To determine the oxidation number of chromium (Cr) in the dichromate ion (Cr₂O₇²⁻), we first need to consider the overall charge of the ion and the oxidation states of the other elements involved. Oxygen usually has an oxidation state of -2. Since there are seven oxygen atoms, their total contribution to the charge is -14 (7 x -2).
Let’s denote the oxidation number of chromium as x. In the Cr₂O₇²⁻ ion, there are two chromium atoms, so we can set up the equation:
x + x + (-14) = -2
Solving for x gives us:
2x – 14 = -2
2x = 12
x = +6
Thus, the oxidation number of Cr in Cr₂O₇²⁻ is +6.
When Cr₂O₇²⁻ is converted to Cr³⁺, the oxidation number changes from +6 to +3. This change signifies a reduction because the oxidation number decreases. In redox reactions, oxidation refers to an increase in oxidation number, while reduction refers to a decrease.
Since Cr₂O₇²⁻ is being reduced in this reaction, it acts as an oxidizing agent, as it causes another substance to be oxidized while it itself is reduced.
In summary, the oxidation number of Cr in Cr₂O₇²⁻ is +6. The conversion to Cr³⁺ is a reduction process, and Cr₂O₇²⁻ acts as an oxidizing agent in this reaction.