Question

Volume $ {V_1}mL $ of $ 0.1M{\text{ }}{K_2}C{r_2}{O_7} $ is needed for complete oxidation of $ 0.678g{\text{ }}{N_2}{H_4} $ in acidic medium. The volume of $ 0.3M{\text{ }}KMn{O_4} $ needed for the same oxidation in acidic medium will be:
A) $ \dfrac{{201}}{5}{V_1} $
B) $ \dfrac{5}{2}{V_1} $
C) $ 113{V_1} $
D) can’t say

Answer 1

Hint: In here we are given two Oxidizing Agents in the same medium (acidic). Remember that the no. of equivalents of Oxidising Agents (in same medium) will always be equal; I.e., the number of equivalents of $ 0.1M{\text{ }}{K_2}C{r_2}{O_7} $ will be equal to $ 0.3M{\text{ }}KMn{O_4} $ .

Complete Step By Step Answer:
For the first reaction the equation can be written as:
$ {K_2}C{r_2}{O_7} + {N_2}{H_4}\xrightarrow{{acidic}}2C{r^{ + 3}} + {N_2} $ -- (1)
Here we can see that the oxidation state of Cr changes from $ + 6 \to + 3 $ and that of N changes from $ - 1({N_2}{H_4}) \to 0({N_2}) $ . To find the no. of equivalents we’ll use the formula:
\[no.of{\text{ }}equivalent = n \times {n_f}\]
It can also be written as: $ no.of{\text{ }}equivalents = M \times V \times {n_f} $ -- (2)
Where n is the no. of moles of Oxidising Agent or Reducing Agent. M is the molarity, V is the Volume, and $ {n_f} $ is the n-factor.
The n-factor can be given by the formula: $ {n_f} = |change{\text{ }}in{\text{ }}Oxidation{\text{ }}State| \times no.of{\text{ }}atoms $
The n-factor of Cr in equation (1) is: $ {n_{f(O.A)}} = | + 3| \times 2 = 6 $ -- (3)
Substituting (3) in (2), the no. of equivalents is: $ no.of{\text{ }}equivalents = 0.1 \times {V_1} \times 6 = 0.6{V_1} $ -- (4)
The second Reaction can be given as:
$ KMn{O_4} + {N_2}{H_4} \to M{n^{ + 2}} + {N_2} $
Here the O.S of Mn changes from $ + 7 \to + 2 $ and that of N changes from $ - 1({N_2}{H_4}) \to 0({N_2}) $ .
The n-factor of O.A can be given as: $ {n_{f(O.A)}} = | + 5| \times 1 = 5 $ -- (5)
Substituting (5) in (2): $ no.of{\text{ }}equivalents = 0.3 \times {V_2} \times 5 $ --- (6)
We have been given that the equivalents of $ 0.1M{\text{ }}{K_2}C{r_2}{O_7} $ and $ 0.3M{\text{ }}KMn{O_4} $ will be equal. Hence equating equation (4) and (6)
$ 0.6{V_1} = 0.3 \times 5 \times {V_2} $
$ {V_2} = \dfrac{{0.6}}{{5 \times 0.3}}{V_1} = \dfrac{{201}}{5}{V_1} $
The correct answer is Option (A).

Note:
The n-factor of certain Oxidizing agents can be easily remembered in different mediums. The n-factor of $ {K_2}C{r_2}{O_6} $ is always 6 in acidic medium. The n-factor of $ KMn{O_4} $ in Basic, Acidic and Neutral Medium is 1,5 and 3 respectively. You can remember this as $ BAN - 153 $ .