# What happens when potassium iodide is added to a copper (II) salt?

$$\ce{Cu^2+ + KI -> ??}$$

I thought of this reaction:

$$\ce{Cu^2+ + 2KI -> CuI2 + 2K+}$$

But since $\ce{CuI2}$ doesn't exist, the $\ce{I-}$ ions must reduce $\ce{Cu^2+}$ to $\ce{CuI}$ and simultaneously get oxidised to $\ce{I2}$ and thus it will proceed the following reaction:-

$$\ce{2Cu^2+ + 4KI → 2CuI + I2 + 2K+}$$

I gave this reaction to my teacher and he gave me an another version of the reaction:

$$\ce{2Cu^2+ + 5KI -> 2CuI + KI3}$$

He says that he performed the reaction experimentally and obtained a white salt ($\ce{CuI}$) in a brown solution ($\ce{KI3}$).

Now, I started wondering what will happen if I alter the reaction and add sodium/lithium iodide instead of potassium iodide. Will it form sodium/lithium triiodide. Definitely not/ Or maybe, it will form.

Two questions:

1. What is the correct reaction?
2. What will happen if we add sodium/lithium iodide instead of potassium iodide?

Related questions:

When writing a reaction, you should take care to keep charges and atoms balanced. So the very first reaction you postulated is good (it has charges and atoms balanced) but then it goes South.

If you are only interested in the redox process, then it makes sense to think only about those reactions that form part of the redox process.

\begin{align}\ce{Cu^2+ + e- & <=> Cu+}\tag{red}\\ \ce{2I- & <=> I2 + 2 e-}\tag{ox}\\ \ce{2Cu^2+ + 2 I- & <=> 2 Cu+ + I2}\tag{redox} \end{align}

The next step, however, is that the $\ce{Cu+}$ ions will combine with iodide to form insoluble $\ce{CuI}$. Thus, we need excess iodide.

$$\ce{2 Cu^2+ + 4 I- -> 2 CuI v + I2}\tag{precipitation}$$

At any of these points, you could decide to include counterions. For example, your iodide could be $\ce{KI}$ and your copper salt could be $\ce{CuSO4}$. That would give:

$$\ce{2 CuSO4 + 4 KI -> 2 CuI v + I2 + 2 K2SO4}\tag{counterions}$$

Or maybe you decide that only $\ce{KI}$ is important, then of course this becomes:

$$\ce{2 Cu^2+ + 4 KI -> 2 CuI v + I2 + 4 K+}\tag{only KI}$$

Finally, you will typically have an excess of $\ce{KI}$, so that that can further react with the iodine generated to redissolve it. (Iodine is poorly soluble in water, $\ce{KI3}$ is well soluble, hence Lugol’s iodine.)

\begin{align}\ce{I2 + KI & -> KI3}\tag{Lugol, balanced}\\ \ce{I2 + I- & -> I3-}\tag{Lugol, ionic}\end{align}

Thus, a full equation will look like any of the following:

\begin{align}\ce{2 Cu^2+ + 5 I- & -> 2 CuI v + I3-}\tag{full, ionic}\\ \ce{2 Cu^2+ + 5 KI & -> 2 CuI v + KI3 + 4 K+}\tag{full, semi-ionic}\\ \ce{2 CuSO4 + 5 KI & -> 2 CuI v + KI3 + 2 K2SO4}\tag{full, balanced}\end{align}

Since the alkaline halides are generally very similar to each other, I think that the reaction should work pretty much identically if potassium is replaced with sodium or lithium. Personally, I didn’t have a chance to perform it, since my inorganic student labs only had pre-mixed potassium iodide solutions.

• So, if replaced with sodium or lithium iodide, will it form sodium/lithium triiodide? I don't think the compound exist. Jun 16, 2016 at 1:52
• @NilayGhosh We are talking about aquaeous solutions. And I have no reason to believe that a sodium cation or a lithium one should cause the breakdown of the triiodide anion. It would be a simple experiment; add $\ce{NaCl}$ to a $\ce{KI3}$ solution and see what happens; likely nothing.
– Jan
Jun 16, 2016 at 10:16
• I've tried this reaction which proceeds immediately, but I am stuck on the redox potentials. I don't understand why I can't calculate a positive electrode potential. Cupric ions reduced to cuprous ions +0.15V Iodide oxidized to Iodine -0.54V It seems to me this reaction should NOT be spontaneous. But it does happen, quickly and readily. Jan 26 at 21:04
• @124c41 Two immediate possibilities: 1) the standard potential of $\ce{Cu^2+/Cu+}$ might be different from $\ce{Cu^2+/CuI}$; or 2) the equilibrium might be driven to the product side through solubility (CuI is very poorly soluble).
– Jan
Jan 27 at 8:52