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)

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the political sponsor of weights and measures reform in the french revolutionary national assembly was the bishop of autun, better known as talleyrand. under his auspices, the french academy appointed several committees to carry out the work of developing a usable system of weights and measures for france. one of the committees recommended a decimalized measurement system based upon a length equal to one ten-millionth of the length of a quadrant of the earth’s meridian (i.e., one ten-millionth of the distance between the equator and the north pole).

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