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, by the equation lnK=nFERT 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/(molK) , and T is the Kelvin temperature.

Reduction half-reaction E (V) Ag+(aq)+eAg(s) 0.80 Cu2+(aq)+2eCu(s) 0.34 Sn4+(aq)+4eSn(s) 0.15 2H+(aq)+2eH2(g) 0 Ni2+(aq)+2eNi(s) 0.26 Fe2+(aq)+2eFe(s) 0.45 Zn2+(aq)+2eZn(s) 0.76 Al3+(aq)+3eAl(s) 1.66 Mg2+(aq)+2eMg(s) 2.37

Part A Use the table of standard reduction potentials given above to calculate the equilibrium constant at standard temperature (25 C) for the following reaction: Fe(s)+Ni2+(aq)Fe2+(aq)+Ni(s) Express your answer numerically.

Part B Calculate the standard cell potential (E) for the reaction X(s)+Y+(aq)X+(aq)+Y(s) if K = 4.30×104. Express your answer to three significant figures and include the appropriate units.

Explanation / Answer

a)

Fe(s)+Ni2+(aq)Fe2+(aq)+Ni(s)

Ni2+(aq)+2eNi(s) 0.26

Fe2+(aq)+2eFe(s) 0.45

E°cell = Ered - Eox = -0.26 --0.45 = 0.19 V

now..

dG = -n*F*E°cell = -2*96500*0.19 = -36670 J/mol

dG = -RT*ln(K)

K = exp(-DG/(RT)) = exp(36670/(8.314*298)) = 2678535.544

B)

ge E°cell given

K = 4.3*10^-4

and e- = 1

dG = -RT*ln(K)

dG = -8.314*298*ln(4.3*10^-4)

dG = 19205.4

dG = -nF*E°cell

E°cell = -dG/(nF) = -19205.4/(1*96500) = -0.19901 V

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