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.
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?
1 answer
1 answer