Exercise 18.67: Problems by Topic - Cell Potential, Free Energy, and the Equilib

Question

Exercise 18.67: Problems by Topic - Cell Potential, Free Energy, and the Equilibrium Constant Calculate the equilibrium constant for each of the reactions at 25 C. Part A Pb2+(aq)+Mg(s)Pb(s)+Mg2+(aq) Express your answer using one significant figure. K = 5×1075 SubmitMy AnswersGive Up All attempts used; correct answer displayed Part B Br2(l)+2Cl(aq)2Br(aq)+Cl2(g) Express your answer using one significant figure. K = 6•1010 SubmitMy AnswersGive Up Incorrect; Try Again; 5 attempts remaining Part C MnO2(s)+4H+(aq)+Cu(s)Mn2+(aq)+2H2O(l)+Cu2+(aq) Express your answer using one significant figure. K = SubmitMy AnswersGive Up

Explanation / Answer

part A

   Pb2+(aq)+Mg(s)Pb(s)+Mg2+(aq)  

E0cell = E0cathode - E0anode

        = (-0.126)-(-2.372)

        = 2.246 v

   DG0 = - nFE0cell

       = -2*96500*2.246

     = -433.478 kjoule

DG0 = - RTlnK

-433478 = -8.314*298lnk

k = 9.65*10^75

part B

Br2(l)+2Cl-(aq) ----> 2Br-(aq)+Cl2(g)

E0cell = E0cathode - E0anode

        = (1.0873)-(1.36)

        = -0.2727 v

   DG0 = - nFE0cell

       = -2*96500*-0.2727

     = 52631.1 joule

DG0 = -RTlnK

52631.1 = -8.314*298lnk

k = 5.95*10^-10

part C

MnO2(s)+4H+(aq)+Cu(s) ----> Mn2+(aq)+2H2O(l)+Cu2+(aq)

E0cell = E0cathode - E0anode

        = (0.337)-(0.95)

        = -0.613 v

   DG0 = - nFE0cell

       = -2*96500*-0.613

     = 118309 joule

DG0 = -RTlnK

118309 = -8.314*298lnk

K = 1.83*10^-21

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