Question

When so2 is passed through potassium dichromate solution the oxidation number of s?

Answer 1

To be an oxidation there must be a reduction.

We look what happen to K2Cr2O7(potassium dichromate)

We can use (Cr2O4)²- as K2Cr2O7

This is the half reaction

(Cr2O4)²- + 8H+ 6e ➡ 2Cr³+ + 4(H20)

It is a oxidation reaction.

It means the S in SO2 becomes reduced

This is the half reaction

SO2 + 2(H20) ➡ (SO4)²- + 4H+ 2e

Here SO2 becomes (SO4)²-

If X is the oxidation number of S atom we can write as follow.

X + 4(-2) = -2

X - 8 = -2

X = +6

Oxidation state of sulphur becomes +6

Answer 2

(Cr2O4)²- + 8H+ 6e ➡ 2Cr³+ + 4(H20)

It is a oxidation reaction.

It means the S in SO2 becomes reduced

This is the half reaction

SO2 + 2(H20) ➡ (SO4)²- + 4H+ 2e

Here SO2 becomes (SO4)²-

If X is the oxidation number of S atom we can write as follow.

X + 4(-2) = -2

X - 8 = -2

X = +6

Oxidation state of sulphur becomes +6
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