Final answer:
Oxidation occurs at the anode with Cr being oxidized from Cr(s) to Cr³+ (aq), losing 3 electrons. Reduction takes place at the cathode with Cu²+ (aq) being reduced to Cu(s), gaining 2 electrons. The overall reaction is balanced by ensuring that the electrons lost equals the electrons gained.
Step-by-step explanation:
Writing Oxidation and Reduction Half-Reactions
Oxidation and reduction are two key concepts in electrochemistry that describe the loss and gain of electrons, respectively. To analyze the changes taking place, we can separate the full reaction into two half-reactions. In the given example of chromium and copper reaction, 2 Cr(s) + 3 Cu²+ (aq) → 2 Cr³+ (aq) + 3 Cu(s), we need to determine which species are oxidized and reduced and then write the individual half-reactions.
The oxidation half-reaction for chromium is Cr(s) → Cr³+ (aq) + 3e⁻. On the other hand, the reduction half-reaction for copper is Cu²+ (aq) + 2e⁻ → Cu(s). The reaction at the anode is where oxidation occurs, and the reaction at the cathode is where reduction takes place.
In an electrochemical cell, electrons flow from the anode to the cathode. Therefore, for this reaction chromium, which loses electrons, is oxidized at the anode, and copper, which gains electrons, is reduced at the cathode. To balance the overall reaction, we need to ensure that the number of electrons lost equals the number gained, necessitating proper coefficients in front of the reactants and products.