a) P4(s) + OH- (aq) → PH3 (g)+ HPO2 (aq) how to balnce it by oxidation no method
Answers
P
0
4
+
O
−2
H
−
+1
⇌
P
−3
H
+1
3
+
H
+1
2
P
+1
O
−2
2
−
⇒ P
4
0
⟶4
H
+1
2
P
+1
O
−2
2
+4e
−
(oxidation)
⇒
P
0
4
+12e
−
⟶4
P
−3
H
+1
3
(Reduction).
Balance the oxygen atoms in oxidation reaction.
P
0
4
+8OH
−
⟶4
H
+1
2
P
+1
O
2
2
+4e
−
3
P
0
4
+24OH
−
⟶12
H
+1
2
P
+1
O
+2
2
+12e
−
(O)
P
0
4
+12e
−
+12H
2
O⟶4
P
−3
H
+1
3
+12OH
−
(R)
4
P
0
4
+24OH
−
+12e
−
+12H
2
O⟶12
H
+1
2
P
+1
O
−2
2
+4
P
−3
H
+1
3
+12e
−
+12OH
−
Simplity the above equation.
P
0
4
+3OH
−
+3H
2
O⟶3
H
+1
2
P
+1
O
−2
2
+
P
−3
H
+1
3
P
4
+3OH
−
+3H
2
O⇌PH
3
+3H
2
PO
2
The equation is already balanced
