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Part A Calculate the equilibrium constant at 25 ∘C for the reaction Fe(s)+2Ag+(aq)→Fe2+(aq)+2Ag(s) Standard Reduction Potentials...

Part A Calculate the equilibrium constant at 25 ∘C for the reaction Fe(s)+2Ag+(aq)→Fe2+(aq)+2Ag(s) Standard Reduction Potentials at 25 ∘C Fe2+(aq)+2e−→Fe(s) E∘= −0.45 V Ag+(aq)+e−→Ag(s) E∘= 0.80 V

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Answer #1

The reactions are Fe+2(aq)+2e- ------>Fe(s), Eo=-0.45V , reversing the reaction, Fe(s)------->Fe+2+2e-, Eo=0.45V (1)

Ag+(aq)+e- ------>Ag, Eo=0.80V (2)

multiplying Eq.2 with 2 and addition with Eq.1 gives

Fe+2Ag+ ------>Fe+2+2Ag, Eo=0.80+0.45= 1.25V

since deltaG= -nFE= -RTlnK, K= equilibrium constant, n= no of exchangeble electrons =2, F= 96500, E= 1.25V, T= 25+273= 298K, R= 8.314 J/mole.K

lnK= nFE/RT = 2* 96500*1.25/ (8.314*298)= 97.4

K= 1.944*1042

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