How It Works: Why More Reactive Metals Always Displace Less Reactive Ones
Part of Displacement Reactions — GCSE Chemistry
This how it works covers How It Works: Why More Reactive Metals Always Displace Less Reactive Ones within Displacement Reactions for GCSE Chemistry. Revise Displacement Reactions in Chemical Changes for GCSE Chemistry with 20 exam-style questions and 20 flashcards. This topic appears regularly enough that it should still be part of a steady revision cycle. It is section 3 of 12 in this topic. Use this how it works to connect the idea to the wider topic before moving on to questions and flashcards.
Topic position
Section 3 of 12
Practice
20 questions
Recall
20 flashcards
⚙️ How It Works: Why More Reactive Metals Always Displace Less Reactive Ones
Displacement reactions are driven by the relative ability of metals to lose electrons. A more reactive metal has a stronger tendency to exist as positive ions — it "wants" to be in ionic form. A less reactive metal has a weaker drive to remain as ions — it "prefers" to be neutral atoms.
When a more reactive metal (e.g., magnesium) is placed in a solution containing less reactive metal ions (e.g., Cu²⁺), the magnesium atoms donate their outer electrons directly to the copper ions. Magnesium atoms lose 2 electrons each (becoming Mg²⁺ ions, which dissolve into solution) while copper ions gain those 2 electrons (becoming neutral Cu atoms, which settle out as brown solid).
This electron transfer is thermodynamically favourable — the system moves to a lower energy state. The magnesium's outermost electrons are less tightly held (higher energy) compared to where they end up on copper. This energy difference is released as heat, which is why displacement reactions are exothermic.
The sulfate ions (SO₄²⁻) are spectator ions: they remain in solution throughout, unchanged. They are not shown in the ionic equation because they play no part in the electron transfer.