Question

Using standard electrode potentials calculate ΔG∘rxn and use its value to estimate the equilibrium constant for each of the reactions at 25 ∘C.

A. Cu2+(aq)+Zn(s)→Cu(s)+Zn2+(aq)

B. Br2(l)+2Cl−(aq)→2Br−(aq)+Cl2(g)

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

Find K. All answers must have one significant figure. Please show all your work and equations used. Thank you!!

Answer 1

We will be calculating ΔG∘rxn for the following reactions.

ΔG∘rxn = nFE° =

Where n is number of moles of electrons. And F is Faradays constant.

A. Cu2+(aq)+Zn(s)→Cu(s)+Zn2+(aq)

This reaction can be divided in two parts oxidation and reduction

Cu2+(aq) + 2e- à Cu(s)    E°red = +0.337V

ΔG∘red = -2 * 96500 * 0.33 = -63690 J

Zn(s) = Zn2+(aq) + 2e-       E°ox = +0.7618V

ΔG∘ox = -2 * 96500 * 0.7618 = -147027.4 J

ΔG∘rxn = ΔG∘red + ΔG∘ox

              = -63.690 + -147.027 = -210.7 kJ

E° = (0.05916/n) log K at 25°C

E° = E°red + E°ox = 0.33 + 0.7618 = 1.0918

Log K = E° * n / 0.05916 = 1.0918 * 4 / 0.05916 = 73.820

K = 6.6 * 10^73

B. Br2(l) + 2Cl−(aq) → 2Br−(aq) + Cl2(g)

This reaction can be divided in two parts oxidation and reduction

Br(l) + 2e-à 2Br-(aq)         E°red = +1.066V

ΔG∘ox = -2 * 96500 * 1.066 = -205738 J

2Cl-(aq) à Cl2(g) + 2e-       E°ox = -1.36 V

ΔG∘ox = -2 * 96500 * -1.36 = 262480 J

ΔG∘rxn = ΔG∘red + ΔG∘ox

              = -205.738 + 262.480 = 56.7 kJ

E° = (0.05916/n) log K at 25°C

E° = E°red + E°ox = 1.066 + -1.36 = -0.294 V

Log K = E° * n / 0.05916 = -0.294 * 4 / 0.05916 = -19.87

K = 1.3 * 10^-20

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

This reaction can be divided in two parts oxidation and reduction

MnO2(s) + 4H+(aq) + 2e- à Mn2+(aq) + 2H2O(l)        E°red = +1.23 V

ΔG∘red = -2 * 96500 * 1.23 = -237390 J

Cu(s) à Cu2+(aq) + 2e-                                                     E°ox = - 0.337V

ΔG∘ox = -2 * 96500 * -0.337 = 65041 J

ΔG∘rxn = ΔG∘red + ΔG∘ox

              = -237.390 + 65.041 = -172.3 kJ

E° = (0.05916/n) log K at 25°C

E° = E°red + E°ox = 1.23 – 0.337 = 0.893 V

Log K = E° * n / 0.05916 = 0.893 * 4 / 0.05916 = 60.37

K = 2.3 * 10^60