Question

What would be the equivalent weight of the reductant in the reaction :
$\left[Fe\right(CN{)}_6{]}^{-3}+H_2{O}_2+2O{H}^{-}\to 2\left[Fe\right(CN{)}_6{]}^{4-}+{2H}_{2}O+O_2$
$[Given:Fe=56,C=12,N=14,O=16,H=1]$

• A. 17
• B. 212
• C. 34
• D. 32

Answer 1

The correct option is A 17

$\left[Fe\right(CN{)}_6{]}^{3-}+H_2{O}_2+2O{H}^{-}\to 2\left[Fe\right(CN{)}_6{]}^4+2H_{2}O+O_2$

$\text{Assigning the oxidation state we get}$

$\left[F{e}^{3+}\right(CN{)}_6{]}^{3-}+H_2{O}_2^{-1}+2O^{-2}{H}^{-}\to 2\left[F{e}^{+2}\right(CN{)}_6{]}^4+2H_{2}O+O_2$

$\begin{array}{c}\text{Reduction reaction is :-}\\ {\text{Fe}}^{3+}\to F{e}^{2+}\\ \text{Oxidation reaction is}\\ 2O^{-}\to O_2\end{array}$

$\begin{array}{c}\text{Since, Fe gets reduced from}+3\text{to}\\ +2\text{oxidation state, It is the oxidant}\\ \text{and}{H}_2{O}_2\text{gets oxidised from}-1\text{to}\\ 0,\text{it is reductant.}\\ \text{considering,}\end{array}$

$\begin{array}{c}2O^{-}⟶O_2\\ \text{change in oxidation no.}\\ =2\times 0-2\times (-1)=2\\ \text{Equivalent mass of reactant,}{H}_2{O}_2=\frac{Molar\text{mass}}{\Delta \text{oxidation no.}}\end{array}$

$=\frac{34}{2}=17$