I'm having trouble balancing this reaction using half reactions: $$\ce{I2 + KOH -> KI + KIO3 + H2O}$$

I so far know that iodine is reduced when it forms $\ce{KI}$ (oxidation noumber decreases from 0 to -1) and oxidised in $\ce{KIO3}$ (the Oxidation number increases from 0 to 5).

Oxidation half: $\ce{I2 -> KIO3}$

Reduction half: $\ce{I2 -> KI}$

The potassium atoms on both halves seem to be the problem because they are only on the product side. When I balance the iodine atoms the $\ce{K}$ atoms aren't.

I've tried this several times and can't work it out.


To make your problem simpler(equivalent of your problem) just balance this reaction: $\ce{I2 -> I^{-} + IO3^{-}}$ in basic medium.

While balancing, the $\ce{OH-}$ on the LHS correspond to the number of moles of $\ce{KOH}$ and the $\ce{IO3-}$ on the RHS correspond to the number of moles of $\ce{KIO3}$ (the $\ce{K}$ is just a distraction).

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1- Determining oxidation number, Oxidation and Reduction: $$ \ce{ \overset{\color{red}{(0)}}{I2}+ \overset{\color{black}{(+1)}}{K} \overset{\color{ black }{(-2)}}{O} \overset{\color{ black }{(+1)}}{H} -> \overset{\color{black}{(+1)}}{K} \overset{\color{green}{(-1)}}{I} +\overset{\color{black}{(+1)}}{K} \overset{\color{ blue }{(+5)}}{I}\overset{\color{ black }{(-2)}}{O3}+\overset{\color{black}{(+1)}}{H2} \overset{\color{black}{(-2)}}{O} }$$ When you look at the reaction: it is an auto-redox reaction, in which:

The atom$\ce{I}$: went from $\color{red}{(0)}$ to $\color{green} {(-1)}$, it gains electrons, so it is reduced.

The atom$\ce{I}$: went from $\color{red}{(0)}$ to $\color{blue}{(+5)}$, it lost electrons, so it is oxidized.

2- Separate the redox reaction into two half-reactions: one for the oxidation, and one for the reduction

$$\text {Reduction half: } \ce{ \overset{\color{red}{(0)}}{I2}+ {KOH} -> {K}\overset{\color{green}{(-1)}}{I} }$$ $$\text {Oxidation half: } \ce{ \overset{\color{red}{(0)}}{I2}+ {KOH} -> {K}\overset{\color{ blue }{(+5)}}{I}{O_3} }$$ 3- Balance all elements except oxygen and hydrogen in both half equations:

$$\text {Reduction half: } \ce{ \overset{\color{red}{(0)}}{I2}+ 2{KOH} -> 2{K}\overset{\color{green}{(-1)}}{I} }$$ $$\text {Oxidation half: } \ce{ \overset{\color{red}{(0)}}{I2}+ 2{KOH} -> 2{K}\overset{\color{ blue }{(+5)}}{I}{O_3} }$$ 4- Add the appropriate number of water molecules to that side of the equation required to balance the oxygen atoms:

$$\text {Reduction half: } \ce{ \overset{\color{red}{(0)}}{I2}+ 2{KOH} -> 2{K}\overset{\color{green}{(-1)}}{I} +2\ce{H2O}} $$ $$\text {Oxidation half: } \ce{ \overset{\color{red}{(0)}}{I2} + 2\ce{KOH} + 4\ce{H2O} -> 2\ce{K}\overset{\color{ blue }{(+5)}}{I}\ce{O3} }$$ 5- Add the appropriate number of hydrogen ions to that side of the equation required to balance the hydrogen atoms: $$\text {Reduction half: } \ce{ \overset{\color{red}{(0)}}{I2} + 2\ce{KOH} + 2\ce{H+} -> 2\ce{K}\overset{\color{green}{(-1)}}{I} +2\ce{H2O}} $$ $$\text {Oxidation half: } \ce{ \overset{\color{red}{(0)}}{I2} + 2\ce{KOH} +4\ce{H2O} -> 2\ce{K}\overset{\color{ blue }{(+5)}}{I}\ce{O3}+ 10\ce{H+} }$$ 6- Use electrons ($\ce{e^-}$) to equalize the net charge on both sides of the equation: $$\text {Reduction half: } \ce{ \overset{\color{red}{(0)}}{I2}+ 2\ce{KOH} +2\ce{H+} + 2\ce{e-} -> 2{K}\overset{\color{green}{(-1)}}{I} + 2\ce{H2O} } $$ $$\text {Oxidation half: } \ce{ \overset{\color{red}{(0)}}{I2} + 2\ce{KOH} +4\ce{H2O} -> 2\ce{K}\overset{\color{ blue }{(+5)}}{I}\ce{O3} + 10\ce{H+} +10\ce{e-} }$$ 7- Make the electrons gained equal to the electrons lost: $$\text{Reduction half: } \ce{ 5\overset{\color{red}{(0)}}{I2}+ 10\ce{KOH} +10\ce{H+} +10\ce{e-} -> 10\ce{K}\overset{\color{green}{(-1)}}{I} + 10\ce{H2O} } $$ $$\text {Oxidation half: } \ce{ \overset{\color{red}{(0)}}{I2} + 2\ce{KOH} +4\ce{H2O} -> 2\ce{K}\overset{\color{ blue }{(+5)}}{I}\ce{O3} + 10\ce{H+} +10\ce{e-} }$$ 8- Combine the two half-reactions:

$ \ce{5\overset{\color{red}{(0)}}{I2} + 10\ce{KOH} +10\ce{H+} +10\ce{e-} + \overset{\color{red}{(0)}}{I2} + 2\ce{KOH} + 4\ce{H2O} -> 10\ce{K}\overset{\color{green}{(-1)}}{I} + 10\ce{H2O} + 2\ce{K}\overset{\color{ blue }{(+5)}}{I}\ce{O3}+ 10\ce{H+} + 10\ce{e-} } $

9- Cancel out the electrons and anything that is the same on both sides then collect the things that is the same on the same side to form one balanced redox equation: $$ \ce{6\overset{\color{red}{(0)}}{I2}+ 12\ce{KOH} -> 10\ce{K}\overset{\color{green}{(-1)}}{I} + 6\ce{H2O} + 2\ce{K}\overset{\color{ blue }{(+5)}}{I}\ce{O3} }$$

10- Shorten the equation in a simplified form: $$ \ce{3\overset{\color{red}{(0)}}{I2}+ 6\ce{KOH} -> 5\ce{K}\overset{\color{green}{(-1)}}{I} + 3\ce{H2O} + \ce{K}\overset{\color{ blue }{(+5)}}{I}\ce{O3} } $$

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Here's a little guide.

  1. Start with determining "relevant" species (those that reduced or oxidized): $$\ce{I2^0 + K+ + OH- -> K+ + I- + K+ + IO3- + H2O}$$

  2. Notice that you have $\ce{OH-}$ here, thus, you have basic conditions.

  3. Half-reactions

    $\ce{I^0 -> I^-}$

    $\ce{I^0 -> IO3-}$

So, you should "compensate" oxygens in $2^{nd}$ reaction (either using water or $\ce{OH-}$, since your conditions are basic, you should use $\ce{OH-}$ and "add" water to the right to compensate for "new" hydrogens).

  1. $\ce{I^0 + 6OH- -> IO3- + 6H2O}$

Usually you simply use twice as many $\ce{OH-}$ as "compensated oxygens". Here you have 3 of 'em, so you should use $\ce{6OH-}$.

  1. New half-reactions, but with electrons so net charge is 0:

$\ce{I^0 + e- -> I^-}$

$\ce{I^0 + 6OH- - 5e- -> IO3- + 6H2O}$

  1. Balance electrons so number of electons for each half-reaction is equal (by multiplication)

$\ce{5I^0 + 5e- -> 5I^-}$

$\ce{I^0 + 6OH- - 5e- -> IO3- + 6H2O}$

then add halves:$$\ce{3I2^0 + 6OH- -> IO3- + 5I^- + 6H2O}$$

  1. Almost there! Don't forget about cations:

$$\ce{3I2 + 6KOH -> KIO3 + 5KI + 6H2O}$$

Equation is balanced :)

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  • 1
    $\begingroup$ Quibbling I'd write the half cell as $\ce{I^0 + 6OH- -> IO3- + 6H2O + 5e- }$ to better conform how it will be written in standard reduction tables. $\endgroup$ – MaxW Dec 10 '18 at 14:15

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