Answer:
[tex]3H_2O+(Bi^{5+}O_3)^-\rightarrow Bi^{3+}+6OH^-[/tex]
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:
[tex]BiO_3^-\rightarrow Bi^{3+}[/tex]
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:
[tex](Bi^{5+}O_3)^-\rightarrow Bi^{3+}[/tex]
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:
[tex]6H^++(Bi^{5+}O_3)^-\rightarrow Bi^{3+}+3H_2O[/tex]
Now, since the balance is is basic media, we add six molecules of hydroxide ions in order to produce water with the hydrogen ones:
[tex]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^-[/tex]
Then, we accommodate the waters to obtain:
[tex]3H_2O+(Bi^{5+}O_3)^-\rightarrow Bi^{3+}+6OH^-[/tex]
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