Question

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

Answer 1

Answer:

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

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.