ElectrochemistryMedium
Electrolysis — Faraday's law
How many grams of copper deposit at the cathode passing 2.50 A for 30.0 min through CuSO₄ solution?
Cu²⁺ + 2e⁻ → Cu(s). F = 96485 C/mol. M(Cu) = 63.55 g/mol.
Cu²⁺ + 2e⁻ → Cu(s). F = 96485 C/mol. M(Cu) = 63.55 g/mol.
Given data
I = 2.50 At = 30.0 min = 1800 sF = 96485 C/molM(Cu) = 63.55 g/moln = 2 e⁻/Cu- Calculate total charge Q = I·t.
- Calculate moles of Cu deposited: n(Cu) = Q/(n·F).
- Calculate mass of Cu = n·M.
Full worked solution
- Calculate total charge Q = I·t.Q = 2.50 × 1800 = 4500 C.
- Calculate moles of Cu deposited: n(Cu) = Q/(n·F).n(Cu) = 4500/(2×96485) = 0.02332 mol.
- Calculate mass of Cu = n·M.m = 0.02332 × 63.55 = 1.482 g.
Result:1.48 g of copper deposited.