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Lec13_drobny_11 - Lecture 13 The Nernst Equation Reading...

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Lecture 13: The Nernst Equation • Reading: Zumdahl 11.4 Recommended Problems:11.47, 11.49, 11.51, 11.55,11.57 • Outline: – Why would concentration matter in electrochem.? – The Nernst equation. – Applications
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Concentration and E cell Consider the following redox reaction: Zn(s) + 2H + (aq) Zn 2+ (aq) + H 2 (g) cell = 0.76 V ! G°= -nFE° cell < 0 (spontaneous) What if [H + ] = 2 M? Expect driving force for product formation to increase. Therefore ! G decreases, and E cell increases How does E cell dependend on concentration?
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Concentration and E cell (cont.) Recall, in general: ! G = ! G° + RTln(Q) • However: ! G = -nFE cell -nFE cell = -nFE° cell + RTln(Q) E cell = E° cell - (RT/nF)ln(Q) E cell = E° cell - (0.0591 V/n)log(Q) The Nernst Equation
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Summary of Relationships: ! G & E cell ! G = ! G° + RTln(Q) ! G = -nFE cell E cell = E° cell - (0.0591/n)log(Q) None of these ideas is separate. They are all connected, and are all derived directly from thermodynamics.
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Concentration and E cell (cont.) With the Nernst Eq., we can determine the effect of concentration on cell potentials. E cell = E° cell - (0.0591 V/n)log(Q) • Example. Calculate the cell potential for the following: Fe(s) + Cu 2+ (aq) Fe 2+ (aq) + Cu(s) Where [Cu 2+ ] = 0.3 M and [Fe 2+ ] = 0.1 M
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Concentration and E cell (cont.) Fe(s) + Cu 2+ (aq) Fe 2+ (aq) + Cu(s) First, need to identify the 1/2 cells Cu 2+ (aq) + 2e- Cu(s) 1/2 = 0.34 V Fe 2+ (aq) + 2e- Fe(s) 1/2 = -0.44 V Fe(s) Fe 2+ (aq) + 2e- 1/2 = +0.44 V Fe(s) + Cu 2+ (aq) Fe 2+ (aq) + Cu(s) cell = +0.78 V
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