balance the following Redox reaction by oxidation number method mno2 + bi R - gives MN 2 + + br2 + H2O acid medium
Answers
Answer:
MnO
4
−
+8H
+
+5
e
−
⟶Mn
2+
+4H
2
O×22Br
−
⟶Br
2
+2
e
−
×5
Explanation:
MnO
4
−
+Br
−
→Mn
2+
+Br
2
(a) First half reaction;
M
+7
nO
4
−
→
M
+2
n
2+
To balance oxygen atoms, H
2
O is added on right side,
MnO
4
−
→Mn
2+
+4H
2
O
(b) To balance hydrogen, H+ ions are added to left side
MnO
4
−
+8H
+
→5e
−
→Mn
2+
+4 H
2
O
Second half reaction
B
−1
r
−
→
B
0
r
2
Or 2Br
−
→Br
2
To balance charge, 2
e
−
are added right hand side
2Br
−
→Br
2
+2
e
−
Now adding the equation (i) (ii)
2MnO
4
−
+16H
+
+10Br
−
⟶2Mn
2+
+8H
2
O+5Br
2
MnO
4
−
+8H
+
+5
e
−
⟶Mn
2+
+4H
2
O×22Br
−
⟶Br
2
+2
e
−
×5