Question

A concentration cell is one in which both the anode and cathode are the same but with different concentrations. Calculate the cell potential with [Zn2+] = 0.10 M for the cathode and the [Zn2+] = 0.010 M for the anode?

A. + 0.06 V
B. - 0.03 V
C. + 0.03 V
D. 0.0 V

Answer 1

Answer: The cell potential of the cell is +0.03 V

Step-by-step explanation:

The half reactions for the cell is:

Oxidation half reaction (anode):
`Zn(s)\rightarrow Zn^(2+)+2e^-`

Reduction half reaction (cathode):
`Zn^(2+)+2e^-\rightarrow Zn(s)`

In this case, the cathode and anode both are same. So,
`E^o_(cell)` will be equal to zero.

To calculate cell potential of the cell, we use the equation given by Nernst, which is:

`E_(cell)=E^o_(cell)-(0.0592)/(n)\log \frac{[Zn^(2+){diluted}}{[Zn^(2+){concentrated}]}`

where,

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

`E_(cell)` = ?

`[Zn^(2+){diluted}]` = 0.010 M

`[Zn^(2+){concentrated}]` = 0.10 M

Putting values in above equation, we get:

`E_(cell)=0-(0.0592)/(2)\log (0.0101M)/(0.10M)`

`E_(cell)=0.03V`

Hence, the cell potential of the cell is +0.03 V

Answer 2

Cell potential is 0.03 V

Step-by-step explanation:

Anode:
`Zn\rightarrow Zn^(2+)(0.01M)+2e^(-)`

Cathode:
`Zn^(2+)(0.10M)+2e^(-)\rightarrow Zn`

Overall:
`Zn^(2+)(0.10M)\rightarrow Zn^(2+)(0.01M)`

Nenrst equation for this cell reaction at
`25^(0)\textrm{C}`:

Cell potential,
`E_(cell)=(-0.059)/(n)log{([Zn^(2+)]_(0.01M))/([Zn^(2+)]_(0.10M))}`

Where n is number of electron exchanged

Here n=2

So,
`E_(cell)=(-0.059)/(2)log{(0.01)/(0.10)}=0.03V`