Chemistry (Electrochemistry)

The standard potential for the following galvanic cell is 1.72V :
Be(s)|Be2+(aq)||Pb2+(aq)|Pb(s)

The standard reduction potential for the Pb2+/Pb half-cell: E = -.13V

Calculate the standard reduction potential for the Be2+/Be half-cell.



What I know:
Be is the anode
Pb is the cathode
I'm confused as to how to approach the problem though. How do you find the other E cell of the half reaction when given one half reaction Ered and the E cell total?

asked by Brian
  1. What I do is write the two half cells.
    Be ==> Be^2+ + 2e....E = x anode
    Pb^2+ + 2e ==> Pb....E = -0.13 cathode
    ---------------------------------
    Be + Pb^2+ ==> Be^2+ + Pb E = 1.72
    The cell consists of an oxidation half cell (Be/Be^2+) and a reduction half cell (Pb^2+/Pb).
    x + (-0.13) = 1.72
    Solve for x which gives you +1.85v AS AN OXIDATION. The problem asks for the reduction of that half cell so change the sign.

    posted by DrBob222

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