For the reaction, $F e S_{2} ⟶ {F e}^{3 +} + S O_{2}$ , the $n$ -factor is :

1

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11

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28

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61

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Solution

The correct option is C $28$
Folowing are the steps to balance the equation.
$F e +2 {2 S}_{2} -1 ⟶ {F e}^{3 +} +3 + S O_{2} +4$
$5 F e S_{2} ⟶ 5 {F e}^{3 +} + 2 S O_{2}$
$5 F e S_{2} + 4 H_{2} O ⟶ 5 {F e}^{3 +} + 2 S O_{2}$
$5 F e S_{2} + 4 H_{2} O ⟶ 5 {F e}^{3 +} + 2 S O_{2} + 8 H^{+}$
To balance the charges add 23 electrons to left hand side.
$5 F e S_{2} + 4 H_{2} O + 23 e^{-} ⟶ 5 {F e}^{3 +} + 2 S O_{2} + 8 H^{+}$
The oxidation number of iron in $F e S_{2}$ is +2.
The oxidation number of iron in $F e^{3 +}$ is +3.
The change in the oxidation number is 1.
Since 23 electrons are involved and change in oxidation number of 5 iron atoms is +5, the n factor is 28.

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