Question

Complete and balance the equation for areaction in acidic medium using the smallest set of whole numbers.

As₂O₃+NO₃^- --> H₃AsO₄ + NO

Hello if you can explain each step that would help alot. I want to be able to relaern the steps.

Answer 1

Oxidation number of each element in As2O3 is
O=-2
As=+3

Oxidation number of each element in NO3-1 is
O=-2
N=+5

Oxidation number of each element in H3AsO4 is
H=+1
O=-2
As=+5

Oxidation number of each element in NO is
O=-2
N=+2

Oxidation number of each element in reactant is
O=-2
As=+3
N=+5

Oxidation number of each element in product is
H=+1
O=-2
As=+5
N=+2

As in As2O3 has oxidation state of +3

As in H3AsO4 has oxidation state of +5

So, As in As2O3 is oxidised to H3AsO4

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:

As2O3 --> 2 H3AsO4 + 4e-

Balance number of electrons to be same in both half reactions

Reduction half cell:

4 NO3- + 12e- --> 4 NO

Oxidation half cell:

3 As2O3 --> 6 H3AsO4 + 12e-

Lets combine both the reactions.

4 NO3- + 3 As2O3 --> 4 NO + 6 H3AsO4

Balance Oxygen by adding water

4 NO3- + 3 As2O3 + 7 H2O --> 4 NO + 6 H3AsO4

Balance Hydrogen by adding H+

4 NO3- + 3 As2O3 + 7 H2O + 4 H+ --> 4 NO + 6 H3AsO4

This is balanced chemical equation in acidic medium

4 NO3- + 3 As2O3 + 7 H2O + 4 H+ --> 4 NO + 6 H3AsO4