction:

AsO₄³¯ ---> As³⁺

Notice that I used the full polyatomic ion for arsenate. This is common practice in going from the molecular equation to the net ionic. Use the full polyatomic ion.

There are three half-reactions and the third (already balanced) is this one:

2e¯ + H⁺ ---> H¯

How do I know this is the third half-reaction? Things that I know: (1) AsH₃ is a hydride, therefore the hydrogen is a -1 oxidation state, (2) it is a product and (3) hydrogen is in a +1 oxidation state as a (4) reactant.

How did I know to split up the AsH₃ into two pieces? I know: (1) you can only have one reduction or one oxidation in a half-reaction and (2) As gets reduced AND H also gets reduced. The solution? Split AsH₃ into two pieces, each with its own half-reaction.

Here is a warning: I will have to recombine the two pieces of AsH₃ in the final answer and there MUST be three H for every one As. Look for how I do that.

Here are the three balanced half-reactions:

Zn ---> Zn²⁺ + 2e¯
2e¯ + 8H⁺ + AsO₄³¯ ---> As³⁺ + 4H₂O
2e¯ + H⁺ ---> H¯

Here's how to equalize the electrons:

4 [Zn ---> Zn²⁺ + 2e¯]
2e¯ + 8H⁺ + AsO₄³¯ ---> As³⁺ + 4H₂O
3 [2e¯ + H⁺ ---> H¯]

The key? There MUST be a three in front of the third half-reaction. This gives me three H¯ to pair up with the one As³⁺. Since that means 8e¯ on the left, I have to put a 4 in front of the first half-reaction, thereby equalizing the electrons. Also, please note that I have to keep the second half-reaction as is because I need to have one As³⁺ for my three H¯.

Add up the three half-reactions to give:

4Zn + 8H⁺ + AsO₄³¯ + 3H⁺ ---> 4Zn²⁺ + AsH₃ + 4H₂O

Notice that I combined the As³¯ and the 3H¯. Also notice that I did not combine the H⁺ on the left-hand side. I shall now combine the 3H⁺ with the arsenate ion:

4Zn + 8H⁺ + H₃AsO₄ ---> 4Zn²⁺ + AsH₃ + 4H₂O

The only remaining task is to re-introduce the nitrate ion; we will need 8 of them:

4Zn + 8HNO₃ + H₃AsO₄ ---> 4Zn(NO₃)₂ + AsH₃ + 4H₂O

Q.E.D.