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Home > Balancing of Disproportionation Redox Reaction: Ion Electrode Method

Balancing of Disproportionation Redox Reaction: Ion Electrode Method - (Concept)

Disproportionation reactions are those reactions in which one species having some oxidation state converts into two different oxidation states, one oxidation state is higher and other is lower.

The balancing of the disproportionation reaction by ion electrode method can be understood by the following example.
The chemical reaction is as follows:

\mathrm{Cl_{2}\: +\: OH^{-}\: \rightarrow \: ClO_{3}^{-}\: +\: Cl^{-}\: +\: H_{2}O}

In this reaction, Cl on the reactant side has zero oxidation state but on the product side, its oxidation states are +5(in ClO_3^-) and -1 in Cl-.

STEP 1: Write oxidation half-reaction 

\mathrm{Cl_{2}\: \rightarrow ClO^{-}_{3}}

Now balance the chlorine atoms on both sides. Thus the balance equation is as follows:

\mathrm{Cl_{2}\: \rightarrow 2ClO^{-}_{3}}

Now chlorine atoms are changing its oxidation states from 0 to 5. Thus, there is a total exchange of 10 electrons. So, write the complete balanced equation as follows:

\mathrm{Cl_{2}\: \rightarrow 2ClO^{-}_{3}\: +\: 10e^{-}\quad\quad\quad\quad\quad\quad............(i)}

STEP 2: Write the reduction half-reaction

\mathrm{Cl_{2}\: \rightarrow \: Cl^{-}}

Now balance the chlorine atoms on both sides. Thus the balance equation is as follows:

\mathrm{Cl_{2}\: \rightarrow \: 2Cl^{-}}

Now in this equation, chlorine atoms are changing its oxidation states from 0 to -1. Thus, there is a total exchange of 2 electrons. So, write the complete balanced equation as follows:

\mathrm{Cl_{2}\: +\: 2e^{-}\: \rightarrow \: 2Cl^{-}\quad\quad\quad\quad\quad\quad............(ii)}

Now balance the electrons exchange of equations (i) and (ii) and then add them both. Thus the final added equation is as follows:

\mathrm{6Cl_{2}\: \rightarrow \: 2ClO^{-}_{3}\: +\: 10Cl^{-}\quad\quad\quad\quad\quad\quad............(iii)}

STEP 3: Balance the charge
In equation(iii), there is a total of -12 charge on the product side and zero charge on the reactant side. Thus, to balance the charge on both sides, add the required number of OH- ions on the deficient side. Thus,

12\mathrm{OH}^{-}+6 \mathrm{Cl}_{2} \longrightarrow 2 \mathrm{ClO}_{3}^{-}+10 \mathrm{Cl}^{-}

STEP 4: Balance the oxygen atoms
To balance the oxygen atoms, add the required number of H2O molecules on the deficient side.

12\mathrm{OH}^{-}+6 \mathrm{Cl}_{2} \longrightarrow 2 \mathrm{ClO}_{3}^{-}+\mathrm{6H_{2}O}\: +\: 10 \mathrm{Cl}^{-}

This is the final balanced equation for the given disproportionation reaction by ion-electrode method.

Exam Chapter
JEE MAIN Redox Reaction and Electrochemistry
Chemistry Part II Textbook for Class XI
Page No. : 275
Line : 38

In this method, the two half equations are balanced separately and then added together to give balanced equation.


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