Balance the reaction step-wise: As2S3(s) + NO3^- + H^+(aq) to AsO4(aq) + S(s) + NO(g) + H2O(l)
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.
Have you made a typo? Is that supposed to be AsO4^3-
To balance the given chemical reaction step-wise, we need to follow a systematic approach by first balancing the atoms in each ionic species or molecule.
Step 1: Break the reaction into two half-reactions, one for oxidation and one for reduction.
Oxidation half-reaction: As2S3(s) → AsO4(aq)
Reduction half-reaction: NO3^-(aq) → S(s) + NO(g) + H2O(l)
Step 2: Balance the atoms other than hydrogen and oxygen in each half-reaction.
Oxidation half-reaction: 2As2S3(s) → As2O8(aq) + 6S(s)
Reduction half-reaction: 2NO3^-(aq) → 2NO(g) + 3S(s) + 8O2-
Step 3: Balance the oxygen atoms by adding water (H2O) to the appropriate side of each half-reaction.
Oxidation half-reaction: 2As2S3(s) + 21H2O(l) → 4As2O8(aq) + 12S(s) + 42H+(aq)
Reduction half-reaction: 2NO3^-(aq) → 2NO(g) + 3S(s) + 8O2-(aq)
Step 4: Balance the hydrogen atoms by adding protons (H+) to the appropriate side of each half-reaction.
Oxidation half-reaction: 2As2S3(s) + 21H2O(l) → 4As2O8(aq) + 12S(s) + 42H+(aq)
Reduction half-reaction: 2NO3^-(aq) + 10H+(aq) → 2NO(g) + 3S(s) + 8O2-(aq) + 5H2O(l)
Step 5: Balance the charges in each half-reaction by adding electrons (e^-) to the appropriate side.
Oxidation half-reaction: 2As2S3(s) + 21H2O(l) → 4As2O8(aq) + 12S(s) + 42H+(aq) + 60e^-
Reduction half-reaction: 2NO3^-(aq) + 10H+(aq) + 60e^- → 2NO(g) + 3S(s) + 8O2-(aq) + 5H2O(l)
Step 6: Multiply each half-reaction by the appropriate number so that the number of electrons gained and lost is equal.
Oxidation half-reaction: 2As2S3(s) + 21H2O(l) + 120e^- → 8As2O8(aq) + 24S(s) + 84H+(aq) + 120e^-
Reduction half-reaction: 120e^- + 20NO3^-(aq) + 100H+(aq) → 40NO(g) + 60S(s) + 160O2-(aq) + 100H2O(l)
Step 7: Combine the balanced half-reactions to give the overall balanced reaction.
2As2S3(s) + 21H2O(l) + 120e^- + 20NO3^-(aq) + 100H+(aq) → 8As2O8(aq) + 24S(s) + 84H+(aq) + 40NO(g) + 60S(s) + 160O2-(aq) + 100H2O(l)
Step 8: Simplify the equation by canceling out any common terms.
2As2S3(s) + 21H2O(l) + 20NO3^-(aq) + 100H+(aq) → 8As2O8(aq) + 84H+(aq) + 24S(s) + 40NO(g) + 160O2-(aq) + 60S(s) + 100H2O(l)
Finally, simplify the equation by cancelling out the excess hydrogen ions (H+) on both sides:
2As2S3(s) + 21H2O(l) + 20NO3^-(aq) → 8As2O8(aq) + 24S(s) + 40NO(g) + 160O2-(aq) + 60S(s) + 100H2O(l)
This is the balanced equation for the reaction step-wise.