A voltaic cell is constructed with two Zn²⁺-Zn electrodes, where the half-reaction is Zn²⁺ + 2e^- → Zn (s) E° = -0.763 V
The concentrations of zinc ion in the two compartments are 5.50 M and 1.11 × 10^-2 M, respectively. The cell emf is __________ V.

Question

Quizplus · Accepted Answer

To calculate the cell EMF, we need to use the Nernst equation:

E = E° - (RT/nF) ln(Q)

where:
E° = -0.763 V (given)
R = 8.314 J/mol*K (gas constant)
T = 298 K (room temperature, 25°C)
n = 2 (because two electrons are transferred in the half-reaction)
F = 96,485 C/mol (Faraday constant)
Q = [Zn2+]cathode / [Zn2+]anode = (5.50 M) / (1.11 × 10^-2 M) = 4.95 × 10^2

Plugging in the values:

E = -0.763 V - (8.314 J/mol*K / (2 * 96,485 C/mol)) ln(4.95 × 10^2)
E ≈ 0.0798 V

Therefore, the answer is C: 0.0798.

A Voltaic Cell Is Constructed with Two Zn2+-Zn Electrodes, Where

A Voltaic Cell Is Constructed with Two Zn²⁺-Zn Electrodes, Where