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=nFE/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/(molK) , and T is the Kelvin temperature.

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)

B) Calculate the standard cell potential (E) for the reaction

X(s)+Y+(aq)X+(aq)+Y(s)

if K = 4.56×103.

Express your answer to three significant figures and include the appropriate units

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

Explanation / Answer

Solution :-

A) Fe(s) + Ni^2+(aq) ----- > Fe^2+(aq) + Ni(s)

Fe is oxidized and Ni is reduced

Fe2+(aq)+2eFe(s)     Eo = -0.45V

Ni2+(aq)+2eNi(s)        Eo = -0.26 V

E cell = E cathode - E anode

         = -0.26 V – (-0.45 V)

         = 0.19 V

Now lets calculate the equilibrium constant

lnK=nFE/RT

n=2

ln K = 2*96500 C *0.19 V /8.314 J per mol K *298 K

ln K = 14.8

K= antiln [14.8]

K= 2.68*10^6

So the equilibrium constant = 2.68*10^6

B) K= 4.56*10^-3

Eo cell = ?

X(s)+Y+(aq)X+(aq)+Y(s)

Number of electrons = n = 1

lnK=nFE/RT

ln 4.56*10^-3 = 1*96500 C * E / 8.314 J per mol K * 298 K

-5.39 = 1*96500 C * E / 8.314 J per mol K * 298 K

-5.39 * 8.314Jper mol K * 298 K / 1 * 96500 C = E

-0.138 V = Eo

So the cell potential is Eo = -0.138 V

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