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

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.

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.

K =

Part B

Calculate the standard cell potential (E) for the reaction

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

if K = 4.04×103.

Express your answer numerically in volts.

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

A.

the potential

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

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

the most negative --> oxidizes

the mos potsivite --> reduces

then

E°cell = Ered - Eox

E°cell = -0.26 +0.45 = 0.19 V

B.

given

nFE°cell = RT*lnK

then

E°cell = RT/(nF)*lnK

n = 1 eelectrons transfer

F = 96500 farady constant

R 0 ideal gas law

T = temperature = 298 K

E°cell = (8.314*298)/(1*96500)*ln(4.04*10^-3) 0 -0.1415

meaning this is not favoured

since K < 1

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