Final answer:
The redox reaction between chromium and copper(II) in acidic solution is balanced by adjusting the half-reactions to make the electrons lost equal the electrons gained, combining them into the overall equation, and finally balancing the hydrogen with water and hydrogen ions.
Step-by-step explanation:
To complete and balance the equation for the redox reaction where chromium (Cr) is oxidized to chromate (CrO42−) and copper(II) (Cu2+) is reduced to copper (Cu), we need to balance each half-reaction and then combine them to make the overall balanced equation. The two half-reactions given in acidic solution are:
- Oxidation: Cr(s) → Cr3+ (aq) + 3e−
- Reduction: Cu2+ (aq) + 2e− → Cu(s)
First, to balance the number of electrons, we multiply the reduction half-reaction by 3 and the oxidation half-reaction by 2, obtaining:
- Oxidation: 2Cr(s) → 2Cr3+ (aq) + 6e−
- Reduction: 3Cu2+ (aq) + 6e− → 3Cu(s)
Next, we combine these balanced half-reactions to form the overall equation:
2Cr(s) + 3Cu2+ (aq) → 2Cr3+ (aq) + 3Cu(s)
Finally, in an acidic solution, we need to balance the hydrogens by adding 7H2O to the left side and 14H+ to the right side, resulting in:
2Cr(s) + 3Cu2+ (aq) + 7H2O(l) → 2CrO42− (aq) + 3Cu(s) + 14H+ (aq)