Question

when SO2 is passed in acidified K2Cr2O7 solution oxidation. state of S changed from
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Answer 1

Answer:

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

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