Chemistry · Redox Equilibrium · NEET
A reducing agent gives away electrons and gets oxidised itself. I⁻ has one extra electron compared to a neutral iodine atom. When I⁻ donates that electron to something else, it reduces that other species and turns back into I₂. So calling iodide a strong reducing agent just means it loses its electron very easily.
Iodine is the largest halogen atom. Its outermost electron sits far from the nucleus and is shielded by many inner shells, so the pull on it is weak. Fluorine is very small, so its electron is held tightly and F⁻ does not want to give it away. Weak grip = easy electron loss = strong reducing power, which is why I⁻ beats Br⁻, Cl⁻ and F⁻.
Reducing power increases down the group: F⁻ < Cl⁻ < Br⁻ < I⁻. So I⁻ is the strongest and F⁻ is the weakest reducing halide. This is the exact opposite of the oxidising power order of the halogens (F₂ > Cl₂ > Br₂ > I₂).
The X₂/X⁻ standard electrode potential drops as you go down the group. I₂ has the lowest (least positive) reduction potential, which means the reverse reaction (I⁻ → I₂ + e⁻) happens most easily. A low reduction potential for the halogen means a high reducing power for its halide ion.
Do not mix up the halogen and the halide ion. Fluorine (F₂) is the strongest oxidising halogen, but fluoride (F⁻) is the weakest reducing halide. F⁻ is tiny and holds its electron so tightly that it almost never donates it. Being a strong oxidiser as a halogen and being a weak reducer as its ion go together.
Try the real previous-year questions from this chapter — each with the answer and a full solution.
Iodide, I⁻. It is the largest halide ion, so its extra electron is loosely held and easily donated, making it the strongest reducing agent among the halides.
I⁻ > Br⁻ > Cl⁻ > F⁻. Reducing power increases down the group as atomic size increases and the outer electron is held less tightly.
Fluoride is very small with a high nuclear pull on its outer electrons. It holds its electron tightly and does not donate it easily, so it is the weakest reducing agent among the halides.
No, they are opposite. Oxidising power of halogens is F₂ > Cl₂ > Br₂ > I₂, while reducing power of halides is I⁻ > Br⁻ > Cl⁻ > F⁻.
Yes. I⁻ is a good enough reducing agent to reduce Fe³⁺ to Fe²⁺ while itself getting oxidised to I₂. This reaction is used in iodometric titrations and shows iodide's strong reducing nature.
