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p-Block Elements – II — Chemistry STD 12 — Question

Tamilnadu BoardEnglish MediumSTD 12Chemistryp-Block Elements – II3 Marks
Question
Give and explain reducing behaviour of sulphur dioxide.
✓

Answer

As Sulphur dioxide can readily be oxidised, it acts as a reducing agent. It reduces chlorine into hydrochloric acid.
$
SO _2+2 H _2 O + Cl _2 \rightarrow H _2 SO _4+2 HCl
$
It also reduces potassium permanganate and dichromate to $Mn ^{2+}$ and $Cr ^{3+}$ respectively.
- $2 KMnO _4+5 SO _2+2 H _2 O \rightarrow K _2 SO _4+2 MnSO _4+2 H _2 SO _4$
- $K _2 Cr _2 O _7+3 SO _2+ H _2 SO _4 \rightarrow K _2 SO _4+ Cr _2\left( SO _4\right)_3+ H _2 O$

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