Posted by Anonymous on Tuesday, March 9, 2010 at 3:21pm.
If the La/La^+3 is the anode, and you remember that the anode is where oxidation occurs, then the half reaction at the anode must be
La ==> La^+3 + 3electrons .... Ehalfcell
2H^+ + 2e ==> H2...............Ehalfcell
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[If La is the anode, then H2 must be the cathode and gain electrons.]
This is a little tough to do the computer but let me recopy the half cell part below.
La/La^+3 half cell = what voltage
H^+/H2 half cell = 0 voltage
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total cell reaction = total voltage.
Now if the total voltage is 2.52 volts and that is composed of ??volts + 0 volts = 2.52 volts, I give you three guess (and the first two don't count) as to what is ?? volts. It MUST be 2.52 volts since 2.52 + 0 = 2.52.
Here is where you need to be careful to make sure what the question is asking.
If the half cell is
La ==>La^+3 + 3e and that is 2.52 volts, what's next? The question is what is the REDUCTION potential. That is
La^+3 + 3e ==> La which is just the reverse of what the lab experiment was. So reverse the sign to make the oxidation potential of +2.52 volts = to -2.52 for the reduction potential.