Question

Compose the balanced reaction equation for the following reaction occuring in acidic aqueous medium.
Reduce all coefficients to the lowest possible integers.
Include H₂O(l) and H⁺(aq) appropriately to complete the equation.

As2S3(s) + NO3−(aq) →H3AsO4(aq) + NO(g) + S8(s)

hint: Identify all atoms which undergo a change in formal oxidation state.
Determine the relative coefficients for all species which are involved in oxidation/reduction.
Determine appropriate coefficients for all components other than H2O, OH⁻ or H⁺.
Next add either OH⁻ (basic solutions) or H⁺ (acidic solutions) to one side only of the reaction to balance ionic charges.
Balance the oxygen atoms by adding H₂O to one side only of the equation.
Check to see that hydrogen atoms balance. If not, you have made a mistake.

Answer 1

As2S3(s) + NO3−(aq) → H3AsO4(aq) + NO(g) + S8(s)
Identify the 2 half reactions
Reduction; N(V) in NO3^- to N(II) in NO (by e gain)
Oxidation; As(III) to As As(V) (by e loss)
AND S(II-) to S(0) (by e loss) [Unusual to find two species oxidised in one reaction]
Work out half equations
Reduction;
e involved N^5+ + 3e ---> N^2+
NO3^- + 3e ---> NO
Balance O's with H2O on appropriate side
NO3^- + 3e ---> NO + 2H2O
Balance H's with H^+ on appropriate side
4H^+ + NO3^- + 3e ---> NO + 2H2O Equation A
Oxidation;
e involved As^3+ ---> As^5+ + 2e
As^3+ ---> AsO4^3- + 2e
AND S^2 ---> S + 2e
Balance O's with H2O on appropriate side
4H2O + As^3+ ---> AsO4^3- + 2e
AND S^2- ---> S + 2e
Balance H's with H^+ on appropriate side AND increase S to form an S8 molecule
4H2O + As^3+ ---> AsO4^3- + 8H^+ + 2e Equation B
AND 8S^2- ---> S8 + 16e Equation C
To combine these two oxidations for As2S3 molecules it is necessary to
(16 x B) + (3 x C)
64H2O + 8As2S3 ---> 16AsO4^3- + 128H^+ + 3S8 + 80e (32+28=80) Equation D

AT THIS STAGE I QUESTION THE EXTREMELY AND UNUSUALLY HIGH "BALANCING
NUMBERS" WHICH ARE ABOUT TO BECOME EVEN HIGHER BUT REALIZE THAT THEY ARE BEING FORCED UPON ME BY THE REQUIREMENT FOR AN OCTA-ATOMIC (S8) MOLECULE ANDTHE LACK OF ANY LOW COMMON MULTIPLE

The two half equations A and D can only be combined with elimination of electrons by
(80 x A) + (3 x D) (giving 240e on both sides) TRY IT. REMEMBERING TO COLLECT LIKE TERMS AT THE SIDE WHERE THERE ARE MOST OF THEM, YOU SHOULD ARRIVEAT

32H2O + 80NO3^- + 24As2S3 ------> 48AsO4^3- + 64H^+ + 9S8 + 80NO
On each side; 64 H; 272 O; 80 N; 48 As; 72 S; charge 80-