A standard electrochemical cell is set up using two standard half-cells, as shown in the diagram below - NSC Physical Sciences - Question 8 - 2019 - Paper 2

Component Q provides a pathway for the movement of ions, ensuring electrical neutrality in the cell and assisting in restoring charge balance.

To identify metal X, we use the cell notation:

$E_{cell} = E_{cathode} - E_{anode}$

Given that the voltmeter initially registers 1.49 V, we can consider the possible half-cell reactions:

• If X is platinum, $E_{anode} = 1.36 ext{ V}$ and
• Therefore, rearranging gives $E_{cell} = E_{anode} = 1.49 + E_{anode}$ and leads us to conclude that X is Lead (Pb), as it has a potential of $-0.13 ext{ V}$. Thus, the calculation confirms that metal X is Pb (Lead).

The reducing agent is the species undergoing oxidation. Therefore, in the context of the cell, the reducing agent is Lead (Pb) or Pb.

The reading on the voltmeter becomes zero when the cell reaches equilibrium, at which point the rate of oxidation is equal to the rate of reduction. This means that the forward and reverse reactions occur at the same rate, leading to no net change in the voltage.

The reading on the voltmeter will DECREASE when AgNO₃ is added due to the formation of a precipitate reducing the availability of reactants in the half-cell.

According to Le Chatelier's principle, when a system at equilibrium is disturbed, the system will adjust to counteract the disturbance. Adding AgNO₃ increases the concentration of Ag⁺ ions in the half-cell, causing the equilibrium to shift towards the precipitate formation. This shift reduces the concentrations of reactants, leading to a decrease in the voltage registered by the voltmeter.