Question

In order to oxidise a mixture one mole of each of $Fe{{C}_{2}}{{O}_{4}}$, $F{{e}_{2}}{{({{C}_{2}}{{O}_{4}})}_{3}}$, $FeS{{O}_{4}}$ and $F{{e}_{2}}{{(S{{O}_{4}})}_{3}}$ in acidic medium, the number of moles of $KMn{{O}_{4}}$ required is:
(A) 3
(B) 2
(C) 1
(D) 1.5

Answer 1

Hint: An oxidizing agent is a reactant that removes electrons from other reactants during a redox reaction. The oxidizing agent typically takes these electrons for itself, thus gaining electrons and being reduced. An oxidizing agent is thus an electron acceptor.

Complete step by step answer: In an acidic solution, permanganate(VII) ion is reduced to the colourless manganese(II) (Mn2+) ion.
\[8{{H}^{+}}+Mn{{O}_{4}}^{-}+5{{e}^{-}}\to M{{n}^{2+}}+4{{H}_{2}}O\]
The manganese atom in this reaction changes from +7 oxidation to +2 oxidation, there is a change of 5 electrons, therefore the n-factor of Permanganate will be 5. Since Permanganate is an oxidizing agent, it will oxidize $Fe{{C}_{2}}{{O}_{4}}$ ,$F{{e}_{2}}{{({{C}_{2}}{{O}_{4}})}_{3}}$, $FeS{{O}_{4}}$ and $F{{e}_{2}}{{(S{{O}_{4}})}_{3}}$ in acidic medium. Now we will see oxidation of each species with change in oxidation number and n-factor.
Oxidation of $Fe{{C}_{2}}{{O}_{4}}$ will give,
\[\begin{align}
  & F{{e}^{+2}}\to F{{e}^{+3}} \\
 & {{C}_{2}}{{O}_{4}}^{-2}\to 2C{{O}_{2}}+2{{e}^{-}} \\
\end{align}\]
There change of 1 electron in iron oxidation state and 2 electron change in oxalic acid, therefore, the n-factor will be 3. Similarly Oxidation of $F{{e}_{2}}{{({{C}_{2}}{{O}_{4}})}_{3}}$ will give,
\[{{C}_{2}}{{O}_{4}}^{-2}\to 2C{{O}_{2}}+2{{e}^{-}}\]
Iron is already in +3 oxidation state, therefore, n-factor is 2x3=6.
Oxidation of $FeS{{O}_{4}}$ will give n-factor 1 and oxidation of $F{{e}_{2}}{{(S{{O}_{4}})}_{3}}$ will give n-factor 0.
Since, Permanganate will oxidize each species therefore, the relation will be,
Moles of permanganate x n-factor= moles of $Fe{{C}_{2}}{{O}_{4}}$ x n-factor+ moles of $F{{e}_{2}}{{({{C}_{2}}{{O}_{4}})}_{3}}$ x n-factor+ moles of $FeS{{O}_{4}}$ x n-factor+ moles of $F{{e}_{2}}{{(S{{O}_{4}})}_{3}}$ x n-factor
\[\begin{align}
  & n\times 5=1\times 3+1\times 6+1\times 1+1\times 0 \\
 & n=\dfrac{10}{5}=2 \\
\end{align}\]
Therefore, 2 moles of permanganate will be needed to oxidize one mole of $Fe{{C}_{2}}{{O}_{4}}$ , $F{{e}_{2}}{{({{C}_{2}}{{O}_{4}})}_{3}}$, $FeS{{O}_{4}}$ and $F{{e}_{2}}{{(S{{O}_{4}})}_{3}}$ in acidic medium.

Hence, the correct option is the B option.

Note: A permanganate is the general name for a chemical compound containing the manganate(VII) ion, $(Mn{{O}_{4}}^{-})$ . Because manganese is in the +7 oxidation state, the permanganate(VII) ion is a strong oxidizing agent.