Chemistry · Redox Equilibrium · NEET
At the cathode, the species with the higher (more positive) reduction potential is discharged. Na+ has E about -2.71 V, while water reducing to H2 (2H2O + 2e- to H2 + 2OH-) is far easier at about -0.83 V. Water is much easier to reduce than Na+, so H2 gas comes out and Na+ stays in solution. This is why you cannot get sodium metal from a water solution; you need MOLTEN NaCl (no water) to deposit sodium.
By pure E values, water oxidising to O2 (about +1.23 V) should beat Cl- oxidising to Cl2 (about +1.36 V). But real electrodes have OVERVOLTAGE (extra voltage needed for a gas to actually form). Oxygen has a high overvoltage on many electrodes, so in concentrated brine, Cl2 is released instead. This is a key exception NEET expects you to know: aqueous NaCl gives H2 at cathode and Cl2 at anode.
Molten NaCl has only Na+ and Cl- (no water), so you get sodium metal at the cathode and chlorine gas at the anode. Aqueous NaCl has water too; water competes and wins at the cathode, giving H2 gas (not Na), while Cl2 comes at the anode due to overvoltage. Same salt, different products, because water changes the game.
Two simple rules. Cathode (reduction): the ion or water with the HIGHER (more positive) reduction potential is discharged first. Anode (oxidation): the species with the LOWER (more negative) reduction potential, meaning easiest to oxidise, is discharged first. Then check if water can beat the ion, and remember overvoltage exceptions like Cl2 over O2 in concentrated solutions.
SO4^2- is very hard to oxidise (its S is already in the highest +6 state), so it is never discharged in water. Instead water is oxidised at the anode to give O2, and H+ (or water) is reduced at the cathode to give H2. The net result is simply the electrolysis of water. Ions like SO4^2- and NO3- are spectators in aqueous electrolysis.
If the metal ion has a reduction potential higher than water (more positive than about -0.83 V), the metal deposits. So Cu2+ (+0.34 V), Ag+ (+0.80 V) and similar unreactive-metal ions plate out as metal. But very reactive metal ions like Na+, K+, Ca2+, Al3+ have E far below water, so water wins and H2 is released instead.
Try the real previous-year questions from this chapter — each with the answer and a full solution.
Yes. NEET tests it as an application of electrode potentials (E values) in the Redox and Electrochemistry units. The classic favourite is aqueous vs molten NaCl, plus electrolysis of water from dilute H2SO4 or Na2SO4. Knowing the cathode/anode rules and the overvoltage exception covers most questions.
At the cathode, the species with the highest (most positive) reduction potential is discharged. Compare the metal ion with water (H2O reducing to H2 at about -0.83 V). Whichever is higher wins. Unreactive metals (Cu, Ag) deposit; reactive-metal ions (Na, K) lose to water and release H2.
At the anode, the species easiest to oxidise (lowest reduction potential) is discharged. Compare the anion with water (H2O oxidising to O2 at about +1.23 V). Ions like SO4^2- and NO3- never discharge (water gives O2), while halide ions like Cl-, Br-, I- do discharge as the free halogen.
Overvoltage is the extra voltage above the theoretical E needed to actually release a gas at an electrode. Oxygen has a high overvoltage, so even though water should give O2 before Cl- gives Cl2, in concentrated NaCl the practical product at the anode is Cl2. This is the key real-world exception NEET expects.
Yes. In concentrated NaCl (brine) the anode gives Cl2, but in very dilute NaCl the anode may give O2 (from water) because there are few chloride ions and the overvoltage effect is reduced. Concentration shifts which species wins the race at the electrode.
