Question

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

Answer 1

Write the half cell reaction for the oxidation of ferrous ion to ferric ion using ceric ion in Ceric ammonium sulphate. Also, write the half cell reactions for the oxidation of ferrous ions to ferric ions using permanganate ions. Using the stoichiometric coefficients of half cell reactions calculate the ratio of the number of moles of ceric ammonium sulphate required per mole of ferrous ammonium sulphate to the number of moles of ${\text{KMn}}{{\text{O}}_{\text{4}}}$ required per mole of ferrous ammonium sulphate.

Complete step by step solution:
The formula of ceric ammonium sulphate is ${{\text{(N}}{{\text{H}}_{\text{4}}}{\text{)}}_{\text{4}}}{\text{Ce(S}}{{\text{O}}_{\text{4}}}{{\text{)}}_{\text{4}}}$. Ammonium ion has +1 charge, sulphate ion has -2 charge hence oxidation state of cerium ion is+4.
The formula of potassium permanganate is ${\text{KMn}}{{\text{O}}_{\text{4}}}$. The oxidation state of ${\text{K}}$ is +1, the oxidation state of ${\text{O}}$ is -2 so oxidation state of ${\text{Mn}}$ is +7.
The formula of ferrous ammonium sulphate is${{\text{(N}}{{\text{H}}_{\text{4}}}{\text{)}}_2}{\text{Fe(S}}{{\text{O}}_{\text{4}}}{{\text{)}}_2} \cdot {\text{6}}{{\text{H}}_{\text{2}}}{\text{O}}$. Here the oxidation state of ferrous ion (${\text{F}}{{\text{e}}^{{\text{2 + }}}}$) is +2.
The formula of ferric sulphate is, ${\text{F}}{{\text{e}}_{\text{2}}}{{\text{(S}}{{\text{O}}_{\text{4}}}{\text{)}}_3}$. Here, the oxidation state of ferric ion (${\text{F}}{{\text{e}}^{{\text{3 + }}}}$) is +3.
Now we will write the half cell for the oxidation of ferrous ion to ferric ion using ceric ion in Ceric ammonium sulphate as follows:
${\text{F}}{{\text{e}}^{{\text{2 + }}}} + {\text{ C}}{{\text{e}}^{{\text{4 + }}}} \to \,{\text{F}}{{\text{e}}^{{\text{3 + }}}} + {\text{ C}}{{\text{e}}^{{\text{3 + }}}}$
Here all atoms are already balanced and the electronic charge is also balanced so we can say that this is a balanced half cell reaction.
From the balanced half cell reaction, we can say that 1 mol ceric ammonium sulphate required per mole of ferrous ammonium sulphate.
Now, similarly, we will write the half cell reaction for the oxidation of ferrous ions to ferric ions using permanganate ions.
${\text{F}}{{\text{e}}^{{\text{2 + }}}} + {\text{ Mn}}{{\text{O}}_{\text{4}}}^{\text{ - }} \to \,{\text{F}}{{\text{e}}^{{\text{3 + }}}} + {\text{ M}}{{\text{n}}^{{\text{2 + }}}}$
Here to balance the oxygen atoms we need to add 4 water molecules on the right side of the equation.
${\text{F}}{{\text{e}}^{{\text{2 + }}}} + {\text{ Mn}}{{\text{O}}_{\text{4}}}^{\text{ - }} \to \,{\text{F}}{{\text{e}}^{{\text{3 + }}}} + {\text{ M}}{{\text{n}}^{{\text{2 + }}}} + {\text{4}}{{\text{H}}_{\text{2}}}{\text{O}}$
Now, to balance the hydrogen atoms we need to add 8 ${{\text{H}}^{\text{ + }}}$ ions on the reactant side.
${\text{F}}{{\text{e}}^{{\text{2 + }}}} + {\text{ Mn}}{{\text{O}}_{\text{4}}}^{\text{ - }} + 8{{\text{H}}^{\text{ + }}} \to \,{\text{F}}{{\text{e}}^{{\text{3 + }}}} + {\text{ M}}{{\text{n}}^{{\text{2 + }}}} + {\text{4}}{{\text{H}}_{\text{2}}}{\text{O}}$.
Now, to balance the charges as follows:
${\text{5F}}{{\text{e}}^{{\text{2 + }}}} + {\text{ Mn}}{{\text{O}}_{\text{4}}}^{\text{ - }} + 8{{\text{H}}^{\text{ + }}} \to \,5{\text{F}}{{\text{e}}^{{\text{3 + }}}} + {\text{ M}}{{\text{n}}^{{\text{2 + }}}} + {\text{4}}{{\text{H}}_{\text{2}}}{\text{O}}$
From the balanced half cell reaction, we can say that $1/5$ moles of ${\text{KMn}}{{\text{O}}_{\text{4}}}$ required per mole of ferrous ammonium sulphate.
So,
$\dfrac{{{\text{Moles of ceric ammonium sulphate}}}}{{{\text{Moles of potassium permanganate}}}} = \dfrac{1}{{1/5}} = 5.0$

Hence, the correct option is (A) 5.0.

Note: Oxidation is the loss of electrons while reduction is the gain of electrons. The oxidising agent is the substance that tends to oxidise other substances and itself get reduced. So in a redox reaction, the oxidising agent accepts the electron from other substances and helps other substances to oxidise. Based on medium used acidic or basic medium there are different methods of balancing the redox reaction.