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Redox Reactions

The equivalen...

Question

The equivalent weight of $F e S_{2}$ in the following reaction is $F e S_{2} + O_{2} \to F e^{+ 3} + S O_{2}$

A

\frac{M o l . w t}{1}

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B

\frac{M o l . w t}{7}

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C

\frac{M o l . w t}{11}

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D

\frac{M o l . w t}{9}

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Solution

The correct option is A $\frac{M o l . w t}{1}$

The equivalent weight of $F e S_{2}$ in the following reaction is $\frac{M o l . w t}{1}$

$F e S_{2} + O_{2} \to F e^{+ 3} + S O_{2}$ .
Becuase in the above equation the oxidation state of $F e$ is changed from $+ 2$ to $+ 3$ .So the no of electrons lost is 1.
$(E q u i v a l e n t w e i g h t) = \frac{M o l . w t}{(N o . O f . e l e t r o n s l o s t o r g a i n e d)}$ .

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