Question

$HBr$ and $HI$ reduce sulphuric acid. $HCl$ can reduce $KMn{O}_{4}$ and $HF$ can reduce:
a) ${H}_{2}S{O}_{4}$
b) $KMn{O}_{4}$
c) ${K}_{2}{Cr}_{2}{O}_{7}$
d) None of these

Answer 1

Hint: Compound X can reduce Y, this statement means that X loses electrons to Y and oxidizes itself and reduces Y. This means that X acts as a reducing agent for Y.

Complete step by step answer:
Reducing agents are the compounds that lose electrons to the fellows compounds and reduces them and itself gets oxidized in a reaction. In this process, the compound which gets oxidized, loses electrons in the reaction.

Now, let us look at the reactions of the given compounds one by one:
1. $HBr$ and $HI$ reduce sulphuric acid. It means that $HBr$ and $HI$ acts as a reducing agent. Let us see the reaction involved.

$2H\overset { -1 }{ Br } /H\overset { -1 }{ I } \quad +\quad H_{ 2 }\overset { +6 }{ S } { O }_{ 4 }\quad \longrightarrow \quad \overset { +4 }{ S } { O }_{ 2 }\quad +\quad 2{ H }_{ 2 }O\quad +\quad \overset { 0 }{ { Br }_{ 2 } } \quad /\quad \overset { 0 }{ { I }_{ 2 }}$

In this reaction $Br /I$ has an oxidation state of -1 and after the reaction they have an oxidation state of 0. This shows that they have lost the electrons and had themselves oxidized. Whereas, $S$ has an oxidation state of +6 and after the reaction it has an oxidation state of +4. This shows that it has gained electrons and therefore got reduced.

2. $HCl$ can reduce $KMn{O}_{4}$. It also means that $HCl$ acts as a reducing agent. Let us see the reaction involved.

$2K\overset { +7 }{ Mn } { O }_{ 4 }\quad +\quad 16H\overset { -1 }{ Cl } \quad \longrightarrow \quad 2KCl\quad +\quad 2\overset { +2 }{ Mn } { Cl }_{ 2 }\quad +\quad 8{ H }_{ 2 }O\quad +\quad 5\overset { 0 }{ { Cl }_{ 2 } }$

In this reaction $Cl$ has an oxidation state of -1 and after the reaction it has an oxidation state of 0. This shows that it has lost the electrons and had itself oxidized. Whereas, $Mn$ has an oxidation state of +7 and after the reaction it has an oxidation state of +2. This shows that it has gained electrons and therefore got reduced.

But in the case of $HF$, $F$ is the most electronegative element and due to its small size, it does not have a tendency to lose electrons any further. And therefore cannot act as a reducing agent and therefore, cannot reduce any of the above compounds.

Hence, option (d) None of the above is the correct answer.

Note: $HF$ is a good oxidizing agent. It has the highest standard reduction potential of ${F}_{2}$ to ${F}^{-}$. Note that $F$ prefers the oxidation state of -1 over the oxidation state of 0.