A voltaic cell is constructed with two -Zn electrodes,where the half-reaction is (aq) + → Zn(s) E° = -0.763 V
The concentrations of zinc ion in the two compartments are 5.50 mol L^-1 and 1.11 ×
Mol L^-1,respectively.The cell emf is ________ V.

Question

A voltaic cell is constructed with two

-Zn electrodes,where the half-reaction is A voltaic cell is constructed with two -Zn electrodes,where the half-reaction is (aq) + → Zn(s) E° = -0.763 V The concentrations of zinc ion in the two compartments are 5.50 mol L-1 and 1.11 × Mol L-1,respectively.The cell emf is ________ V. A) -1.54 × B) -378 C) 0.0798 D) 0.160 E) -0.761

(aq) +

→ Zn(s) E° = -0.763 V
The concentrations of zinc ion in the two compartments are 5.50 mol L^-1 and 1.11 × A voltaic cell is constructed with two -Zn electrodes,where the half-reaction is (aq) + → Zn(s) E° = -0.763 V The concentrations of zinc ion in the two compartments are 5.50 mol L-1 and 1.11 × Mol L-1,respectively.The cell emf is ________ V. A) -1.54 × B) -378 C) 0.0798 D) 0.160 E) -0.761

Mol L^-1,respectively.The cell emf is ________ V.

Quizplus · Accepted Answer

The cell emf can be calculated using the Nernst equation:

Ecell = E°cell - (RT/nF)lnQ

where:
- E°cell is the standard cell potential, given as -0.763 V
- R is the gas constant (8.314 J/K·mol)
- T is the temperature in kelvin (298 K)
- n is the number of electrons transferred in the balanced half-reaction, which is 2 in this case
- F is the Faraday constant (96,485 C/mol)
- Q is the reaction quotient, which can be calculated as:

Q = [Zn2+]anode/[Zn2+]cathode
  = (5.50 mol/L)/(1.11 × 10-3 mol/L)
  = 4.95 × 106

Substituting into the Nernst equation:

Ecell = -0.763 V - (8.314 J/K·mol × 298 K / (2 × 96,485 C/mol))ln(4.95 × 106)
       = 0.0798 V

Therefore, the best choice is C, which is the calculated cell emf.

A Voltaic Cell Is Constructed with Two -Zn Electrodes,where