If 0.025 mol of Cu is produced in the reaction, how many moles of electrons flow through the wire?

As copper(I) ions cannot exist in aqueous solution, the starting material must be copper(II) ions.

Cu²⁺(aq) + 2e⁻ → Cu(s)
OR: Mole ration e⁻ : Cu = 2 : 1

No. of moles of Cu produced = 0.025 mol
No. of moles of e⁻ flow through the wire = (0.025 mol) × 2 = 0.050 mol