Disproportionation of Potassium Chlorate (V)

Disproportionation is a redox reaction where one species undergoes oxidation and reduction to form different products.

Let's consider the disproportionation of potassium chlorate (V) as an example.

Given that potassium chlorate (V) undergoes disproportionation to form chlorate (VII) and chloride.

disproportionation KClO3 products

Chlorate (V), ClO3- is simultaneously oxidised to chlorate (VII), ClO4- and reduced to chloride, Cl- in the same redox reaction.

Let's work out the balanced equation for this disproportionation.

Since disproportionation is just a redox reaction, we can use the half equation method to balance this reaction.

Check out this video to learn how to balance a redox reaction in acidic and alkaline medium.

We can use the following 4 steps to balance the half equation.

disproportionation KClO3 4 steps for balancing half equation

The oxidation of ClO3- to ClO4- is as shown.

disproportionation KClO3 oxidation

Notice the electrons are on the right hand side of the half equation, since oxidation is the loss of electrons.

The reduction of ClO3- to Cl- is shown here.

disproportionation KClO3 reduction

Reduction is the gain of electrons hence the electrons are on the left hand side of the half equation.

We can now combine the 2 half equations together.

The coefficients for the oxidation half equation is multiplied by 3 so that we have the same number of electrons for both oxidation and reduction half equations.

This is to ensure that the electrons will cancel out exactly when the 2 half equations are added together.

disproportionation KClO3 overall balanced equation

Topic: Redox Titration, Physical Chemistry, A Level Chemistry, Singapore

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