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Doss [256]
3 years ago
5

Write a balanced half-reaction for the reduction of bismuth oxide ion to bismuth ion in basic aqueous solution. Be sure to add p

hysical state symbols where appropriate.
Chemistry
1 answer:
Butoxors [25]3 years ago
7 0

Answer:

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

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!

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