Synthesis of organic molecules dates (1800\'s) back to well before the developme
ID: 1052506 • Letter: S
Question
Synthesis of organic molecules dates (1800's) back to well before the development of spectroscopy (early 1900's). Chemists characterized products by a detailed analytical protocol, the "qual scheme". Some tests involved tasting the product. Many tests were based on a colorimetric change for reaction between specific functional groups and a given reagent. Consider some traditional qual scheme tests. Calculate E^0, delta G^0, spontaneous direction, and the balanced chemical reaction. (a) Chlorine Water: Cl_2 gas bubbled through water to saturation (chlorine water) is an indicator for Br^- and I^-. A positive test for Br^- is a yellow wish solution whereas a positive I^- test is a dark red solution. (b) Tollen's Reagent: A reagent that tests for aldehydes, is made by reacting sodium hydroxide and silver nitrate to yield Ag_2O, which is dissolved in aqueous ammonia to form [Ag(NH_3)_2]^+. The sample is then warmed with Tollen's reagent in a test tube. Ketones do not react. Aldehydes yield the positive result that gives the test its other name, silver mirror test. (Use formaldehyde as the analyte.) (c) Milton's Reagent: Milton's reagent tests for albumins and phenols. One part of mercury metal dissolved in one part of cold fuming nitric acid yields a solution of Hg^2+ and Hg_2^2+. The diluted solution is reacted with the analyte. Use dihydroxyquinone as an example phenol. (d) Dichromate Test: Dichromate discriminates between primary and secondary alcohols, as well as between aldehydes and ketones. Dichromate is yellow, while Cr(III) is green and Cr(II) is red. (Note the shift in wavelength on passing from highest to lowest oxidation state.) Use ethanol as a primary alcohol and hydroquinonc as a ersatz secondary alcohol to identify colorimetric result that discriminates between primary and secondary alcohols, as well as between aldehydes and ketones. Two reactions that may be useful are: CH_3CHO + 2e + 2H^+ CH_3CH_2OH, E^0 = -0.197 V CH_3COO^- + 3H^- + +2e CH_3CHO + H_2OExplanation / Answer
2. for the given reactions
(a) balanced equation
Cl2 + 2Br- --> 2Cl- + Br2
Eo = 1.36 - 1.09 = 0.27 V
dGo = -nFEo = -2 x 96485 x 0.27 = -52.102 kJ
Cl2 + 2I- --> 2Cl- + I2
Eo = 1.36 - 0.53 = 0.83 V
dGo = -nFEo = -2 x 96485 x 0.83 = -160.565 kJ
(b) Equation : 2[Ag(NH3)2]+ + HCHO + H2O --> 2Ag + RCOOH + 2H+ + 4NH3
Eo = 0.8 - (-0.03) = 0.83 V
dGo = -2 x 96485 x 0.83 = -160.165 kJ
Similarly given the standard reduction potential values, we can calculate eo and dG for other tests.
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