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Objectives

Objectives. Define redox reactions . Apply Nernst equation . Apply redox titrations . Detect the end point of redox titration using different oxidants and reductants . Identify some important oxidizing agents . Redox reactions. Oxidation:

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Objectives

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  1. Objectives • Define redox reactions . • Apply Nernst equation . • Applyredox titrations . • Detect the end point of redox titration using different oxidants and reductants . • Identify some important oxidizing agents .

  2. Redox reactions • Oxidation: Is a loss of electrons to an oxidizing agent (which gets reduced) to give a higher or more positive oxidation state. • Reduction: Gain of electrons from a reducing agent (which gets oxidized) to give a lower or more negative oxidation state. • A reduction- oxidation reaction, commonly called redox reaction, is one that occurs between a reducing agent and an oxidizing agent: • Ox1 + Red2 Red1 + Ox2 Ox1 is reduced to Red1 and Red2 is oxidized to Ox2. • The oxidizing or reducing tendency of a substance depends mainly on its reduction potential.

  3. We can calculate the potential differences by using the standard electrode potentials for each half cell given in the following table: The one with lower potential acts as the anode (oxidized) and the higher acts as the cathode (reduced)

  4. How to write a cell • Usually the cell is written as follows: Anode/solution/cathode Ecell = Eright-Eleft = Ecathode- Eanode = E+ - E- • Notice that: for voltaic cells, the Ecell is usually positive . • For a reaction to be complete enough to obtain a sharp end point in a titration, there should be at least 0.2 t. 0.3 V difference between the two electrode potentials.

  5. Nernst equationEffect of concentration on Potential • The potential dependence on concentration can be represented by Nernst equation as follows: • aOx + ne bRed Where, E is the reduction potential at the specific concentrations, n is the number of electrons involved in the half-reaction, R is the gas constant (8.3143 coul Deg-1mol-1), T is the absolute temperature, F is Faraday constant (96487 coul eq-1)

  6. Notice the following: • At 25°C (298.16 K), the value of 2.303RT/F = 0.05916. • As usual, the concentration of pure substances such as precipitates and liquids are taken as unity. • The log term of the equation is the ratio of the right side concentrations over the left side concentrations. • Before and after equivalence point, we use the concentration of the component in excess to determine the E of the cell. • At equivalence point, we can apply the following simple equation :

  7. Example 1 • A solution is 10-3M in Cr2O72- and 10-2M in Cr3+. If the pH is 2.0 , what is the potential of this half-reaction? E°Cr2O72-,Cr3+= 1.33 • Cr2O72- + 14H+ + 6e 2Cr3+ + 7H2O E = E°Cr2O72-,Cr3+ - 0.059/6 log [Cr3+]2/[Cr2O72-][H+]14 = 1.33 – 0.0596/6 log (10-2)2/(10-3)(10-2)14 = 1.33 – 0.0596/6 log 1027 = 1.33 – 27(0.0596/6) = 1.06 V

  8. Example 25.0 ml of 0.10M Ce4+ solution is added to 5.0 ml of 0.30M Fe2+ solution . Calculate the potential of a platinum electrode dipped in the solution relative to NHE. • E°Ce4+/Ce3+ = 1.61, E°Fe3+/Fe2+ = 0.771 • We start with 0.30 x 5.0 = 1.5 mmol Fe2+, • and add 0.1 x 5.0 = 0.5mmol Ce4+, thus, • We form 0.50 mmol Fe3+ and Ce3+ and we have 1.0 mmol Fe2+ remaining. • Applying Nernst equation for Fe3+/Fe2+ we will use, • Fe3+ + e Fe2+ (0.5/10ml) (1.0/10ml) = 0.05M = 0.1M E = 0.771 – 0.0596 log [Fe2+]/[Fe3+] = 0.771 – 0.0596 log 0.1/0.05 = 0.753 V

  9. Titration of 100mL of 0.10M Fe2+ solution vis. 0.20 M Ce4+

  10. Visual detection of end point • Obviously, the end point can be determined by measuring potential with an indicator electrode relative to a reference and plotting this against the volume of titrant. • But in other titrations, it is usually more convenient to use a visual indicator. • There are three methods for visual detection : • Self indication. • Starch detection. • Redox indicators.

  11. Visual detection of end point (cont.) • 1. Self indication • If the reducing or oxidizing agent is coloured, the end point can be detected by the appearance or disappearance of the colour of the reagent. • e.g. KMnO4 where the end point can be detected by the pink colour of permanganate. • Specific Indicators • 2. Starch Indicator • This indicator is used for titrations involving iodine. Starch forms a dark blue colour with iodine (I2).

  12. Visual detection of end point (cont.) • 3. Redox indicators • These are highly coloured dyes that are weak reducing or oxidizing agents that can be oxidized or reduced and the colour of both forms are different. • There are not many good redox indicators. • Ferroin [tris(1,10-phenanthroline)iron(II) sulphate] is one of the best indicators that can be used in cerium(IV) titrations. It is oxidized from the red to pale blue at the end point. • Diphenylaminesulphonic acid, is used as an indicator for titrations with dichromate in acid solution. The colour at the end point is purple.

  13. Visual detection of end point (cont.)

  14. Titrations with oxidizing agentsI- Potassium permanganate • permanganateis a widely used oxidizing agent (E°= 1.51V) that can act as a self indicatorin acidic medium MnO4- + 8H+ + 5e Mn2+ + 4H2O • While in basic medium, it will be precipitated as brown MnO2. • Potassium permanganate is standardized by titration with primary standard sodium oxalate 5H2C2O4 + 2MnO4- + 6H+ 10CO2 + 2Mn2+ + 8H2O • The solution must be heated for rapid reaction. The reaction is catalyzed by the Mn2+ product and it goes very slowly until Mn2+ is formed. • Permanganate titration are not possible in the presence of chloride because it will be oxidized to chlorine. • 2MnO4- + 10HCl + 6H+ 5Cl2 + 2Mn2+ + 8H2O

  15. II- potassium dichromate • It is slightly weaker oxidizing agent than potassium permanganate. • The main advantage is its availability as a primary standard material. • It does not react with it HCL so the titrations can be performed in HCl medium. • Cr2O72- + 14H+ + 6e 2Cr3+ + 7H2O • The orange colour of dichromate is not intense to be used to determine the end point, so that is why external indicators should be usede.g diphenylamine sulphonic acid.

  16. Cerium (IV) • Is a powerful oxidizing agent. Its potential depends on the acid in which the reaction takes place. It is 1.44 V on using H2SO4 and 1.70 V in perchloric acid. • It can be used in the same titrations as permanaganate but the oxidation of chloride is slow. • The salt of cerium, ammonium hexanitrocerate is a primary standard material. • The main disadvantage is the increased costcompared to permanagante. • Ferroin is a suitable indicator for such titrations.

  17. Iodimetry • Iodine is a moderately strong oxidizingagent that can be used to titrate reducing agents. Titrations with iodine are called iodimetric titrations. • These reactions are performed in neutral or mildly alkaline pH8 to weakly acid solutions. (GIVE REASON) • If the pH is too alkaline, I2 will disproportionate to hypoiodate and iodide • I2 + 2OH- IO- + I- + H2O • Also in acidic medium, Starch which is used to detect the endpoint of such titration is hydrolyzed, and oxidizing power of iodine will also decrease.

  18. Iodimetry (cont.) • Because I2 is not a strong oxidizing agent this limits the number of reducing agents that can be titrated against iodine and this increases it selectivity. • Although pure iodine can be obtained by sublimation but its solution should be standardized using As2O3. • Iodine has a low solubility in water but the complex I3- is very soluble. So iodine solutions are prepared by dissolving iodine in concentrated solutions of potassium iodide • I2+I- I3- • Therefore, I3- is the actual species used in titration.

  19. Iodometry • Iodide ion is a weak reducing agent and will reduce strong oxidizing agents. It is not used, however, as a titrant mainly because of the lack of convenient visual indicator system, as well as the low speed of the reaction. • When an excess of iodide is added to a solution of an oxidizing agent, I2 is produced in an amount equivalent to the oxidizing agent present. This I2 can be titrated with reducing agents and the result will be the same as if the oxidizing agent was titrated directly. • The titrating agent is sodium thiosulphate. • This method is called Iodometric method.

  20. Iodometry (cont.) • Consider, for example, the determination of dichromate: Cr2O72- + 6I-(excess)+ 14H+ 2Cr3+ + 3I2 + 7H2O I2 + 2S2O32- 2I- + S4O62- tetrathionate • Each Cr2O72- produces 3I2 which in turn react with 6S2O32-. • Thus • 6 millimoles of S2O32- are equivalent to Cr2O72-.

  21. Notice that • We can’t titrate strong oxidizing agents directly with thiosulphate cause it might be oxidized to higher states such as sulphate (SO42-) rather than tetrathionate(S4O62-). • Starch, used to detect the end point, should not be added from the beginning of the titration but near the end of titration where the colour of iodine is pale yellow for two main reasons: • 1. Iodine starch complex is only slowly dissociated and diffuse end point would result if a large amount of iodine is adsorbed on starch. • 2. Iodometric titration usually occur in acid medium which leads to the hydrolysis of starch.

  22. Remember • No of moles can be obtained by MV • Or Weight/Mwt(molecules) • Or Weight/At.wt(atoms) • On solving the redox problems, • Try to find the relation between the No. of moles of your reactants, • Calculate the no. of moles by applying any of the above relations.

  23. Example 1 A 0.2 g sample containing copper is analyzed iodometrically. Copper (II) is reduced to copper(I) by iodide: 2Cu2+ + 4I- 2CuI + I2 What is the percent of copper in the sample if 20 ml of 0.10 M Na2S2O3 is required for the titration of the liberated iodine? I2 + 2S2O32- 2I- + S4O62- • One half of I2 is liberated per mole of Cu2+ and since each I2 reacts with 2S2O32-, thus each mole of Cu2+ is equivalent to one mole S2O32- . • (MV)Cu2+ = (MV)S2O32- • % of Cu2+ = (0.127/0.2) x 100 = 63.5 %

  24. Example 2 • A solution of Na2S2O3 is standardized iodometrically against 0.1262 g of high purity KBrO3, requiring 44.97 ml Na2S2O3. What is the molarity of the Na2S2O3? BrO3- + 6I- + 6H+ Br- + 3I2 + 3H2O 3I2 + 6S2O32- 6I- + 3S4O62- • 6 Each mmol S2O32- = mmol BrO3- • 6 (MS2O32- x 44.97) = (126.2 mg/167.01) • MS2O32- = 0.1008 M

  25. Example 3 • A 0.1809 sample of pure iron wire was dissolved in acid, reduced to the +2 state, and titrated with 31.33 ml of cerium (IV). Calculate the molar concentration of the Ce4+. • Fe2+ Fe3+ + e • Ce4+ + e Ce3+ • 1 mmole of Fe2+ is equivalent to 1 mmole Ce4+ • Wt Femg/At.wt = (MCe4+ x VCe4+) • 180.9 mg/55.84 = MCe4+ x 31.33 • MCe4+ = 0.1034 M

  26. Example 4 • Titration of 0.1467 g of primary standard Na2C2O4 required 28.85 ml of potassium permanganate. Calculate the molar concentration of KMnO4 in this solution . 5H2C2O4 + 2MnO4- + 6H+ 10CO2 + 2Mn2+ + 8H2O From the equation, each 2 mmole of permanganate is equivalent to 5 mmole oxalate, thus 1 mmole of permanganate is equivalent to 5/2 mmole oxalate, • (MV)KMnO4 = (wt/Mwt)oxalate X 5/2 • (M x 28.85)KMnO4 = (146.7mg/133.99)oxalate X 5/2 • MKmnO4 = 0.0151 M

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