A concentration cell is constructed using two Ni electrodes with Ni²⁺ concentrations of 1.0 M and 1.00 $\times$10^-⁴ M in the two half-cells.The reduction potential of Ni²⁺ is -0.23 V.Calculate the potential of the cell at 25°C.

Question

Quizplus · Accepted Answer

The potential of a concentration cell can be calculated using the Nernst equation: E = E° - (RT/nF) * ln(Q), where E° is the standard reduction potential, R is the gas constant, T is the temperature in Kelvin, n is the number of moles of electrons transferred, F is the Faraday constant, and Q is the reaction quotient. For a concentration cell, E° = 0 because both electrodes are of the same material. At 25°C (298 K), and assuming n = 2 for the nickel reaction, the equation simplifies to E = -(0.0592/n) * log([Ni2+]cathode/[Ni2+]anode). Substituting the given concentrations and solving gives a positive potential, indicating option D is correct.

A Concentration Cell Is Constructed Using Two Ni Electrodes with Ni2+

A Concentration Cell Is Constructed Using Two Ni Electrodes with Ni²⁺