Question

The voltage generated by the zinc concentration cell described by the line notation zn(s)|zn²⁺(aq, 0.100M)||zn²⁺(aq, ?M)|zn(s) is 21.0mV at 25 degrees Celsius. Calculate the concentration of the zn²⁺(aq) ion at the cathode. M concentration:

Answer 1

The concentration of Zn²⁺(aq) at the cathode of a 21.0 mV zinc concentration cell at 25°C is approximately 0.0848 M, as determined by applying the Nernst equation.

Step-by-step explanation:

The student is asking to calculate the concentration of Zn²⁺(aq) at the cathode in a zinc concentration cell. To derive this concentration, we use the Nernst equation which relates the cell potential to the standard electrode potential, temperature, and the reaction quotient (Q). The voltage of 21.0 mV (0.021 V) at 25°C that's given indicates that the reaction is at non-standard conditions (since E° for a zinc concentration cell is 0 V). The formula for the Nernst equation in this context is:

E = E° - (0.0591/n)log(Q)

Since E° for a zinc concentration cell is 0 V and n equals 2 (the number of electrons transferred per zinc ion), we can simplify to:

0.021 V = -(0.0591/2)log([Zn²⁺]_cathode/0.100 M)

Solving for the concentration of Zn²⁺ at the cathode:

[Zn²⁺]_cathode = 0.100 M × 10^(-2(n)(0.021 V)/0.0591)

[Zn²⁺]_cathode = 0.100 M × 10^(-2(2)(0.021)/0.0591) = 0.100 M × 10^(-0.0711864

[Zn²⁺]_cathode = 0.100 M × 0.847584

The concentration of Zn²⁺(aq) ion at the cathode in the zinc concentration cell is thus approximately 0.0848 M.