MnO2(s) + 4H+(aq) + Cu(s) → Mn2+(aq) + 2 H2O(l)+ Cu2+(aq)
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
ΔG° = Gibbs Free Energy, J
n = # of e- transferred
F = Faraday’s constant = 96485 J/(mol e-)
E°cell = standard cell potential, V
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)
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