Question

The number of moles of $KMn{{O}_{4}}$ ​reduced by one mole of KI in alkaline medium is:

Answer 1

Hint: First write the equation according to the conditions given in the question then solve it using redox reaction equation balancing concept in basic medium as it is stated in the question that it is in alkaline medium and as you will finish balancing the equation you will get the number of moles required.

Complete step by step answer:
Given reaction:
$K\overset{{}}{\mathop{Mn}}\,{{O}_{4}}+K\overset{{}}{\mathop{I}}\,\to \overset{{}}{\mathop{Mn}}\,{{O}_{2}}+K\overset{{}}{\mathop{I}}\,{{O}_{3}}$
In a balanced chemical reaction the number of all atoms at the right side should be equal to the left side of the reaction.
Balancing a chemical reaction as:
Step 1: Assign oxidation numbers to each of the atoms in the equation and write the numbers above the atom as:
$K\overset{+7}{\mathop{Mn}}\,{{O}_{4}}+K\overset{-1}{\mathop{I}}\,\to \overset{+4}{\mathop{Mn}}\,{{O}_{2}}+K\overset{+5}{\mathop{I}}\,{{O}_{3}}$
Step 2: Identify the atoms that are oxidized and those that are reduced as:
Reduction:
 $K\overset{+7}{\mathop{Mn}}\,{{O}_{4}}\to \overset{+4}{\mathop{Mn}}\,{{O}_{2}}$
\[\] Oxidation: 
$K\overset{-1}{\mathop{I}}\,\to K\overset{+5}{\mathop{I}}\,{{O}_{3}}$
Step 3: oxidation-number change is:
Reduction: 
$K\overset{+7}{\mathop{Mn}}\,{{O}_{4}}\to \overset{+4}{\mathop{Mn}}\,{{O}_{2}}$
​Gain of total 3 electrons
Oxidation: 
$K\overset{-1}{\mathop{I}}\,\to K\overset{+5}{\mathop{I}}\,{{O}_{3}}$
Loss of total 6 electrons
Step 4: Balance the total change in oxidation number as:
Reduction:
$K\overset{+7}{\mathop{Mn}}\,{{O}_{4}}\to \overset{+4}{\mathop{Mn}}\,{{O}_{2}}\times 2$
 Gain of total 6 electrons
Oxidation:
$K\overset{-1}{\mathop{I}}\,\to K\overset{+5}{\mathop{I}}\,{{O}_{3}}\times 1$
Loss of total 6 electrons
Therefore, Reduction: 
$2K\overset{+7}{\mathop{Mn}}\,{{O}_{4}}\to \overset{+4}{\mathop{2Mn}}\,{{O}_{2}}$
Oxidation: 
\[K\overset{-1}{\mathop{I}}\,\to K\overset{+5}{\mathop{I}}\,{{O}_{3}}\]
Step 5: Balance O atoms in reduction reaction by adding ${{H}_{2}}O$ and then balance H by ${{H}^{+}}$ as:
Reduction: 
$2K\overset{+7}{\mathop{Mn}}\,{{O}_{4}}+8{{H}^{+}}\to \overset{+4}{\mathop{2Mn}}\,{{O}_{2}}+4{{H}_{2}}O+2{{K}^{+}}$
Oxidation: 
$K\overset{-1}{\mathop{I}}\,+3{{H}_{2}}O\to K\overset{+5}{\mathop{I}}\,{{O}_{3}}+6{{H}^{+}}$
Step 6: For base catalysed reaction add $O{{H}^{-}}$to both side to neutralize  \[{{H}^{+}}\]as:
$2K\overset{+7}{\mathop{Mn}}\,{{O}_{4}}+8{{H}^{+}}+8O{{H}^{^{-}}}\to \overset{+4}{\mathop{2Mn}}\,{{O}_{2}}+8O{{H}^{-}}+2{{K}^{+}}+4{{H}_{2}}O$
Oxidation: 
$K\overset{-1}{\mathop{I}}\,+6O{{H}^{-}}+3{{H}_{2}}O\to K\overset{+5}{\mathop{I}}\,{{O}_{3}}+6{{H}^{+}}+6O{{H}^{-}}$
Or Reduction: 
$2K\overset{+7}{\mathop{Mn}}\,{{O}_{4}}+4{{H}_{2}}O\to \overset{+4}{\mathop{2Mn}}\,{{O}_{2}}+8O{{H}^{-}}+2{{K}^{+}}$
Oxidation: 
$K\overset{-1}{\mathop{I}}\,+6O{{H}^{-}}\to K\overset{+5}{\mathop{I}}\,{{O}_{3}}+3{{H}_{2}}O$
Thus overall reaction is:
 $2K\overset{{}}{\mathop{Mn}}\,{{O}_{4}}+4{{H}_{2}}O+KI+6O{{H}^{-}}\to \overset{{}}{\mathop{2Mn}}\,{{O}_{2}}+8O{{H}^{-}}+2{{K}^{+}}+KI{{O}_{3}}+3{{H}_{2}}O$
Or 
$2K\overset{{}}{\mathop{Mn}}\,{{O}_{4}}+{{H}_{2}}O+KI\to \overset{{}}{\mathop{2Mn}}\,{{O}_{2}}+2O{{H}^{-}}+2{{K}^{+}}+KI{{O}_{3}}$
The Balanced reaction is:
$2K\overset{{}}{\mathop{Mn}}\,{{O}_{4}}+{{H}_{2}}O+KI\to \overset{{}}{\mathop{2Mn}}\,{{O}_{2}}+2KOH+KI{{O}_{3}}$
Thus, 2 moles of $KMn{{O}_{4}}$ reduced by one mole of KI in alkaline medium.

Note: All steps of balancing equation in acidic and basic medium are the same except one step. In acidic medium oxygen atoms should be balanced with water and hydrogen with${{H}^{+}}$. In basic medium oxygen should be balanced with $O{{H}^{-}}$ and hydrogen with water.