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

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 $BAN - 153$ .