# 14-Redox_4_web - Note some topics of the today lecture were...

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Lecture agenda: The “x   0” approximation (REDOX  TITRATION;  POTENTIOMETRY ) Faraday’s law of electrolysis  Professional electrochemistry,   VOLTAMMETRY COULOMETRIC  TITRATION Note: some topics of the today lecture were advanced; what (the ppts) is indicated below between “advanced” and “the end of advanced” is not for exam # 2)

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The “x   0” approximation  (towards better understanding of redox titration)
Ce 4+ + Fe 2+   Ce 3+ + Fe 3+ (1) Ce 4+ + Fe 2+ Ce 3+ + Fe 3+ (1') Before the equivalence point: [Ce 4+ ] = 0, as the completed reaction 1’   leaves no Ce 4+ product on the left side. Equilibrium concentrations can then be replaced by analytical concentrations: c(Fe 3+ ) is equal to c(Ce 3+ ) c(Fe 2+ ) is equal to c 0 (Fe 2+ ) - c(Fe 3+ ) “x   0” approximation in redox titration

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The reaction is however in  equilibrium: Ce 4+    +   Fe 2+      Ce 3+    +   Fe 3+ (1) If this is so: [Ce 4+  exactly 0, as [Ce 4+ ] is set by reaction 1, which  attains the  equilibrium.    [Ce 4+ ] from the equilibrium is equal to [Fe 2+ ] (from the  equilibrium), a small subset of all Fe 2+ is present in  solution. Because of the Fe 2+  ion is overwhelmingly present, [Ce 4+ is the most practical gauge of equilibrium.
In essence, what we assume is (just for the Fe 3+ form): [Fe 3+ ] = c(Fe 3+ ) – x c(Fe 3+ ), that is, x = 0. where x = [Ce 4+ ]. (Or [Fe 2+ ] from the equilibrium balance only.) This leads to the "x = 0" approximation. Ce 4+ + Fe 2+ Ce 3+ + Fe 3+ (1) Ce 4+ + Fe 2+ Ce 3+ + Fe 3+ (1') How good is the x = 0 approximation?

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] [ ] [ log 0592 . 0 ] [ ] [ log 0592 . 0 2 3 0 3 4 0 + + + + + = + = Fe Fe E Ce Ce E E Fe Ce Calculate the equilibrium constant K eq  for: Ce 4+    +   Fe 2+       Ce 3+    +   Fe 3+ (1) ] ][ [ ] ][ [ 4 2 3 3 + + + + = Ce Fe Ce Fe eq K ] ][ [ ] ][ [ log 0592 . 0 4 2 3 3 0 0 + + + + - = - Ce Fe Ce Fe E E Ce Fe -0.84 = -0.0592 ∙ log K eq ; K eq  = 1.55 x 10 14 , a very high  value!    Notice that K eq  is proportional to the  E 0   difference. The master equation:
] [ ] [ log 0592 . 0 2 3 0 + + + = Fe Fe E E Fe Fe ] [ ) /( ) ( ] [ ) /( log 0592 . 0 4 , 0 , 0 , 0 4 , 0 0 + + + + - - + + = Ce V V V c V c Ce V V V c E E Ce Fe Ce Ce Fe Fe Ce Fe Ce Ce Fe Fe Before the equivalence point.   The correct equations should contain the term [Ce 4+ ] from  the equilibrium 1  (as there is certain concentration of  [Ce 4+

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## This note was uploaded on 12/28/2010 for the course CHEMISTRY 222 taught by Professor Wieckowski during the Spring '10 term at University of Illinois at Urbana–Champaign.

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14-Redox_4_web - Note some topics of the today lecture were...

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