Why does bromine react with iron(II) but not iodine?

My question comes from the answer (C) to this question:

Firstly, (correct me if I'm wrong please) I thought that the answer could either be C or D since the other answers are negative ions which can't react with iodine or bromine since they (as diatomic molecules) have a full shell of outer electrons. Now, from looking at the question, I knew it had something to do with oxidation and reduction as well as displacement reactions, however, as iodine and bromine aren't on the reactivity series, we can't determine which would react with which metal. From there, I just had no more ideas. Could anyone provide me with a hint or a guide as to how to do this ?

All four of the ions are reducing agents. But usually halogen gases' tendency to undergo reduction to become halide ions decreases down the group (along with their reactivity). Out of the $4$ ions, only $\ce{Fe^{2+}}$ has a mild enough reducing power.

$$\ce{2Fe^{2+} +Br2 -> 2Br^- +2Fe^{3+}}$$

But $\ce{Fe^{2+}}$ is not strong enough to reduce $\ce{I2}$ as it is relatively stable.

In fact, $\ce{SO3^{2-}}$ has the strongest reducing power among the four options, as it can reduce both $\ce{I2}$ and $\ce{Br2}$.

$\ce{OH^{-}}$ is a mildly strong reducing agent, it can react with both $\ce{I2}$ and $\ce{Br2}$, and its typical reaction is

$$\ce{4OH^- -> 2H2O +O2 + 4e^-}$$

$\ce{Mn^{2+}}$ is the least reducing, because it is the product of a strong oxidizing agent that has been reduced:

$$\ce{MnO4^{-} + 8H+ +5e^- -> Mn^{2+} +4H2O}$$

Here, permanganate ion $\left(\ce{MnO4^-}\right)$ is a very strong oxidizing agent; the product thus would be weakly reducing.

Actually, a similar idea can be seen with $\ce{SO3^{2-}}$ (sulphite) ions. Since it is a strong reducing agent, we can expect the product formed after it is oxidized is a weak oxidizing agent.

$$\ce{SO3^{2-} + H2O -> SO4^{2-} + 2H+ + 2e^-}$$

Indeed, sulphate ions alone are weak oxidizers, and stable.

I suggest you refer to E.C.S. (electrochemical series) to know whether a reaction would occur. It is easy to find online; e.g., at Wikipedia.

• Nice answer! Thanks for wandering over here from Math.SE. :-) Apr 2, 2016 at 12:30
• @Brian Thanks for the warm welcome! I had some free time so I decided to come over see what I can learn here :)
– lEm
Apr 2, 2016 at 13:05