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Potassium Dichromate

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Question

What happens when $S O_{2}$ is passed through acidified $K_{2} C r_{2} O_{7}$ solution?

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Solution

When ${S O}_{2}$ is passed through acidified $K_{2} {C r}_{2} O_{7}$ solution. ${S O}_{2}$ gas turns acidified potassium dichromate solution from orange to green reduced chromium $+ 4$ to $+ 3$ .
$K_{2} {C r}_{2} O_{7} + 2 H_{2} {S O}_{4} + 3 {S O}_{2} \to 2 {C r}_{2} {({S O}_{4})}_{3} + K_{2} {S O}_{4} + H_{2} O$

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S O_{2}

is passed through acidified

K_{2} C r_{2} O_{7}

S O_{2}

is passed through acidified solution of

H_{2} S

:

{S O}_{2}

is passed through acidified

K_{2} {C r}_{2} O_{7}

Q. When SO₂ is passed through acidified K₂Cr₂O₇ solution.
(a) The solution turns green
(b) The solution is decolourized
(c) Reduction of SO₂ takes place
(d) Green ppt of Cr₂(SO₄)₃ is formed

S O_{2}

is passed in acidified

K_{2} C r_{2} O_{7}

solution oxidation state of

S

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