JEE Chemistry Redox Reactions Notes Set 07

Equivalent Weight and n-Factor

The equivalent weight of a substance is the mass that combines with or displaces 1.008 g of hydrogen, 8 g of oxygen or 35.5 g of chlorine. In calculations it is \( E=\dfrac{M}{n} \), where M is the molar (or atomic) mass and n is the n-factor, the number of moles of electrons, protons or charges exchanged per mole of the substance in the reaction considered. The value is therefore reaction-dependent.

Acids and bases: n is the number of H⁺ (or OH^-) actually replaced. \( \ce{H3PO4} \) has n = 1, 2 or 3 depending on whether it forms \( \ce{H2PO4^-} \), \( \ce{HPO4^{2-}} \) or \( \ce{PO4^{3-}} \). Hypophosphorous acid is monobasic and phosphorous acid dibasic. Salts in precipitation or double decomposition: n is the total charge on the cations (for \( \ce{Ca3(PO4)2} \), n = 6). Oxidants and reductants: n is the total change in oxidation number per formula unit, which is the number of electrons gained or lost.

Important redox n-factors: \( \ce{KMnO4} \) is 5 in acid (\( \ce{Mn^{2+}} \)), 3 in neutral or weakly basic medium (\( \ce{MnO2} \)) and 1 in strongly basic medium (\( \ce{MnO4^{2-}} \)); \( \ce{K2Cr2O7} \) is 6; \( \ce{H2O2} \) is 2 either as oxidant or reductant; \( \ce{FeC2O4} \) is 3 (Fe(II) gives 1 and oxalate 2); \( \ce{Cu2S} \) is 8 when it goes to \( \ce{Cu^{2+}} \) and \( \ce{SO2} \); \( \ce{Na2S2O3} \) is 1 against iodine (to tetrathionate) but 8 against chlorine (to sulphate). In a disproportionation, n is found from the electrons gained or lost in one direction only; for \( \ce{3Cl2 + 6NaOH -> 5NaCl + NaClO3 + 3H2O} \), n per \( \ce{Cl2} \) = 5/3.

Normality and the Law of Equivalence

Normality is equivalents of solute per litre of solution: \( N=M\times n \) (M = molarity). Mixtures of acids add up as milliequivalents: \( N_{mix}=\dfrac{N_1V_1+N_2V_2}{V_1+V_2} \). Dilution keeps equivalents constant, \( N_1V_1=N_2V_2 \). The law of equivalence states that at the end point, equivalents (or milliequivalents) of one reactant equal those of the other, whatever the reaction type; this is the only equation needed in most titration problems.

Example 1. 70 percent (w/w) phosphoric acid of density 1.54 g/mL has \( 1000\times1.54\times0.70/98=11 \) mol per litre, so \( N=11\times3=33 \). Example 2. Mixing 100 mL of 0.2 M \( \ce{H2SO4} \) with 200 mL of 0.2 M HCl gives \( (40+40)/300=0.267\ \text{N} \).

Redox Titrations

Permanganate is its own indicator (the pink colour persists at the end point) and must be used only with dilute sulphuric acid; hydrochloric acid would be oxidised to chlorine and nitric acid would oxidise the reductant. Typical reactions in acid:

\( \ce{MnO4^- + 5Fe^{2+} + 8H+ -> Mn^{2+} + 5Fe^{3+} + 4H2O} \)
\( \ce{2MnO4^- + 5C2O4^{2-} + 16H+ -> 2Mn^{2+} + 10CO2 + 8H2O} \)
\( \ce{Cr2O7^{2-} + 6Fe^{2+} + 14H+ -> 2Cr^{3+} + 6Fe^{3+} + 7H2O} \)
\( \ce{2MnO4^- + 5H2O2 + 6H+ -> 2Mn^{2+} + 5O2 + 8H2O} \)

Oxalic acid and its salts behave both as acids (n = 2 for \( \ce{H2C2O4} \)) and as reductants (n = 2 as well). Ferrous oxalate has n = 3. In the reaction of peroxide with permanganate, the oxygen gas comes from the peroxide, and an isotopic label on peroxide appears in the oxygen, not in water.

Example. 25 mL of 0.0208 M oxalic acid needs \( 25\times0.0208\times\tfrac25/0.0154=13.5 \) mL of 0.0154 M permanganate, because 5 mol of oxalic acid react with 2 mol of permanganate.

Iodometry and Iodimetry

In iodimetry a reducing agent is titrated directly with standard iodine. In iodometry an oxidising agent is allowed to react with excess iodide, and the iodine liberated is titrated with hypo (sodium thiosulphate): \( \ce{I2 + 2S2O3^{2-} -> 2I^- + S4O6^{2-}} \). Millimoles of hypo therefore equal milliequivalents of the oxidant. Starch added near the end point gives a blue colour that disappears at the end point.

Common iodometric relations: \( \ce{2Cu^{2+} + 4I^- -> 2CuI + I2} \) (n = 1 for copper); \( \ce{Cr2O7^{2-}} \) liberates 3 mol of iodine; \( \ce{IO3^-} \) and \( \ce{BrO3^-} \) take 6 electrons to give the halide; ozone gives one mole of iodine per mole, with oxygen unchanged; \( \ce{AsO4^{3-} + 2H+ + 2I^- -> AsO3^{3-} + H2O + I2} \) (arsenate takes 2 electrons).

Hydrogen Peroxide, Bleaching Powder and Hardness

Volume strength of hydrogen peroxide is the volume of oxygen at STP released by one volume of the solution on decomposition. From \( \ce{2H2O2 -> 2H2O + O2} \): 68 g gives 22.4 L, so volume strength \( =5.6\times N=11.2\times M \) and strength in g/L \( =\text{vol. strength}\times 3.04 \). A 11.2 V solution is 1 M, 2 N and 3.4 percent (w/v).

Available chlorine in bleaching powder is the chlorine released by dilute acid, expressed as a percentage of the sample. Taking the formula as \( \ce{CaOCl2} \) (127 g/mol) the maximum is 71/127, or 55.9 percent. It is estimated iodometrically: the powder is treated with excess KI and acetic acid, and the iodine is titrated with hypo; each mole of hypo corresponds to 35.5 g of available chlorine.

Hardness is due to calcium and magnesium salts. Temporary hardness (bicarbonates) is removed by boiling or by Clark's method, in which lime reacts 1 : 1 with calcium bicarbonate: \( \ce{Ca(HCO3)2 + Ca(OH)2 -> 2CaCO3 + 2H2O} \). Permanent hardness (chlorides and sulphates) is removed by washing soda, which precipitates the carbonate, or by ion-exchange resins that swap the metal ions for H⁺ or Na⁺. Hardness is expressed in ppm as grams of calcium carbonate equivalent per million grams of water.

Acid-Base Titrations with Mixed Alkalis

Phenolphthalein changes colour near pH 8-9, so it marks neutralisation of NaOH and conversion of carbonate to bicarbonate only. Methyl orange changes near pH 3-4 and marks complete neutralisation to carbonic acid. If phenolphthalein uses volume x and methyl orange uses total volume y of the same acid: NaOH (mmol-equivalent) = x - (y - x), carbonate = 2(y - x) equivalents, and for a bicarbonate-carbonate mixture the carbonate is 2x and bicarbonate y - 2x in equivalents of acid.

Example. 40 mL of 0.05 M sodium sesquicarbonate (\( \ce{Na2CO3.NaHCO3.2H2O} \)) contains 2 mmol of each ion. Phenolphthalein uses 40 mL of 0.05 M HCl (2 mmol), methyl orange 120 mL (6 mmol), so the difference is 80 mL.

Kjeldahl's method converts nitrogen in an organic compound to ammonia, which is absorbed in a known excess of standard acid and back-titrated: percent N = \( \dfrac{1.4\times(\text{meq of acid used by } \ce{NH3})}{\text{mass of sample in g}} \).