Question

Consider a redox reaction:

$Fe{S}_2+KMn{O}_4+H^{+}\to F{e}^{3+}+S{O}_4^{2-}+M{n}^{2+}+H_{2}O$.

If the molar mass of $Fe{S}_2$ is M, then equivalent mass of $Fe{S}_2$ would be equal to:

• A. $M$
• B. $\frac{M}{5}$
• C. $\frac{M}{10}$
• D. $\frac{M}{15}$

Answer 1

The correct option is D $\frac{M}{15}$

If in a reaction, two elements of same reactant molecule are undergoing either oxidation or reduction then the n-factor of the molecule will be the sum of the individual n-factors of both the elements.

$\stackrel{+2-1}{Fe{S}_2}\to \stackrel{+6}{S}{O}_4^{2-}+\stackrel{+3}{F}{e}^{3+}\text{oxidation half reaction}$

$n_f=[1\times (3-2)+2(6-(-1)]=15$

($\therefore$ n factor of a compound undergoing redox change is equal to total number of moles of electrons lost, gained or exchange by 1 mole of the compound.)

So, equivalent mass of $Fe{S}_2=\frac{M}{15}$