Question

complete and balance the following redox reaction in acidic solution. be sure to include the proper phases for all species within the reaction.
clo₃⁻(aq) + i₂(s) → clo₂(g) + io₃⁻(aq)

Explanation:

Step1: Assign oxidation numbers

• For \(ClO_{3}^{-}\): \(Cl\) has an oxidation number of \(+5\) (since \(O\) is \(- 2\), let \(x\) be \(Cl\)'s oxidation number: \(x+3\times(-2)=-1\), \(x = +5\)).
• For \(ClO_{2}\): \(Cl\) has an oxidation number of \(+4\) (since \(O\) is \(-2\), let \(x\) be \(Cl\)'s oxidation number: \(x + 2\times(-2)=0\), \(x=+4\)). So \(Cl\) is reduced (\(ClO_{3}^{-}\to ClO_{2}\)).
• For \(I_{2}\): \(I\) has an oxidation number of \(0\). For \(IO_{3}^{-}\): \(I\) has an oxidation number of \(+5\) (since \(O\) is \(-2\), let \(x\) be \(I\)'s oxidation number: \(x+3\times(-2)=-1\), \(x = +5\)). So \(I\) is oxidized (\(I_{2}\to IO_{3}^{-}\)).

Step2: Write half - reactions

• Reduction half - reaction (\(ClO_{3}^{-}\to ClO_{2}\)):
• Balance atoms (except \(O\) and \(H\)): Already balanced (\(1\) \(Cl\) on each side).
• Balance \(O\) atoms: Add \(H_{2}O\). \(ClO_{3}^{-}\to ClO_{2}+H_{2}O\).
• Balance \(H\) atoms (in acidic solution, add \(H^{+}\)): \(ClO_{3}^{-}+2H^{+}\to ClO_{2}+H_{2}O\).
• Balance charge: \(ClO_{3}^{-}+2H^{+}+e^{-}\to ClO_{2}+H_{2}O\).
• Oxidation half - reaction (\(I_{2}\to IO_{3}^{-}\)):
• Balance \(I\) atoms: \(I_{2}\to 2IO_{3}^{-}\).
• Balance \(O\) atoms: Add \(H_{2}O\). \(I_{2}+6H_{2}O\to 2IO_{3}^{-}\).
• Balance \(H\) atoms: \(I_{2}+6H_{2}O\to 2IO_{3}^{-}+12H^{+}\).
• Balance charge: \(I_{2}+6H_{2}O\to 2IO_{3}^{-}+12H^{+}+10e^{-}\).

Step3: Make electron loss equal to electron gain

• Multiply the reduction half - reaction by \(10\) (since the oxidation half - reaction has a loss of \(10e^{-}\)):
• \(10ClO_{3}^{-}+20H^{+}+10e^{-}\to 10ClO_{2}+10H_{2}O\).
• The oxidation half - reaction remains: \(I_{2}+6H_{2}O\to 2IO_{3}^{-}+12H^{+}+10e^{-}\).

Step4: Add the two half - reactions

• \(10ClO_{3}^{-}+20H^{+}+I_{2}+6H_{2}O\to 10ClO_{2}+10H_{2}O+2IO_{3}^{-}+12H^{+}\).
• Simplify (cancel out \(H^{+}\) and \(H_{2}O\) on both sides):
• \(10ClO_{3}^{-}+(20 - 12)H^{+}+I_{2}\to 10ClO_{2}+(10 - 6)H_{2}O+2IO_{3}^{-}\).
• \(10ClO_{3}^{-}+8H^{+}+I_{2}\to 10ClO_{2}+4H_{2}O+2IO_{3}^{-}\).

Answer:

\(10ClO_{3}^{-}(aq)+8H^{+}(aq)+I_{2}(s)\to 10ClO_{2}(g)+4H_{2}O(l)+2IO_{3}^{-}(aq)\)