Complete the following questions based on this reaction:

MnO4 - (aq) + Al (s) ⟶ Mn2 + (aq) + Al3 + (aq)

Balance the reaction in an acidic solution.

What is being oxidized?

What is being reduced?

Identify the oxidizing agent.

Identify the reducing agent.

Calculate the Standard Cell Potential for this reaction.

Is this reaction spontaneous as written?

To balance a redox reaction in we use the ion-electron method. In acidic solution, it proposes the following steps:

Identify and write separately half-reactions of reduction and of oxidation. To balance masses, add as many H⁺ on the side that is lacking. In case there are missing oxygen atoms, add water molecules on that side and the double of H⁺ on the other side. Add electrons to the proper side of the half-reaction so the charges are the same on both sides. Multiply both half-reactions by proper numbers so that the number of electrons gained is the same that the number of electrons lost. Use the numbers obtained to balance the equation.

In the reaction:

MnO₄⁻ (aq) + Al (s) ⇄ Mn²⁺ (aq) + Al³⁺ (aq)

We identify these half-reactions:

MnO₄⁻ (aq) ⇄ Mn²⁺ (aq) Reduction (the species gains electrons)

Al (s) ⇄ Al³⁺ (aq) Oxidation (the species loses electrons)

Let's use the ion-electron method for both half-reactions.

In the reduction, we have to add 4 molecules of H₂O to the right and 8 atoms of H⁺ to the left.

8H⁺ + MnO₄⁻ (aq) ⇄ Mn²⁺ (aq) + 4 H₂O

Now masses are balanced. With respect to the charges, there is a total charge of + 7 in the left and + 2 in the right, so we need to add 5 electrons (negative charges) to the left side.

5e⁻ + 8H⁺ + MnO₄⁻ (aq) ⇄ Mn²⁺ (aq) + 4 H₂O

Since the species gains electrons, we can confirm it is a reduction.

Regarding the oxidation half-reaction, masses are balanced, so we just have to add 3 electrons to the right to balance charges.

Al (s) ⇄ Al³⁺ (aq) + 3e⁻

Since the species loses electrons, we can confirm it is an oxidation.

Now, let's put together both results.

5e⁻ + 8H⁺ + MnO₄⁻ (aq) ⇄ Mn²⁺ (aq) + 4 H₂O

Al (s) ⇄ Al³⁺ (aq) + 3e⁻

We have to multiply the first reaction by 3, and the second by 5, so the number of electrons gained and lost is the same (15 electrons). The result would be:

24H⁺ + 3MnO₄⁻ (aq) + 5 Al (s) ⇄ 3Mn²⁺ (aq) + 12 H₂O + 5 Al³⁺ (aq)

This is the balanced equation.

What is being oxidized?

The species that undergoes oxidation is Al (s) since it loses electrons.

What is being reduced?

The species that undergoes reduction is MnO₄⁻ (aq) since it gains electrons.

Identify the oxidizing agent.

The oxidizing agent is the one that reduces, therefore making the other oxidize. The oxidizing agent is MnO₄⁻ (aq).

Identify the reducing agent.

The reducing agent is the one that oxidizes, therefore making the other reduce. The reducing agent is Al (s).

Calculate the Standard Cell Potential for this reaction.

The Standard Cell Potential (E°) is equal to the difference between the reduction potential of the reduction reaction and the reduction potential of the oxidation reaction. The reduction potentials can be found in tables and in this case are:

5e⁻ + 8H⁺ + MnO₄⁻ (aq) ⇄ Mn²⁺ (aq) + 4 H₂O E° = 1.51 V

Al³⁺ (aq) + 3e⁻ ⇄ Al (s) E° = - 0.66 V

E° = 1.51V - (-0.66V) = 2.17 V

Is this reaction spontaneous as written?

By convention, when E° is positive (2.17 V in this case), the reaction is spontaneous in the way it is written.