Respuesta :
Answer : The [tex]\Delta G[/tex] for this reaction is, -88780 J/mole.
Solution :
The balanced cell reaction will be,
[tex]Cu(s)+2Ag^+(aq)\rightarrow Cu^{2+}(aq)+2Ag(s)[/tex]
Here, magnesium (Cu) undergoes oxidation by loss of electrons, thus act as anode. silver (Ag) undergoes reduction by gain of electrons and thus act as cathode.
The half oxidation-reduction reaction will be :
Oxidation : [tex]Cu\rightarrow Cu^{2+}+2e^-[/tex]
Reduction : [tex]2Ag^++2e^-\rightarrow 2Ag[/tex]
Now we have to calculate the Gibbs free energy.
Formula used :
[tex]\Delta G^o=-nFE^o[/tex]
where,
[tex]\Delta G^o[/tex] = Gibbs free energy = ?
n = number of electrons to balance the reaction = 2
F = Faraday constant = 96500 C/mole
[tex]E^o[/tex] = standard e.m.f of cell = 0.46 V
Now put all the given values in this formula, we get the Gibbs free energy.
[tex]\Delta G^o=-(2\times 96500\times 0.46)=-88780J/mole[/tex]
Therefore, the [tex]\Delta G[/tex] for this reaction is, -88780 J/mole.
