(D) When $SO_{2}$ is passed through an acidified $K_{2}Cr_{2}O_{7}$ solution,the orange dichromate ion $(Cr_{2}O_{7}^{2-})$ is reduced to the green ch

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green $Cr_{2}(SO_{4})_{3}$ is formed

(D) When $SO_{2}$ is passed through an acidified $K_{2}Cr_{2}O_{7}$ solution,the orange dichromate ion $(Cr_{2}O_{7}^{2-})$ is reduced to the green chromium$(III)$ ion $(Cr^{3+})$.The balanced chemical equation is:$K_{2}Cr_{2}O_{7} + H_{2}SO_{4} + 3SO_{2} \rightarrow K_{2}SO_{4} + Cr_{2}(SO_{4})_{3} + H_{2}O$As a result,the solution turns green due to the formation of $Cr_{2}(SO_{4})_{3}$.

Class 11 Chemistry Redox Reactions Redox reaction and Method for balancing Redox reaction

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When $SO_{2}$ is passed through acidified $K_{2}Cr_{2}O_{7}$,the process that takes place is

MHT CET 2020 All MHT CET

A
the solution turns blue
B
the solution is decolourised
C
$SO_{2}$ is reduced
D
green $Cr_{2}(SO_{4})_{3}$ is formed

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Redox Reactions

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Redox reaction and Method for balancing Redox reaction

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When $SO_{2}$ is passed through acidified $K_{2}Cr_{2}O_{7}$,the process that takes place is

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