1. Balance the following in acidic solution.
Solution
Oxidation number of each element in reactant is
P=0
O=-2
N=+5
Oxidation number of each element in product is
H=+1
O=-2
P=+5
N=+2
P in P4 has oxidation state of 0
P in H2PO4- has oxidation state of +5
So, P in P4 is oxidised to H2PO4-
N in NO3- has oxidation state of +5
N in NO has oxidation state of +2
So, N in NO3- is reduced to NO
Reduction half cell:

2. NO3- + 3e- --> NO
Oxidation half cell:
P4 --> 4 H2PO4- + 20e-
Balance number of electrons to be same in both half reactions
Reduction half cell:
20 NO3- + 60e- --> 20 NO
Oxidation half cell:
3 P4 --> 12 H2PO4- + 60e-
Lets combine both the reactions.
20 NO3- + 3 P4 --> 20 NO + 12 H2PO4-
Balance Oxygen by adding water
20 NO3- + 3 P4 + 8 H2O --> 20 NO + 12 H2PO4-
Balance Hydrogen by adding H+
20 NO3- + 3 P4 + 8 H2O + 8 H+ --> 20 NO + 12 H2PO4-
This is balanced chemical equation in acidic medium
Answer:
20 NO 3
-
+ 3 P 4 + 8 H 2 O + 8 H +
--> 20 NO + 12 H 2 PO 4
-