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3rd lecture Reduction- Oxidation Reactions. Learning Objectives What are some of the key things we learned from this lecture? Calculating redox titration.

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Presentation on theme: "3rd lecture Reduction- Oxidation Reactions. Learning Objectives What are some of the key things we learned from this lecture? Calculating redox titration."— Presentation transcript:

1 3rd lecture Reduction- Oxidation Reactions

2 Learning Objectives What are some of the key things we learned from this lecture? Calculating redox titration curves Detection of end point in redox titration Redox indicators. -Requirements of a redox indicator -Transition range of redox indicator - Examples of redox indicators - Determination of equilibrium constant (K eq ) for redox reactions

3 Redox titration curves It is a plot of the number of mls of titrant aganist potential in volts. These curves show the change in potential during the progress of a redox titration. Suppose the titration of 100 ml of 0.1 N solu. of ferrous sulphate with 0.1 N ceric sulphate. 1- Upon addition of 10ml ceric sulphate the ratio of Fe 3+ / Fe 2+ becomes 10/90 (10 ml ceric reacts with 10 ml ferrous to give 10 ml ferric) E= /1 log 90/10 = 0.69v 2- Addition of 50 ml ceric, E= /1 log 50/50 = 0.77v 90 ml Fe ml Fe ml Ce ml Fe ml Fe ml Ce 3+ 2

4 3- Addition of 99.9 ml ceric, E= /1 log 0.1/99.9 = 0.93v 4- Addition of 100 ml ceric At the equivalence point, Ee.p. = n 1 E n 2 E 2 0 n 1 + n 2 E 1 0 Standard oxidation potential of Fe 3+ / Fe 2+ E 2 0 Standard oxidation potential of Ce 4+ / Ce 3+ n 1 number of electrons lost or gained by Fe 3+ / Fe 2+ n 2 number of electrons lost or gained by Ce 4+ / Ce 3+ E e.p. = 0.77 X x 1 / 1+1 = 1.1v 0.1 ml Fe ml Fe ml Ce ml Fe ml Fe ml Ce 3+ 4

5 5- After the equivalence point Addition of ml ceric E= /1 log 100 /0.1 = 1.27v It is clear from the curve that there is a sudden change in potential at the equivalence point. E ml of titrant e.p. 0.1 ml Ce ml Fe ml Ce 3+ 5 New system

6 Detection of end point in redox titration 1- Potentiometric method: The potential of a cell involving the solution titrated is followed. The end point is known from the inflection point in the titration curve. 2- Miscellaneous methods: A- Specific indicators: substances which react specifically with one of the reagents in a titration to produce a color. e.g. starch gives blue color with iodine SCN - ion gives red color with Fe 3+

7 B- No (self) indicators: If the color of the titrating agent undergoes a sharp enough change in color at the equivalence point. E.g. KMnO 4, I 2 1- When KMnO 4, is used as atitrant for reducing agents, during titration the purple color of permenganate disappears due to formation of colorless Mn 2+ When all the reducing agent has been oxidized a slight excess of KMnO 4 colors the solution pink. 2- Similarly, titration with iodine, at the e.p. There should be the brown color of exx iodine. But the color of I 2 is unstable, so, we add starch which forms an intense blue color (adsorption compound) with exx free I 2.

8 C- Irreversible indicators (acid-base indicators): These are highly colored organic compounds which undergo irreversible oxidation or reduction with little excess of a titrant. e.g. Methyl orange and methyl red in titrations with BrO 3 -  M.O. and M.R. are decolorised irreversibly (destroyed) by first exx of strong oxidants e.g. BrO 3 -.  Another example is naphthol blue black which changes from pink to colorless by a slight excess of BrO 3 -  The irrevesible indicator must be added near the end point.

9 D- External indicators: E.g.1- Spot test method: for the determination of Fe 2+ with K 2 Cr 2 O 7 Near the equivalence point, drops of ferrous sample are removed and brought in contact with freshly prepared solu of potassium ferricyanide on a spot plate.  Normally Fe 2+ forms a blue color with ferricyanide.  When all Fe 2+ is oxidized by K 2 Cr 2 O 7, no more Fe 2+ ions are available and the sample will not give a blue color with ferricyanide. BB Ferri cyanide Ferri cyanide Before e.p. After e.p. Sample solu

10 E- Internal redox indicators: Redox indicator changes its color when the oxidation potential of the titrated solution reaches a definite value. Requirement of a redox indicator: 1- The color change should be intense so that small amount of titrant gives the color change. 2- The indicator action should be reversible so that back titration can be performed. 3- The 2 colors of the indicator (reduced form, and oxidised form) should be sufficiently different so that the color change would be sharp. 4- The transition potential of the indicator should be in between the E o values of the 2 systems used in the titration. i.e. Lower than the titrant but higher than the sample WHY? 5- The transition potential of the indicator should not be affected by changes in pH, other wise the pH must be controlled.

11 Transition range of redox indicator  Redox indicators are typical redox systems.  They can be represented by the general half reaction as follows: In ox + n H + + ne  In red color A color B  At a potential E, the ratio of the 2 forms of the indicator are determined by the Nernst equation as follows: E ind = E 0ind -  A good redox indicator must show the colour of its oxidized form when the ratio of [In red ] / [In oxd ] is not more than 1/10. At this limit the potential becomes: E ind = E o ind - E ind = E o ind +

12  And shows the colour of its reduced form when the ratio of [In red ] / [In oxd ] is not more than 10/1. At this limit the potential becomes: E ind = E 0 ind - E ind = E 0 ind –  In other words a good redox indicator, must change its color at a potential expressed by the following formula: E ind = E 0 ind ±

13 Examples or redox indicators: Diphenylamine:1 % solu in conc H 2 SO 4 is used (E 0 = 0.76, n = 2)  The range of diphenylamine E ind = 0.76 ± 0.059/2 = v.  At potential below 0.73 v the color of the rduced form predominates (colorless).  At potential above 0.79v the color of the oxidized form predomintes (blue-violet).  Between v the color of the solution changed gradually from colorless to blue-violet.

14  The first reaction involving the formation of diphenylbenzidine is non-reversible, the second gives violet product which can be reversed and constitutes the actual indicator reaction.  It is prepared in sulphuric acid because it is insoluble in water  sulphonic acid derivative of diphenylamine (diphenylamine p-sulphonic acid) has the same mechanism of action as diphenylamine.  It is water soluble with sharp color change during oxidation (Transition potential 0.8V), and is independent on pH. colorless

15  Diphenylamine is unsuitable indicator for the determination of ferrous with dichromate WHY? Because E 0 of Fe 3 /Fe and E 0 of diphenylamine 0.76 which are very close. So we have to lower the oxidation potential of Fe 3+ /Fe 2+ by adding PO 4 3. Phenylanthranilic acid is suitable indicator for the determination of ferrous with dichromate WHY? Because E 0 of phenylanthranilic acid 1.08 which is intermediate between E 0 Fe 3+ /Fe and E 0 Cr 2 O 7 2- /Cr

16 Chelate of ferrous with 1,10 ortho phenanthroline (Ferroin)  It is intensely red and is converted by oxidation into the pale blue ferric complex (Ferrin). (Ph) 3 Fe 3+ + e  (Ph) 3 Fe 2+ (E 0 = V) pale blueRed (Ferrin) (Ferroin)  It is an excellent indicator for Ce 4+.

17 Determination of equilibrium constant (K eq ) for redox reactions a A ox + n e a A red E 0 A cathode B red b B ox + n e E 0 A anode By combining the two equations: a A ox + b B red a A red + b B ox Keq = [ A red ] a [ B ox ] b / [A ox ] a [ B red ] b E cell = E cathode - E anode Equilibrium constant, indicates reaction completeness. High value of Keq indicate complete reaction while low value indicate incomplete reaction.

18 At equilibrium the current stops E cathode = E anode E A = E B s E 0 A - 0.o59/ n log [A red ] a / [A oxd ] a = E 0 B - 0.o59/ n log [B red ] b / [B oxd ] b E 0 A - E 0 B = 0.o59/ n log Keq Log Keq = n (E 0 A - E 0 B ) / 0.059


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