Question

$S{{O}_{2}}$ acts as both oxidant and reductant. Explain

Answer 1

Oxidant is defined as those, whose oxidation number decreases. The reductant is defined as those, whose oxidation number increases. The reducing nature of sulfur dioxide is due to the evolution of nascent hydrogen. The oxidising nature appears when it reacts with strong oxidising agents.

Complete step by step answer:
Sulfur is the element of group 16 and forms a dioxide. It is called sulfur dioxide and has a formula $S{{O}_{2}}$.
The sulfur dioxide acts as a reducing agent as well as an oxidizing agent.
First, let us study the reducing property of $S{{O}_{2}}$(sulfur dioxide).
In the presence of moisture, $S{{O}_{2}}$acts as a good reducing agent. Its reducing character is due to the evolution of nascent hydrogen. The reaction is given below:
$S{{O}_{2}}+{{H}_{2}}O\to {{H}_{2}}S{{O}_{4}}+2[H]$

Some important reducing properties of $S{{O}_{2}}$are discussed below:
(i)- $S{{O}_{2}}$reduces halogens to halogen acids. When $S{{O}_{2}}$reacts with $C{{l}_{2}}$ it forms hydrochloric acid.
$C{{l}_{2}}+S{{O}_{2}}+2{{H}_{2}}O\to 2HCl+{{H}_{2}}S{{O}_{4}}$

(ii)- It decolorizes the pink violet color of acidified $KMn{{O}_{4}}$ solution. When $S{{O}_{2}}$reacts with pink violet $KMn{{O}_{4}}$it forms colorless $MnS{{O}_{4}}$
$2KMn{{O}_{4}}(pink\text{ }violet)+5S{{O}_{2}}+2{{H}_{2}}O\to {{K}_{2}}S{{O}_{4}}+2MnS{{O}_{4}}(colorless)+2{{H}_{2}}S{{O}_{4}}$

(iii)- It turns orange colored potassium dichromate solution green. When $S{{O}_{2}}$reacts with orange ${{K}_{2}}C{{r}_{2}}{{O}_{7}}$ it forms green $C{{r}_{2}}{{(S{{O}_{4}})}_{3}}$
${{K}_{2}}C{{r}_{2}}{{O}_{7}}(orange)+{{H}_{2}}S{{O}_{4}}+3S{{O}_{2}}\to {{K}_{2}}S{{O}_{4}}+C{{r}_{2}}{{(S{{O}_{4}})}_{3}}(green)+{{H}_{2}}O$

(iv)- It reduces ferric to ferrous salts. When $S{{O}_{2}}$reacts with $F{{e}_{2}}{{(S{{O}_{4}})}_{3}}$ it forms $FeS{{O}_{4}}$.
$S{{O}_{2}}+F{{e}_{2}}{{(S{{O}_{4}})}_{3}}+2{{H}_{2}}O\to 2FeS{{O}_{4}}+2{{H}_{2}}S{{O}_{4}}$

(v)- It reduces acidified potassium iodate to iodine. When $S{{O}_{2}}$reacts with $KI{{O}_{3}}$it form ${{I}_{2}}$
$2KI{{O}_{3}}+4S{{O}_{2}}+4{{H}_{2}}O\to 4KHS{{O}_{4}}+3{{H}_{2}}S{{O}_{4}}+{{I}_{2}}$

Now let us study the oxidizing property of $S{{O}_{2}}$. It acts as a mild oxidizing agent particularly when it reacts with strong reducing agents.
Some examples are given below:
(i)- It oxidizes hydrogen sulfide to sulfur.
$2{{H}_{2}}S+S{{O}_{2}}\to 2{{H}_{2}}O+3S$

(ii)- It oxidizes active metals like magnesium, potassium, and iron. The reactions are given below:
For magnesium- $3Mg+S{{O}_{2}}\to 2MgO+MgS$
For potassium- $4K+3S{{O}_{2}}\to {{K}_{2}}S{{O}_{3}}+{{K}_{2}}{{S}_{2}}{{O}_{3}}$
For iron- $3Fe+S{{O}_{2}}\to 2FeO+FeS$

(iii)- It oxidises carbon from carbon monoxide to carbon dioxide.
$2CO+S{{O}_{2}}\to 2C{{O}_{2}}+S$

Note: Besides being a reducing as well as an oxidizing agent $S{{O}_{2}}$is used in the refining of petroleum and bleaching of sugarcane juice. It is also used for a bleaching agent for delicate articles like wool and silk.