1.

When sulphur dioxide is passed through an acidified K_(2)Cr_(2)O_(7) solution, the oxidation state of sulphur is changed from

Answer»

`+4` to `+6`
`+6` to `+4`
`+4` to 0
`+4` to `+2`

SOLUTION :`ul(K_(2) Cr_(2)O_(7) +4H(2) SO_(4)rarr K_(2)SO_(4) + Cr_(2)(SO_(4))_(3) + 7H_(2)O+ 3(O))`
`SO_(2) + (O) + H_(2) O rarr H_(2) SO_(4) ] xx3`
`K_(2)Cr_(2)O_(7) + H_(2)SO_(4) + 3SO_(2) rarr K_(2)SO_(4) + Cr_(2)(SO_(4))_(3) + H_(2)O`
In `SO_(2)` , OX. state of `S= + 4 `. In `SO_(4)^(2-)`,ox.state of `S= +6`


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