# What is reduced and oxidized when potassium bromate reacts to oxygen and postassium bromide?

I'm trying to help lead my daughter to a solution and my high school chemistry is almost 30 years ago.

For the following reaction, indicate the species being oxidezed and reduced and show the oxidation numbers above these symbols. Lastly, indicate the number of electrons that have been lost or gained.

$$\ce{2 KBrO3 -> 2 KBr + 3 O2}$$

What is being oxidized, and what is being reduced? My daughter tells me that since the charge on the two components doesn't change there isn't reduction or oxidation. You could almost say the $\ce{KBr}$ is being 'de-oxidized' because it's losing oxygen, but that would mean the oxygen is being oxidized, which in this case seems nonsensical.

If someone can explain to me which it is and why, so I can walk her through it, or direct me to a place where this is done, I'd appreciate it. This is beyond my knowledge.

• One would hope that a concept that fundamental is explained at a high school level in the high school level textbook, which the school is using to teach a high school chemistry class. – Martin - マーチン Aug 20 '18 at 14:52
• She's taking the course online, and has talked about some of the oversights and misplaced information she's encountered. I have the sense this course was not constructed well. – Keith Davies Aug 20 '18 at 19:31

1. Defining oxidation number (O.N.), Oxidation and Reduction

• Oxidation number, also called oxidation state, is an indicator of the degree of oxidation (loss of electrons) of an atom in a chemical compound

• Oxidation is the loss of electrons or an increase in oxidation number by a molecule, atom, or ion.

• Reduction is the gain of electrons or a decrease in oxidation number by a molecule, atom, or ion.

2. To calculate the oxidation number of $\ce{Br}$ in $\ce{KBrO3}$:

$\ce{O}$ is the most electronegative element in the compound, assume it gets all the electrons and it therefore has the oxidation number $-2$. $\ce{K}$ has $+1$, so $\ce{Br}$ gets $+5$ according to the following calculation: \begin{align} 1 \times (\text{O.N. of K}) + 1 \times (\text{O.N. of Br}) + 3 \times{(\text{O.N. of O})} &= 0\\ 1 \times (+1) + 1 \times (\text{O.N. of Br}) + 3 \times (-2) &= 0\\ \text{O.N. of Br} &= (+5) \end{align}

3. To calculate the oxidation number of $\ce{Br}$ in $\ce{KBr}$: \begin{align} 1 \times (\text{O.N. of K}) + 1 \times (\text{O.N. of Br}) &= 0 \\ 1 \times (+1) + 1 \times(\text{O.N. of Br}) &= 0\\ \text{O.N. of Br} &= (-1) \end{align}

4. Look at $\ce{O2}$, it gets $0$.

In total that will give you: $$\ce{ 2\overset{\color{black}{(+1)}}{K} \overset{\color{red}{(+5)}}{Br} \overset{\color{green}{(-2)}}{O_3} -> 2\overset{\color{black}{(+1)}}{K} \overset{\color{red}{(-1)}}{Br} +3 \overset{\color{green}{(0)}}{O2}}$$

When you look at the reaction: it is an auto-redox reaction, in which

• $\ce{Br}$ went from $+5$ to $-1$, it gains electrons, so it is reduced.
• $\ce{O}$ went from $-2$ to $0$, it lost electrons, so it is oxidized.
• Ah ha. Of course. The reaction does not necessarily involve changing the charge on each molecule, but can rearrange how they are 'on the atoms' of each molecule. Thank you for explaining this so simply, my high school chemistry was long, long ago. – Keith Davies Aug 20 '18 at 4:58

This is not a reaction where ions exchange electrons, but the formal oxidation numbers do change definitively.

Electronegativity of O is 3.5, Br 2.7. That's a rather large difference, so in $\ce{KBrO3}$, that gives O its preferred formal oxidation state -2, K has +1, so Br gets the difference +5. I think you can take it from here.

And yes, the oxygen gets "oxidised", nothing wrong with that! Look at the reverse reaction: Oxygen oxidises KBr to the Bromate, so it obviously itself gets reduced.