QuestionMay 11, 2025

Write balanced half-reactions for the following redox reaction: BiO_(3)^-(aq)+3H_(2)O(l)+2I^-(aq)arrow Bi^3+(aq)+6OH^-(aq)+I_(2)(s) reduction: square oxidation: square

Write balanced half-reactions for the following redox reaction: BiO_(3)^-(aq)+3H_(2)O(l)+2I^-(aq)arrow Bi^3+(aq)+6OH^-(aq)+I_(2)(s) reduction: square oxidation: square
Write balanced half-reactions for the following redox reaction: BiO_(3)^-(aq)+3H_(2)O(l)+2I^-(aq)arrow Bi^3+(aq)+6OH^-(aq)+I_(2)(s) reduction: square oxidation: square

Solution
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Answer

Reduction: BiO_3^- + 3H_2O + 6e^- \rightarrow Bi^{3+} + 6OH^- ### Oxidation: 2I^- \rightarrow I_2 + 2e^- Explanation 1. Write the reduction half-reaction Reduction involves BiO_3^- being reduced to Bi^{3+}. Balance Bi first, then oxygen using water, hydrogen using H^+, and charge using electrons: BiO_3^- + 3H_2O + 6e^- \rightarrow Bi^{3+} + 6OH^- 2. Write the oxidation half-reaction Oxidation involves I^- being oxidized to I_2. Balance iodine atoms first, then charge using electrons: 2I^- \rightarrow I_2 + 2e^-

Explanation

1. Write the reduction half-reaction<br /> Reduction involves $BiO_3^-$ being reduced to $Bi^{3+}$. Balance Bi first, then oxygen using water, hydrogen using $H^+$, and charge using electrons:<br />$BiO_3^- + 3H_2O + 6e^- \rightarrow Bi^{3+} + 6OH^-$<br /><br />2. Write the oxidation half-reaction<br /> Oxidation involves $I^-$ being oxidized to $I_2$. Balance iodine atoms first, then charge using electrons:<br />$2I^- \rightarrow I_2 + 2e^-$
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