# Problem: Using standard electrode potentials calculate ΔG° and use its value to estimate the equilibrium constant for each of the reactions at 25 °C.Pb2+(aq) + Mg(s) → Pb(s) + Mg2+(aq)

###### FREE Expert Solution

Pb2+(aq) + Mg(s) → Pb(s) + Mg2+(aq)

Recall:

Lose               Gain
Electron         Electrons
Oxidation       Reduction

cathode → reduction → oxidation number decreases

anode → oxidation → oxidation number increases

For ions: charge of the ion = oxidation number
For neutral atoms/compound: oxidation number = 0

For Pb: There is a decrease in oxidation number, therefore, it undergoes reduction and is the cathode.

Pb2+(aq) + 2e- → Pb(s) E° = -0.13 V

For Mg: There is an increase in oxidation number, therefore, it undergoes oxidation and is the anode.

Mg2+(aq) + 2e- → Mg(s)  E° = -2.37 V

E°cell = 2.24 V

$\overline{){\mathbf{∆}}{\mathbf{G}}{\mathbf{°}}{\mathbf{=}}{\mathbf{-}}{\mathbf{nFE}}{{\mathbf{°}}}_{{\mathbf{cell}}}}$

ΔG° = Gibbs Free Energy, J
n = # of e- transferred
F = Faraday’s constant = 96485 J/(mol e-)
cell = standard cell potential, V

90% (116 ratings) ###### Problem Details

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

Pb2+(aq) + Mg(s) → Pb(s) + Mg2+(aq)