Respuesta :
Answer : The cell potential for this cell 0.434 V
Solution :
The balanced cell reaction will be,
[tex]Cu(s)+2Ag^{+}(aq)\rightarrow Cu^{2+}(aq)+2Ag(s)[/tex]
Here copper (Cu) undergoes oxidation by loss of electrons, thus act as anode. silver (Ag) undergoes reduction by gain of electrons and thus act as cathode.
First we have to calculate the standard electrode potential of the cell.
[tex]E^o_{[Cu^{2+}/Cu]}=0.34V[/tex]
[tex]E^o_{[Ag^{+}/Ag]}=0.80V[/tex]
[tex]E^o=E^o_{[Ag^{+}/Ag]}-E^o_{[Cu^{2+}/Cu]}[/tex]
[tex]E^o=0.80V-(0.34V)=0.46V[/tex]
Now we have to calculate the concentration of cell potential for this cell.
Using Nernest equation :
[tex]E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Cu^{2+}][Ag]^2}{[Cu][Ag^+]^2}[/tex]
where,
n = number of electrons in oxidation-reduction reaction = 2
[tex]E_{cell}[/tex] = ?
Now put all the given values in the above equation, we get:
[tex]E_{cell}=0.46-\frac{0.0592}{2}\log \frac{(1.14)\times (1)^2}{(1)\times (0.15)}[/tex]
[tex]E_{cell}=0.434V[/tex]
Therefore, the cell potential for this cell 0.434 V
