Question of 96
Q.How does SO2 react with the acidic solution of the following ?
(i) KMnO4
(ii) K2Cr2O7
Bihar BsebBihar Board Intermediate 2022Subjective· 5mImportance★★★★★
0% · 0/96 Questions
You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.
Start your 14-day free trial to unlock the full solution →SO2 is a reducing agent: it decolourises acidic KMnO4 (Mn goes +7 → +2) and turns acidic K2Cr2O7 orange→green (Cr goes +6 → +3), while SO2 itself is oxidised from S(+4) to sulphate S(+6).
Sulphur dioxide (SO2) contains sulphur in the +4 oxidation state, which can go up to +6 (as sulphate/sulphuric acid). Hence SO2 readily acts as a reducing agent and reduces strong oxidising agents.
(i) Reaction with acidic KMnO4:
- Acidified potassium permanganate is purple (Mn in +7 state). SO2 reduces Mn(+7) to Mn(+2), so the purple colour is discharged (decolourised).
- Balanced equation: 2KMnO4 + 5SO2 + 2H2O → K2SO4 + 2MnSO4 + 2H2SO4
- Observation: the pink/purple colour of KMnO4 disappears — a test that shows SO2 is a reducing gas.
(ii) Reaction with acidic K2Cr2O7:
- Acidified potassium dichromate is orange (Cr in +6 state). SO2 reduces Cr(+6) to Cr(+3), giving green chromium(III) sulphate.
- Balanced equation: K2Cr2O7 + 3SO2 + H2SO4 → K2SO4 + Cr2(SO4)3 + H2O …
Unlock everything free for 14 days
- Full step-by-step solutions
- Concept-first explanations
- Methods, shortcuts & mistakes
- PYQ mapping + timed mock tests
Full access for 14 days. No credit card required.