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

$$\begin{align}\ce{Cu^2+ + e- &amp; &lt;=&gt; Cu+}\tag{red}\\
\ce{2I- &amp; &lt;=&gt; I2 + 2 e-}\tag{ox}\\
\ce{2Cu^2+ + 2 I- &amp; &lt;=&gt; 2 Cu+ + I2}\tag{redox}
\end{align}$$

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

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

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:

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

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

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

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.)

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

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

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

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.