Find the n-factor in the following chemical changes :

$C_{2} O_{4}^{2 -} \to C O_{2}$

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none of these

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Solution

The correct option is B 2
Given chemical change:
$C_{2} O_{4}^{2 -} ⟶ {C O}_{2}$
Here , oxidation state of $C$ in reactant is +3 while
that in product is +4 .
In this way, two carbons are going from +3 to +4 (oxidized),
hence, $n -$ factor $= 2$
so correct option is B

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