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 = ?
Answer:
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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