Displacement as Redox

A highly reactive metal has a much stronger tendency to lose its outer electrons than a less reactive metal. When placed in a mixture, the more reactive metal atom throws away its electrons to form a positive ion, meaning it undergoes oxidation.

The less reactive metal starts off as a positive ion trapped in the compound. It is forced to gain these transferred electrons, turning back into a neutral, solid atom through reduction.

In Higher Tier papers, you must also identify the agents causing this change. The more reactive metal atom acts as a reducing agent because it forces the other substance to take its discarded electrons. The less reactive metal ion acts as the oxidising agent because it accepts electrons, forcing the other substance to oxidise.

Cutting Out the Bystanders

If you watch a football match, you focus on the players moving the ball, not the thousands of people sitting in the stands. In chemistry, spectator ions are exactly like the crowd.

These are ions (such as $SO_4^{2-}$ or $NO_3^-$) that do not change their charge, state, or participate in the electron transfer during a reaction.

To show the true redox mechanism, chemists write an ionic equation. This strips away the spectator ions to reveal only the particles that actually gain or lose electrons. We can break this down even further into a half-equation, which explicitly shows the exact number of electrons (written as $e^-$) moving to or from a specific element.

Worked Example: Metal Displacement

Explain the redox processes occurring when solid zinc is added to copper(II) sulfate solution.

Step 1: Write the full chemical equation.

$$Zn(s) + CuSO_4(aq) \rightarrow ZnSO_4(aq) + Cu(s)$$

Step 2: Remove the spectator ions to form the ionic equation.

• The sulfate ion ($SO_4^{2-}$) remains unchanged in the solution.

$$Zn(s) + Cu^{2+}(aq) \rightarrow Zn^{2+}(aq) + Cu(s)$$

Step 3: Write the oxidation half-equation.

• Zinc atoms are more reactive, so they lose two electrons to form zinc ions.

$$Zn \rightarrow Zn^{2+} + 2e^-$$

Step 4: Write the reduction half-equation.

• Copper(II) ions are less reactive, so they gain the two electrons to form neutral solid copper atoms.

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

(Practical Observations: The blue $Cu^{2+}$ solution would fade, and a reddish-brown solid of copper would form on the dissolving zinc).

Worked Example: Halogen Displacement

Halogens follow the exact same redox principles, but because they are non-metals, the more reactive element gains electrons instead of losing them.

Explain why the reaction between chlorine gas and sodium iodide solution is a redox reaction.

Step 1: Write the full chemical equation.

$$Cl_2(g) + 2NaI(aq) \rightarrow 2NaCl(aq) + I_2(aq)$$

Step 2: Form the ionic equation by removing the sodium spectator ions ($Na^+$).

$$Cl_2(g) + 2I^-(aq) \rightarrow 2Cl^-(aq) + I_2(aq)$$

Step 3: Identify the reduction process.

• The more reactive chlorine molecules gain electrons to form chloride ions.

$$Cl_2 + 2e^- \rightarrow 2Cl^-$$

Step 4: Identify the oxidation process.

• The less reactive iodide ions lose electrons to form iodine molecules. Notice how the total charge is balanced on both sides of the arrow.

$$2I^- \rightarrow I_2 + 2e^-$$