Calculate the n-factor of Zn and HNO3 in the following reaction:
Dear student
In the given reaction, following oxidation and reduction takes place
OXIDATION: 4 Zn + 8HNO3 → 4Zn(No3)2 + 8e- + 8H+
Reduction : HNO3 + HNO3 + 8e- + 8H+ → NH4NO3 + 3H2O
NET reaction : 4Zn + 10HNO3 + 8e- + 8H+ → 4Zn(NO3)2 + NH4NO3 + 8e- + 3H2O + 8H+
In the given reaction, following oxidation and reduction takes place
OXIDATION: 4 Zn + 8HNO3 → 4Zn(No3)2 + 8e- + 8H+
Reduction : HNO3 + HNO3 + 8e- + 8H+ → NH4NO3 + 3H2O
NET reaction : 4Zn + 10HNO3 + 8e- + 8H+ → 4Zn(NO3)2 + NH4NO3 + 8e- + 3H2O + 8H+
We can see that 8 electrons have been exchanged in the reaction therefore n-factor=8 Regards