Question

A certain amount of a reducing agent reduces x mole of ${\text{KMn}}{{\text{O}}_4}$ and y mole of ${{\text{K}}_{\text{2}}}{\text{C}}{{\text{r}}_{\text{2}}}{{\text{O}}_{\text{7}}}$ ​ in different experiments in acidic medium. If the change in oxidation state in reducing agent is the same in both experiments, x : y
A.5 : 3
B.3 : 5
C.5 : 6
D.6 : 5

Answer 1

We shall write the equation of reduction for both the compounds and find the electrons exchanged. The electrons exchanged will help determine the moles reacted by the law of equivalence.

Complete step by step answer:
From the question, we know that the amount of reducing agent is the same in both the experiments.
$\therefore$ No. of equivalents of the reducing agent will be equal.
From the law of equivalence, we can say that:
no. of the equivalent of ${\text{KMn}}{{\text{O}}_4}$ ​ = no. of the equivalent of reducing agent = no. of the equivalent of ${{\text{K}}_{\text{2}}}{\text{C}}{{\text{r}}_{\text{2}}}{{\text{O}}_{\text{7}}}$ ​
To use the above relation we need to find the n-factor or the number of electrons exchanged during the reduction reaction.
For ${\text{KMn}}{{\text{O}}_4}$ reduction in an acidic medium, we can write the equation as: $8{H^ + } + MnO{4^ - } + 5{e^ - } \to M{n^{2 + }} + 4{H_2}O$.
We can see that the n-factor of ${\text{KMn}}{{\text{O}}_4}$ in an acidic medium is 5.
Now ${{\text{K}}_{\text{2}}}{\text{C}}{{\text{r}}_{\text{2}}}{{\text{O}}_{\text{7}}}$ , we can write the reaction as:
${K_2}C{r_2}{O_7}{\text{ + }}14{H^ + } + 6{e^ - } \to 2{K^ + } + 2C{r^{3 + }}{\text{ + }}7{H_2}O$.
Here, we can see the n-factor of ${{\text{K}}_{\text{2}}}{\text{C}}{{\text{r}}_{\text{2}}}{{\text{O}}_{\text{7}}}$ in acidic the medium ​ is 6.
We know that , no. of equivalent = ${\text{no}}{\text{. of moles\; \times \;n\; - factor}}$
$\Rightarrow {{\text{n}}_{{\text{KMn}}{{\text{O}}_{\text{4}}}}} \times 5 = {{\text{n}}_{{{\text{K}}_{\text{2}}}{\text{C}}{{\text{r}}_{\text{2}}}{{\text{O}}_{\text{7}}}}} \times 6$
Rearranging:
$\Rightarrow x \times 5 = y \times 6$
$\Rightarrow \dfrac{x}{y} = \dfrac{6}{5}$

Therefore, we can conclude that the correct answer to this question is option D.

Note:
Equivalent weight is defined as the amount of substance that produces 1-mole electrons or 96500 C charge. In any reaction, the number of electrons gained by one component equals the number of electrons lost by another component. So, the concept of the number of gram equivalents is based upon how electrons are transferred in reaction. And because of this number of grams equivalent of each component involved in the reaction are equal. This principle can be used to solve various kinds of problems.