Oxidation of Sulphides by Acidified Potassium Dichromate

Acidified potassium dichromate ($K_2Cr_2O_7$/$H^+$) is a powerful oxidizing agent used in various chemical reactions. When it reacts with sulphides, specifically the sulfide ion ($S^{2-}$), it causes the oxidation of the sulfur atom.

Understanding the Reaction

The sulfide ion ($S^{2-}$) contains sulfur in its lowest common oxidation state, which is -2. Oxidizing agents like acidified potassium dichromate tend to increase this oxidation state.

In this specific reaction, the sulfide ion ($S^{2-}$) is oxidized to elemental sulphur (S). In elemental sulfur, the oxidation state of sulfur is 0.

The overall reaction can be represented as:

$3S^{2-} + Cr_2O_7^{2-} + 8H^+ \rightarrow 3S + 2Cr^{3+} + 7H_2O$

In this redox reaction:

• Sulfide ($S^{2-}$) is oxidized (oxidation state changes from -2 to 0).
• Dichromate ($Cr_2O_7^{2-}$) is reduced (the chromium changes oxidation state from +6 to +3).

Analysis of Options

Let's look at why elemental sulphur (S) is the product:

• Sulphur (S): This corresponds to the oxidation state 0. This is a common outcome when sulfides are oxidized, especially under controlled conditions.
• $SO_4^{2-}$ (Sulfate): Here, sulfur has an oxidation state of +6. While possible with very strong oxidizing conditions, it's not the typical primary product in many sulfide oxidations by dichromate.
• $SO_3^{2-}$ (Sulfite): Here, sulfur has an oxidation state of +4. This represents a significant oxidation from -2.
• $SO_2$ (Sulfur Dioxide): This also represents sulfur in the +4 oxidation state.

While further oxidation to sulfite or sulfate is possible depending on the reaction conditions (concentration, temperature, reaction time), the oxidation of sulphides ($S^{2-}$) to elemental sulphur (S) is a well-documented reaction pathway with acidified potassium dichromate.