The reaction of sulphur in alkaline medium is given below:
S₈₍ₛ₎ + a OH⁻_((aq)) → b S²⁻_((aq)) + c S₂O₃²⁻_((aq)) + d H₂O₍ₗ₎
The values of 'a' is _______. (Integer answer)

Solution

Chemistry ma’am · 4 steps

Sulphur at 0 goes both ways here, down to -2 and up to +2, so balance the electrons first.

1. Spot the disproportionation. Sulphur starts at oxidation state 0. In S²⁻ it is -2, so each S gains 2 electrons. In S₂O₃²⁻ each S is +2, so one thiosulphate ion loses 2 × 2 = 4 electrons.
2. Match the electrons. One S₂O₃²⁻ gives 4 electrons, which is enough for two S²⁻. So b : c = 2 : 1, and one such set uses 2 + 2 = 4 sulphur atoms. S₈ has 8, so take two sets: b = 4, c = 2.
3. Balance charge. The right side carries 4(-2) + 2(-2) = -12. Only OH⁻ is charged on the left, so a = 12. Then O gives 12 = 2 × 3 + d, so d = 6, and H checks out as 12 = 6 × 2.

   S₈ + 12OH⁻ → 4S²⁻ + 2S₂O₃²⁻ + 6H₂O

4. So. The coefficient of OH⁻ is a = 12.

The common slip: 6. Balancing with only 4 sulphur atoms, as if the reactant were S₄, gives 6. S₈ needs two full sets of products, which doubles every coefficient.

The idea to keep. When one element is both oxidised and reduced, balance electron gain against electron loss first. In basic medium, finish with charge using OH⁻, then O and H with water.

Answer from the NTA final answer key. Solution by PrepWiser.