To get the quantity of electricity for discharge at an electrode for each ions, we multiply the number of electrons by the number of moles, since 1 mole = 1F

• 4.0 moles of Cl\(^-\)  → 1e\(^-\) x  4 = 4F
• 3.0 moles of Na\(^+\)  → 3e\(^-\) x 1 = 3F
• 2.5 moles of Cu\(^{2+}\) → 2e\(^-\) x 2.5 = 5F
• 2.0 moles of Al\(^{3+}\)  → 3e\(^-\) x 2  = 6F.  The Ions that require the largest quantity of electricity for discharge at an electrode is the Al\(^{3+}\).