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Lesson_25_-_Example_Solutions-2009

# Lesson_25_-_Example_Solutions-2009 - Lesson 25 Example 25-1...

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Lesson 25 Example 25-1 Is Cu + a stable ion in aqueous solution? To decide this, first determine the equilibrium constant for the reaction 2 Cu + = Cu 2+ + Cu Use two different pairs of half-reactions in your calculation. Note that they have different ∆ε o values but the same K value. ___________________________________________________________________________________________________________ _ Solution: (1) Cu + + e - Cu ε o = 0.518 (2) Cu 2+ + 2 e - Cu ε o = 0.339 (3) Cu 2+ + e - Cu + ε o = 0.160 1

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Lesson 25 1 st pair (1) Cu + + e - Cu ε o = 0.518 -(3) Cu + Cu 2+ + e - ε o = - 0.160 2 Cu + Cu 2+ + Cu ∆ ε o = 0.358 ( 29 ( 29 6 059 . 0 358 . 0 1 059 . 0 o 10 x 17 . 1 10 10 K = = = ε n Since K is very large, Cu + is not stable in solution. Let’s check the calculations with another pair of half-reactions. 2
Lesson 25 2 nd pair (2) Cu 2+ + 2 e - Cu ε o = 0.339 -2x(3) 2 Cu + 2 Cu 2+ + 2 e - ε o = -0.160 2 Cu + Cu 2+ + Cu ∆ ε o = 0.179 ( 29 ( 29 6 059 . 0 179 . 0 2 059 . 0 o 10 x 17 . 1 10 10 K = = = ε n As expected, the same result is obtained. Cu + is not stable in aqueous solution. 3

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Lesson 25 Example 25-2 Knowing that Fe 3+ + e - = Fe 2+ ε 1 o = 0.77 V Fe 2+ + 2 e - = Fe ε 1 o = -0.44 V Calculate the value of ε o for Fe 3+ + 3 e - = Fe ____________________________________________________________________________________________ Solution: Can we simply add the two half-cell reactions?
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Lesson_25_-_Example_Solutions-2009 - Lesson 25 Example 25-1...

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