Calculate the equilibrium constant for each of the reactions at 25 C. Standard E

Question

Calculate the equilibrium constant for each of the reactions at 25 C. Standard Electrode Potentials at 25 C Reduction Half-Reaction E(V) Pb2+(aq)+2e Pb(s) -0.13 Mg2+(aq)+2e Mg(s) -2.37 I2(s)+2e 2I(aq) 0.54 Cl2(g)+2e 2Cl(aq) 1.36 MnO2(s)+4H+(aq)+2e Mn2+(aq)+2H2O(l) 1.21 Cu2+(aq)+2e Cu(s) 0.16 Part A Pb2+(aq)+Mg(s)Pb(s)+Mg2+(aq) Express your answer using three significant figures.

I2(s)+2Cl(aq)2I(aq)+Cl2(g)

Express your answer using two significant figures.

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

Express your answer using two significant figures.

Explanation / Answer

Part A)

Pb2+(aq)+Mg(s)Pb(s)+Mg2+(aq)

Eocell = - 0.13 - (- 2.38)

          = 2.24 V

- n F Eocell = -R T ln Keq

- 2 x 96485 x 2.24 = - 8.314 x 298 x ln Keq

Keq = 5.91 x 10^75

equilibrium constant = 5.91 x 10^75

part B)

I2(s)+2Cl(aq)2I(aq)+Cl2(g)

Eocell = 0.54 - 1.36 = - 0.82

- 2 x 96485 x - 0.82 = - 8.314 x 298 x ln K

K = 1.83 x 10^-28

part C)

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

Eocell = 1.21 - 0.16 = 1.05

equilibrium constant = 3.29 x 10^35

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