The equilibrium constant, K, for a redox reaction is related to the standard pot

Question

The equilibrium constant, K, for a redox reaction is related to the standard potential, E degree, by the equation ln K = nFE degree / RT where n is the number of moles of electrons transferred, F(the Faraday constant) is equal to 96,500 C/(mol e^-), R (the gas constant) is equal to 8.314 J/(mol K), and T is the Kelvin temperature. Fe(s) + Ni^2+ (aq) rightarrow Fe^2+ (aq) + Ni(s) Express your answer numerically. Calculate the standard cell potential (E degree) for the reaction X(s) + Y^+ (aq) rightarrow X^+ (aq) + Y(s) if K = 6.91 Times 10^-3. Express your answer to three significant figures and include the appropriate units.

Explanation / Answer

nF*E°cell = RT*lnK

E°cell = RT/(nF)*lnK

Then

E°cell = 8.314*298/(1*96500) * ln(6.91*10^-3)

Solve

E°cell = -0.12772 V

This is not favoured since K is too small (reagents are favoured, therefore current of electorns go backward, i.e. negative V)

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