Michael Faraday was the first scientist who described the quantitative aspects of electrolysis. Now Faraday’s laws also flow from what has been discussed earlier. Faraday’s Laws of Electrolysis After his extensive investigations on electrolysis of solutions and melts of electrolytes, Faraday published his results during 1833-34 in the form of the following well known Faraday’s two laws of electrolysis:
(i) First Law: The amount of chemical reaction which occurs at any electrode during electrolysis by a current is proportional to the quantity of electricity passed through the electrolyte (solution or melt).
(ii) Second Law: The amounts of different substances liberated by the same quantity of electricity passing through the electrolytic solution are proportional to their chemical equivalent weights (Atomic Mass of Metal ÷ Number of electrons required to reduce the cation).
Impure copper is converted into copper of high purity by electrolysis, using an aqueous solution of copper sulphate. Which of the following correctly describes this process?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → Impure copper is made the anode; it dissolves as Cu2+ ions, and pure copper is deposited at the cathode.
In an electrolytic cell with copper electrodes in copper sulphate solution, copper is oxidised at the anode and reduced at the cathode. So the impure copper is made the anode; it dissolves as Cu2+ ions, and these ions are discharged as pure copper on the cathode:
Anode: \( \displaystyle \ce{Cu(s) -> Cu^2+(aq) + 2e^-} \)
Cathode: \( \displaystyle \ce{Cu^2+(aq) + 2e^- -> Cu(s)} \)
Why the other options are incorrect
Option (1) Impure copper is made the cathode, and pure copper from the solution is deposited on it. — Incorrect. Copper deposited on an impure cathode would stay mixed with the impurities. The impure metal must dissolve, and dissolving (oxidation) takes place only at the anode.
Option (3) Impure copper is made the anode; oxygen is evolved at it, and pure copper is deposited at the cathode. — Incorrect. A copper anode is a reactive electrode: it takes part in the electrode reaction and is itself oxidised to Cu2+. Oxygen would come from water only at an inert anode such as platinum.
Option (4) Impure copper is made the anode; it dissolves as Cu2+ ions, and hydrogen is evolved at the cathode. — Incorrect. The Cu2+ ions in the solution are reduced at the cathode (E° of Cu2+/Cu is +0.34 V, higher than 0.00 V for H+/H2), so copper, not hydrogen, is deposited there.
NCERT (2026-27 edition):
2.5 Electrolytic Cells and Electrolysis – “…One of the simplest electrolytic cell consists of two copper strips dipping in an aqueous solution of copper sulphate. If a DC voltage is applied to the two electrodes, then Cu2+ ions discharge at the cathode (negatively charged) … Thus copper is dissolved (oxidised) at anode and deposited (reduced) at cathode. This is the basis for an industrial process in which impure copper is converted into copper of high purity. The impure copper is made an anode that dissolves on passing current and pure copper is deposited at the cathode.”