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

Explanation:

For the given chemical reaction :

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

Using Nernst equation :

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

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 = ?

$E_{cell}=0-\frac{2.303\times 8.314\times 298}{2\times 96500}\log\frac{0.010}{0.10}$

$E_{cell}=0.029V$

Thus the cell potential will be 0.029 V