Redox ReactionsMedium

Redox balancing — half-reaction method

Balance the redox reaction in acidic medium: MnO₄⁻ + Fe²⁺ → Mn²⁺ + Fe³⁺.

Given data

MnO₄⁻ + Fe²⁺ → Mn²⁺ + Fe³⁺

Steps

0 / 4

Write the reduction half-reaction: MnO₄⁻ → Mn²⁺. How many e⁻ to reduce Mn(VII) to Mn(II)?
Write your working — the numerical result must appear in the text e⁻
Write the oxidation half-reaction: Fe²⁺ → Fe³⁺. How many e⁻ are lost?
To balance electrons, multiply Fe by 5. What is the Fe²⁺:MnO₄⁻ ratio?
Balance O and H in acidic medium. How many H₂O on the right?

Full worked solution

Write the reduction half-reaction: MnO₄⁻ → Mn²⁺. How many e⁻ to reduce Mn(VII) to Mn(II)?
$7 - 2$
$7 - 2 = 5 = \mathbf{5\,e^-}$ . Mn(VII) reduces to Mn(II).
Write the oxidation half-reaction: Fe²⁺ → Fe³⁺. How many e⁻ are lost?
$3 - 2$
$3 - 2 = 1 = \mathbf{1\,e^-}$ . Fe²⁺ oxidizes to Fe³⁺.
To balance electrons, multiply Fe by 5. What is the Fe²⁺:MnO₄⁻ ratio?
$5 : 1$
$5:1 = 5 = \mathbf{5}$ . Fe²⁺ : MnO₄⁻ ratio.
Balance O and H in acidic medium. How many H₂O on the right?
$4 = \mathbf{4\,H_2O}$ . MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O.

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

Keep studying