MHT CET · Chemistry · p Block Elements (Group 15, 16, 17 & 18)
When \(\mathrm{SO}_{2}\) is passed through acidified \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) solution
- A reduction of \(\mathrm{SO}_{2}\) takes place
- B the solution turns orange
- C the solution turns blue
- D the solution turns green
Answer & Solution
Correct Answer
(D) the solution turns green
Step-by-step Solution
Detailed explanation
Action of acidified potassium dichromate on \(\mathrm{SO}_2\):
Sulphur dioxide gas is oxidized to sulphuric acid when passed through acidified potassium dichromate solution. The colour of the solution changes from orange to green because potassium dichromate is reduced to chromic sulphate.
\(\underset{\substack{\text { Potassium } \\ \text { dichromate }}}{ \text{K} _2 \text{Cr} _2 \text{O} _7}+\underset{\substack{\text { Sulphur } \\ \text { dioxide }}}{3 \text{SO} _2}+\underset{\substack{\text { Sulphuric } \\ \text { acid }}}{ \text{H} _2 \text{SO} _4} \longrightarrow \underset{\substack{\text { Potassium } \\ \text { sulphate }}}{ \text{K} _2 \text{SO} _4}\) \(+~\underset{\substack{\text { Chromic } \\ \text { sulphate }}}{ \text{Cr} _2\left( \text{SO} _4\right)_3}+\underset{\text {Water }}{ \text{H} _2 \text{O }}\)
Sulphur dioxide gas is oxidized to sulphuric acid when passed through acidified potassium dichromate solution. The colour of the solution changes from orange to green because potassium dichromate is reduced to chromic sulphate.
\(\underset{\substack{\text { Potassium } \\ \text { dichromate }}}{ \text{K} _2 \text{Cr} _2 \text{O} _7}+\underset{\substack{\text { Sulphur } \\ \text { dioxide }}}{3 \text{SO} _2}+\underset{\substack{\text { Sulphuric } \\ \text { acid }}}{ \text{H} _2 \text{SO} _4} \longrightarrow \underset{\substack{\text { Potassium } \\ \text { sulphate }}}{ \text{K} _2 \text{SO} _4}\) \(+~\underset{\substack{\text { Chromic } \\ \text { sulphate }}}{ \text{Cr} _2\left( \text{SO} _4\right)_3}+\underset{\text {Water }}{ \text{H} _2 \text{O }}\)
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