What are the half reactions for the redox reaction between SO2 and iodine?

Question

What are the half reactions for the redox reaction between $\ce{SO2}$ and iodine? The relevant chemical equation: $$\ce{ SO2 + H2O + I2 -> H2SO4 + 2HI}$$

This is what I think it is, not sure if it's correct.

Oxidation: $\ce{SO2 + 2H2O -> SO4^2- + 4H+ + 2e-}$

Reduction: $\ce{I2 + 2e- -> 2I-}$

Answer 1

Your initial comment exchange made me realize that you have some difficulties in balancing redox equations. Thus, I decided to include some clues for your benefit.

An important part of writing half-reactions is making sure they're balanced by mass and charge. Since most redox reactions are done in aqueous medium, you can always balance $\ce{O}$ by $\ce{H2O}$. By doing so, you contribute extra $\ce{H}$ to the equation so you can balance them by $\ce{H+}$ since it is mostly an acid medium reactions (if they are in base or neutral conditions, you may cancel $\ce{H+}$ by adding the same amount of $\ce{OH-}$ to both sides of the reaction). Finally, cancel the net plus charge by adding $\ce{e-}$s to the appropriate side.

Let's see the easy one first, in your case reduction half reaction where $\ce{I2}$ reduce to $\ce{I-}$: $\ce{I2 -> I-}$. Balance its mass, which gives you: $\ce{I2 -> 2I-}$. Now, balance the negative charges by $\ce{e-}$s. So, you got balanced reduction half-reaction ($\ce{e-}$s are in LHS): $$\ce{I2 + 2e- -> 2I-} \qquad \mathrm{E^\circ = \pu{0.536 V}} \qquad \text{(1)}$$

Now, see the more difficult second equation, the oxidation half reaction where $\ce{SO2}$ oxidizes to $\ce{SO4^2-}$: $\ce{SO2 -> SO4^2-}$. Its $\ce{S}$ is already balanced, but $\ce{O}$ is not. So, balance it with $\ce{H2O}$, which gives you: $\ce{SO2 + 2H2O -> SO4^2-}$. Now, balance additional $\ce{H}$ by $\ce{H+}$, which gives you a mass-balanced equation: $\ce{SO2 + 2H2O -> SO4^2- + 4H+}$. Now, balance the negative charges by $\ce{e-}$s. So, you got balanced oxidation half-reaction ($\ce{e-}$s are in RHS): $$\ce{SO2 + 2H2O -> SO4^2- + 4H+ + 2e-} \qquad \mathrm{E^\circ = \pu{0.157 V}} \qquad \text{(2)}$$ If you add (1) and (2) together in order to cancel $\ce{e-}$s, you get the redox reaction you are looking for: $$\ce{SO2 + 2H2O + I2 -> SO4^2- + 4H+ + 2I-} \qquad \mathrm{E^\circ_{cell} = \pu{0.693 V}} \qquad \text{(3)}$$

The value of $\mathrm{E^\circ_{cell}}$ is positive means the reaction is spontaneous.

(Note: The value of $\mathrm{E^\circ_{\ce{SO2/SO4^2-}}}$ is from Ref.1)

Reference:

1. J. A. O’Brien, J. T. Hinkley, S. W. Donne, “Electrochemical Oxidation of Aqueous Sulfur Dioxide II. Comparative Studies on Platinum and Gold Electrodes,” J. Electrochem. Soc. 2012, 159(9), F585–F593 (DOI: 10.1149/2.060209jes).