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14_REDOX_titrations

14_REDOX_titrations - Both oxidation and reduction occur...

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Both oxidation and reduction occur during a titration. The equivalence point is based on the concentration of the oxidized and reduced form of all species involved. A ox + B red = A red + B ox Ce 4+ + Fe 2+ = Ce 3+ + Fe 3+ For a REDOX titration, the equivalence point is the point where E forward = E reverse . E cell = 0 Since the E values are concentration dependent, we must rely on the Nernst equation E = E o - log 0.0592 n [A red ] [A ox ] At the equivalence point: E A o - 0.0592 n A log [A red ] [A ox ] E B o - 0.0592 n B log [B red ] [B ox ] = Since A red =B ox and B red =A ox at the equivalence point, we can reduce this to: E eq = n A E A o + n B E B o n A + n B E eq = n A E A o + n B E B o n A + n B Determine the E eq for the following reaction: Fe 2+ + Ce 4+ = Fe 3+ + Ce 3+ E o Fe 3+ /Fe 2+ = +0.771V E o Ce 4+ /Ce 3+ = +1.70 V n Fe = 1, n Ce = 1 E eq = = 1.24 V 1.70V + 0.771V 2

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Determining E eq becomes much more complicated for more complex systems. The inclusion of one additional species, like H + is relatively common. Lets look at another example. 6Fe 2+ + Cr 2 O 7 2- + 14H + = 6Fe 3+ + 2Cr 3+ + 7H 2 O First, obtain both half reactions: E eq = E o Fe - E eq = E o Cr - 0.0592 1 log [ Fe 2+ ] [ Fe
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14_REDOX_titrations - Both oxidation and reduction occur...

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