Question

Ceric ammonium sulphate and potassium permanganate are used as oxidizing agents in acidic medium for oxidation of ferrous ammonium sulphate to ferric sulphate. The ratio, of number of moles of ceric ammonium sulphate required per mole of ferrous ammonium sulphate to the number of moles of ferrous sulphate to the number of moles of $\text{KMn}{{\text{O}}_{4}}$ required per mole of ferrous ammonium sulphate is:
A. 5.0
B. 0.2
C. 0.6
D. 2.0

Answer 1

This question is based on balancing the redox reaction and the change in the oxidative and reductive state of an element. The change in the oxidation state of an element can be obtained by finding the number of electrons involved in the reaction for oxidation and reduction.

Complete answer:
Let us solve this question step by step;

Step (1)- Write the molecular formula for the chemical names of the compounds given:
Ceric ammonium sulphate- ${{(\text{N}{{\text{H}}_{4}}\text{)}}_{4}}\text{Ce(S}{{\text{O}}_{4}}{{\text{)}}_{4}}$
Potassium permanganate- $\text{KMn}{{\text{O}}_{4}}$
Ferrous ammonium sulphate- ${{\text{(N}{{\text{H}}_{4}}\text{)}}_{2}}\text{Fe(S}{{\text{O}}_{4}}{{\text{)}}_{2}}\text{.6}{{\text{H}}_{2}}\text{O}$

Step (2)- Write the reaction involved in acidic medium- The reaction of $\text{KMn}{{\text{O}}_{4}}$ in acidic medium is $\text{Mn}{{\text{O}}_{4}}^{-}+8{{\text{H}}^{+}}+5{{\text{e}}^{-}}\ to \text{M}{{\text{n}}^{+2}}+4{{\text{H}}_{2}}\text{O}$.

Step (3)- Write the oxidation states- The oxidation state of $\text{Ce}$ in ceric ammonium sulphate ${{(\text{N}{{\text{H}}_{4}}\text{)}}_{4}}\text{Ce(S}{{\text{O}}_{4}}{{\text{)}}_{4}}$ is +4 and of $\text{Fe}$ in ferrous ammonium sulphate ${{\text{(N}{{\text{H}}_{4}}\text{)}}_{2}}\text{Fe(S}{{\text{O}}_{4}}{{\text{)}}_{2}}\text{.6}{{\text{H}}_{2}}\text{O}$ is +2.

Step (4)- Write the oxidation reactions- The reactions involved are being oxidizing agent $\text{Ce}$ in ceric ammonium sulphate ${{(\text{N}{{\text{H}}_{4}}\text{)}}_{4}}\text{Ce(S}{{\text{O}}_{4}}{{\text{)}}_{4}}$ will undergo reduction and $\text{Fe}$ in ${{\text{(N}{{\text{H}}_{4}}\text{)}}_{2}}\text{Fe(S}{{\text{O}}_{4}}{{\text{)}}_{2}}\text{.6}{{\text{H}}_{2}}\text{O}$ will oxidise; $\text{C}{{\text{e}}^{+4}}+{{\text{e}}^{-}}\ to \text{C}{{\text{e}}^{+3}}$and$\text{F}{{\text{e}}^{+2}}\ to \text{F}{{\text{e}}^{+3}}+{{\text{e}}^{-}}$.

Step (5)- The change in reduction of $\text{KMn}{{\text{O}}_{4}}$ is $5{{\text{e}}^{-}}$ and change in oxidation of ${{\text{(N}{{\text{H}}_{4}}\text{)}}_{2}}\text{Fe(S}{{\text{O}}_{4}}{{\text{)}}_{2}}\text{.6}{{\text{H}}_{2}}\text{O}$ is $\text{1}{{\text{e}}^{-}}$. Therefore, the oxidation of 1 mole of ${{\text{(N}{{\text{H}}_{4}}\text{)}}_{2}}\text{Fe(S}{{\text{O}}_{4}}{{\text{)}}_{2}}\text{.6}{{\text{H}}_{2}}\text{O}$ will require moles $\frac{1}{5}$ of $\text{KMn}{{\text{O}}_{4}}$.

Step (6)- The ceric ammonium sulphate and ferrous ammonium sulphate; the change in oxidation states are the reduction of ${{(\text{N}{{\text{H}}_{4}}\text{)}}_{4}}\text{Ce(S}{{\text{O}}_{4}}{{\text{)}}_{4}}$ is $\text{1}{{\text{e}}^{-}}$ change and that of ${{\text{(N}{{\text{H}}_{4}}\text{)}}_{2}}\text{Fe(S}{{\text{O}}_{4}}{{\text{)}}_{2}}\text{.6}{{\text{H}}_{2}}\text{O}$ is also found to be $\text{1}{{\text{e}}^{-}}$. So, 1 mole of ${{\text{(N}{{\text{H}}_{4}}\text{)}}_{2}}\text{Fe(S}{{\text{O}}_{4}}{{\text{)}}_{2}}\text{.6}{{\text{H}}_{2}}\text{O}$ requires 1 mole of ${{(\text{N}{{\text{H}}_{4}}\text{)}}_{4}}\text{Ce(S}{{\text{O}}_{4}}{{\text{)}}_{4}}$ for oxidation.
Thus, the ratio of number of moles of ${{(\text{N}{{\text{H}}_{4}}\text{)}}_{4}}\text{Ce(S}{{\text{O}}_{4}}{{\text{)}}_{4}}$ and $\text{KMn}{{\text{O}}_{4}}$ needed for oxidation of 1 mole of ${{\text{(N}{{\text{H}}_{4}}\text{)}}_{2}}\text{Fe(S}{{\text{O}}_{4}}{{\text{)}}_{2}}\text{.6}{{\text{H}}_{2}}\text{O}$ is $1:\frac{1}{5}$ ; is also 5.

So, the correct answer is “Option A”.

Note:
Utmost care is needed while finding the change in oxidation state. The molecular formula of the compounds must be known to avoid errors. The reactions need to be correctly balanced, so that the change in element’s oxidation state can be found easily and the number of moles of any element can be found directly.