1. Identify the reaction involved

When a salt containing $Cu^{2+}$ reacts with potassium iodide, iodide ion $I^-$ acts as a reducing agent.

So, $Cu^{2+}$ is reduced to $Cu^+$ and iodide is oxidized to iodine.

2. Write the ionic change

The redox process is:

$2Cu^{2+} + 4I^- \rightarrow 2CuI + I_2$

3. Nature of the copper product

The $Cu^+$ formed combines with $I^-$ to give cuprous iodide:

$Cu^+ + I^- \rightarrow CuI$

Cuprous iodide is insoluble and precipitates as a white solid.

4. Check the options

• A: $Cu_2I_2$
  This is just the dimeric representation of $CuI$, but the standard formula of cuprous iodide is $CuI$.
• B: $Cu_2I_3$
  Incorrect formula.
• C: $CuI$
  Correct. This is cuprous iodide formed in the reaction.
• D: $Cu(I_3)_2$
  Incorrect in this context.

5. Final answer

The product formed is:

$\boxed{CuI}$

So, the correct option is C.