Question

For the reaction,
$Fe{S}_2+KMn{O}_4+H^{+}\to F{e}^{3+}+H_{2}S{O}_4+M{n}^{2+}+H_{2}O$.
If the molar mass of $Fe{S}_2$ is M, then equivalent mass of $Fe{S}_2$ would be equal to:

• A. $M$
• B. $\frac{M}{10}$
• C. $\frac{M}{11}$
• D. $\frac{M}{15}$

Answer 1

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

$\stackrel{+2-1}{Fe{S}_2}\to 2H_2\stackrel{+6}{S}{O}_4+\stackrel{+3}{F}{e}^{3+}\text{oxidation half reaction}$
$n-factor=[1\times (3-2)+2(6-(-1)]=15$
($\therefore$ n factor of a compound undergoing redox change is equal to no. of moles of electrons lost, gained or exchange by 1 mole of the compound.)
So, eq.wt. of $Fe{S}_2=\frac{M}{15}$
Hence, (d) is the correct answer.