Question

When $S{O_2}$ is passed in acidified potassium dichromate solution, the oxidation number of S is changed from:
A) + 4 to zero
B) + 4 to + 2
C) + 4 to + 6
D) + 6 to + 4

Answer 1

Hint: Oxidation involves loss of electrons whereas reduction involves gain of electrons. Any of these two reactions cannot take place independently. So, they are half reactions. Collectively such reactions are called redox reactions.

Complete step by step answer:
Potassium dichromate is a strong oxidising agent which oxidised in acidic, basic as well as in neutral medium. Potassium dichromate in acidic medium acts as strong oxidizing agent. It oxidises sulphur dioxide to sulphuric acid in an acidic medium.
The reaction of potassium dichromate and sulphur dioxide in acidic medium can be written as,
Reduction half reaction:
${K_2}C{r_2}{O_7} + 4{H_2}S{O_4}\xrightarrow{{}}{K_2}S{O_4} + C{r_2}{\left( {S{O_4}} \right)_3} + 4{H_2}O + 3\left[ O \right]$
Oxidation half reaction:
\[\left[ {S{O_2} + {H_2}O + \left[ O \right]\xrightarrow{{}}{H_2}S{O_4}} \right] \times 3\]
Overall reaction:
${K_2}C{r_2}{O_7} + {H_2}S{O_4} + 3S{O_2}\xrightarrow{{}}{K_2}S{O_4} + C{r_2}{\left( {S{O_4}} \right)_3} + 1{H_2}O$
 In the form of charge transfer the reaction can be written as,
\[\mathop {C{r_2}}\limits^{ + 7} \mathop {{O_7}}\limits^{2 - } + 3{\mathop {S}\limits^{ + 4} O_2} + 2{H^ + }\xrightarrow{{}}2C{r^{3 + }} + 3\mathop S\limits^{ + 6} O_4^{2 - } + {H_2}O\]
Now change in the oxidation state of sulphur from $S{O_2}$ to $SO_4^{2 - }$ is \[\mathop S\limits^{ + 4}{O_2} \to\mathop S\limits^{ + 6} O_4^{2 - }\]
Change in oxidation state of sulphur = 6 - 4 = 2
So, oxidation state of sulphur in $S{O_2}$ is $ + 4$ which is changed to $ + 6$ in $SO_4^{2 - }$

Hence, the correct option is (C).

Note: Potassium dichromate is an oxidising agent which is used in various oxidation reactions as well as in volumetric analysis. Sulphur dioxide is a reducing agent and is used in various compound formations like sulphuric acid. It is also used as a bleaching agent but its bleaching action is temporary.
Redox reactions are very important as they are also used in electrochemistry. These reactions involve loss and gain of electrons so these reactions are used for various electrochemical processes.