Question

Sodium thiosulfate, Na 2S 2O 3, is used as a "fixer" in black and white photography. Identify the reducing agent in the reaction of thiosulfate with iodine. 2S 2O 3 2–( aq) + I 2( aq) → S 4O 6 2–( aq) + 2I –( aq)

Answer 1

Answer: Reducing agent in the given reaction is $S_{2}O^{2-}_{3}$.

Explanation:

A reducing agent is defined as an element which tends to lose electrons to other element leading to an increase in its oxidation number.

In the given reaction, oxidation state of sulfur in $S_{2}O^{2-}_{3}$ is +2 and $I_{2}(aq)$ has 0 oxidation state.

In $S_{4}O^{2-}_{6}(aq)$ oxidation state of S is 2.5 and in $2I^{-}(aq)$ oxidation state of I is -1.

Since, an increase in oxidation state of S is occurring from +2 to +2.5. Hence, it is acting as a reducing agent.

Thus, we can conclude that reducing agent in the given reaction is $S_{2}O^{2-}_{3}$.