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jpodolski jpodolski
wrote...
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11 years ago
A Daniell cell uses Zinc and Copper electrodes. ie, for every Zn atom oxidised, 2 electrons are released and exactly 1 Cu atom is reduced. What happens when electrodes are chosen such that the anode creates more or less electrons per oxidised atom than the cathode uses per reduced atom?
For Example, Aluminum (3 valence electrons) and Copper (2 valence electrons).
How will this effect the E.M.F. (voltage) and current of the electron flow between the two half cells?
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wrote...
11 years ago
2 Al >> 2 Al3+ + 6e-  E° = + 1.676 V
3 Cu >> 3 Cu2+ + 6e- E° = - 0.337 V
The reaction that occurs is
2 Al + 3 Cu2+ >> 2 Al3+ + 3 Cu

for 2 Al 6 electrons are released and 3 Cu2+ are oxidized
wrote...
11 years ago
The EMF of the cell is based upon the energy per unit of charge and is a function of the two half reactions.  Changing one of the electrode will typically change one of the half reactions and thus the voltage of the cell.

The current is a function of the number of electrons which must move from the anode to the cathode to balance the half reactions.
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