Question

Use the Nernst equation to calculate the cell potentials of the following cells at 298 K (use data in Appendix E of the textbook).

2 Ag+ (aq) (0.50 M) + Ni(s) ? 2 Ag(s) + Ni2+(aq) (0.20 M)

Answer 1

To calculate the cell potential for the given reaction at 298 K, we apply the Nernst equation using the provided concentrations for Ag+ and Ni2+. The specific formula includes constants for R, T, n, and F, and the equation simplifies due to the standard cell potential being zero.

Step-by-step explanation:

The question is about calculating the cell potentials for a chemical reaction using the Nernst equation. Given the half-cell reactions of silver and nickel, we need to determine the cell potential at a specified temperature of 298 K and given concentrations.

To calculate the cell potential under non-standard conditions, we utilize the Nernst equation:

E = E° - (RT/nF) × ln(Q)

Where E is the cell potential, E° is the standard cell potential, R is the universal gas constant, T is the temperature in Kelvins, n is the number of moles of electrons transferred, F is Faraday's constant, and Q is the reaction quotient.

For this reaction, since E°cell = 0, as Ag is both the anode and cathode, the Nernst equation simplifies to:

E = -(RT/nF) × ln(Q)

We can calculate Q from the individual concentrations of Ag+ and Ni2+ ions provided:

Q = ([Ni2+]/[Ag+]2) = (0.20 M)/(0.50 M)2

Now we can insert the values along with R = 8.314 J/(mol·K), T = 298 K, n = 2, and F = 96485 C/mol to find the cell potential. After calculations, we will obtain a numerical value which will be our final answer. The sign of this number will indicate whether the cell reaction is spontaneous or nonspontaneous.