A student set up the cell shown in Figure 2 - AQA - A-Level Chemistry - Question 6 - 2018 - Paper 1

To calculate the electrode potential of the left-hand electrode, we need to consider the concentration of Cu²⁺ in that cell, which is 0.15 mol dm⁻³. Using the Nernst equation:

$E = E^0 - \frac{RT}{nF} \ln Q$

For the reaction:

$Cu^{2+} + 2e^- \leftrightarrow Cu$

Here, (R) is the gas constant, (T) is the temperature in Kelvin, (n) is the number of moles of electrons (2), and (F) is Faraday's constant. Plugging values into the equation:

At 25 °C (298 K):

$E = 0.34 - \frac{0.0592}{2} \log \left(\frac{1}{0.15^2}\right)$

Calculating this will yield the electrode potential for the left-hand electrode.

The left-hand electrode does not have an electrode potential of +0.34 V because the concentration of Cu²⁺ ions is not 1.0 mol dm⁻³. Since the electrode potential is dependent on the concentration of the ions, a lower concentration results in a lower electrode potential.

The conventional representation for the cell in Figure 2 is:

$Cu | Cu^{2+}(0.15 mol dm^{-3}) || Cu^{2+}(1.0 mol dm^{-3}) | Cu$

The concentration of copper(II) ions in the left-hand electrode increases as the reaction proceeds. The reduction of Cu²⁺ ions to form solid copper at the right-hand electrode causes diffusion from the left-hand solution to maintain the dynamic equilibrium.

The EMF decreases to 0 V because the concentrations of Cu²⁺ ions in both half-cells become equal over time. As a result, there is no longer a concentration gradient to drive the redox reactions, leading to the cessation of electron flow.