number of electrons required to reduce 1 mole of KMnO4 in acidic medium
Answers
Answer:
Given reaction: KMnO
4
+KI→MnO
2
+KIO
3
In a balanced chemical reaction number of all atoms at 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
Mn
+7
O
4
+K
I
−1
→
Mn
+4
O
2
+K
I
+5
O
3
Step 2: Identify the atoms that are oxidized and those that are reduced as:
Reduction: K
Mn
+7
O
4
→
Mn
+4
O
2
Oxidation: K
I
−1
→K
I
+5
O
3
Step 3: oxidation-number change is:
Reduction: K
Mn
+7
O
4
→
Mn
+4
O
2
:gain of total 3 electrons
Oxidation: K
I
−1
→K
I
+5
O
3
Loss of total 6 electrons
Step 4: Balance the total change in oxidation number as:
Reduction: K
Mn
+7
O
4
→
Mn
+4
O
2
×2: gain of total 6 electrons
Oxidation: K
I
−1
→K
I
+5
O
3
×1 Loss of total 6 electrons
∴ Reduction: 2K
Mn
+7
O
4
→2
Mn
+4
O
2
Oxidation: K
I
−1
→K
I
+5
O
3
Step 5: Balance O atoms in reduction reaction by adding H
2
O and then balance H by H
+
as:
Reduction: 2K
Mn
+7
O
4
+8H
+
→2
Mn
+4
O
2
+4H
2
O+2K
+
Oxidation: K
I
−1
+3H
2
O→K
I
+5
O
3
+6H
+
Step 6: For base catalysed reaction add OH
−
to both side to neutralize H
+
as:
2K
Mn
+7
O
4
+8H
+
+8OH
−
→2
Mn
+4
O
2
+8OH
−
+2K
+
+4H
2
O
Oxidation: K
I
−1
+6OH
−
+3H
2
O→K
I
+5
O
3
+6H
+
+6OH
−
or Reduction: 2K
Mn
+7
O
4
+4H
2
O→2
Mn
+4
O
2
+8OH
−
+2K
+
Oxidation: K
I
−1
+6OH
−
→K
I
+5
O
3
+3H
2
O
Thus overall reaction is:
2KMnO
4
+4H
2
O+KI+6OH
−
→2MnO
2
+8OH
−
+2K
+
+KIO
3
+3H
2
O
or 2KMnO
4
+H
2
O+KI→2MnO
2
+2OH
−
+2K
+
+KIO
3
the Balanced reaction is:
2KMnO
4
+H
2
O+KI→2MnO
2
+2KOH+KIO
3
Thus 2 moles of KMnO
4
reduced by one mole of KI in alkaline medium.
Explanation: