Question

The voltage generated by the zinc concentration cell described by Zn(s)|Zn2 (aq, 0.100 M)||Zn2 (aq, ? M)|Zn(s), is 14.0 mV at 25 °C. Calculate the concentration of the Zn2 (aq) ion at the cathode.

Answer 1

Ans- Zn(s) | Zn2+ (0.100 M ) || Zn2+ (? M) | Zn(s)
Anode .............................. ....................... cathode
Oxidation ......................... ....................... reduction

This is what is called a "concentration cell." It produces a voltage because of the two different concentrations of zinc ions.

Zn(s) --> Zn2+ + 2e- ....... E = +0.763V
Zn2+ + 2e- --> Zn2+ ........E = -0.763V
The standard cell potential must be zero. But since the zinc ions are each at different concentrations, and neither is 1M, then there is a potential difference between the anode and cathode, albeit, rather small.

The Nernst equation...

.............. RT..... [C]^c [D]^d
E = E° - ----- ln -------------
.............. nF..... [A]^a [B]^b

Combine RT/F and ln(x) to log(x) conversion all into 0.0592

Let x = [Zn2+] at the cathode

0.014V = 0 - 0.0592 / 2 (log(x / 0.100))
sove thisequation and evaluate value of x

that is the [Zn2+] at cathode is in M