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?
To find the standard reduction potential for the Be2+/Be half-cell, you can use the relationship between the standard reduction potentials and the standard cell potential.
The standard cell potential (E°cell) can be calculated by subtracting the reduction potential of the anode (E°anode) from the reduction potential of the cathode (E°cathode):
E°cell = E°cathode - E°anode
In this case, you are given the standard cell potential (E°cell) as 1.72 V and the standard reduction potential for the Pb2+/Pb half-cell (E°cathode) as -0.13 V.
Now, rearranging the equation, you can find the standard reduction potential for the Be2+/Be half-cell (E°anode):
E°anode = E°cathode - E°cell
Substituting the given values:
E°anode = (-0.13 V) - (1.72 V)
Simplifying the equation:
E°anode = -1.85 V
Therefore, the standard reduction potential for the Be2+/Be half-cell is -1.85 V.
To calculate the standard reduction potential for the Be2+/Be half-cell, you can use the equation:
Ecell = Ered(cathode) - Ered(anode)
In this case, the half-cell reaction for the cathode (Pb2+/Pb) is given with a standard reduction potential of -0.13 V. Now, we need to find the standard reduction potential for the anode (Be2+/Be).
Since the overall cell potential is given as 1.72 V, we can rearrange the equation to solve for Ered(anode):
Ered(anode) = Ered(cathode) - Ecell
Substituting the given values:
Ered(anode) = -0.13 V - 1.72 V
Ered(anode) = -1.85 V
Therefore, the standard reduction potential for the Be2+/Be half-cell is -1.85 V.
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.