# Problem: Using standard electrode potentials calculate ΔG° and use its value to estimate the equilibrium constant for each of the reactions at 25 °C.MnO2(s) + 4H+(aq) + Cu(s) → Mn2+(aq)+2 H2O(l) + Cu2+(aq)

###### FREE Expert Solution

MnO2(s) + 4H+(aq) + Cu(s) → Mn2+(aq) + 2 H2O(l)+ Cu2+(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 Mn: There is a decrease in oxidation number, therefore, it undergoes reduction and is the cathode.

MnO2(s) + 4H+ + 2e ⇌ Mn2+ + 2H2O(l)  E° = +1.23 V

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

Cu2+(aq)+2e- → Cu(s)  E° = +0.34 V

E°cell = 0.89 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

97% (192 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.

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