Question

The concentration of zinc is 0.5M and copper is 0.05M at 25 0C. Using the Nernst equation calculate the cell potential of Zn/Zn2+//Cu2+/Cu.

Answer 1

The concentration of zinc is 0.5M and copper is 0.05M at 25 0C. Using the Nernst equation calculate the cell potential of Zn/Zn²⁺//Cu²⁺/Cu.

The equations of reactions taking place in the cell are the following:

Cu²⁺ + 2 e^– ⇄ Cu(s)     E⁰_Cu²⁺_/Cu = 0,34 V: cathode

Zn(s) ⇄ Zn²⁺ + 2 e^–      E⁰_Zn²⁺_/Zn = -0,76 V: anode

Cu²⁺(aq) + Zn(s) → Cu(s) + Zn²⁺(aq).

The Nernst equation for the potential of the cell is:

$E = E^0 - \frac{RT}{nF}\text{ln}Q$ , where:

• n is the number of the electrons taking part in the reaction, 2 in our case
• R = 8,314 J.mol^–1.K^–1 : ideal gas constant
• T : absolute temperature in Kelvin
• F = N_A × e = 6.022 141 79x10²³ × 1.602 176 487x10^–19 = 96485.3399 C : Faraday constant (charge of 1 mole of electrons)
• $Q$ is the reaction quotient, $\frac{c(Zn^{2+})}{c(Cu^{2+})}$ .

Therefore, the potential of the cell is:

$E = 0.34 - (-0.76) )- \frac{8.314\cdot298}{2\cdot96485.3399}\cdot\text{ln}\frac{0.5}{0.05}= 1.07$ V.

Answer: the cell potential is 1.07 V.