Question

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

• A. -4
• B. 6
• C. 2
• D. 4

Answer 1

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 $|O.S{.}_{Product}-O.S{.}_{Reactant}|\times \text{number of atom}$ for the valency factor
$S{O}_{3\left(aq\right)}^{2-}+H_{2}O\to S{O}_{4\left(aq\right)}^{2-}+2H_{\left(aq\right)}^{+}+2e^{-}$
O.S. of S in reactant side
$=x+3\times (-2)=-2$
$x=+4$
O.S. of S in product side
$=x+4\times (-2)=-2$
$x=+6$
$\text{valency factor}=\text{number of S atom}|O.S{.}_{Product}-O.S{.}_{Reactant}|$
$\text{valency factor}=|+6-4|$
$\text{valency factor}=2$