How many total electrons are transferred in the redox reaction? Al (s) + Cr(NO3)2 (aq) – Cr (s) + Al(NO3)3 (aq)

A. 2
B. 3
C. 5
D. 6

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 6 electrons are transferred in the redox reaction.

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