Br2(l) + 2 Cl-(aq) → 2 Br-(aq) + Cl2(g)
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 Br: There is a decrease in oxidation number, therefore, it undergoes reduction and is the cathode.
Br2(l) + 2e- → 2 Br-(aq) E° = 1.09 V
For Cl: There is an increase in oxidation number, therefore, it undergoes oxidation and is the anode.
Cl2(g) + 2e- → 2 Cl-(aq) E° = 1.36 V
E°cell = -0.27 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.
Br2(l) +2 Cl-(aq)→2 Br-(aq) + Cl2(g)
Frequently Asked Questions
What scientific concept do you need to know in order to solve this problem?
Our tutors have indicated that to solve this problem you will need to apply the Cell Potential concept. If you need more Cell Potential practice, you can also practice Cell Potential practice problems.