Answer to Question #248032 in General Chemistry for Aria

Question

Write the cell notation, half reactions and overall reactions of a Galvanic
cell diagram that consists of one half-cell with a silver bar in a 5 x 10 - 2 M
silver nitrate solution (AgNO3
) and another half-cell with a copper bar in a
3.5 x 10 – 4 M copper (II) nitrate solution, Cu(NO3
)2
and a potassium
nitrate,KNO3 use as a salt bridge solution. Measurements indicate that the
copper electrode is negative in relative to the silver electrode. Calculate the
following:
A) Gibbs Free – Energy
B) Standard Gibbs Free – Energy
C) Electrode potential of the cell.

Answer 1

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⁺:

$Cu|Cu^{2+} \, (3.5 \times 10^{-4}\, M)||Ag^+\,(5\times 10^{-2}M)|Ag$

Then, the half reactions that occur at each side of the galvanic cell are:

Anode/ reduction: $Cu^{2+}_{(aq)}+ 2 e^- \to Cu_{(s)}$

Cathode / oxidation: $Ag_{(s)} \to Ag^{+}_{(aq)}+e^-$

The overall reaction is: $Cu^{2+}_{(aq)}+ 2 Ag_{(s)} \to 2 Ag^{+}_{(aq)}+Cu_{(s)}$

The electrode potential of the cell can be calculated with the Nernst equation and the standard reduction potentials:

$\Delta E = \Delta E^{\circ}-\frac{0.06}{2}\log{\cfrac {[Cu^{2+}]}{[Ag^{+}]^2}} \\ \Delta E^{\circ} =E_{Cu^{2+}/Cu}^{\circ}-E_{Ag^{+}/Ag}^{\circ}=(0.80-0.34)V \\ \Delta E^{\circ} =0.46\,V \\ \therefore \Delta E = 0.34\,V- (0.03 \,V)\log {\frac {3.5 \times 10^{-4}\, M}{{[5\times 10^{-2}M]}^2 }} = 0.3989\,V$

The Gibbs and the Standard Gibbs Free Energy can be found with the relation $\Delta G^{(\circ)}=-nF \Delta E^{(\circ)}$:

$\Delta G^{\circ}=-nF \Delta E^{\circ}=-(96485 \,C/mol)(2)(0.46\,V)=-88766.2\frac{J}{mol} \\ \text{On the other hand, } \\\Delta G=-nF \Delta E=-(96485 \,C/mol)(2)(0.3989\,V)=-76975.733\frac{J}{mol}$

In conclusion,

A) Gibbs Free – Energy: $\Delta G=-76975.733\frac{J}{mol}$

B) Standard Gibbs Free – Energy: $\Delta G°=-88766.2\frac{J}{mol}$

C) Electrode potential of the cell: $\Delta E = 0.3989\,V$

Reference

• Chang, R., & Goldsby, K. A. (2010). Chemistry. Chemistry, 10th ed.; McGraw-Hill Education: New York, NY, USA.