(Solved): Use the Nernst equation to calculate the cell potentials (in V) of the following cells at 298.15 K. ...

Expert Answer

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The Nernst equation relates the cell potential of an electrochemical cell to the standard electrode potential and the activities (or concentrations) of the species involved:




where E is the cell potential, E° is the standard electrode potential, R is the gas constant, T is the temperature in Kelvin, n is the number of electrons transferred in the reaction, F is the Faraday constant, and Q is the reaction quotient.
Using this equation, we can calculate the cell potentials of the given cells at 298.15 K as follows:

(a)




The half-reactions involved are:







The overall reaction is the sum of the two half-reactions:




The reaction quotient Q is:




At equilibrium, the cell potential E is equal to zero, so we can solve for E°:







Substituting the given values, we get:
E° = (8.314 J/K/mol) * (298.15 K) / (2 * 96485 C/mol) * ln (0.20 M / 0.65 M)^2 E° = -0.068 V
Thus, the cell potential is:
E = -0.068 V - (RT/2F) ln [Co2+]/[Ag+]^2
Substituting the given values, we get:
E = -0.068 V - (8.314 J/K/mol * 298.15 K / (2 * 96485 C/mol)) * ln (0.65 M / 0.20 M)^2 E = -0.068 V - 0.127 V E = -0.195 V
Therefore, the cell potential of the given cell is -0.195 V.