The Activity Series and Displacement Reactions

Some metals give up electrons more easily than others. This single fact is arranged into a table called the activity series. It lets you predict whether one metal will displace another metal — or hydrogen gas — from an aqueous solution. You can do this without running the experiment first. The activity series is a ranked list of metals by how readily they lose electrons. It is built from Oxidation-Reduction Reactions. A metal higher on the list will reduce the cation of any metal below it. This happens because it is the stronger reducing agent. The same ranking tells you which metals dissolve in acid to release hydrogen gas. It also shows which are so unreactive (gold, platinum) that they resist almost everything. This article shows how to read the series. It also covers how to write and balance single-displacement equations. You'll learn to check predictions using Assigning Oxidation Numbers. The reasoning is entirely qualitative-turned-quantitative. A ranking becomes a yes/no prediction. A balanced equation becomes a mole-ratio calculation.

Key takeaways

The activity series ranks metals from most easily oxidized (most reactive) to least easily oxidized (least reactive, "noble").
A metal displaces a less active metal's cation from solution; it cannot displace a more active metal's cation.
Metals above hydrogen in the series react with acids (or water, for the most active ones) to release H₂ gas; metals below hydrogen do not.
Single-displacement reactions are redox reactions: one element is oxidized, another is reduced, verified with oxidation numbers.
Reaction feasibility is predicted from position in the series alone — no thermodynamic data are needed at this level.
Key idea: if the reactant metal sits higher in the activity series than the metal (or hydrogen) it would replace, the displacement occurs; otherwise nothing happens.

The core insight

A single-displacement reaction proceeds only when the free element is a stronger reducing agent than the element it would replace — read directly from the activity series.

The physical picture

Picture a strip of solid zinc metal dropped into a blue solution of copper(II) sulfate, $\text{CuSO}_4(aq)$ . Almost immediately, the surface of the zinc darkens as reddish-brown copper metal plates out onto it, and the blue color of the solution fades as $\text{Cu}^{2+}$ ions are consumed. The reaction is

\text{Zn}(s) + \text{CuSO}_4(aq) \rightarrow \text{ZnSO}_4(aq) + \text{Cu}(s)

Zinc atoms give up two electrons each, becoming $\text{Zn}^{2+}(aq)$ ; those electrons are accepted by $\text{Cu}^{2+}(aq)$ ions, which become neutral copper atoms. Zinc is oxidized, copper is reduced — a textbook redox process wearing a simple, visible disguise: a solid dissolves while another plates out. Now run the reverse experiment: dip a strip of copper into a solution of zinc sulfate. Nothing happens, no matter how long you wait. Copper metal cannot give up its electrons to $\text{Zn}^{2+}$ ions, because copper is the weaker reducing agent of the two. The asymmetry is the entire content of the activity series: it is simply a ranked list recording which metal "wins" this electron tug-of-war against which other metal, and against $\text{H}^+$ itself. The series also governs metal–acid reactions. Magnesium ribbon dropped into dilute hydrochloric acid fizzes vigorously as hydrogen gas escapes:

\text{Mg}(s) + 2\text{HCl}(aq) \rightarrow \text{MgCl}_2(aq) + \text{H}_2(g)

Copper wire in the same acid produces nothing — copper cannot reduce $\text{H}^+$ to $\text{H}_2$ , because copper lies below hydrogen in the series.

Figure

The activity series of common metals, most active (top) to least active (bottom); the H₂ line marks the acid-reactivity cutoff

The core relations

The predictive rule for a single-displacement reaction between a metal $A$ and a salt solution of metal $B$ 's cation is:

A(s) + B^{n+}(aq) \rightarrow A^{n+}(aq) + B(s) \quad \text{occurs only if } A \text{ is above } B \text{ in the activity series}

where:

$A(s)$ is the free (elemental) metal added to solution,
$B^{n+}(aq)$ is the cation already dissolved,
$n$ is the charge transferred (must balance electrons, not necessarily equal for $A$ and $B$ ),
"above" means higher position, i.e. greater tendency to be oxidized.

For reactions with acid, hydrogen occupies a fixed slot in the same ranking:

\text{M}(s) + n\,\text{H}^+(aq) \rightarrow \text{M}^{n+}(aq) + \tfrac{n}{2}\text{H}_2(g) \quad \text{occurs only if M is above H in the series}

where:

M is the metal, with characteristic charge $n$ in its common compounds,
$n\,\text{H}^+$ supplies $n$ electrons' worth of oxidizing power,
$\tfrac{n}{2}\text{H}_2$ balances hydrogen atoms and charge simultaneously.

Once feasibility is established, ordinary stoichiometry from the balanced equation gives quantitative yields:

\text{mol of product} = \text{mol of limiting reactant} \times \frac{\text{coefficient of product}}{\text{coefficient of limiting reactant}}

where:

the coefficients come from the balanced chemical equation,
"limiting reactant" is whichever reactant runs out first (see Limiting Reactants for the general method, foreshadowed only lightly here since displacement stoichiometry uses the identical logic).

Derivation ABalancing a displacement equation by electron balance

Goal. Given that aluminum metal displaces silver from silver nitrate solution, write and balance the equation, and show why the electron count — not the atom count alone — dictates the coefficients.

Step 1. Write the skeletal (unbalanced) equation. Aluminum is above silver in the activity series, so the reaction is feasible:

\text{Al}(s) + \text{AgNO}_3(aq) \rightarrow \text{Al(NO}_3)_3(aq) + \text{Ag}(s)

Step 2. Assign oxidation numbers to identify the redox pair. Aluminum goes from $0$ in $\text{Al}(s)$ to $+3$ in $\text{Al(NO}_3)_3$ ; silver goes from $+1$ in $\text{AgNO}_3$ to $0$ in $\text{Ag}(s)$ . Nitrate is a spectator polyatomic ion, unchanged throughout (Assigning Oxidation Numbers confirms N stays $+5$ and O stays $-2$ on both sides).

Step 3. Write the two half-reactions and their electron counts.

\text{Al} \rightarrow \text{Al}^{3+} + 3e^-

\text{Ag}^+ + e^- \rightarrow \text{Ag}

Aluminum releases 3 electrons per atom; silver ions each accept only 1 electron.

Step 4. Multiply the silver half-reaction so electrons lost equal electrons gained.

3 \times (\text{Ag}^+ + e^- \rightarrow \text{Ag}) \;\Rightarrow\; 3\text{Ag}^+ + 3e^- \rightarrow 3\text{Ag}

Now both half-reactions transfer exactly 3 electrons.

Step 5. Recombine and carry the coefficient through to the full equation. Since 3 $\text{Ag}^+$ ions are needed, 3 $\text{NO}_3^-$ spectators travel with them, giving 3 $\text{AgNO}_3$ on the reactant side and forcing $\text{Al(NO}_3)_3$ (not $3\,\text{AlNO}_3$ ) on the product side to keep aluminum's charge at $+3$ :

\text{Al}(s) + 3\text{AgNO}_3(aq) \rightarrow \text{Al(NO}_3)_3(aq) + 3\text{Ag}(s)

Step 6. Verify atom and charge balance. Al: 1 = 1. Ag: 3 = 3. N: 3 = 3. O: 9 = 9. Charge: both sides are neutral overall, and the electron transfer (3 electrons) matches Step 4. The equation is balanced.

Worked examples

Example 1 — Predicting and quantifying a metal–metal displacement. A student places a 2.50 g strip of zinc metal into 150.0 mL of 0.400 M copper(II) sulfate solution. Predict whether a reaction occurs, and if so, find the mass of copper deposited and which reactant (if either) is left over. Step 1: Check feasibility. Zinc lies above copper in the activity series (see the figure above), so zinc displaces $\text{Cu}^{2+}$ :

\text{Zn}(s) + \text{CuSO}_4(aq) \rightarrow \text{ZnSO}_4(aq) + \text{Cu}(s)

The mole ratio is 1:1:1:1. Step 2: Convert to moles.

n_{\text{Zn}} = \frac{2.50\ \text{g}}{65.38\ \text{g/mol}} = 0.03824\ \text{mol}

n_{\text{CuSO}_4} = 0.1500\ \text{L} \times 0.400\ \text{mol/L} = 0.06000\ \text{mol}

Step 3: Identify the limiting reactant. The 1:1 ratio means whichever has fewer moles limits the reaction. Since $0.03824 < 0.06000$ , zinc is limiting. Step 4: Compute copper produced.

n_{\text{Cu}} = n_{\text{Zn (reacted)}} \times \frac{1\ \text{mol Cu}}{1\ \text{mol Zn}} = 0.03824\ \text{mol}

m_{\text{Cu}} = 0.03824\ \text{mol} \times 63.55\ \text{g/mol} = 2.430\ \text{g}

Step 5: Leftover reactant.

n_{\text{CuSO}_4,\ \text{remaining}} = 0.06000 - 0.03824 = 0.02176\ \text{mol}

about 0.0218 mol of $\text{Cu}^{2+}$ (and its accompanying sulfate) remains in solution — the blue color fades but does not disappear entirely. Limiting-case check: if the CuSO₄ concentration were instead so dilute that $n_{\text{CuSO}_4} < n_{\text{Zn}}$ , the roles reverse and copper(II) becomes limiting; the calculation method is identical, only the identity of the limiting reactant changes. This confirms the stoichiometric logic does not depend on which reactant happens to be scarcer.

Example 2 — Metal–acid reaction with gas evolution. A 1.20 g sample of aluminum foil reacts completely with excess hydrochloric acid. Find the volume of hydrogen gas produced at STP (0 °C, 1 atm, where 1 mol of ideal gas occupies 22.4 L). Step 1: Feasibility and balanced equation. Aluminum lies well above hydrogen in the activity series, so the reaction proceeds:

2\text{Al}(s) + 6\text{HCl}(aq) \rightarrow 2\text{AlCl}_3(aq) + 3\text{H}_2(g)

(Balanced by the same electron-counting method: each Al loses 3 e⁻, each H⁺ gains 1 e⁻, so 2 Al: 6 H⁺ equalizes 6 electrons transferred.) Step 2: Moles of aluminum.

n_{\text{Al}} = \frac{1.20\ \text{g}}{26.98\ \text{g/mol}} = 0.04448\ \text{mol}

Step 3: Moles of hydrogen gas via the mole ratio.

n_{\text{H}_2} = n_{\text{Al}} \times \frac{3\ \text{mol H}_2}{2\ \text{mol Al}} = 0.04448 \times 1.500 = 0.06672\ \text{mol}

Step 4: Convert to volume at STP.

V_{\text{H}_2} = 0.06672\ \text{mol} \times 22.4\ \text{L/mol} = 1.494\ \text{L}

So roughly 1.49 L of hydrogen gas bubbles off — easily collected and tested with the classic "pop" of the burning-splint test. Limiting-case check: if instead a metal below hydrogen (say copper) were used, $n_{\text{H}_2}$ would be exactly zero regardless of mass, since no reaction occurs — the formula correctly predicts a null result only through the feasibility check in Step 1, not through the stoichiometry alone.

Limiting cases & common traps

Situation	What actually happens	Common mistake
Metal above target cation	Displacement proceeds spontaneously	Assuming any metal displaces any salt
Metal below target cation	No visible reaction at all	Writing a balanced equation for a reaction that cannot occur
Metal above H, added to acid	H₂ gas evolves; metal dissolves	Forgetting to balance charge, using wrong subscript on H₂
Metal below H (Cu, Ag, Au, Pt) in acid	No reaction with dilute non-oxidizing acids	Confusing this with reaction in oxidizing acids like conc. HNO₃, a different mechanism entirely
Very reactive metals (K, Na, Ca) near water	React directly with water, not just acid	Applying acid-only reasoning; these react even without added acid

The chemist behind it

The systematic ordering of metals by "displacing power" grew out of nineteenth-century electrochemical work relating metal reactivity to voltage, most directly traceable to the studies of galvanic cells that quantified how strongly different metals tend to lose electrons. The modern activity series is essentially a simplified, qualitative shadow of the standard reduction potential table, reordered so the strongest reducing agents (most negative reduction potentials) sit at the top.

Origins · 1 minAlessandro Volta and how this was discovered

Alessandro Volta

1745–1827 · Italy, invention of the voltaic pile

Volta's stacked-disc battery of 1800, built from alternating zinc and copper (or silver) discs separated by brine-soaked cloth, was the first device to generate a sustained electric current from a chemical reaction. The very pairing he chose — zinc oxidizing while copper or silver is protected — is a direct experimental ancestor of the activity series: zinc's willingness to lose electrons to copper is precisely the displacement reaction that powers the pile. Volta's work gave chemists their first quantitative handle on relative "electromotive" strength between metals, decades before the language of oxidation numbers or standard electrode potentials existed.

Ask the tutor: If I drop a coil of iron wire into a solution containing both silver nitrate and copper(II) nitrate at the same concentration, which metal plates out first, and why?

Why does copper not react with hydrochloric acid, but it does react with concentrated nitric acid?

Copper sits below hydrogen in the activity series, so it cannot reduce H⁺ to H₂ — that displacement mechanism genuinely fails. Concentrated nitric acid dissolves copper through a completely different pathway: the nitrate ion itself acts as the oxidizing agent, being reduced to NO₂ or NO while copper is oxidized to Cu²⁺. No H₂ gas forms in that case, which is a clue that a different mechanism is operating.

Does a higher position in the activity series always mean a faster reaction?

Not necessarily. The series predicts thermodynamic feasibility — whether a reaction can occur at all — not its rate. Aluminum, for example, is quite high in the series but reacts sluggishly with many acids at room temperature because a thin, tough oxide layer (Al₂O₃) coats its surface and must be breached first. Once that layer is disrupted, the reaction proceeds essentially as predicted.

Can I use the activity series to predict reactions between two different metal salts, with no free metal involved?

No. The activity series specifically governs reactions where a free, uncombined metal (oxidation state 0) meets a dissolved cation. If both species are already ions in solution, the relevant tool is a solubility-based Precipitation Reactions analysis instead, since no elemental metal is present to act as a reducing agent.

Why does the equation for aluminum plus silver nitrate need a coefficient of 3, not something simpler?

Aluminum forms a $3+$ ion and silver forms a $1+$ ion, so one aluminum atom must supply enough electrons for three silver ions. Balancing electrons transferred (3 lost by Al, 1 gained by each Ag⁺) forces the ratio 1 Al : 3 AgNO₃ : 1 Al(NO₃)₃ : 3 Ag, exactly as derived by half-reaction balancing.

Where does hydrogen "sit" if it isn't actually a metal?

Hydrogen is included in the activity series purely as a reference point, because so many metal reactions of interest involve acids. Its position marks the dividing line: metals above it can reduce H⁺ to H₂ gas, and metals below it cannot, even though hydrogen itself is a nonmetal under ordinary conditions.