A concentration cell consists of two Zn/Zn²⁺ electrodes. The electrolyte in compartment A is 0.10 M Zn(NO₃)₂and in compartment B is 0.60 M Zn(NO₃)₂. What is the voltage of the cell at 25°C?

Question

Quizplus · Accepted Answer

The equation for the half-reaction is:
ZnS1U1P12 -> Zn2+ + 2e-
The E° value for this reaction is -0.76 V.
In compartment A, the anode, zinc is being oxidized to Zn2+ ions, so the Ecell of the half-reaction is E°cell = -0.76V.
In compartment B, the cathode, Zn2+ ions are being reduced to zinc metal, so the Ecell of that half-reaction is E°cell = +0.76V.
We can use the Nernst equation to calculate the Ecell of the concentration cell at 25°C.
Ecell = E°cell - (RT/nF) ln Q
where R is the gas constant (8.31 J/mol*K), T is the temperature in Kelvin (298 K), n is the number of electrons involved in the reaction (2 electrons in this case), F is Faraday's constant (96500 C/mol), and Q is the reaction quotient.
At equilibrium, the concentrations of the two half-cells will be equal, so Q = 1.
Plugging in the values:
Ecell = 0.76 V - (8.31 J/mol*K * 298 K / (2 * 96500 C/mol) ) * ln 1/(0.10/0.60)
Ecell = 0.76 V - (0.0257 V) * ln 6
Ecell = 0.76 V - 0.0257 V * 1.79
Ecell = 0.76 V - 0.046 V
Ecell = 0.714 V
Therefore, the voltage of the cell is 0.714 V at 25°C. The closest answer choice is C, which is 0.023 V off the actual value.

A Concentration Cell Consists of Two Zn/Zn2+ Electrodes

A Concentration Cell Consists of Two Zn/Zn²⁺ Electrodes