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

Question

he 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 to96,500 C/(mol e?) , R (the gas constant) is equal to 8.314 J/(mol?K) , and T is the Kelvin temperature.

Standard reduction potentials

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

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

if K = 1.02

Reduction half-reaction E? (V) Ag+(aq)+e??Ag(s) 0.80 Cu2+(aq)+2e??Cu(s) 0.34 Sn4+(aq)+4e??Sn(s) 0.15 2H+(aq)+2e??H2(g) 0 Ni2+(aq)+2e??Ni(s) ?0.26 Fe2+(aq)+2e??Fe(s) ?0.45 Zn2+(aq)+2e??Zn(s) ?0.76 Al3+(aq)+3e??Al(s) ?1.66 Mg2+(aq)+2e??Mg(s) ?2.37

Explanation / Answer

For the reaction:

X(s) +Y+(aq)   ---- > X+(aq) +Y(S)

K=1.02*10-3

ln K= nFE0/RT

E0=(RT/nF)lnK

       =(8.314*298/1*96485) ln(1.02*10-3)

       =-0.177V

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