Question

Study this chemical reaction: 2Cu+Cl2 → 2 CuCl Then, write balanced half-reactions describing the oxidation and reduction that happen in this reaction oxidation: reduction

Answer 1

For the chemical reaction 2Cr(s) + 3Cu²⁺ (aq) → 2Cr³⁺ (aq) + 3Cu(s), the oxidation half-reaction at the anode is Cr(s) → Cr³⁺ (aq) + 3e⁻, and the reduction half-reaction at the cathode is Cu²⁺ (aq) + 2e⁻ → Cu(s). The overall reaction does not change as it is already balanced.

Step-by-step explanation:

Writing Balanced Half-Reactions

To write balanced half-reactions for the given chemical reaction 2Cr(s) + 3Cu²⁺ (aq) → 2Cr³⁺ (aq) + 3Cu(s), we need to separate the two processes of oxidation and reduction that occur at the anode and the cathode, respectively.

Oxidation (anode): The loss of electrons. For chromium:
Cr(s) → Cr³⁺ (aq) + 3e⁻

Reduction (cathode): The gain of electrons. For copper:
Cu²⁺ (aq) + 2e⁻ → Cu(s)

Balance the half-reactions for electrons lost and gained. Then combine the balanced half-reactions, ensuring the number of electrons lost in oxidation is equal to those gained in reduction.

Adding the half-reaction equations and simplifying yields the overall reaction without any change as the reaction is already balanced:

2Cr(s) + 3Cu²⁺ (aq) → 2Cr³⁺ (aq) + 3Cu(s)