JEE Chemistry Redox Reactions Notes Set 01

Equivalent Concept

The equivalent weight of a substance is its molar mass divided by the n-factor. For acids and bases, n-factor is the number of H+ ions replaced or OH- ions neutralised per formula unit. For oxidising and reducing agents, it is the number of electrons gained or lost per formula unit in the given reaction. Equivalents = mass / equivalent weight = moles x n-factor. Normality = equivalents per litre = molarity x n-factor.
The key rule of volumetric analysis: at the end point, equivalents (or milliequivalents) of the two reactants are equal. For titrations written as N1V1 = N2V2, a back titration subtracts the leftover equivalents from the equivalents originally added.

n-factors to Remember

KMnO4: 5 in acid (to Mn2+), 3 in neutral or weakly basic medium (to MnO2), 1 in strongly basic medium (to MnO4 2-).
K2Cr2O7: 6 in acid (to Cr3+).
H2C2O4 and oxalates: 2 per C2O4 unit (to CO2).
FeSO4, Fe2+ to Fe3+: 1. Na2S2O3 to Na2S4O6: 1 per Na2S2O3, so equivalent weight equals the molar mass.
H2O2: 2 as a reducing agent (to O2) and 2 as an oxidising agent (to H2O).
I2 + 2 S2O3 2- gives 2 I- + S4O6 2-: iodine and hypo react in a 1 : 2 mole ratio.

Acid-Base Titrations Involving Carbonates

Na2CO3 reacts in two stages: with phenolphthalein, CO3 2- becomes HCO3- (half the total acid), and with methyl orange, HCO3- becomes CO2 (the other half). NaOH reacts fully at the phenolphthalein stage. So:
Mixture of Na2CO3 and NaOH: phenolphthalein titre = NaOH + half Na2CO3; the extra methyl orange titre equals half Na2CO3.
Mixture of Na2CO3 and NaHCO3: phenolphthalein titre = half Na2CO3; the extra methyl orange titre = half Na2CO3 + NaHCO3.
Remember that titre values are the volumes of the same acid solution, so they compare directly.

Redox Titrations

Common titrants: KMnO4 (self-indicator, used in acid), K2Cr2O7 (needs an indicator), iodine and hypo in iodometry. In iodometry, an oxidant such as Cu2+ or Cl2 releases iodine from iodide, and the iodine is titrated with hypo; the equivalents of oxidant equal the equivalents of hypo used. A double titration is used when two components react with the same titrant in different ways.

Worked example 1. How many mL of 0.1 N KMnO4 (in acid) are needed to oxidise 0.63 g of oxalic acid dihydrate (molar mass 126)?
Answer:
Equivalents of oxalic acid = 0.63/63 = 0.01 (n-factor 2).
Volume = 0.01/0.1 = 0.1 L = 100 mL.

Worked example 2. A 1 g sample of impure CaCO3 requires 18 mL of 1 N HCl for complete reaction. What is its percentage purity?
Answer:
Equivalents = 0.018, equivalent weight of CaCO3 = 50.
Mass = 0.9 g, so purity = 90%.