E goes up due to forward reaction favored… Summary and Extension…Galvanic(Voltaic)Cells Ce4+ + 1e- → Ce3+ E ° = 1.70 V Au3+ + 3e- → Au E ° = 1.50 V C → ← Au The push and pull from the anode to the cathode of electrons ←anode ←cathode Standard Cell Potential = E ° = 0.20 V Three times 1M Ce4+, Ce3+ 1M Au3+ EMF ( ) in volts Electro-motive force Ce4+ + 1e- → Ce3+ E ° = 1.70 V Au → 3e- + Au3+ E º° = -1.50 V Overall reaction: __________________________ 3Ce4+ + Au → Au3+ + 3Ce3+ -work charge -w q defined as: J/C E = ________ = _____(symbolically) 96, 500 C or w = -E q = -E n F 1 mole of electrons charge would be: _______________ (this is called a ,F ) Faraday or w = DG DG° = -E °nF or since we know another name for w→ DG = -E nF (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 forward reaction favored… Recall…DG = DG° + RT lnQ Substituting in… -E nF = -E °nF + RT lnQ or ( ) Dividing by -nF We get……E = E ° - RT lnQ called: _______________________ The Nernst Equation nF E = 0.20 V – (0.0592/3) (log [0.500][0.300]3/[0.200]3) E = 0.196 V = 0.20V At 25.0°C ( a common temp for running these cells) 0.0592 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 n 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