Clear Notion
Login

Concepts

Read through concept notes with worked explanations and diagrams, chapter by chapter.

Concept 04 / 04Redox Reactions

Redox Titrations and the n-Factor of Oxidising and Reducing Agents

Using redox reactions in volumetric analysis: n-factor (valence factor) of oxidants and reductants, KMnO4 in acidic, neutral and basic media, dichromate titrations, iodometry and iodimetry with thiosulfate, the law of equivalence, and choice of acid and indicator.

Figure 1
Redox titrations determine the amount of an oxidant or reductant by titrating it against a standard solution of the other. The key idea is the law of equivalence: at the end point, the number of electrons given by the reductant equals the number accepted by the oxidant. n-factor (valence factor). For an oxidant or reductant, the n-factor is the number of electrons gained or lost per formula unit (the total change in ON per formula unit). Then equivalents == moles ×\times n-factor, equivalent weight =Mn= \frac{M}{n}, normality == molarity ×\times n, and at the equivalence point N1V1=N2V2N_{1}V_{1} = N_{2}V_{2}, or equivalently n1M1V1=n2M2V2n_{1}M_{1}V_{1} = n_{2}M_{2}V_{2}. Common n-factors. KMnO4\mathrm{KMnO_{4}}: acidic 55 (to Mn2+\mathrm{Mn^{2+}}), neutral or weakly basic 33 (to MnO2\mathrm{MnO_{2}}), strongly basic 11 (to MnO42−\mathrm{MnO_{4}^{2-}}) (see figure). K2Cr2O7\mathrm{K_{2}Cr_{2}O_{7}} (acid): 66. Fe2+\mathrm{Fe^{2+}}: 11. C2O42−\mathrm{C_{2}O_{4}^{2-}} (oxalate to CO2\mathrm{CO_{2}}): 22. H2O2\mathrm{H_{2}O_{2}}: 22 (as oxidant or reductant). S2O32−\mathrm{S_{2}O_{3}^{2-}} to S4O62−\mathrm{S_{4}O_{6}^{2-}}: 11. I2\mathrm{I_{2}} to I−\mathrm{I^{-}}: 22. For a species where two elements change (as in FeC2O4\mathrm{FeC_{2}O_{4}} oxidised by KMnO4\mathrm{KMnO_{4}}: Fe +2→+3+2 \to +3 and C2\mathrm{C_{2}} +6→+8+6 \to +8), add the changes: n =1+2=3= 1 + 2 = 3. Permanganometry. KMnO4\mathrm{KMnO_{4}} is used in dilute H2SO4\mathrm{H_{2}SO_{4}}. It acts as its own indicator: the first excess drop gives a permanent pink colour. HCl is avoided because permanganate oxidises Cl−\mathrm{Cl^{-}} to Cl2\mathrm{Cl_{2}}; HNO3\mathrm{HNO_{3}} is avoided because it is itself an oxidant. Oxalic acid titrations are warmed to about 60∘C60^{\circ}\mathrm{C} because the reaction is slow until Mn2+\mathrm{Mn^{2+}} (an autocatalyst) forms. KMnO4\mathrm{KMnO_{4}} is not a primary standard and is standardised against oxalic acid or Mohr's salt. Dichromatometry. K2Cr2O7\mathrm{K_{2}Cr_{2}O_{7}} is a primary standard (pure, stable) and can be used in the presence of HCl. An external or internal indicator (diphenylamine) is needed, because the orange-to-green change is not sharp. Iodometry and iodimetry. Iodimetry: direct titration with standard I2\mathrm{I_{2}} (for reductants such as SO32−\mathrm{SO_{3}^{2-}}, As2O3\mathrm{As_{2}O_{3}}). Iodometry: an oxidant liberates I2\mathrm{I_{2}} from excess KI, and the I2\mathrm{I_{2}} is titrated with standard sodium thiosulfate using starch (blue to colourless): 2Cu2++4I−→2CuI+I2\mathrm{2Cu^{2+} + 4I^{-} \rightarrow 2CuI + I_{2}}; I2+2S2O32−→2I−+S4O62−\mathrm{I_{2} + 2S_{2}O_{3}^{2-} \rightarrow 2I^{-} + S_{4}O_{6}^{2-}}. So 2 mol2\,\mathrm{mol} Cu2+\mathrm{Cu^{2+}} correspond to 2 mol2\,\mathrm{mol} thiosulfate. Key Takeaways: - n-factor == electrons transferred per formula unit; equivalents == moles ×\times n. - At equivalence: N1V1=N2V2N_{1}V_{1} = N_{2}V_{2}. - KMnO4\mathrm{KMnO_{4}}: n =5= 5, 33, 11 in acidic, neutral, strongly basic media. - KMnO4\mathrm{KMnO_{4}} is a self-indicator; use H2SO4\mathrm{H_{2}SO_{4}}, not HCl or HNO3\mathrm{HNO_{3}}. - Pitfall: in iodometry, relate the oxidant to thiosulfate through the liberated I2\mathrm{I_{2}}, not directly. Brain-Teaser 1: What volume of 0.02 M0.02\,\mathrm{M} KMnO4\mathrm{KMnO_{4}} is needed to oxidise 20 mL20\,\mathrm{mL} of 0.1 M0.1\,\mathrm{M} FeSO4\mathrm{FeSO_{4}} in acid? Solution: Equivalents of Fe2+\mathrm{Fe^{2+}}: 20×0.1×1=2 meq20 \times 0.1 \times 1 = 2\,\mathrm{meq}. KMnO4\mathrm{KMnO_{4}} normality =0.02×5=0.1 N= 0.02 \times 5 = 0.1\,\mathrm{N}. Volume =20.1=20 mL= \frac{2}{0.1} = 20\,\mathrm{mL}. Brain-Teaser 2: How many moles of KMnO4\mathrm{KMnO_{4}} (acidic) are needed to oxidise one mole of ferrous oxalate FeC2O4\mathrm{FeC_{2}O_{4}} completely? Solution: Ferrous oxalate loses 1e−1e^{-} from iron and 2e−2e^{-} from oxalate: n =3= 3, so 3 mol3\,\mathrm{mol} of electrons. Each MnO4−\mathrm{MnO_{4}^{-}} accepts 55: 35=0.6 mol\frac{3}{5} = 0.6\,\mathrm{mol} KMnO4\mathrm{KMnO_{4}}. Brain-Teaser 3: A sample of CuSO4\mathrm{CuSO_{4}} is treated with excess KI, and the liberated iodine needs 25 mL25\,\mathrm{mL} of 0.1 M0.1\,\mathrm{M} Na2S2O3\mathrm{Na_{2}S_{2}O_{3}}. Find the mass of Cu2+\mathrm{Cu^{2+}} in the sample. Solution: Thiosulfate: 2.5 mmol2.5\,\mathrm{mmol}, which reacts with 1.25 mmol1.25\,\mathrm{mmol} I2\mathrm{I_{2}}, liberated by 2.5 mmol2.5\,\mathrm{mmol} Cu2+\mathrm{Cu^{2+}}. Mass =2.5×10−3×63.5=0.159 g= 2.5 \times 10^{-3} \times 63.5 = 0.159\,\mathrm{g}.

Practice this concept

Free · no account required · instant solutions

Redox Titrations and the n-Factor of Oxidising and Reducing Agents — DPP

10 questionsSee the solution the moment you answer

Practice