Pharmaceutical Analytical Chemistry PHCM223 Lecture 9 REDOX REACTIONS (I) Dr. Nesrine El Gohary 9 th lecture
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1 Pharmaceutical Analytical Chemistry PHCM223 Lecture 9 REDOX REACTIONS (I) Dr. Nesrine El Gohary 9 th lecture
2 Learning outcomes Define redox reactions. Apply redox titrations. Detect the end point of redox titration using different indicators. Identify some important oxidizing agents. 2
3 Redox reactions Redox reactions are oxidation-reduction reactions. Oxidation: It is the loss of electrons by a reagent (Reducing agent). Reduction: It is the gain of electrons by a reagent (Oxidizing agent). Redox reactions takes place when an oxidizing agent reacts with a reducing agent. Ox 1 is reduced to Red 1 and Red Ox Red 2 Red 1 + Ox 2 is oxidized to Ox 2. The oxidizing or reducing tendency of a substance depends mainly on its reduction potential. This table shows the values of standard reduction potentials for common half reactions measured against standard hydrogen electrode. The one with higher potential acts as the cathode (reduced) and the lower acts as the anode (oxidized) 3
4 The potential (E) is dependent on the concentrations of the species involved in the half reaction. The potential could be calculated theoretically by applying Nernst equation: Where: E= E n [reduced]b log [oxidized] a E = Standard reduction potential n= Number of electrons [reduced]= reduced form concentration [oxidized] = oxidized form concentration a and b are coefficients in the balanced equation Example: Determine the potential of a Pt indicator electrode if dipped in a solution containing 0.1M Sn 4+ and 0.01M Sn 2+. Sn e - Sn +2 E o = 0.15V E= Nernst Equation aox + ne bred log [0.01] [0.1] = 0.18 V Equilibrium constant K 4
5 Reducing agent Oxidizing agent Redox titration It is a volumetric method of analysis which relies on oxidation or reduction of the analyte. OR Oxidizing agent Reducing agent Difference in E should be at least 0.2 V for a complete reaction to take place. The reduction potential of the two reacting substances will determine which one of them will act as a reducing agent and which will act as an oxidizing agent. The endpoint can be determined by measuring the potential developed using a voltammeter or visually by using certain indicators. 5
6 Redox Reaction Titration Curve It is a plot of the potential measured against volume of titrant added. The potential can be determined practically by recording the potential of an indicator electrode relative to a reference electrode. Theoretically the potential could be calculated by applying Nernst 6 equation.
7 Calculating the potential in a redox titration Before the endpoint Apply Nernst equation To the sample half reaction to calculate the potential At the endpoint Apply the following equation E n 1 E n 1 1 n n 2 2 E 2 After the endpoint Apply Nernst equation To the titrant half reaction to calculate the potential 7
8 Example: On titrating 5.0 ml of 0.30M Fe 2+ solution, calculate the potential of a platinum electrode dipped in the solution relative to NHE when: a) 5.0 ml of 0.1 M Ce 4+ is added. 1-Write the balanced redox reaction that takes place. Fe 2+ Fe 3+ + e Ce 4+ + e Ce 3+ Fe 2+ + Ce 4+ Fe 3+ + Ce Perform the necessary stoichiometric calculations. Before reaction: Fe 2+ + Ce 4+ Fe 3+ + Ce 3+ 5 ml x 0.3 M = 1.5 mmol 5 ml x 0.1 M = 0.5 mmol 0 mmol 0 mmol After reaction: 1 mmol 0 mmol 0.5 mmol 0.5 mmol E Ce4+/Ce3+ = 1.70, E Fe3+/Fe2+ = BEFORE THE END POINT Determine the volume of titrant needed to reach the endpoint: (MVn) titrant = (MVn) sample n is the number of electrons involved in each half reaction (0.1 x V x 1) = (0.3 x 5 x 1) V = (0.3 x 5 x 1)/(0.1x1)= 15 ml 3- Apply Nernst equation for Fe 3+ /Fe 2+ E= log 1 10 ( ) = V 8
9 Example cont.: b) 7.5 ml of 0.1 M Ce 4+ is added. HALF THE END POINT Fe 2+ + Ce 4+ Fe 3+ + Ce 3+ Before 5 ml x 0.3 M 7.5 ml x 0.1 M 0 mmol 0 mmol reaction: = 1.5 mmol = 0.75 mmol After reaction: 0.75 mmol 0 mmol 0.75 mmol 0.75mmol Apply Nernst equation for Fe 3+ /Fe 2+ E= log ( ) = V E = E Fe3+/Fe2+ = V 9
10 Example cont.: c) 15.0 ml of 0.1 M Ce 4+ is added. END POINT Before reaction: Fe 2+ + Ce 4+ Fe 3+ + Ce 3+ 5 ml x 0.3 M = 1.5 mmol 15 ml x 0.1 M = 1.5 mmol 0 mmol After reaction: 0 mmol 0 mmol 1.5 mmol E n 1 E n 1 1 n n 2 2 E 2 0 mmol 1.5 mmol Apply the following equation to calculate the potential:??? This equation is derived from the addition of Nernst equation for the two half reactions. E= (1 x 1.7) + (1 x 0.771) = V The potential at the endpoint is independent of the volume or concentration of both the sample and the titrant. 10
11 Example cont.: d) 20.0 ml of 0.1 M Ce 4+ is added. AFTER THE END POINT Fe 2+ + Ce 4+ Fe 3+ + Ce 3+ Before reaction: 5 ml x 0.3 M = 1.5 mmol 20 ml x 0.1 M = 2 mmol 0 mmol 0 mmol After reaction: 0 mmol 0.5 mmol 1.5 mmol 1.5 mmol Apply Nernst equation for Ce 4+ /Ce 3+ E= log ( ) = V 11
12 Example cont.: e) Sketch the titration curve Note that the initial potential cannot be calculated because we don t how much Fe 3+ is present. But always there must be some Fe 3+ present initially either as impurity or from oxidation of Fe 2+ by atmospheric oxygen. 12
13 Visual Detection of the endpoint Redox indicators Specific indicators Self indicators 13
14 Visual Detection of the endpoint Redox indicators Redox indicators are weak reducing or oxidizing agents that undergo a color change as they go from their oxidized form to their reduced form and vice-versa. Example: Ferroin is a redox indicator commonly used with Ce 4+ titrations. Ce 4+ Sample + Ferroin Its color changes from red to pale blue at the endpoint of the titration. Diphenylamine sulphonic acid is another redox indicator commonly used with dichromate (Cr 2 O 7 ) 2- titrations. Its color changes from colorless to purple at the end point of the titration. 14
15 Visual Detection of the endpoint Redox indicators Like acid-base indicators there is a range in which the indicator changes its color POTENTIAL RANGE This potential range is determined by applying Nernst equation for the indicator In(ox) + ne - In(red) E= E [In(red)] log [In(ox)] n As with acid-base indicators, the color change is observed when one form of the indicator is 10 times the other. Applying in Nernst equation shows us that the color change will occur over the range: E= (E ± n ) Volts The potential at the endpoint should lay within this potential range for 15 the indicator to be useful.
16 Some common redox indicators 16
17 Test Yourself: Which indicator should you choose for the following redox titration? Methylene blue: E = 0.53 OR Diphenylamine: E =
18 Visual Detection of the endpoint Specific indicators Starch Indicator This indicator is used for titrations involving iodine. Starch forms a dark blue color with iodine (I 2 ). Starch is not a redox indicator because it responds specifically to the presence of iodine not to potential change. Starch easily undergoes decomposition so it should be freshly prepared. Glucose is a product of its decomposition, glucose acts as a reducing agent so will lead to errors in the titration. 18
19 Visual Detection of the endpoint Self indicators If the sample or titrant is colored, the end point can be detected by the appearance or disappearance of the colour of the reagent. e.g. KMnO 4 where the end point can be detected by the pink color of permanganate. START POINT END POINT 19
20 Common Oxidizing agents used as titrants Potassium Permanganate KMNO 4 Cerium (IV) Ce 4+ Potassium dichromate K 2 Cr 2 O 7 Iodine I 2 20
21 Potassium Permanganate KMNO 4 It is a widely used oxidizing agent (E = 1.51V). It is reduced to different forms depending on the ph of the medium. In acidic ph it is reduced to the colorless Mn 2+ ion. MnO H + + 5e Mn H 2 O In neutral or alkaline ph it is converted to a brown precipitate of MnO 2 so it is not used in these phs. It is used as a self indicator in acidic ph. It is not a primary standard so solutions of permanganate are standardized against primary standard sodium oxalate. 5H 2 C 2 O 4 + 2MnO H + 10CO 2 + 2Mn H 2 O The reaction between permanganate and oxalate is slow at room temperature so must be heated to fasten the reaction. The reaction is autocatalyzed by the Mn 2+ product and it goes very slowly until Mn 2+ is formed. Permanganate titration are not possible in the presence of chloride because it will be oxidized to chlorine so HCl is not a suitable acid to 21 be used to adjust ph. Usually H 2 SO 4 is used instead.
22 Potassium dichromate K 2 Cr 2 O 7 It is slightly weaker oxidizing agent permanganate (E = 1.36V). Cr 2 O H + + 6e 2Cr H 2 O than potassium The main advantage is its availability as a primary standard material. It is used in acidic medium where it is reduced to the green Cr 3+ ion. In basic solutions it is converted to CrO 2-4 which has no oxidizing properties. It does not react with HCl so the titrations can be performed in HCl medium. The orange color of dichromate is not intense to be used to determine the end point, so that is why external indicators should be used e.g diphenylamine sulphonic acid. 22
23 Cerium (IV) Ce 4+ Like permanganate it is a powerful oxidizing agent. It is used in acidic medium where it is reduced to colorless Ce 3+ ion. Ce 4+ + e Ce 3+ yellow colorless Its potential depends on the acid in which the reaction takes place. It is 1.44 V on using H 2 SO 4 and 1.70 V in perchloric acid (HClO 4 ). It not used in basic solutions since it is precipitated. The yellow color or Ce 4+ at the endpoint is not clear to be used as a self indicator so ferroin is used as an indicator with Ce 4+ titrations. It can be used in the same titrations as permanganate but the oxidation of chloride is slow so could be used with HCl solutions. The salt of cerium, ammonium hexanitrocerate, (NH 4 ) 2 Ce(NO 3 ) 6 is a primary standard material. 23
24 Iodine I 2 Iodine is a weak oxidizing agent ((E = 0.536V). It is used to titrate only strong reducing agents, thus this increases its selectivity where it is possible to titrate strong reducing agents in the presence of weak ones. Titrations performed with I 2 are called Iodimetric titrations These titrations are performed in neutral or mildly alkaline (ph8) to weakly acid solutions. If the ph is too alkaline: I 2 will disproportionate (undergo oxidation and reduction reaction at the same time) to hypoiodate and iodide I 2 + 2OH - IO - + I - + H 2 O If the ph is too acidic: Starch the indicator used in Iodimetric titrations is hydrolyzed. I 2 Reducing agent + Starch Start point: Colorless End point: Blue color 24
25 Iodine I 2 Cont. Iodine has a low solubility in water, so the actual titrant is I 3-. I 3- is prepared by dissolving iodine in concentrated solutions of potassium iodide. I 2 +I - I 3 - Triiodide Although Pure iodine is available but its solution should be standardized using As 2 O 3. It should be standardized because it is highly volatile. 25
26 References D. A. Skoog, D.A. West, F.J. Holler, S.R. Crouch, Analytical Chemistry, an introduction, 7th Edition, ISBN (Chapter 17and 18). Lecture 9 by Prof. Rasha ElNashar, GUC, SS
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