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SO(2) reduces acidified K(2)Cr(2)O(7) to...

`SO_(2)` reduces acidified `K_(2)Cr_(2)O_(7)` to

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Acidified K_(2)Cr_(2)O_(7) turns green by

STATEMENT - 1 : SO_(2) turns acidified K_(2)Cr_(2)O_(7) green. And STATEMENT - 2 : SO_(2) converts Cr_(2)O_(7)^(2-) ion to Cr^(+3) which gives green colour.

STATEMENT - 1 : SO_(2) turns acidified K_(2)Cr_(2)O_(7) green. And STATEMENT - 2 : SO_(2) converts Cr_(2)O_(7)^(2-) ion to Cr^(+3) which gives green colour.

When SO_(2) is passed in acidified K_(2)Cr_(2)O_(7) solution oxidation state of S changed from

When SO_(2) is passed into acidified K_(2)Cr_(2)O_(7) solution, oxidation state of sulphur changes from

If(X) turns acidified K_(2)Cr_(2)O_(7) solution green , then X may be