A galvanic cell is composed of these two half-cells, with the standard reduction potentials shown:
Cu2+(aq) + 2 e- ---->Cu(s)
+0.34volt
Cd2+(aq) + 2 e- ---->Cd(s)
-0.40volt
What is the standard free energy change for the cell reaction of this galvanic cell?
Reverse the Cd half cell and add to the Cu half cell as written. Calculate Eo cell.
Then dG = -nFEcell.
n is 2 and F is 96,485.
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To find the standard free energy change (ΔG°) for the cell reaction of this galvanic cell, you can use the formula:
ΔG° = -nFΔE°
Where:
ΔG° = standard free energy change for the cell reaction
n = number of electrons transferred in the balanced redox equation
F = Faraday's constant (96,485 C/mol)
ΔE° = standard cell potential (also known as the standard electromotive force or standard cell voltage), which is equal to the difference between the reduction potentials of the two half-cells.
In this case, since the reduction potentials are given, and the balanced redox equation shows that 2 electrons are transferred in both half-reactions, you can directly substitute the values into the formula:
n = 2 (since 2 electrons are transferred in both half-reactions)
F = 96,485 C/mol
ΔE° = E°(cathode) - E°(anode)
ΔE° = (+0.34 V) - (-0.40 V)
= +0.74 V
Now, substitute the values into the formula to find ΔG°:
ΔG° = -nFΔE°
= -(2)(96,485 C/mol)(0.74 V)
≈ -142,168 J/mol
Therefore, the standard free energy change for the cell reaction of this galvanic cell is approximately -142,168 J/mol.