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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: 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 - .

Educação

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
-

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