Titrating Iodine with SO2

To be honest I'm not really sure what is going on in this reaction.

It says the SO2 is oxidised by 1 mole of I2. Thus, I2 should be reduced and hence gain electrons from SO2 :confused:

Thus, for an ion of iodine to be formed it needs to receive one electron from SO2 but since its 1 mole of a molecule of Iodine 2 electrons will be donated by SO2 :confused:

Is my thinking correct ?

Original post by Ari Ben Canaan

To be honest I'm not really sure what is going on in this reaction.

It says the SO2 is oxidised by 1 mole of I2. Thus, I2 should be reduced and hence gain electrons from SO2 :confused:

Thus, for an ion of iodine to be formed it needs to receive one electron from SO2 but since its 1 mole of a molecule of Iodine 2 electrons will be donated by SO2 :confused:

Is my thinking correct ?

Pretty much:

half equation 1: I₂ + 2e --> 2I^-

so the iodine must absorb 2 electrons, hence the SO₂ must provide these 2 electrons:

half equation 2: SO₂ + 2H₂O --> SO₄^2- + 4H⁺ + 2e

Sulphur changes its oxidation state from IV to VI

I₂ + SO₂ + 2H₂O --> 2I^- + SO₄^2- + 4H⁺

(edited 13 years ago)

Reply 2

ahmed123453

how did you figure out the half equation 0f SO2 ?

Reply 3

username3249896

Original post by ahmed123453

how did you figure out the half equation 0f SO2 ?

SO₂ has oxidation state +4

Iodine in I₂ has oxidation state 0
I^-has oxidation state -1

Reduction half eqn: I₂ + 2e^- -> 2I^-
we are given 1 mole of I₂ reacts with 1 mole of SO₂.

Since 1 mole of I₂ accepts 2 moles of electrons, SO₂ must donate 2 electrons. The oxidation state must increase by 2. So oxidation state of sulfur increases from +4 to +6. (Ans D)

SO₂ is oxidised to SO₄^2-
SO₂ + 2H₂O - > SO₄^2- + 2e^- +4H⁺

Reply 4