Question

when Sulphur dioxide is passed through an acidified K2Cr2O7 solution ,the oxidation state of sulphur changes from
1.) +4 to 0
2.) +4 to +2
3) +4 to +6
4) +6 to +4

Answer 1

Answer:

When Sulphur dioxide is passed through an acidified K2Cr2O7 solution ,the oxidation state of sulphur changes to (3) +4 to +6.

Explanation:

For occurance oxidation, the occurance of reduction is also important. (Cr2O4) -2 {valency is -2} is used as K2Cr2O7.

The half reaction which is also the oxidation reaction is performed.

(Cr2O4) -2+8H +6e-->2Cr+3 + 4H2O

S reduces I SO2. The half reaction which is also reduced reaction is performed.

SO2+2H2O-->SO4-2 +4H+ 2e

SO 2 becomes SO4-2

Let oxidation state of S be X then,

X+4(-2) =-2

X=6

Oxidation state of sulphur is +6.

Answer 2

Answer:

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