Question

Find the potentials of the following electrochemical cell:

Cd | Cd2+, M = 0.10 ‖ Ni2+, M = 0.50 | Ni

Answer 1

It is 0.167V!!!

Answer 2

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 :

where,

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

`E^o_(cell)` = standard electrode potential = 0.16 V

`E_(cell)=0.16-(0.0592)/(2)\log ([0.10])/([0.5])`

`E_(cell)=0.18V`

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