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Chapter 18 Lecture 5

Chapter 18 Lecture 5 - 2 in one liter at equilibrium Ag Cu...

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Chapter 18 Lecture 5 Effects of concentration: What if you measured the potential of a cell where all the concentrations were not ONE? You just have to enter a cheat term to compensate. ΔG = ΔG o + RTlnQ Because: ΔG o = ─nFE o cell Q nF RT E E o ln - = Nernst Equation For all concentrations: ν ϑ ] tan [Re ] [Pr log 0591 . 0 ts ac oducts n E E o - = For a half reaction: O + ne - R ] [ ] [ log 0591 . 0 O R n E E o - = Write both half reactions as reductions when using this formula
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At equilibrium: ν ϑ ] tan [Re ] [Pr log 0591 . 0 0 ts ac oducts n E o - = Keq n ts ac oducts n E o log 0591 . 0 ] tan [Re ] [Pr log 0591 . 0 = = ν ϑ Example: Determine Keq for the following reaction: O 2 + 4I + 4H + 2I 2 (s) + 2H 2 O From Appendix C: O 2 + 4H + + 4e 2H 2 O E o = 1.229 V I 2 + 2e 2I E o = 0.535 V
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Example 2: Consider a piece of copper placed in a 1.0 M Ag+ solution. Consider the redox reaction: 2Ag + + Cu → 2Ag(s) + Cu 2+ Will a reaction take place? From Appendix C: Ag + + 1e Ag(s) E o = 0.800 V Cu 2+ + 2e Cu(s) E o = 0.340 V What will the equilibrium concentrations be? Find K.
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Equilibrium Concentrations can be found by the “x” method Let x = the moles of Cu
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Unformatted text preview: 2+ in one liter at equilibrium. Ag + Cu 2+ I C Eq 2 2 ] [ ] [ + + = Ag Cu K How to find E for a cell where concentrations are not 1.0 M Given cell notation: 1. Look up half reactions 2. Calculate E for each half-reaction using Nernst Equation At this point you should write BOTH half reactions as reductions and use the sign listed in the E o table when entering voltages into the equation. 3. Calculate Ecell from: E o meter or cell = E o red or right - E o black or left Example: Find Ecell for: Pt│H 2 (1 atm), 0.0675 M HCl ││ 0.025 M KOH │H 2 (1 atm)│Pt You can also calculate solution concentrations: If Ecell = ─0.015 V, Find the Ag + concentration in the right half cell. Pt│Fe 2+ (0.125 M), Fe 3+ (0.068 M)││ Ag + (? M )│Ag From Appendix C: Ag + + 1e ─ → Ag(s) E o = 0.800 V Fe 3+ + 1e ─ → Fe 2+ E o = 0.771 V...
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