The diagram of the cell will be starting with Cu/Cu2+ because that is the negative pole or anode, while the oxidation occurs to the half that involves Ag/Ag+:
Cu∣Cu2+(3.5×10−4M)∣∣Ag+(5×10−2M)∣Ag
Then, the half reactions that occur at each side of the galvanic cell are:
Anode/ reduction: Cu(aq)2++2e−→Cu(s)
Cathode / oxidation: Ag(s)→Ag(aq)++e−
The overall reaction is: Cu(aq)2++2Ag(s)→2Ag(aq)++Cu(s)
The electrode potential of the cell can be calculated with the Nernst equation and the standard reduction potentials:
ΔE=ΔE∘−20.06log[Ag+]2[Cu2+]ΔE∘=ECu2+/Cu∘−EAg+/Ag∘=(0.80−0.34)VΔE∘=0.46V∴ΔE=0.34V−(0.03V)log[5×10−2M]23.5×10−4M=0.3989V
The Gibbs and the Standard Gibbs Free Energy can be found with the relation ΔG(∘)=−nFΔE(∘):
ΔG∘=−nFΔE∘=−(96485C/mol)(2)(0.46V)=−88766.2molJOn the other hand, ΔG=−nFΔE=−(96485C/mol)(2)(0.3989V)=−76975.733molJ
In conclusion,
A) Gibbs Free – Energy: ΔG=−76975.733molJ
B) Standard Gibbs Free – Energy: ΔG°=−88766.2molJ
C) Electrode potential of the cell: ΔE=0.3989V
Reference
- Chang, R., & Goldsby, K. A. (2010). Chemistry. Chemistry, 10th ed.; McGraw-Hill Education: New York, NY, USA.
Comments