Question #248032
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.
1
Expert's answer
2021-10-08T02:08:57-0400

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+:


CuCu2+(3.5×104M)Ag+(5×102M)AgCu|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(aq)2++2eCu(s)Cu^{2+}_{(aq)}+ 2 e^- \to Cu_{(s)}


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


The overall reaction is: Cu(aq)2++2Ag(s)2Ag(aq)++Cu(s)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:


ΔE=ΔE0.062log[Cu2+][Ag+]2ΔE=ECu2+/CuEAg+/Ag=(0.800.34)VΔE=0.46VΔE=0.34V(0.03V)log3.5×104M[5×102M]2=0.3989V\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 ΔG()=nFΔE()\Delta G^{(\circ)}=-nF \Delta E^{(\circ)}:


ΔG=nFΔE=(96485C/mol)(2)(0.46V)=88766.2JmolOn the other hand, ΔG=nFΔE=(96485C/mol)(2)(0.3989V)=76975.733Jmol\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: ΔG=76975.733Jmol\Delta G=-76975.733\frac{J}{mol}

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

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


Reference

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

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