# Video: Predicting the Products and Their Amounts for the Reaction between Bromine and Potassium Iodide Solution

One mole of bromine gas reacts completely with excess potassium iodide solution, according to the given equation. Br₂ + 2I⁻ (excess) ⟶ ＿. Which of the following will be obtained? [A] 1 mole of Br⁻ and 2 moles of I⁻ [B] 1 mole of Br⁻ and o1ne mole of I⁻ [C] 1 mole of Br⁻ and 1 mole of I₂ [D] 2 moles of Br⁻ and 1 mole of I₂ [E] 2 moles of Br⁻ and 2 moles of I₂

04:01

### Video Transcript

One mole of bromine gas reacts completely with excess potassium iodide solution, according to the given equation. Br₂ plus 2I⁻ in excess react to form blank. Which of the following will be obtained? A) One mole of Br⁻ and two moles of I⁻. B) One mole of Br⁻ and one mole of I⁻. C) One mole of Br⁻ and one mole of I₂. D) Two moles of Br⁻ and one mole of I₂. Or E) Two moles of Br⁻ and two moles of I₂.

Bromine is a halogen. A halogen is an element in group 17, sometimes known as group seven, of the periodic table. At the top of group 17, we have fluorine, followed by chlorine, followed by bromine. The other component of our reaction is potassium iodide. The potassium is a spectator ion. So we only have iodide in our equation. The potassium ion doesn’t react with anything in this equation. So it can be ignored. The name iodide derives from the name for the element, iodine. Iodine is the fourth element in group 17, just below bromine. Atoms of group 17 elements have seven valence electrons each and have a tendency to absorb an extra electron to form halides, x⁻ ions.

As we move down the group, we’re adding one shell each time. So the valence shell for iodine is much, much further from the nucleus than the valence shell for fluorine. This means as you go down the group, the attraction felt by that extra electron in the halide gets weaker. This means that as you go up the halogens, they become progressively better oxidizing agents. They release more energy when they oxidize something else and gain electrons themselves. Meanwhile, the halides, the negative ions of the halogens, become better reducing agents as they get bigger. The large iodide ion gives up its electron much more readily than the compact fluoride ion does.

So in our reaction, we expect that better oxidizing agent, Br₂, to take electrons from the better reducing agent, I⁻. So we’re going to form Br⁻ ions. In turn, the I⁻ ions will give up their electrons to Br₂, forming I₂. Remember, the natural state for any halogen is a diatom, where two individual atoms share an electron each to form a single bond. This fulfills their octets. So the products for this reaction are bromide and iodine.

The next thing we have to do is balance the equation. We have two bromine atoms on the left, but only one bromide ion on the right. So we need two bromide ions. However, the iodines are already balanced with two on both sides. The charge is balanced with minus two on both sides, meaning that we have a balanced equation. The equation tells us that we have one mole of bromine gas reacting completely. Br₂ and Br⁻ in this equation are in a ratio of one to two. So for one mole of Br₂ in, we should get two moles of Br⁻ out. Meanwhile, bromine and iodine are in a ratio of one to one. So one mole of bromine in means one mole of iodine out.

This output corresponds with option D, two moles of bromide and one mole of iodine.