S8-bOH- cS2-sS2O2-3-H2O- Find the value of c-

Let us look at the half reactions i.e. oxidation and reduction separately. Oxidation: S_8 → nbsp; S_2O^2 _3 S_8 + 24OH^ → nbsp; 4S_2O^2 _3 + 12H_2O +16e^ ...

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Byju's Answer

Standard VII

Chemistry

Balancing Redox Reaction

S8+bOH-→ cS2-...

Question

$S_{8} + b O H^{-} \to c S^{2 -} + s S_{2} O_{3}^{2 -} + H_{2} O$ . Find the value of c.

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Solution

Let us look at the half reactions i.e. oxidation and reduction separately.
Oxidation:
$S_{8} \to S_{2} O_{3}^{2 -}$
$S_{8} + 24 O H^{-} \to 4 S_{2} O_{3}^{2 -} + 12 H_{2} O + 16 e^{-} \dots . (1)$
Reduction:
$S_{8} \to S^{2 -}$
$S_{8} + 16 e^{-} \to 8 S^{2 -}$ .......(2)
Adding both the reactions (1) and (2),
$2 S_{8} + 24 O H^{-} \to 4 S_{2} O_{3}^{2 -} + 8 S^{2 -} + 12 H_{2} O$
Dividing the whole equation by 2,
$S_{8} + 12 O H^{-} \to 2 S_{2} O_{3}^{2 -} + 4 S^{2 -} + 6 H_{2} O$
So, c = 4

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S8+bOH−→cS2−+sS2O2−3+H2O. Find the value of c.

$S_{8} + b O H^{-} \to c S^{2 -} + s S_{2} O_{3}^{2 -} + H_{2} O$ . Find the value of c.