1.

When sulphur dioxide is passed in 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 :`K_(2)Cr_(2)O_(7)` acts as a STRONG OXIDISING agent . Itoxidises other compounds and GETS itself reduced . In `SO_(2)`, S is in +4 state so it gets oxidised to +6 state which is maximum group oxidation state.
`(K_(2)Cr_(2)O_(7)+4H_(2)SO_(4)toK_(2)SO_(4)+Cr_(2)(SO_(4))_(3)+7H_(2)O+3S[SO_(2)+[O]+H_(2)OtoH_(2)SO_(4)]xx3)/(K_(2)Cr_(2)+[O]+H_(2)SO_(4)+2SO_(2)toK_(2)SO_(4)+Cr_(2)(SO_(4))_(3)+H_(2)O)`


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