# Exercise 18.67: Problems by To

Exercise 18.67: Problems by Topic – Cell Potential, Free Energy,and the Equilibrium Constant Calculate the equilibrium constant foreach of the reactions at 25 ∘C. Part APb2+(aq)+Mg(s)→Pb(s)+Mg2+(aq) Express your answer using onesignificant figure. K = ? Part B Br2(l)+2Cl−(aq)→2Br−(aq)+Cl2(g)Express your answer using one significant figure. K = ? Part CMnO2(s)+4H+(aq)+Cu(s)→Mn2+(aq)+2H2O(l)+Cu2+(aq) Express your answerusing one significant figure. K = ?

Part A

Pb2+(aq)+Mg(s) —>Pb(s)+Mg2+(aq)

E0cell = E0cathode – E0anode

=(-0.126)-(-2.372)

= 2.246v

DG0 = – nFE0cell

=-2*96500*2.246

= -433478 joule

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)

= -0.2727v

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)

= -0.613v

DG0 = – nFE0cell

=-2*96500*-0.613

= 118309 joule

DG0 = -RTlnK

118309 = -8.314*298lnk

K = 1.83*10^-21

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