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


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)


  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).

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