Which one of the following statements is correct when $S O_{2}$ is passed through acidified $K_{2} C r_{2} O_{7}$ solution?

Green

C r_{2} (S O_{4})_{3}

is formed

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The solution turns blue

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The solution is decolourized

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S O_{2}

is reduced

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Solution

The correct option is A Green $C r_{2} (S O_{4})_{3}$ is formed
$K_{2} C r_{2} O_{7} \to C r_{2} (S O_{4})_{3}$ green solution is obtained where as $S O_{2}$ oxidises into sulphate $S O {_{4}}^{2 -}$
$(K_{2} C r_{2} O_{7} + 3 S O_{2} + H_{2} S O_{4} \to C r_{2} (S O_{4})_{3} + H_{2} O + K_{2} S O_{4})$

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Similar questions

S O_{2}

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

S O_{2}

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

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