Question

The observed cell potential for a voltaic cell is 2.067 V when the temperature is 298 K and the concentration of copper(ll) ions is 1.07 M. What is the concentration of aluminum ions in this cell?

Al(s)|A13+ (aq, ?M)||Cu2+(aq, 1.07M)|Cu(s)
Cu2+(aq) + 2 e- → Cu(s) 0.337 V
Al3+(aq) + 3 e- → Al(s) -1.66 V

Answer 1

3.22 * 10^-4 M

Step-by-step explanation:

The cell reaction equation is;

2Al(s) + 3Cu^2+(aq) ------> 2Al^3+(aq) + 3Cu(s)

E° = 0.337 V - (-1.66 V)

E° = 1.997 V

Given that the cell potential is 2.067 V, from Nernst's equation;

E = E° - 0.0592/n log Q

Substituting values;

2.067 = 1.997 - 0.0592/6 log [Al^3+]^2/[1.07]^3

0.07 = - 0.0592/6 log [Al^3+]^2/[1.07]^3

- 7.07 = log [Al^3+]^2/[1.07]^3

Antilog (- 7.07) = [Al^3+]^2/[1.07] ^3

8.5 * 10^-8 = [Al^3+]^2/[1.07]^3

[Al^3+]^2 = 8.5 * 10^8 * 1.07 ^3

[Al^3+]^2 = 1.04 * 10^-7

[Al^3+] = √1.04 * 10^-7

[Al^3+] = 3.22 * 10^-4 M