What is E° of the cell if an electrochemical cell consisted of:
Pb|Pb2+ (0.50M)||Au3+(0.10M| Au°
I think you meant "what is the Ecell).Write the equation.
Pb==> Pb^2+ + 2e Eo = ? look up this oxidation E as E1
Au^3+ + 3e ==> Au Eo =? look up this reduction E as E2
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3Pb + 2Au^3+ ==> 3Au + 2Pb^2+
Calculate Ecell by Ecell = E1 + E2 from above. but this is Ecell for 1 M soltuiion of eacxh and you don't have that. Now convert to the concentrations listed in the problem with this equation.
Ecell = Eo - (0.059/n) log Q and I write Q because I can't get all of that on one line. So n is 6 for total electrons and
Q = concentrations of right side/left side; i.e.,
[(Pb^2+)^3*(Au)^2]/[(Au^3+)^2(Pb)^3. You are given (Pb^2+) and (Au^3+) and (Pb) = 1 and (Au) = 1 by definition.
Post your work if you get stuck.
To find the standard cell potential (E°) of the electrochemical cell, you need to use the Standard Reduction Potentials table. The E° values in the table represent the reduction potentials for different half-cell reactions.
Given the following cell notation: Pb|Pb2+(0.50M)||Au3+(0.10M)|Au°
1. Identify the half-reactions for each electrode:
- The left electrode consists of the Pb metal and Pb2+ ions: Pb → Pb2+ + 2e-
- The right electrode consists of the Au3+ ions and Au metal: Au3+ + 3e- → Au
2. Look up the Standard Reduction Potentials for each half-reaction in the table:
- Pb2+ + 2e- → Pb (E° = -0.13 V)
- Au3+ + 3e- → Au (E° = 1.50 V)
3. Add the reduction potentials of the half-reactions to get the overall E° of the cell:
E°cell = E°right - E°left
E°cell = 1.50 V - (-0.13 V)
E°cell = 1.63 V
Therefore, the standard cell potential (E°) of the given electrochemical cell is 1.63 V.