IB Chemistry - Oxidation

IB Chemistry home > Syllabus 2016 > Redox processes > Half-equations involving hydrogen ions

Syllabus ref: 9.1

Most redox processes happen in aqueous solution. Hydrogen ions or water are often needed to allow the reduction or oxidation half-equation to occur.

Reactions in acidic solution

When potassium manganate(VII) acts as an oxidising agent, the manganate(VII) ion is reduced in (acidic solution) according to the half-equation:

MnO4- + 8H+ + 5e Mn2+ + 4H2O

The eight hydrogen ions are needed to absorb the four oxygen atoms from the manganate(VII) ion. Notice that the half-equation is balanced electronically. On the left hand side there are 8 positive charges and 6 negative charges making an overall 2+ chage. On the right hand side, there is also a 2+ charge, the equation is balanced both in terms of atoms and electrical charges.
Balance potasium manganate(VII) reactions

To find the number hydrogen ions required is usually a matter of seeing how many water molecules must be created from any oxygen in the oxyions.

Example: Construct the half-equation for the reduction of the dichromate(VI) ion Cr2O72- in acidic solution, if the product is the Cr3+ ion.

The dichromate(VI) ion contains 7 oxygen atoms wheras the product chromium (III) ion contains none. This means that all of the oxygen atoms must go to produce water. This requires the use of 14 hydrogen ions.

Cr2O72- + 14H+ 2Cr3+ + 7H2O

However, the equation is not balanced electronically. The two chromium atoms in the dichromate ion are in the (VI) oxidation state and they both end up in the (III) oxidation state; they have absorbed between them 6 electrons. These must be added to the equation:

Cr2O72- + 14H+ + 6e 2Cr3+ + 7H2O

The equation is now balanced electronically.

  • LHS = -2 + 14 + (-6) = +6
  • RHS = 2 x (+3) = +6


Use of water

sulfate(IV) ions can be oxidised to form sulfate(VI) in acidic solution. In this case the sulfate(IV) must pick up an oxygen from the aqueous medium. It used a water molecule.

SO32- + H2O SO42- + 2H+

The equation needs to be balanced in terms of charges by adding electrons to the appropriate side. One sulfur atom changes from the (IV) oxidation state to the (VI) oxidation state, thus releasing two electrons. The electrons must be added to the right hand side.

SO32- + H2O SO42- + 2H+ +2e

Sulfate(IV) is therefore a reducing agent.


Reactions in alkali solution

Redox behaviour is pH sensitive and products are often different in alkali solution. The IB does not require a knowledge of specific half-equations in alkali medium.