# The halide ions as reducing agents

So I'm currently an A-level student, and we learnt at some point last year that iodide ions were (at least compared to the halides above it) strong reducing agents - strong enough such that if sodium iodide reacted with concentrated sulphuric acid, it would be able to lower the oxidation state of the sulphur atom all the way from +6 to -2 (hydrogen sulfide).

One of the possible redox reactions goes as follows:

$$\ce{2I–(aq) -> I2(g) + 2 e–}$$

$$\ce{SO4^{2–}(aq) + 4 H+(aq) + 2 e– -> SO2(g) + 2 H2O}$$

Overall: $$\ce{2I–(aq) + SO4^{2–}(aq) + 4 H+(aq) -> I2(g) + SO2(g) + 2 H2O}$$

However, my problem is as follows: When I looked up the redox potentials of the two half equations above, the redox potential of the I2/I- equation was actually higher (more positive) than the redox potential of the $\ce{SO4^{2–}|SO2}$ equation.

By my understanding, that should mean that the reaction is unfeasible : the EMF of the proposed reaction (calculated by subtracting the potential at the anode from the cathode) is negative.

So why does this reaction occur?

• There's standard potential and there are actual potentials depending on situation. Why do you think conc. H2SO4 oxidises things diluted does not? – Mithoron May 22 '18 at 19:08
• So if H2SO4 is concentrated, the redox potential of this half equation would become more positive - such that it would exceed the potential of the I2/I-? – HunnyBunch May 22 '18 at 20:13