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 is decolourised
B.
$$S{O_2}$$ is reduced
C.
Green $$C{r_2}{\left( {S{O_4}} \right)_3}$$ is formed
D.
The solution turns blue
Answer :
Green $$C{r_2}{\left( {S{O_4}} \right)_3}$$ is formed
Solution :
When $$S{O_2}$$ is passed through acidified $${K_2}C{r_2}{O_7}$$ solution, green chromium sulphate is formed. In this
reaction, oxidation state of $$Cr$$ changes from + 6 to + 3.
$$\mathop {{K_2}C{r_2}{O_7}}\limits_{OS\,\,{\text{of}}\,\,Cr = + 6} + {H_2}S{O_4} + 3S{O_2} \to $$ \[{{K}_{2}}S{{O}_{4}}+\underset{\begin{smallmatrix}
OS\,\,\text{of}\,Cr=+3 \\
\text{Green}
\end{smallmatrix}}{\mathop{C{{r}_{2}}{{\left( S{{O}_{4}} \right)}_{3}}}}\,+{{H}_{2}}O\]
The appearance of green colour is due to the reduction of chromium metal.