balance mno4- + c2h5oh gives mn2+ + ch3cooh by oxidation number method
Answers
Answer:
The unbalanced redox equation is as follows:
C
2
H
5
OH+MnO
4
−
→C
2
H
3
O
−
+MnO
2
(s)+H
2
O
All atoms other than H and O are balanced.
The oxidation number of C changes from -2 to -1. The change in the oxidation number of 1 C atom is 1. The change in the oxidation number of 2 C atoms is 2.
The oxidation number of Mn changes from +7 to 4. The change in the oxidation number of Mn is 3.
The increase in the oxidation number is balanced with a decrease in the oxidation number by multiplying C
2
H
5
OH and C
2
H
3
O
−
with 3 and by multiplying MnO
4
−
and MnO
2
(s) with 2.
3C
2
H
5
OH+2MnO
4
−
→3C
2
H
3
O
−
+2MnO
2
(s)+H
2
O
O atoms are balanced by adding 3 water molecules on RHS..
3C
2
H
5
OH+2MnO
4
−
→3C
2
H
3
O
−
+2MnO
2
(s)+4H
2
O
Hydrogen atoms are balanced by adding 1 H
+
atoms on the RHS.
3C
2
H
5
OH+2MnO
4
−
→3C
2
H
3
O
−
+2MnO
2
(s)+4H
2
O+H
+
Since the reaction occurs in basic medium, add 1 hydroxide ion on both sides of the equation.
3C
2
H
5
OH+2MnO
4
−
+OH
−
→3C
2
H
3
O
−
+2MnO
2
(s)+4H
2
O+H
+
+OH
−
On RHS, 1 H
+
ion combine with 1 OH
−
ion to form 1 water molecule.
3C
2
H
5
OH+2MnO
4
−
+OH
−
→3C
2
H
3
O
−
+2MnO
2
(s)+5H
2
O
This is the balanced chemical equation.
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