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:
3rd lecture Reduction- Oxidation Reactions
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
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
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- 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
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+
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.
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.
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
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.
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 +
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 ±
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.
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
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
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+.
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.
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