What is my answer for the second part? The equilibrium constant, K, for a redox

Question

What is my answer for the second part?

The equilibrium constant, K, for a redox reaction Is related to the standard potential, E degree, by the equation In 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 middot K), and T is the Kelvin temperature. Use the table of standard reduction potentials given above to calculate the equilibrium constant at standard temperature (25 degree C) for the following reaction: 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 = 1.92 times 10^-3. Express your answer to three significant figures and include the appropriate units.

Explanation / Answer

B) Given that K = 1.92 x 10-3

In the given reaction,

no of electrons transferred n = 1

We know that Go = -RT In K and Go = -nFEo

                 Then,

                 -RT In Keq = -nFEo

                      Eo = RT In Keq / nF

                           = [8.314 J/K/mol x 298 K x In (1.92 x 10-3) ] / [ 1 x 96500 C/mol]

                           = -0.16 V

                     Eo = -0.16 V

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