Final answer:
The net redox reaction when Cu reacts with acidic K₂Cr₂O₇ is the oxidation of chromium (anode) and reduction of copper (cathode), balancing to 2Cr(s) + 3Cu²⁺ (aq) → 2Cr³⁺ (aq) + 3Cu(s).
Step-by-step explanation:
When Cu comes into contact with acidic K₂Cr₂O₇, a redox reaction takes place. The reaction involves the oxidation of chromium (Cr) and the reduction of copper (Cu). In the half-reaction at the anode (oxidation), chromium goes from being a solid metal to a trivalent ion in solution:
Cr(s) → Cr³⁺ (aq) + 3e⁻
In the half-reaction at the cathode (reduction), copper ion in solution is reduced to solid copper metal:
Cu²⁺ (aq) + 2e⁻ → Cu(s)
To balance the electrons between the oxidation and the reduction half-reactions, we need to multiply the chromium reaction by 2 and the copper reaction by 3, yielding:
2Cr(s) + 3Cu²⁺ (aq) → 2Cr³⁺ (aq) + 3Cu(s)
This balanced equation represents the overall redox reaction.