Question

Balance the following equation :
MnO4- + I- --> MnO2 + IO3- (basic medium) in half reaction method not oxidation.

Answer 1

The given equation is :

MnO⁴⁻ + I⁻ ====== > MnO₂ + IO³⁻

The complete equation will be :

MnO⁴⁻ + I⁻ + H₂O  ====== > MnO₂ + IO³⁻ + OH⁻

In order to balance the equation proceed these steps :

Add 2 beside MnO⁴⁻ .

Add 2 beside MnO₂ so as to balance the Mn .

Add 2 beside OH⁻ .

Count and see whether they are balanced or not .

Mn = 2

O = 8 + 1 = 9

I = 1

H = 2

This is balanced on both sides .

BALANCED EQUATION :

2 MnO⁴⁻ + I⁻ + H₂O  ====== > 2 MnO₂ + IO³⁻ + 2 OH⁻

Answer 2

Answer:

The balanced chemical equation in the basic medium is

$\[\begin{gathered} \hfill \\ 2Mn{O_4}^ - + {I^ - } + {H_2}O \to 2Mn{O_2} + IO{3^ - } + 2O{H^ - } \hfill \\ \end{gathered} \]$

Explanation:

As per the question we have given the reaction that is

$\[Mn{O_4}^ - + {I^ - } \to Mn{O_2} + IO{3^ - }\]$

This reaction is taking place in a basic medium

In the basic medium we will add water on the left side and hydroxide ion on the right side then the reaction is to be balanced algebraically.

Therefore,

$\[\begin{gathered} Mn{O_4}^ - + {I^ - } + {H_2}O \to Mn{O_2} + IO{3^ - } + O{H^ - } \hfill \\ 2Mn{O_4}^ - + {I^ - } + {H_2}O \to 2Mn{O_2} + IO{3^ - } + 2O{H^ - } \hfill \\ \end{gathered} \]$

Hence balanced chemical equation in the basic medium is

$\[\begin{gathered} \hfill \\ 2Mn{O_4}^ - + {I^ - } + {H_2}O \to 2Mn{O_2} + IO{3^ - } + 2O{H^ - } \hfill \\ \end{gathered} \]$