Question

Find the potentials of the following electrochemical cell:

Answer 1

Answer: 0.18 V

Explanation:-

$Cd/Cd^{2+}(0.10M)//Ni^{2+}(0.50M)?Ni$

Here Cd undergoes oxidation by loss of electrons, thus act as anode. nickel undergoes reduction by gain of electrons and thus act as cathode.

$E^0_(Cd^{2+}/Cd)$=-0.40V[/tex]

$E^0_(Ni^{2+}/Ni)$=-0.24V[/tex]

$Cd+Ni^{2+}\rightarrow Cd^{2+}+Ni$

Here Cd undergoes oxidation by loss of electrons, thus act as anode. nickel undergoes reduction by gain of electrons and thus act as cathode.

$E^0=E^0_{cathode}- E^0_{anode}$

Where both $E^0$ are standard reduction potentials.

$E^0=E^0_{[Ni^{2+}/Ni]}- E^0_{[Cd^{2+}/Cd]}$

$E^0=-0.24-(-0.40)=0.16V$

Using Nernst equation :

$E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Cd^{2+}]}{[Ni^{2+]}$

where,

n = number of electrons in oxidation-reduction reaction = 2

$E^o_{cell}$ = standard electrode potential = 0.16 V

$E_{cell}=0.16-\frac{0.0592}{2}\log \frac{[0.10]}{[0.5]}$

$E_{cell}=0.18V$

Thus the potential of the following electrochemical cell is 0.18 V.