For the reaction of thiosulfate anion with iodine, the thiosulfate anion acts as the reducing agent according to the oxidation half-reaction,

2S2O32-(aq) -> S4O62-(aq) + 2e-.

Which of the following reducing half-reactions is correct to give the overall reaction of thiosulfate anion with iodine?

a. I2(aq) + 2e- -> 2I-(aq)

b. I2(aq) + e- -> 2I-(aq)

c. I2(aq) + 2e- -> 2IO3-(aq)

Explanation:

1. Oxidation half-reaction (given)

2. Reduction half-reaction:

Reduction is the gain of electrons with the consequent reduction in the number of oxidation.

The starting reactant is iodine which is diatomic; thus, it is I₂.

Each iodine atom gains one electron; thus, in total 2 electrons are gained and each I atom in solution will become an I⁻ anion.

Half-reaction:

The overall reaction of thiosulfate anion with iodine is obtained when you add the two half-reactions:

        (the electrons are canceled)

it is to be noted that an alpha particle is an helium nucleus. so it contains 2 protons 2 neutrons and has a mass number of 4. when alpha decay particles are ejected from the nucleus it gets transformed into an atom which has atomic number less than 2 and a mass number less than 4.

an equation that could represent alpha decay is radium 226 decay of alpha particle forms radon 222.

→