Balance the following equations in basic medium by ion-electron method

The Oxidation no. of Cldecreases from +7 in \( Cl_{ 2 }O_{ 7 }\;to +3\) in \( ClO_{ 2 }^{ – }\)

The oxidation no. of Oincreases from \( -1\; in\; H_{ 2 }O_{ 2 }\)to 0 in \( O_{ 2 }\)

Therefore, \( H_{ 2 }O_{ 2 }\)​behaves as a reducing agent while \( Cl_{ 2 }O_{ 7 }\)​ behaves as an oxidizing agent in the reaction. Ion – electron method:

 – The oxidation half reaction:

\( H_{ 2 }O_{ 2 \; (aq) } \; \rightarrow \; O_{ 2 \; (g) } \)

By adding two electrons to the oxidation number, the oxidation number is balanced as follows:

2OH - ions are added to balance the charge as follows:

By adding \( 2H_2 O_2\) as follows, the oxygen atoms are balanced.

The half-reduction equation is as follows: The Cl atoms are balanced in the following way:

\( Cl_{ 2 }O_{ 7 \; (g) } \; \rightarrow \; ClO^{ – }_{ 2 \; (aq) }\)

By adding 8 electrons to the oxidation number, the oxidation number is balanced:

\( Cl_{ 2 }O_{ 7 \; (g) } \; + \; 8\; e^{ – } \; \rightarrow \; 2 \; ClO^{ – }_{ 2 \; (aq) }\)

6OH' is added to balance the charge as follows:

By adding \( 3H_2O\) as follows, the oxygen atoms are balanced.

By multiplying equation (i) by 4 and adding equation (ii) toit, you can get the balanced equation.

Oxidation number method:

 – Reduction in the oxidation no. of \( Cl_{ 2 }O_{ 7 }\) = 4× 2 = 8

 – Increment in the oxidation no. of \( H_{ 2 }O_{ 2 }\)​ = 2× 1 = 2

Multiply \( H_{ 2 }O_{ 2 }\)​ by 4 and \( O_{ 2 }\)by 4 to balance the reduction and increment of the oxidation no. :

The Cl atoms are balanced in the following way:

The O atoms are balanced by adding \(3 H_2O\) in the following way:

\( 2OH^{-}\)and \( 2H_2O\)are used to balance the H atoms as follows:

This is the equation that must be balanced.