Cl2O7+H2O2 gives out ClO2^-+O2 balance the equation in basic medium by ion electron method
Answers
Answer:
Balance the given reaction by oxidation number method a.Cl2O7 ... Cl2O7 + H2O2 ClO2- + O2 + H+ (basic medium) b. MnO4- + SO2 Mn2+ + HSO4- (acidic medium) The oxidation number of chlorine decreases from +7 to +3 and the oxidation number of O increases from -1 to 0.SO CI2O7 is oxidizing agent and H2O2 is reducing agent .
The oxidation number of chlorine decreases from +7 to +3 and the oxidation number of O increases from -1 to zero. Thus, Cl
2
O
7
is oxidizing agent and H
2
O
2
is the reducing agent.
Ion electron method:
The oxidation half equation is
H
2
O
2
(aq)→O
2
(g)
To balance oxidation number, 2 electrons are added.
H
2
O
2
(aq)→O
2
(g)+2e
−
2 hydroxide ions are added to balance the charge.
H
2
O
2
(aq)+2OH
−
→O
2
(g)+2e
−
2 water molecules are added to balance the O atoms.
H
2
O
2
(aq)+2OH
−
→O
2
(g)+2H
2
O(aq)+2e
−
The reduction half reaction is Cl
2
O
7
→ClO
2
−
(aq)
The Cl atoms are balanced
Cl
2
O
7
→ClO
2
−
(aq)
8 electrons are added to balance the oxidation number.
Cl
2
O
7
+8e
−
→ClO
2
−
(aq)
6 hydroxide ions are added to balance the charge.
Cl
2
O
7
+8e
−
→ClO
2
−
(aq)+6OH
−
(aq)
The oxidation half equation is multiplied with 4 and added to reduction
half equation.
CI
2
O
7
(g)+4H
2
O
2
(aq)+2OH
−
→CIO
2
−
(aq)+4O
2
(g)+5H
2
O(l)
Oxidation number method:
Total decrease in oxidation number of Cl
2
O
7
is 8.
Total increase in oxidation number of H
2
O
2
is 2.
H
2
O
2
and O
2
are multiplied with 4
Cl
2
O
7
(g)+4H
2
O
2
(aq)→ClO
2
−
(aq)+4O
2
(g)
Chlorine atoms are balanced
Cl
2
O
7
(g)+4H
2
O
2
(aq)→2ClO
2
−
(aq)+4O
2
(g)
O atoms are balanced by adding 3 water molecules.
Cl
2
O
7
(g)+4H
2
O
2
(aq)→2ClO
2
−
(aq)+4O
2
(g)+3H
2
O(l)
H atoms are balanced by adding 2 hydroxide ions and 2 water molecules.
CI
2
O
7
(g)+4H
2
O
2
(aq)+2OH
−
→CIO
2
−
(aq)+4O
2
(g)+5H
2
O(l)