Question

The following skeletal oxidation-reduction reaction occurs under basic conditions. Write the balanced OXIDATION half reaction.

Bi(OH)3 + NO2 → Bi + NO3-

Answer 1

`N^(4+)O_2+2OH^-\rightarrow (N^(5+)O_3)^-+1e^-+H_2O`

Step-by-step explanation:

Hello,

In this case, for the given reaction, we first start by the writing of the oxidation states of all the involved elements:

`Bi^(3+)(OH)^-+N^(4+)O^(2-)_2\rightarrow Bi^0+(N^(5+)O^(2-)_3)^-`

In such a way, we are noticing nitrogen is undergoing an increase in its oxidation state, therefore it is being the oxidized species, for which the oxidation half reaction, should be (considering basic conditions):

`N^(4+)O_2+H_2O+2OH^-\rightarrow (N^(5+)O_3)^-+1e^-+2H^++2OH^-\\\\N^(4+)O_2+H_2O+2OH^-\rightarrow (N^(5+)O_3)^-+1e^-+2H_2O\\\\N^(4+)O_2+2OH^-\rightarrow (N^(5+)O_3)^-+1e^-+H_2O`

Best regards.