Consider the following reduction potentials: Cu²⁺ + 2e^- $\to$ Cu $\varepsilon$ ^$\circ$ = 0.342 V
Pb²⁺ + 2e^- $\to$ Pb $\varepsilon$ ^$\circ$ = -0.130 V
For a galvanic cell employing the Cu,Cu²⁺and Pb,Pb²⁺ couples,calculate the maximum amount of work that would accompany the reaction of one mole of lead under standard conditions.

Question

Consider the following reduction potentials: Cu²⁺ + 2e^-

\to

Cu

\varepsilon

^$\circ$ = 0.342 V
Pb²⁺ + 2e^-

\to

Pb

\varepsilon

^$\circ$ = -0.130 V
For a galvanic cell employing the Cu,Cu²⁺and Pb,Pb²⁺ couples,calculate the maximum amount of work that would accompany the reaction of one mole of lead under standard conditions.

Quizplus · Accepted Answer

The maximum work done by a galvanic cell is given by the formula: W_max = -nFE_cell. Here, n = 2 (moles of electrons), F = 96485 C/mol (Faraday's constant), and E_cell = E_cathode - E_anode = 0.342 V - (-0.130 V) = 0.472 V. Thus, W_max = -2 * 96485 * 0.472 = -91.1 kJ.

Consider the Following Reduction Potentials: Cu2+ + 2e- →\to→ Cu ε\varepsilonε ∘\circ∘

Consider the Following Reduction Potentials: Cu²⁺ + 2e^- $\to$ Cu $\varepsilon$ ^$\circ$