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 number of moles of cerium ammonium sulphate required per mole of ferrous ammonium sulphate to the number of moles of $KMn{O}_4$ required per mole of ferrous ammonium sulphate is

• A. 5.0
• B. 0.2
• C. 0.6
• D. 2.0

Answer 1

Answer: (A)

$Mn{O}_4^{-}+8H^{+}\to M{n}^{2+}+4H_{2}O+5e^{-}$
$(N{H}_4{)}_{2}Ce(S{O}_4{)}_2\equiv C{e}^{4+}$ and
$(N{H}_4{)}_{2}FeS{O}_{4}6H_{2}O=F{e}^{2+}$
Now $C{e}^{4+}+e^{-}\to C{e}^{3+}$
and $F{e}^{2+}\to F{e}^{3+}+e^{-}$
Since reduction of $KMn{O}_4$ is a $5e^{-}$ change and oxidation of $(N{H}_4{)}_{2}S{O}_4$ is a $1e^{-}$ change, therefore, oxidation of 1 mole of $(N{H}_4{)}_{2}S{O}_4.FeS{O}_4.6H_{2}O$ will require $1/5$ mole of $KMn{O}_4$
Further, since reduction of $(N{H}_4{)}_{2}S{O}_4.Ce(S{O}_4{)}_2$ is a $1e^{-}$ change and that of oxidation of $(N{H}_4{)}_{2}S{O}_4.FeS{O}_4.6H_{2}O$ is also also change, therefore, oxidation of $1$ mole of $(N{H}_4{)}_{2}S{O}_4.FeS{O}_4.6H_{2}O$ will require one mole of $(N{H}_4{)}_{2}S{O}_4.C{e}_2(S{O}_4{)}_2$. Thus, the ratio of the number of moles of $(N{H}_4{)}_{2}S{O}_4.C{e}_2(S{O}_4{)}_2$ and $KMn{O}_4$ required to oxidise $1$ mole of $(N{H}_4{)}_{2}S{O}_4.FeS{O}_4.6H_{2}O$ is $1:1/5$ or $5:1$.