Chemistry · Electrochemistry · NEET
In MOLTEN NaCl there is no water. Only Na+ and Cl- are present. So sodium metal (Na) forms at the cathode and chlorine gas (Cl2) forms at the anode. In AQUEOUS NaCl, water is also present, giving H+ and OH- ions too. Now four ions compete. At the cathode H+ is discharged before Na+ (because Na is very hard to reduce), so hydrogen gas forms, not sodium. At the anode Cl- is discharged, giving chlorine gas. So aqueous NaCl gives H2 + Cl2 + NaOH, while molten NaCl gives Na + Cl2. This is why we melt the salt when we actually want the pure metal.
Dilute H2SO4 in water contains H+, OH- (from water) and SO4^2- ions. At the anode, negative ions are attracted. The SO4^2- ion is very stable and is NOT easily oxidised, so it stays in solution. Instead the OH- ions from water are discharged, releasing oxygen gas: 2H2O -> O2 + 4H+ + 4e-. This is a fixed NEET answer: the anode product with dilute H2SO4 on Pt electrodes is oxygen gas (O2), never SO2 or H2S.
At the CATHODE (reduction, positive ions come here), the ion with the HIGHER reduction potential is discharged first. That is why H+ (or a less reactive metal like Cu) is discharged before very reactive metals like Na+, K+, Ca^2+. At the ANODE (oxidation, negative ions come here), OH- is usually discharged before oxo-anions like SO4^2- and NO3-, but halides (Cl-, Br-, I-) are discharged before OH- when concentrated. Simple rule to remember: reactive metals and oxo-anions stay in solution; water's ions or easy ions leave as gas.
Sodium (Na+) has a very negative reduction potential, meaning it is extremely hard to reduce. Water and H+ are much easier to reduce. So the cathode reaction is 2H2O + 2e- -> H2 + 2OH- (or 2H+ + 2e- -> H2). Hydrogen bubbles off and OH- is left behind, which is why aqueous NaCl electrolysis makes NaOH (this is the chlor-alkali process). To actually get sodium metal you must use MOLTEN NaCl where no water is present.
Count the electrons in the electrode half-reaction. For Ca^2+ + 2e- -> Ca, one mole of Ca needs 2 Faradays. For 2Cl- -> Cl2 + 2e-, one mole of Cl2 gas needs 2 Faradays. For H+ + e- -> H (half of H2), one mole of H2 needs 2 F. Always write the balanced half-reaction, look at how many electrons appear, and that number is the Faradays per mole. This is the most tested calculation in NEET electrolysis questions.
On electrolysis of dilute sulphuric acid using platinum (Pt) electrodes, the product obtained at the anode will be:
The number of Faradays (F) required to produce 20 g of calcium from molten CaCl2 (atomic mass of Ca = 40 g/mol) is:
During the electrolysis of molten sodium chloride, the time required to produce 0.10 mol of chlorine gas using a current of 3 amperes is:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
When more than one ion can be discharged at an electrode, only one is chosen. The ion that is easier to discharge (higher reduction potential at the cathode, or easier to oxidise at the anode) is released first. This choice is called preferential discharge and it decides the final products of electrolysis.
Hydrogen gas (H2) at the cathode, chlorine gas (Cl2) at the anode, and sodium hydroxide (NaOH) left in the solution. This is called the chlor-alkali process. Note you do NOT get sodium metal from aqueous NaCl.
These metals are too reactive to be reduced from a water solution, because hydrogen from water is discharged first at the cathode. Melting the salt removes water, so only the metal ion can gain electrons and the pure metal is deposited.
Yes. Inert electrodes like Pt or graphite do not react, so only the ions in solution are discharged. Reactive electrodes (like a copper anode in CuSO4) can themselves dissolve, changing the products. NEET usually specifies Pt or inert electrodes.
Electrolysis product questions and Faraday calculations appear almost every year. Knowing the molten-vs-aqueous rule and the oxo-anion (O2 at anode) trap lets you answer both theory MCQs and numerical questions quickly and correctly.
