Find the n-factor the half equation for the oxidation of sulphite ion :
$S O_{3 (a q)}^{2 -} + H_{2} O \to S O_{4 (a q)}^{2 -} + 2 H_{(a q)}^{+} + 2 e^{-}$

-4

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Solution

The correct option is C 2
Steps for finding valency factor/n-factor:
1. Find the oxidation state (O.S.) of metal in reactant side.
2. Find the oxidation state of metal in product side.
3. Take the magnitude of difference of two oxidation state $| O . S ._{P r o d u c t} - O . S ._{R e a c t a n t} | \times number of atom$ for the valency factor
$S O_{3 (a q)}^{2 -} + H_{2} O \to S O_{4 (a q)}^{2 -} + 2 H_{(a q)}^{+} + 2 e^{-}$
O.S. of S in reactant side
$= x + 3 \times (- 2) = - 2$
$x = + 4$
O.S. of S in product side
$= x + 4 \times (- 2) = - 2$
$x = + 6$
$valency factor = number of S atom | O . S ._{P r o d u c t} - O . S ._{R e a c t a n t} |$
$valency factor = | + 6 - 4 |$
$valency factor = 2$

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