Reaction between Copper(II) ions and Iodide Ions

If a solution containing Cu2+ ions and aqueous iodide ions
- Iodide ions are oxidised to iodine 2I− -----> I2 +2e-

- Copper (II) ions are reduced to copper (I) ions: Cu2+ + e- -----> Cu+

The final equation is
2Cu2+ + 4I- ---------> CuI(s) + I2


I dont understand how they got this final equation as I tried to balance the electrons and adding half equations but I don not end up with this equation.

Help?
Thank you x

They arrived at this answer (although it should have 2CuI(s) on the right side) by reasoning that when Cu2+ ions and I- ions are mixed, you get a precipitate of copper(II) iodide, CuI2. But CuI2 is unstable and breaks into copper(I) iodide (CuI) and iodine (I2).

Cu2+(aq) + 2I-(aq) ----> CuI2(s) ----> CuI(s) + 1/2 I2(aq) . . .or double everything to get
2Cu2+(aq) + 4I-(aq) ----> 2CuI2(s) ----> 2CuI(s) + I2(aq)