Question

The ratio of number of moles $KMn{O_4}$​ and ${K_{2}}C{r_2}{O_7}$​ required to oxidise 0.1mol $S{n^{ + 4}}$ in acidic medium:
A.6:5
B.5:6
C.1:2
D.2:1

Answer 1

During the reaction $KMn{O_4}$ and ${K_{2}}C{r_2}{O_7}$ are getting 5 and 6 electrons, respectively. At the same time $S{n^{ + 4}}$ loses 2 electrons and becomes $S{n^{ + 2}}$. Thus we have to equate the equivalent moles of both potassium permanganate and potassium dichromate to the amount of stannous ions. This will give us the ratio of both these reagents to oxidise the reaction.

Complete answer:
The reaction of $S{n^{ + 4}}$ in acidic medium is given as below:

$\Rightarrow S{n^{ + 4}}\xrightarrow{{reducing{\text{ }}agent}}S{n^{ + 2}}$
Thus this reduction requires 2 electrons. Thus the n factor of $S{n^{ + 4}}$ is 2
The reaction of $KMn{O_4}$ in an acidic medium is given as below:

$\Rightarrow KMn{O_4}\xrightarrow{{acidic{\text{ }}medium}}M{n^{ + 2}}$
Here the oxidation number of Manganese will change from +7 to +2 and thus we can say that this reduction can be done by using 5 electrons. Thus n factor of $KMn{O_4}$ is 5
Thus we can equate the number of equivalents of $KMn{O_4}$ and $S{n^{ + 4}}$
[Number of equivalents = moles $\times$ n factor]
n factor of $KMn{O_4}$ is 5 and that of $S{n^{ + 4}}$ is 2
Thus we can say that:
$\Rightarrow {n_{KMn{O_4}}} \times n{\text{ }}facto{r_{KMn{O_4}}} = {n_{S{n^{ + 4}}}} \times n{\text{ }}facto{r_{S{n^{ + 4}}}}$
We can substitute the values of all the above terms and thus obtain the equation below:
$\Rightarrow {n_{KMn{O_4}}} \times 5 = 0.1 \times 2$
By taking 5 to the other side, we get:
$\Rightarrow {n_{KMn{O_4}}} = \dfrac{{0.2}}{5}$
Simplifying this, we get:
$\Rightarrow {n_{KMn{O_4}}} = 0.04$
Now we can take a look at the reaction of potassium dichromate in acidic medium

$\Rightarrow {K_2}C{r_2}{O_7}\xrightarrow{{acidic{\text{ }}medium}}C{r^{ + 3}}$
Thus we can say that the oxidation state of chromium changes from +6 to +3 indicating that this requires 3 electrons. Thus the n factor of this reaction is 6 (Since there are 2 chromium ions per molecule).
Now we can equate the number of equivalents as we have done in the above case and thus we have the equation:
$\Rightarrow {n_{{K_{2}}C{r_2}{O_7}}} \times n{\text{ }}facto{r_{{K_{2}}C{r_2}{O_7}}} = {n_{S{n^{ + 4}}}} \times n{\text{ }}facto{r_{S{n^{ + 4}}}}$
Substituting the values as mentioned above, we get:
$\Rightarrow {n_{{K_{2}}C{r_2}{O_7}}} \times 6 = 0.1 \times 2$
Simplifying this we get,
$\Rightarrow {n_{{K_{2}}C{r_2}{O_7}}} = \dfrac{{0.2}}{6}$
$\Rightarrow {n_{{K_{2}}C{r_2}{O_7}}} = 0.03333$
Now we need to find the ratio of the number moles of $KMn{O_4}$​ and ${K_{2}}C{r_2}{O_7}$​. To do that we need to divide the value of the number of moles of both $KMn{O_4}$ and ${K_{2}}C{r_2}{O_7}$.
$\Rightarrow \dfrac{{{n_{KMn{O_4}}}}}{{{n_{{K_{2}}C{r_2}{O_7}}}}} = \dfrac{{0.04}}{{0.0333}}$
This can be written as:
$\Rightarrow \dfrac{{{n_{KMn{O_4}}}}}{{{n_{{K_{2}}C{r_2}{O_7}}}}} = \dfrac{6}{5}$
Thus the correct answer for this question is option (A).

Note:
While calculating the n factor of a molecule make sure u have taken into account the number of atoms into consideration. In this question we know that one molecule of potassium dichromate will use 3 electrons to change to chromium (+3) ion. But there are 2 chromium ions in the molecule and therefore 6 electrons are required for the reduction and hence the n factor is 6.