Question

What is the correct relation between normality $\left( {\text{N}} \right)$ ${\text{&}}$ molarity $\left( {\text{M}} \right)$ of ${K_2}C{r_2}{O_7}$ acting as an oxidizing agent in acidic medium.
A) $N = 6M$
B) $M = 6N$
C) $N = 3M$
D) $M = 3N$

Answer 1

Hint: The reagent ${K_2}C{r_2}{O_7}$ acts as an oxidizing agent that means it itself gets reduced which will lead to loss of electrons. One can write a reaction equation for this and try to put that relationship in the equation of relation between normality and molarity.

Complete step by step answer:
1) First of all we will discuss the formula which gives a relation between normality and molarity.
$Normality{\text{ }} = {\text{ }}molarity{\text{ }} \times {\text{ }}n{\text{ factor}}$
2) Now let us write the chemical conversion equation for ${K_2}C{r_2}{O_7}$ when it oxidizes other molecules and gets reduced itself.
${K_2}C{r_2}{O_7}\xrightarrow{{}}C{r^{ + 3}}$
The $C{r^{ + 3}}$ form is a reduced form where the ${K_2}C{r_2}{O_7}$ loses electrons to other molecules which means it oxidizes other molecules. The ${\text{ + 3}}$ charge shows that electrons have been donated or lost which shows oxidation has been done and it also indicates the oxidation state of the Cr ion.
3) Now let's calculate the oxidation state of Cr in the molecule ${K_2}C{r_2}{O_7}$ as below,
In ${K_2}C{r_2}{O_7}$ molecule, the potassium has +1 oxidation state and the oxygen has -2 oxidation state and to find out the oxidation state of Cr lets take that value as x,
$\left( {2 \times + 1} \right) + \left( {2 \times x} \right) + \left( {7 \times - 2} \right) = 0$
Now let's calculate the equation,
$2 + 2x - 14 = 0$
By doing the subtraction we get,
$2x - 12 = 0$
By changing side we get,
$2x = + 12$
By changing the side the numerator 2 will go to the denominator,
$x = \dfrac{{ + 12}}{2}$
$x = + 6$
Therefore, the oxidation state of the Cr in ${K_2}C{r_2}{O_7}$ is ${\text{ + 6}}$ and it changes to ${\text{ + 3}}$ in the product.
Hence, we can say that the change of oxidation state of Cr is ${\text{ + 3}}$.
4) Now the change of oxidation state for one Cr is +3 and the molecule ${K_2}C{r_2}{O_7}$ has two Cr atoms hence the total change of oxidation state of Cr is ${\text{ + 6}}$.
5) Now the value we got ${\text{ + 6}}$ is the value of n factor and lets put this value in the normality-molarity relation equation we get,
$Normality{\text{ }} = {\text{ }}molarity{\text{ }} \times {\text{ }}n{\text{ factor}}$
$N = 6 \times M$
6) Therefore, the correct relation between normality $\left( {\text{N}} \right)$ ${\text{&}}$ molarity $\left( {\text{M}} \right)$ of ${K_2}C{r_2}{O_7}$ acting as an oxidizing agent in acidic medium is $N = 6M$ which shows option A as the correct choice.
Therefore, the correct option is A.

Note:
The gain of electrons and loss of oxygen atoms is called reduction. The loss of electrons and gain of oxygen atom is called oxidation. An oxidizing agent oxidizes others and gets reduced itself.