Question

4) Balance the following redox reaction in an acidic solution. What are the coefficients in front of H⁺ and Fe3+ in the balanced reaction? How many moles of electrons are transferred? Fe2+(aq) + MnO4⁻(aq) → Fe3+(aq) + Mn2+(aq)

Answer 1

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