During the electrolysis of 200 ml of a solution containing both 0.1 M AgNO₃ and 0.2 M Cu(NO₃)₂, a current of 1.93 A was passed for 20 minutes. Calculate the mass of the metals deposited at the cathode.
- 2.16 g Ag and 1.28 g Cu
- 2.16 g Ag and 0.127 g Cu
- 1.08 g Ag and 0.64 g Cu
- 2.16 g Ag only
📝 Detailed explanation
Total charge Q = 1.93 × 1200 = 2316 C.Total Faradays = 2316 / 96500 = 0.024 F.
Moles Ag⁺ = 0.2 × 0.1 = 0.02 mol.
Moles Cu²⁺ = 0.2 × 0.2 = 0.04 mol.
Silver deposits first: 0.02 mol Ag requires 0.02 F.
Remaining Faradays = 0.024 - 0.02 = 0.004 F.
Copper requires 2 F per mole: moles Cu = 0.004 / 2 = 0.002 mol.
Mass Ag = 0.02 × 108 = 2.16 g.
Mass Cu = 0.002 × 63.5 = 0.127 g.