Question 28(b): Explain why copper displaces Q from its nitrate solution, given the standard electrode potentials.

$E^\circ_{Cu^{2+}/Cu} = +0.34 \text{ V}$
$E^\circ_{Q^+/Q} = +0.80 \text{ V}$

Since $E^\circ_{Cu^{2+}/Cu}$ is less positive than $E^\circ_{Q^+/Q}$, copper metal is a stronger reducing agent (more easily oxidized) than metal Q.

Because copper is a stronger reducing agent, it will spontaneously reduce the $Q^+$ ions in the solution to solid Q, while the copper metal itself gets oxidized to $Cu^{2+}$ ions. This means copper displaces Q from its nitrate solution.

Reason: Copper is a stronger reducing agent (more reactive) than metal Q. Therefore, copper metal is oxidized to $Cu^{2+}$ ions, causing the copper strip to dissolve, while $Q^+$ ions are reduced to solid Q, which deposits on the copper strip.