Question

What would be the equivalent weight of the reductant in the given reaction?
 ${\left[ {{\text{Fe}}{{\left( {{\text{CN}}} \right)}_{\text{6}}}} \right]^{3 - }}\, + \,{{\text{H}}_{\text{2}}}{{\text{O}}_2}\, + \,2{\text{O}}{{\text{H}}^ - }\, \to 2\,\,{\left[ {{\text{Fe}}{{\left( {{\text{CN}}} \right)}_{\text{6}}}} \right]^{4 - }} + \,2\,{{\text{H}}_{\text{2}}}{\text{O}}\, + \,{{\text{O}}_2}$
(Given Fe = $56$, C = $12$, N = $14$, O = $16$, H = $1$)
A. $17$
B. $212$
C. $34$
D. $32$

Answer 1

Hint: To answer we should know what equivalent weight is. First, we will determine the change in oxidation state of each to identify the reductant. Then by diving the molecular mass of reductant with its valence factor we will determine the equivalent weight.
\[{\text{equivalent weight}}\,{\text{ = }}\,\dfrac{{{\text{Molecular weight}}}}{{{\text{valance}}\,{\text{factor}}}}\]


Complete step by step solution:
The formula to determine the equivalent weight is as follows:
\[{\text{equivalent weight}}\,{\text{ = }}\,\dfrac{{{\text{Molecular weight}}}}{{{\text{valance}}\,{\text{factor}}}}\]
The given reaction is as follows:
${\left[ {{\text{Fe}}{{\left( {{\text{CN}}} \right)}_{\text{6}}}} \right]^{3 - }}\, + \,{{\text{H}}_{\text{2}}}{{\text{O}}_2}\, + \,2{\text{O}}{{\text{H}}^ - }\, \to 2\,\,{\left[ {{\text{Fe}}{{\left( {{\text{CN}}} \right)}_{\text{6}}}} \right]^{4 - }} + \,2\,{{\text{H}}_{\text{2}}}{\text{O}}\, + \,{{\text{O}}_2}$
The oxidation of Fe in ${\left[ {{\text{Fe}}{{\left( {{\text{CN}}} \right)}_{\text{6}}}} \right]^{3 - }}$ is as follows:
The oxidation state of ${\text{C}}{{\text{N}}^ - }$ is $ - 1$so,
$x\, + \,\left( { - 1 \times 6} \right)\, = \, - 3$
$x\, = \, - 3 + 6$
$x\, = \, + 3$
So, the oxidation state of Fe in ${\left[ {{\text{Fe}}{{\left( {{\text{CN}}} \right)}_{\text{6}}}} \right]^{3 - }}$is $ + 3$.
The oxidation of Fe in ${\left[ {{\text{Fe}}{{\left( {{\text{CN}}} \right)}_{\text{6}}}} \right]^{4 - }}$ is as follows:
$x\, + \,\left( { - 1 \times 6} \right)\, = \, - 4$
$x\, = \, - 4 + 6$
$x\, = \, + 2$
So, the oxidation state of Fe in ${\left[ {{\text{Fe}}{{\left( {{\text{CN}}} \right)}_{\text{6}}}} \right]^{4 - }}$is$ + 2$.
So, the oxidation state of Fe in changing from $ + 3$to $ + 2$ means it Fe is getting reduced means it is the oxidizing agent or oxidant.
Reduction reaction: ${\left[ {{\text{Fe}}{{\left( {{\text{CN}}} \right)}_{\text{6}}}} \right]^{3 - }}\, \to 2\,\,{\left[ {{\text{Fe}}{{\left( {{\text{CN}}} \right)}_{\text{6}}}} \right]^{4 - }}$
The oxidation of oxygen in ${\text{O}}{{\text{H}}^ - }$ is as follows:
$\left( { + 1\, \times 1} \right)\, + \,\left( x \right)\, = \, - 1$
$x\, = \, - 1 - 1$
$x\, = \, - 2$
So, the oxidation state of oxygen in ${\text{O}}{{\text{H}}^ - }$is$ - 2$.
The oxidation of oxygen ${{\text{H}}_{\text{2}}}{\text{O}}$ is as follows:
$\left( { + 1\, \times 2} \right)\, + \,\left( x \right)\, = \,0$
$x\, = \, - 2$
So, the oxidation state of oxygen in ${{\text{H}}_{\text{2}}}{\text{O}}$is$ - 2$.
So, the oxidation state of oxygen does not change.
No change in oxidation state: $2{\text{O}}{{\text{H}}^ - }\, \to \,2\,{{\text{H}}_{\text{2}}}{\text{O}}$
The oxidation of oxygen in ${{\text{H}}_{\text{2}}}{{\text{O}}_2}$ is as follows:
The oxidation state of hydrogen is $ + 1$so,
$\left( { + 1\, \times 2} \right)\, + \,\left( {x \times 2} \right)\, = \,0$
$2\,x\, = \, - 2$
$x\, = \, - 1$
So, the oxidation state of oxygen in ${{\text{H}}_{\text{2}}}{{\text{O}}_2}$is$ - 1$.
The oxidation state of ${{\text{O}}_2}$ zero.
So, the oxidation state of oxygen in changing from $ - 1$to $0$ means it ${{\text{H}}_{\text{2}}}{{\text{O}}_2}$is getting oxidised means it is the reducing agent or reductant.
Oxidation reaction: ${{\text{H}}_{\text{2}}}{\mathop {\text{O}}\limits^{ - 1} _2}\, \to \,{\mathop {\text{O}}\limits^0 _2}$
So, the oxidation state of oxygen in ${{\text{H}}_{\text{2}}}{{\text{O}}_2}$ is changing from $ - 1$to $0$ and here, two oxygen atoms are present in ${{\text{H}}_{\text{2}}}{{\text{O}}_2}$ so, the total change in oxidation state is of $2$. So, the valence factor for the ${{\text{H}}_{\text{2}}}{{\text{O}}_2}$is $2$. The molecular weight of ${{\text{H}}_{\text{2}}}{{\text{O}}_2}$is$34$.
So, on substituting $2$ for valance factor and ${{\text{H}}_{\text{2}}}{{\text{O}}_2}$ for molecular weight in equivalent weight formula,
\[{\text{equivalent weight}}\,{\text{ = }}\,\dfrac{{34}}{2}\]
\[{\text{equivalent weight}}\,{\text{ = }}\,17\]
So, the equivalent weight of the reductant in the given reaction is\[17\].

Thus, the correct option is (A).


Note: Molecular weight is the product of equivalent weight and balance factor. Molecular weight is determined by adding the atomic mass of each constituting atom. The valence factor is also known as n-factor. The valence factor is the number of electrons gained or lost by a species or the oxidation number or charge of the species In the case of acid, the valance factor is determined as the number of protons donated by the species.