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A galvanic cell consists of La^3+/La half-cell and a standard hydrogen electrode. If the La^3+/La half cell functions as the anode and the cell potential is 2.52 V, what is the reduction potential for the La^3+/La half cell?

the choices are -2.52V,-0.84V, +0.84V,+2.52V

  • Chemistry -

    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
    [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
    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.

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