Chemistry · Redox Equilibrium · NEET
Find the oxidation number of the changing element before and after the reaction. The n-factor is the total change in oxidation number per one formula unit (multiplied by the number of such atoms). For a reductant it is the electrons lost; for an oxidant it is the electrons gained. Example: in KMnO4 (acidic) Mn changes +7 to +2, a change of 5, so n = 5.
Because Mn ends up at a different oxidation state. In acid Mn goes +7 to +2 (n = 5). In neutral or faintly alkaline medium it goes +7 to +4, forming MnO2 (n = 3). In strongly basic medium it goes +7 to +6, forming MnO4^2- (n = 1). Same molecule, but n-factor and equivalent weight change with the medium.
Oxalic acid (H2C2O4) has two carbon atoms. Each carbon is at +3 and each goes to +4 in CO2, losing 1 electron. Two carbons lose 2 electrons in total, so n-factor = 2. Do not confuse this with its acidic n-factor of 2 (it has 2 replaceable H) — here we count electrons because it is acting as a reductant.
Equivalent weight = molar mass / n-factor. For KMnO4 (molar mass 158) in acid, equivalent weight = 158 / 5 = 31.6 g. For oxalic acid dihydrate (molar mass 126) in redox, equivalent weight = 126 / 2 = 63 g. Milli-equivalents = normality x volume(mL) = molarity x n-factor x volume.
Total electrons lost must equal total electrons gained. So multiply each species by a number that makes the electron counts equal. For MnO4^- (n = 5) reacting with C2O4^2- (n = 2), take 2 MnO4^- and 5 C2O4^2- because 2 x 5 = 5 x 2 = 10 electrons. The mole ratio is the inverse of the n-factors.
Valency is a fixed combining capacity of an element. n-factor depends on the specific reaction: for acids it is the number of H+ given, for bases the number of OH-, and for redox species it is the electrons transferred. The same substance can have different n-factors in different reactions, but its valency does not change with the reaction.
For the redox reaction MnO4^- + C2O4^2- + H^+ -> Mn^2+ + CO2 + H2O, the correct coefficients of the reactants (MnO4^-, C2O4^2-, H^+ respectively) are:
In acidic medium, 10 mL of 0.25 M oxalic acid is titrated with KMnO4 solution. If the volume of KMnO4 required to reach the end point is 10 mL, the strength of the KMnO4 solution is:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
In K2Cr2O7 there are 2 chromium atoms, each going from +6 to +3 (a change of 3). Total electrons gained = 2 x 3 = 6, so the n-factor is 6 and the equivalent weight is 294 / 6 = 49 g.
For most NEET redox species yes, because electrons are counted as whole numbers per atom. It can appear fractional per formula unit in unusual disproportionation cases, but for standard titration species (KMnO4, K2Cr2O7, oxalic acid, Mohr salt) it is a small whole number.
Fe^2+ is oxidised to Fe^3+, a change of 1 electron, so its n-factor is 1. Against KMnO4 (n = 5 in acid), 5 moles of Fe^2+ react with 1 mole of KMnO4.
At the end point the equivalents of oxidant equal the equivalents of reductant, regardless of their mole ratio. Using equivalents (normality x volume) lets you solve the titration in one line without first balancing the whole equation.
Its numeric value happens to be 2 in both roles, but for a different reason: 2 replaceable H+ in acid-base, and 2 electrons lost by its two carbons in redox. Always identify the role first.
