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Summary and Extension…Galvanic (Voltaic Cells) PowerPoint Presentation
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Summary and Extension…Galvanic (Voltaic Cells)

Summary and Extension…Galvanic (Voltaic Cells)

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Summary and Extension…Galvanic (Voltaic Cells)

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  1. Summary and Extension…Galvanic (Voltaic Cells) Ce4+ + 1e-→ Ce3+ E ° = 1.70 V Au3+ + 3e-→ Au E ° = 1.50 V The push and pull from the anode to the cathode of electrons C → ← Au ←anode ←cathode Standard Cell Potential = E ° = 0.20 V 1M Ce4+, Ce3+ 1M Au3+ EMF ( ) in volts Electro-motive force Three times Ce4+ + 1e-→ Ce3+ E °= 1.70 V Au → 3e- + Au3+E °= -1.50 V 3Ce4+ + Au → Au3+ + 3Ce3+ Overall reaction: __________________________ defined as: J/C -w q -work charge E = _________ = _____(symbolically) 96, 500 C 1 mole of electrons charge would be: _______________ -E q = -E nF or w = Faraday (this is called a ,F) or w = DG DG° = -E °nF DG = -E nF or since we know another name for w→ (or rewritten under standard form) But what happens to the voltage when the concentration is changed??? For example: __________ …. Explain __________________________________ [Au3+] ↓ E goes up due to LCP Recall…DG = DG° + RT lnQ Substituting in… -E nF = -E °nF + RT lnQ or ( ) Dividing by -nF The Nernst Equation We get……E = E ° - RT lnQ called: _______________________ nF E = 0.20 V – (0.0592/3) (log [0.500][0.300]3/[0.200]3) At 25.0°C ( a common temp for running these cells) What if the concentrations of the ions were [Ce4+] = 0.200 M, [Ce3+] = 0.300 M and [Au3+] = 0.500 M E = E ° - ______log Q 0.0592 n E = 0.196 V = 0.20V For the process we just looked at, find the conc. of the Au3+ ( with all other [ ]’s standard) if the voltage of the cell at 25.0°C is 0.30 V. E = 0.30 V = 0.20 V – 0.0592/3 (log [Au3+]) [Au3+] = 8.56 x 10-6 M