Answer:
Option D.
Explanation:
Let's balance the hole reaction by the ion electron method.
Al (s) + Cr(NO₃)₂ (aq) – Cr (s) + Al(NO₃)₃ (aq)
We know that oxidation state at ground state is 0.
We determine the half reactions:
Al ⇆ Al³⁺ + 3e⁻
Aluminum increase the oxidation state. This is the oxidation.
Cr²⁺ + 2e⁻ ⇄ Cr
Chromium decrease the oxidation state. It changed from +2 to 0.
This is the reduction. In order to balance the half reactions, we need to multiply by 2 and by 3. In that way, we can cancel the electrons:
(Al ⇆ Al³⁺ + 3e⁻) . 2
(Cr²⁺ + 2e⁻ ⇄ Cr) . 3
We sum the half reactions:
2 Al + 3 Cr²⁺ + 6e⁻ ⇄ 2Al³⁺ + 6e⁻ + 3Cr
We cancel the electrons, so now the balanced reaction is:
2Al + 3Cr(NO₃)₂ → 3Cr + 2Al(NO₃)₃
Final answer: 6 electrons are transferred in the redox reaction.
