Final answer:
For the given equation, the oxidation states in the reactants are +6 for Cr in dichromate, +3 for C in oxalate, and -2 for O in both. In the products, Cr has an oxidation state of +3, and C in carbon dioxide has +4, while the O in all compounds is -2.
Step-by-step explanation:
The assignment of oxidation states for the elements in the chemical equation involves determining the charge on an atom in a compound. The oxidation state is a hypothetical charge on an atom if all bonds were ionic. For the reactant dichromate ion (Cr₂O₇²⁻), chromium (Cr) typically has an oxidation state of +6 because the oxygen (O) has a consistent oxidation state of -2. In oxalate ion (C₂O₄²⁻), carbon (C) is typically assigned an oxidation state of +3 while O remains at -2. In the products, chromium ions (Cr³⁻) have an oxidation state of +3, and in carbon dioxide (CO₂), carbon is +4 and the oxygens remain at -2. Let's assign the requested oxidation states:
- Cr₂O₇²⁻(aq): Cr = +6, O = -2
- C₂O₄²⁻(aq): C = +3, O = -2
- Products: Cr = +3, C = +4, O = -2
It's important to note the consistency of oxygen at -2 across different compounds unless it's in peroxides or with the presence of fluorine. The overall equation must be balanced not only for atoms but also for the charges by ensuring the number of electrons gained equals the number of electrons lost, which illustrates the key principle of preserving mass and charge in a chemical reaction.