Question AD.2: A Reverse Disproportionation Now try the next reaction, whic...
A Reverse Disproportionation
Now try the next reaction, which represents the reverse of a disproportionation. (In a disproportionation, an element in one oxidation state reacts to give the same element in higher and lower oxidation states.)
\underset{\begin{matrix} 0 \\ Iodine \end{matrix} }{I_{2}} + \underset{\begin{matrix} +5 \\ Iodate \end{matrix} }{IO_{3}^{−}} + \underset{- 1}{Cl^{−}} \rightleftharpoons \underset{+1 – 1}{ICl_{2}^{−}}
- The oxidation numbers are assigned above. Note that chlorine has an oxidation number of −1 on both sides of the equation. Only iodine is involved in electron transfer.
- Oxidation half-reaction: \underset{0}{I_{2}} \rightleftharpoons \underset{+1}{ICl_{2}^{−}}
Reduction half-reaction: \underset{+5}{IO_{3}^{−}} \rightleftharpoons \underset{+1}{ICl_{2}^{−}}
3. We need to balance I atoms in the first reaction and add Cl^{−} to each reaction to balance Cl.
I_{2} + 4Cl^{−} \rightleftharpoons 2ICl_{2}^{−}
IO_{3}^{−} + 2Cl^{−} \rightleftharpoons ICl_{2}^{−}
4. Now add electrons to each.
I_{2} + 4Cl^{−} \rightleftharpoons 2ICl_{2}^{−} + 2e^{−}
IO_{3}^{−} + 2Cl^{−} + 4e^{−} \rightleftharpoons ICl_{2}^{−}
The first reaction needs 2e^{−} because there are two I atoms, each of which changes from 0 to +1.
5. The second reaction needs 3H_{2}O on the right side to balance
oxygen atoms.
IO_{3}^{−} + 2Cl^{−} + 4e^{−} \rightleftharpoons ICl_{2}^{−} + 3H_{2}O
6. The first reaction is balanced, but the second needs 6H^{+} on the left.
IO_{3}^{−} + 2Cl^{−} + 4e^{−} + 6H^{+} \rightleftharpoons ICl_{2}^{−} + 3H_{2}O
As a check, the charge on each side of this half-reaction is −1, and
all atoms are balanced.
7. Multiply and add.
\begin{matrix} 2(I_{2} + 4Cl^{−} \quad \quad \quad \quad \quad \quad \rightleftharpoons 2ICl_{2}^{−} + \cancel{2e}^{−}) \\ IO_{3}^{−} + 2Cl^{−} + \cancel{4e}^{−} + 6H^{+} \rightleftharpoons ICl_{2}^{−} + 3H_{2}O \\ \hline 2I_{2} + IO_{3}^{−} + 10Cl^{−} + 6H^{+} \rightleftharpoons 5ICl_{2}^{−} + 3H_{2}O \end{matrix} (D-1)
We multiplied the first reaction by 2 so that there would be the same number of electrons in each half-reaction. You could have multiplied the first reaction by 4 and the second by 2, but then all coefficients would simply be doubled. We customarily write the smallest coefficients.