Under identical conditions, which halide ion is the strongest reducing agent?

• A. Iodide
• B. Chloride
• C. Fluoride
• D. Bromide

Answer: (A)

The reducing power of halide ions comes from how readily they are oxidized to the corresponding halogen. In standard redox terms, a halide ion X− will reduce something by giving up electrons to form X2, so the easier this oxidation is, the stronger the reducing agent it is under identical conditions. This is quantified by the X2/X− reduction potential: the lower this potential, the easier the oxidation and the stronger the reducing agent.

For the halides, the approximate reduction potentials are:

I2 + 2e− → 2 I− around +0.54 V

Br2 + 2e− → 2 Br− around +1.07 V

Cl2 + 2e− → 2 Cl− around +1.36 V

F2 + 2e− → 2 F− around +2.87 V

Because iodide has the smallest reduction potential, it is the easiest to oxidize to I2, making iodide the strongest reducing agent under the same conditions. The order of reducing power follows the down-the-group trend: iodide > bromide > chloride > fluoride.