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).
An aqueous solution of sodium chloride is electrolysed using inert electrodes. Which of the following is the net reaction at the cathode?
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( \ce{H2O(l) + e^- -> 1/2H2(g) + OH^-(aq)} \)
The cathode is where reduction takes place. In aqueous NaCl there are H2O molecules and H+ ions besides Na+, so two reductions compete at the cathode:
\( \displaystyle \ce{Na+(aq) + e^- -> Na(s)} \qquad E^\circ = -2.71\ \mathrm{V} \)
\( \displaystyle \ce{H+(aq) + e^- -> 1/2H2(g)} \qquad E^\circ = 0.00\ \mathrm{V} \)
The reaction with the higher E° is preferred, so hydrogen is set free, not sodium. The H+ ions come from water, \( \ce{H2O(l) -> H+(aq) + OH^-(aq)} \), and adding the two steps gives the net cathode reaction:
\( \displaystyle \ce{H2O(l) + e^- -> 1/2H2(g) + OH^-(aq)} \)
The OH− ions formed stay in the solution with Na+, which is why NaOH is one of the products.
Why the other options are incorrect
Option (1) \( \ce{Na+(aq) + e^- -> Na(s)} \) — Incorrect. This reduction has the much lower E° (−2.71 V), so it loses to the reduction of hydrogen. Sodium is obtained at the cathode only from molten NaCl, where there is no water.
Option (2) \( \ce{Cl^-(aq) -> 1/2Cl2(g) + e^-} \) — Incorrect. Electrons are given out here, so this is an oxidation. It can take place only at the anode.
Option (3) \( \ce{2H2O(l) -> O2(g) + 4H+(aq) + 4e^-} \) — Incorrect. This too is an oxidation of water, one of the reactions possible at the anode, not at the cathode.
NCERT (2026-27 edition):
2.5.1 Products of Electrolysis – “At the cathode there is competition between the following reduction reactions:”
\( \displaystyle \ce{Na+(aq) + e^- -> Na(s)} \qquad E^\circ = -2.71\ \mathrm{V} \)
\( \displaystyle \ce{H+(aq) + e^- -> 1/2H2(g)} \qquad E^\circ = 0.00\ \mathrm{V} \)
“The reaction with higher value of E° is preferred and therefore, the reaction at the cathode during electrolysis is:
H+(aq) + e– → ½ H2(g) (2.33)
but H+(aq) is produced by the dissociation of H2O, i.e.,
H2O(l) → H+(aq) + OH–(aq) (2.34).
Therefore, the net reaction at the cathode may be written as the sum of (2.33) and (2.34) and we have
H2O(l) + e– → ½H2(g) + OH– (2.35)”