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(AKA Electrochemical Cell or Voltaic Cell)
What you need to know:
Oxidation Occurs at the Anode, reduction occurs at the Cathode.
Electrons flow through the wire (from lost to gained) Oxidation to reduction or Anode to Cathode.
Red Cat , An Ox
Electrodes must be metal (no ions), if there are only ions present an inert electrode (ex. platinum) should be used.
The salt bridge or porous membrane is used to allow "Ions Flow". This allows electrons to continuously move through the wire while replacing these lost negatives charges.
In other words an electron goes left and a negative ion goes right. Everything is still neutral.
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How do you use the Reduction table? When determining oxidation and reduction reactions find both possible reduction reactions. Which ever is high is the actual reduction reaction that will take place. The lower reaction is actually oxidation and must be reversed. To find the voltage of the cell add the 2 voltages together, remembering that you reverse the lower reaction so the Eo must also be reversed (+ becomes - or vice versa). Example Determining Oxidation and Reduction Here are the 2 possible reduction reactions (reduction charge is goes down) Cu2+ + 2e- --> Cu Zn2+ + 2e- --> Zn Now look for them on the table. Since Cu2+ + 2e- --> Cu is higher on the table, the other reaction is reversed. So the reactions are written like this: Cu2+ + 2e- --> Cu (Red.) Cu is the cathode Zn--> Zn2+ + 2e- (Ox.) Zn is the anode Determining Voltages Cu2+ + 2e- --> Cu Eo= +0.34v Zn--> Zn2+ + 2e- Eo= -(-0.76v) Note- since we reversed Zn we reverse the sign Now just add them up 0.34v + 0.76v=1.10v ***Never to be multiply cell potential (voltages) even when you balance you reaction*** See below for explanation |
| ***The volt has the dimensions of joules/coulomb– the energy produced per quantity of charge passing through the cell. Because voltage is the quotient of two extensive quantities, it is itself intensive. When we multiply the anodic and cathodic half-reactions by the stoichiometric factors required to ensure that each involves the same quantity of charge, the free energy change and the number of coulombs both increase by the same factor, leaving the potential (voltage) unchanged. This explains why we do not have to multiply the E°s of the anode and cathode reactions by stoichiometric factors when we are finding the potential of a complete cell. |
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