Summary and Extension…Galvanic (Voltaic Cells)
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Summary and Extension…Galvanic(Voltaic)Cells 1M Ce 4+ , Ce 3+ 1M Au 3+ EMF ( ) in volts defined as: Overall reaction: __________________________ E = ________ = _____(symbolically) or w = 1 mole of electrons charge would be: _______________ or w = (this is called a ,F ) 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 ______________________________________________ Recall…G = Substituting in… or ( ) We get……E = E ° - RT lnQ called: _______________________ nF At 25.0°C ( a common temp for running these cells) E = E ° - ______log Q For the process we just looked at, find the conc. of the Au 3+ ( with all other [ ]’s standard) if the voltage of the cell at 25.0°C is 0.30 V. Ce 4+ + 1e - → Ce 3+ E ° = 1.70 V Au 3+ + 3e - → Au E ° = 1.50 V ←anode C → ← Au ←cathode Ce 4+ + 1e - → Ce 3+ E ° = 1.70 V Au → 3e - + Au 3+ E º ° = -1.50 V 3Ce 4+ + Au → Au 3+ + 3Ce 3+ Standard Cell Potential = E ° = 0.20 V The push and pull from the anode to the cathode of electrons Electro-motive force Three times J/C -work charg e -w q -E q 96, 500 C Faraday = -E n F G G = -E nF G° = -E °nF [Au 3+ ] ↓ ↓ E goes up due to forward reaction favored… G° + RT lnQ -E nF = -E °nF + RT lnQ Dividing by -nF The Nernst Equation 0.0592 What if the concentrations of the ions were [Ce 4+ ] = 0.200 M, [Ce 3+ ] = 0.300 M and [Au 3+ ] = 0.500 M E = 0.196 V = 0.20V E = 0.30 V = 0.20 V – 0.0592/3 (log [Au 3+ ]) [Au 3+ ] = 8.56 x 10 -6 M n E = 0.20 V – (0.0592/3) (log [0.500][0.300] 3 /[0.200] 3 )
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Summary and Extension…Galvanic (Voltaic Cells). Ce 4+ + 1e - → Ce 3+ E ° = 1.70 V. Au 3+ + 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 Ce 4+ , Ce 3+. - PowerPoint PPT Presentation
Transcript of Summary and Extension…Galvanic (Voltaic Cells)
Summary and Extension…Galvanic(Voltaic)Cells
1M Ce4+, Ce3+ 1M Au3+
EMF ( )
in volts defined as:
Overall reaction: __________________________
E = ________ = _____(symbolically)
or w = 1 mole of electrons charge would be: _______________
or w = (this is called a ,F )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 ______________________________________________
Recall…G =
Substituting in…or ( )We get……E = E ° - RT lnQ called: _______________________
nF
At 25.0°C ( a common temp for running these cells)
E = E ° - ______log Q
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.
Ce4+ + 1e- → Ce3+ E ° = 1.70 VAu3+ + 3e- → Au E ° = 1.50 V
←anodeC → ← Au
←cathode
Ce4+ + 1e- → Ce3+
E ° = 1.70 V
Au → 3e- + Au3+
E º° = -1.50 V3Ce4+ + Au → Au3+ + 3Ce3+
Standard Cell Potential = E ° = 0.20 V
The push and pull from the anode to the cathode of electrons
Electro-motive force
Th
ree
tim
es
J/C-work charge
-w q
-E q96, 500 C
Faraday = -E n F
G
G = -E nFG° = -E °nF
[Au3+] ↓↓ E goes up due to forward reaction favored…G° + RT lnQ
-E nF = -E °nF + RT lnQDividing by -nFThe Nernst Equation
0.0592 What if the concentrations of the ions were [Ce4+] = 0.200 M, [Ce3+] = 0.300 M and [Au3+] = 0.500 M
E = 0.196 V = 0.20V
E = 0.30 V = 0.20 V – 0.0592/3 (log [Au3+]) [Au3+] = 8.56 x 10-6 M
n
E = 0.20 V – (0.0592/3) (log [0.500][0.300]3/[0.200]3)
