MHT CET · Chemistry · Redox Reactions
When \(\mathrm{SO}_{2}\) is passed through acidified \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{0}\), the process that takes place is
- A the solution turns blue
- B the solution is decolourised
- C \(\mathrm{SO}_{2}\) is reduced
- D green \(\mathrm{Cr}_{2}\left(\mathrm{SO}_{4}\right)_{3}\) is formed
Answer & Solution
Correct Answer
(D) green \(\mathrm{Cr}_{2}\left(\mathrm{SO}_{4}\right)_{3}\) is formed
Step-by-step Solution
Detailed explanation
When \(\mathrm{SO}_{2}\) is passed through acidified \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) solution. \(\mathrm{SO}_{2}\) gas turns acidified potassium dichromate solution from orange to green reduced chromium \(+4\) to \(+3\).
\(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}+2 \mathrm{H}_{2} \mathrm{SO}_{4}+3 \mathrm{SO}_{2} \rightarrow 2 \mathrm{Cr}_{2}\left(\mathrm{SO}_{4}\right)_{3}\) \(+~\mathrm{K}_{2} \mathrm{SO}_{4}+\mathrm{H}_{2} \mathrm{O}\)
\(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}+2 \mathrm{H}_{2} \mathrm{SO}_{4}+3 \mathrm{SO}_{2} \rightarrow 2 \mathrm{Cr}_{2}\left(\mathrm{SO}_{4}\right)_{3}\) \(+~\mathrm{K}_{2} \mathrm{SO}_{4}+\mathrm{H}_{2} \mathrm{O}\)
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