Balance the reaction step-wise: As2S3(s) + NO3^- + H^+(aq) to AsO4(aq) + S(s) + NO(g) + H2O(l)

Answer 1

Here is the NO3^- half.

1. Write it down.
NO3^- ==> NO

2. Oxidation state of N on left is +5; on the right is +2. Change in oxidation state is gain of 3e. Do that as step 3.[5+ +(-3) = 2+]

3. NO3^- +3e ==> NO

4. Count up the charge on both sides. I see -4 on the left and zero on the right.
a. if in acid solution add H^+ to balance the change.
b. if in basic solution, add OH^- to balance the charge.
c. so it looks this way.
NO3^- + 3e + 4H^+ ==> NO

5. Add H2O to the appropriate side to balanced the H^+.
NO3^- + 3e + 4H^+ ==> NO + 2H2O

You should check it but this half equation is balanced.

Answer 2

Have you made a typo? Is that supposed to be AsO4^3-