1. 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- + Au E º° = V
Overall reaction: __________________________
3Ce4+ + Au → Au Ce3+
-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 = 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+] = M, [Ce3+] = M and [Au3+] = 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 – /3 (log [Au3+])
[Au3+] = 8.56 x 10-6 M