Answer:
- The coefficients in front of H⁺ and Fe³⁺ are 8 and 5
- There are transferred 5 moles of e-
Explanation:
This is the reaction:
Fe²⁺(aq) + MnO₄⁻(aq) → Fe³⁺(aq) + Mn²⁺(aq)
Let's think the oxidation numbers:
Fe2+ changed to Fe3+
It has increased the oxidation number → OXIDATION
Mn in MnO₄⁻ acts with +7 and it changed to Mn²⁺
It has decreased the oxidation number → REDUCTION
Let's make the half reactions:
Fe²⁺ → Fe³⁺ + 1e⁻ (it has lost 1 mol of e⁻)
MnO₄⁻ + 5e⁻ → Mn²⁺ (it has gained 5 mol of e⁻)
Now we have to ballance the O. In acidic medium we complete with water as many oxygens we have, in the opposite side. We have 4 O in reactant side, so we fill up with 4 H2O in products side.
MnO₄⁻ + 5e⁻ → Mn²⁺ + 4H₂O
Now we have to ballance the H, so as we have 8 H in products side, we complete with 8H⁺ in reactants, this is the complete half reaction:
8H⁺ + MnO₄⁻ + 5e⁻ → Mn²⁺ + 4H₂O
Notice that have 1e⁻ in oxidation and 5e⁻ in reduction. We must multiply (x5) the half reaction of oxidation, so the electrons can be cancelled.
(Fe²⁺ → Fe³⁺ + 1e⁻ ) .5
5Fe²⁺ → 5Fe³⁺ + 5e⁻
8H⁺ + MnO₄⁻ + 5e⁻ → Mn²⁺ + 4H₂O
We sum both half reactions:
5Fe²⁺ + 8H⁺ + MnO₄⁻ + 5e⁻ → 5Fe³⁺ + 5e⁻ + Mn²⁺ + 4H₂O
The electrons are cancelled, so the ballanced reaction is this:
5Fe²⁺ + 8H⁺ + MnO₄⁻ → 5Fe³⁺ + Mn²⁺ + 4H₂O
