Question

Write a balanced half-reaction for the reduction of bismuth oxide ion to bismuth ion in basic aqueous solution. Be sure to add physical state symbols where appropriate.

Answer 1

`3H_2O+(Bi^(5+)O_3)^-\rightarrow Bi^(3+)+6OH^-`

Step-by-step explanation:

Hello there!

In this case, according to the required half-reaction, we start by setting it up from bismuth (V) oxide ion to bismuth (III) ion:

`BiO_3^-\rightarrow Bi^(3+)`

Thus, next realize that the oxidation state of Bi in BiO3^- is 5+ because oxygen is 2- (-2*3+x=-1;x=-1+6;x=+5), so we obtain:

`(Bi^(5+)O_3)^-\rightarrow Bi^(3+)`

Thereafter, we realize three water molecules are needed on the right in order to balance the oxygens and consequently 6 hydrogen atoms on the left to balance hydrogen:

`6H^++(Bi^(5+)O_3)^-\rightarrow Bi^(3+)+3H_2O`

Now, since the balance is is basic media, we add six molecules of hydroxide ions in order to produce water with the hydrogen ones:

`6OH^-+6H^++(Bi^(5+)O_3)^-\rightarrow Bi^(3+)+3H_2O+6OH^-\\\\6H_2O+(Bi^(5+)O_3)^-\rightarrow Bi^(3+)+3H_2O+6OH^-\\\\6H_2O-3H_2O+(Bi^(5+)O_3)^-\rightarrow Bi^(3+)+6OH^-`

Then, we accommodate the waters to obtain:

`3H_2O+(Bi^(5+)O_3)^-\rightarrow Bi^(3+)+6OH^-`

Best regards!