Chemistry · Redox Equilibrium · NEET
A half reaction is one part of a redox reaction that shows only the electron loss OR only the electron gain, not both. A full redox reaction always has two half reactions happening together: one oxidation half and one reduction half. We write them separately so we can clearly see how many electrons move. For NEET this makes balancing and electrode-potential problems much easier.
In the OXIDATION half, the species loses electrons, so electrons (e-) go on the RIGHT (product side). Example: Zn -> Zn2+ + 2e-. In the REDUCTION half, the species gains electrons, so electrons go on the LEFT (reactant side). Example: Cu2+ + 2e- -> Cu. Remember: reduction PULLS electrons IN, so they sit on the left with the reactants.
Step 1: Find the oxidation number of each atom on both sides. Step 2: The element whose oxidation number goes UP is oxidised — write its half reaction with electrons on the right. Step 3: The element whose oxidation number goes DOWN is reduced — write its half with electrons on the left. Step 4: The number of electrons lost = number gained. For Zn + Cu2+ -> Zn2+ + Cu, the halves are Zn -> Zn2+ + 2e- and Cu2+ + 2e- -> Cu.
Electrons are never created or destroyed. Every electron lost by the oxidised species is picked up by the reduced species. So if you balance both halves to the SAME number of electrons and add them, the e- terms appear equally on both sides and cancel out. This is exactly why the total electrons lost must equal total electrons gained — a key idea NEET tests in balancing questions.
The oxidation half loses electrons; its atom's oxidation number increases (goes more positive). The reduction half gains electrons; its atom's oxidation number decreases (goes more negative). The species that is oxidised is the reducing agent, and the species that is reduced is the oxidising agent. Both halves always occur together — you cannot have one without the other.
Yes, when the reaction happens in water. In acidic medium you balance oxygen using H2O and hydrogen using H+. In basic medium you use OH- and H2O. You add electrons to balance the charge. This is the half-reaction (ion-electron) method of balancing, which NEET uses for KMnO4 and K2Cr2O7 reactions.
Given below are half-cell reactions: MnO4- + 8H+ + 5e- -> Mn2+ + 4H2O, E°(MnO4-/Mn2+) = +1.510 V; ½O2 + 2H+ + 2e- -> H2O, E°(O2/H2O) = +1.223 V. Will the permanganate ion MnO4- liberate O2 from water in the presence of an acid?
For the redox reaction MnO4- + C2O4^2- + H+ -> Mn2+ + CO2 + H2O, the correct coefficients of the reactants (MnO4-, C2O4^2-, H+ respectively) are:
On balancing the redox reaction, the coefficients a, b, c are: a Cr2O7^2- + b SO3^2- + c H+ -> 2Cr3+ + b SO4^2- + (c/2) H2O
Try the real previous-year questions from this chapter — each with the answer and a full solution.
Yes, in electrochemistry each electrode carries out one half reaction. Oxidation happens at the anode and reduction happens at the cathode. So an electrode reaction is just a half reaction taking place at one electrode.
No. Electrons cannot simply disappear or appear. An oxidation half must always be paired with a reduction half so the electrons lost are taken up by another species. That is why redox reactions always come in pairs of halves.
Check the oxidation number. If it increases (becomes more positive), that half is oxidation and electrons are on the right. If it decreases (becomes more negative), that half is reduction and electrons are on the left.
They make balancing tough redox equations (like KMnO4 and K2Cr2O7 reactions) simple and systematic, and they are the basis of standard electrode potential and cell EMF questions, which appear almost every year.
The n-factor is the number of electrons transferred in that half reaction. For MnO4- to Mn2+ it is 5, and for oxalate to CO2 it is 2 per oxalate. NEET uses n-factor for equivalents in redox titration calculations.
