4) Balance the following redox reaction in an acidic solution. What are the coefficients in front of H⁺ and Fe3+ in the balanced

Answer:

- The coefficients in front of H⁺ and Fe³⁺ are 8 and 5 respectively.

- A total of 5 moles of electrons are exchanged.

Explanation:

This reaction is represented as:

Fe²⁺(aq) + MnO₄⁻(aq) → Fe³⁺(aq) + Mn²⁺(aq)

Analyzing the oxidation states:

Fe²⁺ transitions to Fe³⁺

This indicates an increase in oxidation state → OXIDATION

Meanwhile, Mn in MnO₄⁻ starts with +7 and transforms into Mn²⁺

This suggests a decrease in oxidation state → REDUCTION

Let's formulate the half reactions:

Fe²⁺ → Fe³⁺  +  1e⁻    (it loses 1 mole of electrons)

MnO₄⁻ + 5e⁻ →  Mn²⁺  (it gains 5 moles of electrons)

Next, we will balance the oxygen atoms. In an acidic environment, water is added to balance the oxygens on the opposite side. Since there are 4 oxygens on the reactant side, we add 4 H₂O to the product side.

MnO₄⁻ + 5e⁻ →  Mn²⁺  + 4H₂O

Now, to balance the hydrogen atoms, we have 8 hydrogens in the products, necessitating the inclusion of 8H⁺ in the reactants, yielding the complete half-reaction:

8H⁺  + MnO₄⁻ + 5e⁻ →  Mn²⁺  + 4H₂O

Notably, there's 1e⁻ in the oxidation and 5e⁻ in the reduction. To cancel electrons, we must multiply the oxidation half-reaction by 5.

(Fe²⁺ → Fe³⁺  +  1e⁻) x 5

5Fe²⁺ → 5Fe³⁺  +  5e⁻  

8H⁺  + MnO₄⁻ + 5e⁻ →  Mn²⁺  + 4H₂O

By adding both half reactions, we have:

5Fe²⁺  + 8H⁺  + MnO₄⁻ + 5e⁻ →  5Fe³⁺  +  5e⁻   + Mn²⁺  + 4H₂O

The electrons cancel out, resulting in the balanced equation:

5Fe²⁺  + 8H⁺  + MnO₄⁻  →  5Fe³⁺  + Mn²⁺  + 4H₂O