Question

Which one of the following statements is correct when $S{O_2}$ is passed through
acidified ${K_2}C{r_2}{O_7}$ solution?
A. The solution turns blue
B. The solution is decolourized
C. $S{O_2}$ is reduced
D. Green $C{r_2}{(S{O_4})_3}$ is formed

Answer 1

Hint: When this reaction takes place, the oxidation state of $Cr$ reduces from +6 to +3. Also, the orange colour changes and chromium sulphate is formed.

Step by step answer:Before moving on to the reaction between $S{O_2}$ and ${K_2}C{r_2}{O_7}$ we need to have information about the two compounds.
${K_2}C{r_2}{O_7}$ is also called Potassium dichromate and is a common inorganic chemical reagent and is most commonly used as an oxidizing agent in various laboratory and industrial applications. It is acutely and chronically harmful to health. It is a crystalline ionic solid with a very bright and red-orange colour. This salt is popular in the laboratory because it is not deliquescent (tendency to become liquid), in contrast to the more industrially relevant salt sodium dichromate.
$S{O_2}$ is also called sulphur dioxide is the chemical compound with the formula $S{O_2}$ It is a colourless gas or liquid with a strong, choking odour. It is a toxic gas responsible for the smell of burnt matches. It is released naturally by volcanic activity and is produced as a by-product of copper extraction and the burning of fossil fuels contaminated with sulphur compounds. The sulphur atom has an oxidation state of +4 and a formal charge of +1. Also, some dried fruits are preserved using $S{O_2}$ to prevent discoloration of the fruit.
When $S{O_2}$ is passed through the acidified ${K_2}C{r_2}{O_7}$ solution, the orange colour of potassium dichromate solutions turns to clear green due to the formation of chromium sulphate. In this reaction, the oxidation state of $Cr$ changes from +6 to +3. The appearance of green colour is due to the reduction of chromium metal.
\[{K_2}C{r_2}{O_7} + {H_2}S{O_4} + 3S{O_2} \to {K_2}S{O_4} + C{r_2}{(S{O_4})_3} + {H_2}O\]

Therefore, the correct statement is option D.

Note: When sulphur dioxide reacts with Potassium Dichromate, the colour of ${K_2}C{r_2}{O_7}$ changes from orange to green and $C{r_2}{(S{O_4})_3}$ is formed and oxidation state of $Cr$ changes from +6 to +3.