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[Zn2+] = 0.10 M for the cathode and the [Zn2+] = 0.010 M[Zn2+] = 0.010 M for the anode?

Answer 1

Answer: 0.029 V

Step-by-step explanation:

For the given chemical reaction :

`Zn^(2+)(0.10M)(aq)+Zn(s)\rightarrow Zn^2+{0.010M)(aq)+Zn(s)`

Using Nernst equation :

`E_(cell)=E^0_(cell)-(2.303RT)/(nF)\log\frac{\text {anodic ion concentration}}{\text {cathodic ion concentration}}`

where,

F = Faraday constant = 96500 C

R = gas constant = 8.314 J/mol.K

T = room temperature = 298 K

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

`E^0_(cell)` = standard electrode potential of the cell = 0 (as both metals are same )

`E_(cell)` = emf of the cell = ?

`E_(cell)=0-(2.303* 8.314* 298)/(2* 96500)\log(0.010)/(0.10)`

`E_(cell)=0.029V`

Thus the cell potential will be 0.029 V