# Question #e32d1

Sep 15, 2015

${\text{IO"_text(3(aq])^(-) + 5"I"_text((aq])^(-) + 6"H"_text((aq])^(+) -> 3"I"_text(2(aq]) + 3"H"_2"O}}_{\textrm{\left(l\right]}}$

#### Explanation:

The iodate, ${\text{IO}}_{3}^{-}$, and iodide, ${\text{I}}^{-}$, ions will react in acidic medium to form iodine, ${\text{I}}_{2}$.

${\text{IO"_text(3(aq])^(-) + "I"_text((aq])^(-) + "H"_text((aq])^(+) -> "I"_text(2(aq]) + "H"_2"O}}_{\textrm{\left(l\right]}}$

Right from the start, you can probably tell that this is a disproportionation raction, which a redox reaction in which the same chemical species is being reduced and oxidized at the same time.

Assign oxidation numbers to all the atoms that take part in the reaction

$\stackrel{\textcolor{b l u e}{+ 5}}{\text{I") stackrel(color(blue)(-2))("O"_3^(-)) + stackrel(color(blue)(-1))("I"^(-)) + stackrel(color(blue)(+1))("H"^(+)) -> stackrel(color(blue)(0))("I"_2) + stackrel(color(blue)(+1))("H"_2) stackrel(color(blue)(-2))("O}}$

Some of the iodine atoms are being reduced from an oxidation state of +5 to an oxidation state of 0, while other are being oxidized from an oxidation state of -1 to an oxidation state of 0.

The two half-reactions will be

• oxidation half-reaction

$2 \stackrel{\textcolor{b l u e}{- 1}}{{\text{I"^(-)) -> stackrel(color(blue)(0))("I}}_{2}} + 2 {e}^{-}$

Each iodine atom loses one electrons, so two iodine atoms will lose 2 electrons.

• reduction half-reaction

$2 \stackrel{\textcolor{b l u e}{+ 5}}{{\text{I")"O"_3^(-) + 10e^(-) -> stackrel(color(blue)(0))("I}}_{2}}$

Each iodine atom will gain 5 electrons, which means that two iodine atoms will gain a total of 10 electrons* to form ${\text{I}}_{2}$.

Since you're in acidic solution, use water to balance the oxygen atoms and protons to balance the hydrogen atoms

$12 \text{H"^(+) + 2stackrel(color(blue)(+5))("I")"O"_3^(-) + 10e^(-) -> stackrel(color(blue)(0))("I"_2) + 6"H"_2"O}$

In any redox reaction, the number of electrons lost in the oxidation half-reaction must be equal to the number of electrons gained in the reduction half-reaction.

This means that you need to multiply the oxidation half-reaction by $5$ to get

$10 {\text{I"^(-) -> 5"I}}_{2} + 10 {e}^{-}$

The two half-reactions will now be

$\left\{\begin{matrix}10 \text{I"^(-) -> 5"I"_2 + 10e^(-) \\ 12"H"^(+) + 2"IO"_3^(-) + 10e^(-) -> "I"_2 + 6"H"_2"O}\end{matrix}\right.$

Add the two half-reaction to get

$10 \text{I"^(-) + 12"H"^(+) + 2"IO"_3^(-) + color(red)(cancel(color(black)(10e^(-)))) -> 5"I"_2 + color(red)(cancel(color(black)(10e^(-)))) + "I"_2 + 6"H"_2"O}$

$2 {\text{IO"_text(3(aq])^(-) + 10"I"_text((aq])^(-) + 12"H"_text((aq])^(+) -> 6"I"_text(2(aq]) + 6"H"_2"O}}_{\textrm{\left(l\right]}}$

The balanced net ionic equation will thus be

${\text{IO"_text(3(aq])^(-) + 5"I"_text((aq])^(-) + 6"H"_text((aq])^(+) -> 3"I"_text(2(aq]) + 3"H"_2"O}}_{\textrm{\left(l\right]}}$