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
