Respuesta :
Answer: 0.18 V
Explanation:-
[tex]Cd/Cd^{2+}(0.10M)//Ni^{2+}(0.50M)?Ni[/tex]
Here Cd undergoes oxidation by loss of electrons, thus act as anode. nickel undergoes reduction by gain of electrons and thus act as cathode.
[tex]E^0_(Cd^{2+}/Cd)[/tex]=-0.40V[/tex]
[tex]E^0_(Ni^{2+}/Ni)[/tex]=-0.24V[/tex]
[tex]Cd+Ni^{2+}\rightarrow Cd^{2+}+Ni[/tex]
Here Cd undergoes oxidation by loss of electrons, thus act as anode. nickel undergoes reduction by gain of electrons and thus act as cathode.
[tex]E^0=E^0_{cathode}- E^0_{anode}[/tex]
Where both [tex]E^0[/tex] are standard reduction potentials.
[tex]E^0=E^0_{[Ni^{2+}/Ni]}- E^0_{[Cd^{2+}/Cd]}[/tex]
[tex]E^0=-0.24-(-0.40)=0.16V[/tex]
Using Nernst equation :
[tex]E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Cd^{2+}]}{[Ni^{2+]}[/tex]
where,
n = number of electrons in oxidation-reduction reaction = 2
[tex]E^o_{cell}[/tex] = standard electrode potential = 0.16 V
[tex]E_{cell}=0.16-\frac{0.0592}{2}\log \frac{[0.10]}{[0.5]}[/tex]
[tex]E_{cell}=0.18V[/tex]
Thus the potential of the following electrochemical cell is 0.18 V.
