The reaction of sulphur in alkaline medium is given below: $\mathrm{S}_{8(\mathrm{~s})}+a\mathrm{OH}^{-}_{(\mathrm{aq})} \longrightarrow b\mathrm{S}^{2-}_{(\mathrm{aq})}+c\mathrm{S}_{2}\mathrm{O}_{3(\mathrm{aq})}^{2-}+d\mathrm{H}_{2}\mathrm{O}_{(\ell)}$ The value of 'a' is ____. (Integer answer)

JEE Main · 2021 · 24 Feb 2021 · Shift 1 · Redox Reactions and Electrochemistry

The reaction of sulphur in alkaline medium is given below:

S_{8 (s)} + a OH_{(aq)}^{-} ⟶ b S_{(aq)}^{2 -} + c S_{2} O_{3 (aq)}^{2 -} + d H_{2} O_{(ℓ)}

The value of 'a' is ____. (Integer answer)

Answer: 12

This is a disproportionation reaction where sulphur is both oxidized and reduced.

Balancing the equation:

In

S_{8}

, sulphur is in 0 oxidation state.

In $S^{2 -}$ , sulphur is in -2 oxidation state (reduction)
In $S_{2} O_{3}^{2 -}$ (thiosulfate), the average oxidation state of S is +2 (oxidation)

Step 1: Balance sulphur atoms

S_{8}

has 8 sulphur atoms.
Let's assume:

b = 4

(4 S²⁻ ions) and

c = 2

(2 S₂O₃²⁻ ions, containing 4 S atoms)
Total S on right = 4 + 4 = 8 ✓

Step 2: Balance oxygen atoms
Right side has

2 \times 3 = 6

oxygen atoms from

S_{2} O_{3}^{2 -}

These come from

OH^{-}

and form

H_{2} O

Step 3: Balance charge and hydrogen
Left side charge:

- a

(from OH⁻)
Right side charge:

- 8

(from 4S²⁻) +

- 4

(from 2S₂O₃²⁻) =

- 12

Therefore:

a = 12

Step 4: Balance hydrogen
12 OH⁻ provides 12 H atoms, forming 6 H₂O:

d = 6

Balanced equation:

S_{8} + 12 OH^{-} ⟶ 4 S^{2 -} + 2 S_{2} O_{3}^{2 -} + 6 H_{2} O

Answer: 12

JEE Main · 2021 · 24 Feb 2021 · Shift 1 · Integer answer