Salt preparation by displacement

Not every salt can be made by simply mixing an acid and a base. Some metals are reactive enough to shove a less reactive species clean out of its compound, and that shoving match — displacement — gives chemists a direct, one-step route to certain salts. Drop a strip of zinc into dilute sulphuric acid and it fizzes as hydrogen gas bubbles away, leaving zinc sulphate behind. Drop the same zinc into blue copper sulphate solution and something even more visible happens: the blue fades, a reddish crust of copper grows on the zinc, and a new colourless salt, zinc sulphate, forms in solution. Both are displacement, and both are governed by a single ranking of metals called the activity series. This article builds the two displacement routes, explains why the activity series decides who wins, and works through the practical steps — excess metal, filtration, evaporation, crystallisation — used to isolate a soluble salt.

The core insight

A metal higher in the activity series displaces hydrogen from a dilute acid or a lower metal from its salt solution, forming a new salt, and this reactivity ranking alone predicts which displacement reactions happen and which do not.

The core idea

Displacement, in the salt-preparation context, means a more reactive metal pushing out a less reactive species — either hydrogen from a dilute acid, or a less reactive metal from a solution of its salt — and taking that species' place in a new compound.

Route 1 — metal + dilute acid. This is the standard method for preparing a soluble salt of a moderately reactive metal, one that sits above hydrogen in the activity series but is not so violently reactive that it is unsafe to handle this way (zinc, iron, magnesium and aluminium are typical; sodium and potassium are excluded because they react far too vigorously with acids):

\text{Metal} + \text{Dilute acid} \rightarrow \text{Salt} + \text{H}_2\uparrow

where: the metal must lie above hydrogen in the activity series for the reaction to proceed at a usable rate, the acid must be dilute (concentrated acids bring side reactions and safety hazards), and the hydrogen gas escapes as bubbles, confirmed by the characteristic "pop" when a lit splint is held to it.

Route 2 — metal + salt solution. A more reactive metal displaces a less reactive metal from a solution of the less reactive metal's salt:

\text{More reactive metal} + \text{Salt of less reactive metal} \rightarrow \text{New salt} + \text{Less reactive metal}

where: the reaction proceeds only if the added metal sits above the metal already in the salt, read off the activity series $\text{K} > \text{Ca} > \text{Na} > \text{Mg} > \text{Al} > \text{Zn} > \text{Fe} > \text{Pb} > (\text{H}) > \text{Cu} > \text{Ag} > \text{Au}$ ; if the added metal sits below, there is simply no reaction. The figure below lays out that ranking as a ladder, with the two displacement routes marked against it: a metal reaching down past hydrogen displaces it from an acid, and a metal reaching down past another metal displaces that metal from its salt.

Figure

A vertical activity series ladder from potassium at the top to gold at the bottom with hydrogen marked partway down, and two arrows showing zinc displacing hydrogen from an acid and iron displacing copper from a salt solution

Why it works

Displacement is not arbitrary — it follows directly from how eagerly each metal loses electrons to become a positive ion, which is exactly what the activity series ranks.

A more reactive metal loses electrons more readily than a less reactive one. When zinc metal touches a solution containing $\text{Cu}^{2+}$ ions, zinc atoms give up electrons to become $\text{Zn}^{2+}$ far more readily than copper atoms would give up electrons to become $\text{Cu}^{2+}$ . Those freed electrons are picked up by the $\text{Cu}^{2+}$ ions already in solution, turning them into neutral copper atoms that deposit on the zinc surface. The zinc has, in effect, traded electrons for the copper's place in solution.

The reverse never happens on its own. Copper metal placed in zinc sulphate solution cannot push electrons onto $\text{Zn}^{2+}$ ions, because copper holds onto its own electrons more tightly than zinc does — there is no driving force for the swap, so nothing is observed. This is why checking the activity series before predicting a reaction is not optional: the ranking is the entire basis for whether displacement occurs at all.

The same logic explains metal + acid. A dilute acid supplies $\text{H}^+$ ions in solution. A metal above hydrogen in the series loses electrons more readily than hydrogen "would," so it reduces $\text{H}^+$ to $\text{H}_2$ gas while itself going into solution as a positive ion — exactly the same electron hand-off, just with hydrogen ions as the recipient instead of a second metal.

Why excess metal, not excess acid, is used. After the reaction, any leftover solid metal is filtered off with a simple filter paper, leaving a clean salt solution behind. Leftover acid, by contrast, stays fully dissolved and cannot be filtered out — it would remain in the crystals as an impurity. So the metal is always added in excess and the acid is the limiting reagent, ensuring every drop of acid reacts and only solid metal (easily removed) remains unreacted.

Worked example

Question: Describe, with balanced equations, how solid magnesium chloride crystals are prepared by displacement, and how a strip of zinc behaves when placed in copper sulphate solution, in terms of the activity series.

Step 1. Locate magnesium and hydrogen on the activity series. Magnesium sits well above hydrogen, so it reacts readily and safely with dilute hydrochloric acid:

\text{Mg} + 2\text{HCl} \rightarrow \text{MgCl}_2 + \text{H}_2\uparrow

Step 2. Carry out the preparation. A measured volume of dilute HCl is taken in a beaker, and magnesium ribbon is added in small pieces, in excess, so that all the acid is consumed. Effervescence (hydrogen, confirmed by the pop test) continues as long as acid remains.

Step 3. Isolate the product. Once effervescence stops, the excess unreacted magnesium is filtered off. The clear filtrate — a solution of magnesium chloride — is evaporated to concentrate it, then cooled to crystallise, giving solid $\text{MgCl}_2$ crystals.

Step 4. Now consider zinc in copper sulphate solution. Locating both metals on the series, zinc sits above copper, so zinc displaces copper:

\text{Zn} + \text{CuSO}_4 \rightarrow \text{ZnSO}_4 + \text{Cu}

The blue colour of the copper sulphate solution fades toward colourless as $\text{Cu}^{2+}$ ions are replaced by $\text{Zn}^{2+}$ ions, a reddish-brown deposit of copper forms on the zinc strip, and the zinc strip itself gradually thins as it dissolves. Answer: $\text{MgCl}_2$ crystals form via Mg + 2HCl to MgCl2 + H2, and zinc visibly displaces copper from CuSO4 solution because Zn lies above Cu in the activity series.

Competitive stretch

If a strip of copper were placed in zinc sulphate solution instead, predict the outcome before reading on: copper sits below zinc in the activity series, so copper cannot lose electrons to $\text{Zn}^{2+}$ , and no reaction is observed at all — the solution stays blue-free (it was never blue) and the copper strip is unchanged. This asymmetry — reaction one way, nothing the other way — is exactly what an activity series is for, and it is a favourite way examiners test whether the ranking has actually been understood rather than memorised as isolated equations.

Common traps

Trying to prepare a salt using a highly reactive metal such as sodium or potassium in a metal + acid reaction — these react far too violently and dangerously with dilute acids and are never used this way.
Using excess acid instead of excess metal — leftover acid contaminates the salt solution and, unlike leftover solid metal, cannot be removed by simple filtration.
Forgetting to check the activity series first — assuming any metal will displace any other metal from its salt, when in fact a less reactive metal (e.g. copper) shows no reaction at all with a more reactive metal's salt solution (e.g. zinc sulphate).
Assuming hydrogen displacement only works with hydrochloric or sulphuric acid — the same idea applies to any dilute acid, provided the metal is reactive enough and lies above hydrogen.

Ask the tutor: want to practise predicting whether a given metal-and-salt-solution pair reacts, or run through more worked examples of the excess-metal filtration method?

Why can zinc displace copper from copper sulphate solution but copper cannot displace zinc from zinc sulphate solution?

It comes down to how readily each metal gives up electrons to form a positive ion. Zinc loses electrons more easily than copper, so when zinc atoms meet $\text{Cu}^{2+}$ ions in solution, zinc hands over electrons and goes into solution as $\text{Zn}^{2+}$ , while the $\text{Cu}^{2+}$ ions pick up those electrons and deposit as copper metal. Copper cannot do the reverse, because it holds its electrons more tightly than zinc does, so there is no driving force for copper to push zinc ions out of solution. The activity series is simply this electron-losing tendency written as a ranked list, and it predicts the direction of every such reaction.

Could I use dilute nitric acid instead of hydrochloric or sulphuric acid for this kind of preparation?

Not usually, for salt preparation by simple metal displacement. Dilute nitric acid tends to act as an oxidising agent rather than reacting by the clean metal-displaces-hydrogen route, so it often produces oxides of nitrogen instead of a steady stream of hydrogen gas, and the reaction is less predictable. That is why ICSE practical work sticks to dilute hydrochloric acid or dilute sulphuric acid for this method — they react cleanly with suitable metals to give hydrogen gas and a soluble salt, exactly as the general equation describes.

Why filter off the excess metal before evaporating, instead of just evaporating everything together?

Any unreacted solid metal left in the flask would stay behind as an insoluble residue when the water evaporates, mixing with and contaminating the salt crystals. Filtering first removes that solid metal cleanly, leaving only the dissolved salt in the filtrate. Evaporating and crystallising this clear filtrate then gives pure salt crystals, free of leftover metal — which is exactly why excess metal (easy to filter) is deliberately preferred over excess acid (impossible to filter) in this method.