Consider the following reduction potentials: For a galvanic cell employing the Cu, Cu²⁺ and Pb, Pb²⁺ couples, calculate the maximum amount of work that would accompany the reaction of 1 mol of lead under standard conditions.

Question

Quizplus · Accepted Answer

To calculate the maximum amount of work that can be done, we can use the formula:
ΔG° = -nFE°
where ΔG° is the change in Gibbs free energy, n is the number of moles of electrons transferred, F is the Faraday constant, and E° is the standard reduction potential.
The balanced half-cell reactions and their reduction potentials are:
Cu2+(aq) + 2e- → Cu(s) - E° = +0.34 V
Pb2+(aq) + 2e- → Pb(s) - E° = -0.13 V
CuS1U1P12+(aq) + e- → Cu(s) + S1U1P12-(aq) - E° = -0.16 V
PbS1U1P12+(aq) + e- → Pb(s) + S1U1P12-(aq) - E° = -0.22 V
The overall cell reaction is:
Pb(s) + CuS1U1P12+(aq) → PbS1U1P12+(aq) + Cu(s)
The standard electromotive force (EMF) of the cell, E°cell, can be calculated by the formula:
E°cell = E°reduction (cathode) - E°reduction (anode)
E°cell = (+0.34 V) - (-0.22 V) = +0.56 V
The number of moles of electrons transferred is 2, since two electrons are transferred in each half-reaction. Therefore, n = 2.
Substituting the values into the formula, we get:
ΔG° = -nFE° = -(2 mol)(96,485 C/mol)(+0.56 V) = -90,706 J/mol
Converting the unit to kJ/mol, we get:
ΔG° = -90.7 kJ/mol
Therefore, the maximum amount of work that would accompany the reaction of 1 mol of lead under standard conditions is -90.7 kJ.

Consider the Following Reduction Potentials: for a Galvanic Cell