Question

Assuming complete ionization, same moles of which of the following compounds will require the least amount of acidified$KMn{{O}_{4}}$ for complete oxidation?
A. $Fe{{C}_{2}}{{O}_{4}}$
B. $Fe{{\left( N{{O}_{2}} \right)}_{2}}$
C. $FeS{{O}_{4}}$
D. $FeS{{O}_{3}}$

Answer 1

Hint: All of these given options contain ferrous ions and they get oxidized into ferric ions when potassium permanganate is reacted with them. But they have different radicals attached to them and so the permanganate will form different compounds with the different radicals which will decide the amount of permanganate needed for the reaction.

Complete answer:
-Based on the property of the compounds being able to dissociate into their ions, they are classified in 2 groups- electrolytes and non-electrolytes. Electrolytes are further classified as strong and weak electrolytes.
-Electrolyte is a compound which dissociates into its constituent ions, cation and anion in presence of DC current under the process of electrolysis.
-Many types of compound can be used as electrolytes but the most preferred compounds are acids, bases and salts. The electrolytes give redox reactions for dissociation into their ions.
-Electrolysis is a technique used to separate a compound in its different constituents with the help of DC current. It is used in non-spontaneous chemical reactions.
-The amount of voltage to be given depends on the type of the electrolyte used. The voltage is called decomposition potential.
-There are two electrodes, cathode and anode dipped in the electrolyte. They are connected to the battery. In acidified water, water is mixed with dilute sulfuric acid. This mixture acts as electrolyte.
-Cathode is the electrode where the cations are attracted. Reduction takes place at cathode. Anode is the electrode where the anions are attracted. Oxidation takes place at anode.
-In all the given compounds, iron gets oxidized from iron(II) to iron(III). The reaction can be shown as $F{{e}^{2+}}-1{{e}^{-}}\to F{{e}^{3+}}$ . Potassium permanganate has variable valencies. Here it has a valency of +7 and it can reduce to many valencies which decide the moles of the compound needed.
-1 mole of the compound needs $\dfrac{1}{5}$ moles of electrons to reduce potassium permanganate. This is because 5 electrons are needed for the process and the reaction can be shown as
     \[Mn{{O}_{4}}^{-}+8{{H}^{+}}+5{{e}^{-}}\to M{{n}^{2+}}+4{{H}_{2}}O\]
-Now looking at the options we see that the option A needs 3 moles of electrons for oxidation and so the moles of permanganate needed would be $\dfrac{3}{5}$ moles. Option B needs 5 moles for oxidation and so 1 mole of permanganate. The option C needs 1 mole for oxidation and so $\dfrac{1}{5}$ moles of permanganate. Option D needs 3 moles of electrons and so $\dfrac{3}{5}$ moles of permanganate.

Thus we see that $FeS{{O}_{4}}$ needs the least moles of permanganate and the correct option is C.

Note:
We should always write the separate reactions for the oxidation and the reduction process and the reaction should be balanced. Only then we can get the correct number of electrons as well as the correct number of the ions. If the reaction is not balanced, we may get different answers if the moles of the reducing and oxidizing agents are not same.