Can sodium pull copper out of copper(II) sulfate?
In 1.8.3 you learned that a metal can replace any metal below it in the activity series from a solution of its salt. Sodium is far above copper, so dropping a small piece of sodium into copper(II) sulfate solution should deposit red copper, right? In fact, the sodium skates around on the surface and gives off gas, and a blue precipitate appears, but no copper at all. This lesson uses the activity series more completely: with water as well as acids and salt solutions, and for halogens as well as metals.
The activity series of metals
Li > K > Ba > Ca > Na > Mg > Al > Zn > Fe > Ni > Sn > Pb > (H) > Cu > Hg > Ag > Pt > Au
This is a typical US activity series金属活动性顺序, with the most reactive metal on the left. Textbooks differ slightly: the Chinese textbook list leaves out Li, Ba, and Ni, and some US books include more metals. The common metals are always in the same order, because the list is based on how easily each metal gives up electrons (its standard electrode potential). In a single replacement置换反应 reaction, a more active element pushes a less active one out of its compound.
Metals and water.
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Li, K, Ba, Ca, and Na react with cold water to form a hydroxide and hydrogen gas:
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Mg barely reacts with cold water but slowly reacts with hot water.
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Fe does not react with cold water, but at high temperature it reacts with steam:
Metals and acids. Metals above hydrogen replace hydrogen in hydrochloric acid or dilute sulfuric acid. Iron forms iron(II) salts, whose solutions are pale green; aluminum forms Al³⁺ salts.
Metals and salt solutions. A metal replaces any metal below it from a solution of its salt, with one exception: K, Ca, and Na react with the water first. They form a hydroxide, which then reacts with the salt. Sodium in copper(II) sulfate solution does this:
Watch the charges when you write products. The ion charges decide the formula of the new compound: in 2Al + 3CuSO₄ → Al₂(SO₄)₃ + 3Cu, Al³⁺ and SO₄²⁻ combine as Al₂(SO₄)₃. When no reaction happens, write NR (no reaction).
Halogens replace halogens
Nonmetals can do single replacement too. The halogen activity series is
As in 2.3.6, reactivity decreases down the group. A more active halogen replaces a less active one from its salt:
The solution turns orange-yellow when bromine forms and brown when iodine forms. The reverse does not happen: I₂ cannot replace Br₂ or Cl₂.
Chinese textbooks teach the same rules. They write reaction conditions over the arrow and ↑ or ↓ for gases and precipitates instead of state symbols.
Pick an element and a solution, predict whether they react, and then check. Switch to Halogens to try nonmetal replacement:
Example 1: will it react?
Predict whether each pair reacts. If it does, write the balanced equation.
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The rule. A metal replaces any metal below it in the activity series from a solution of its salt: Li > K > Ba > Ca > Na > Mg > Al > Zn > Fe > Ni > Sn > Pb > (H) > Cu > Hg > Ag > Pt > Au.
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Sanded aluminum in copper(II) sulfate solution. Al is above Cu, so it reacts. Aluminum forms Al³⁺, so the product is Al₂(SO₄)₃. Balance SO₄ first (3CuSO₄ for one Al₂(SO₄)₃), then Al and Cu:
Red copper coats the aluminum, and the blue color fades. Sanding removes the tight Al₂O₃ coating; without it, the reaction is hard to see.
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Copper in zinc sulfate solution. Cu is below Zn: NR.
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Iron wire in silver nitrate solution. Fe is above Ag, so it reacts. In single-replacement reactions iron usually forms Fe²⁺, so the product is iron(II) nitrate:
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Check. Each equation balances atoms and charge: in the last one, Fe²⁺ needs two NO₃⁻, which is why 2AgNO₃ and 2Ag appear.
Example 2: halogens and the exception
Predict whether each pair reacts. If it does, write the balanced equation.
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The halogen series. F₂ > Cl₂ > Br₂ > I₂. A more active halogen replaces a less active one from a solution of its salt.
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Bromine water added to sodium chloride solution. Br₂ is less active than Cl₂, so it cannot replace chlorine: NR.
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Chlorine bubbled into sodium iodide solution. Cl₂ is more active than I₂, so it reacts. The I₂ formed turns the solution brown (and starch turns blue-black):
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Potassium dropped into silver nitrate solution. K is near the top of the series and reacts violently with cold water. It reacts with the water first, forming KOH and H₂:
The KOH then reacts with AgNO₃ to form a precipitate, but no silver metal is deposited.
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The pattern. Metals that react with cold water (Li, K, Ba, Ca, Na) can’t be used to replace metals from salt solutions. The “higher replaces lower” rule works from Mg down.
- Zinc in iron(II) sulfate solution
Zn is above Fe: Zn + FeSO₄ → ZnSO₄ + Fe, and the pale green color fades.
- Iron in zinc sulfate solution
Fe is below Zn, so it cannot replace zinc.
- Copper in hydrochloric acid
Cu is below hydrogen, so it cannot replace hydrogen in the acid.
- Tin in hydrochloric acid
Sn is above hydrogen and slowly releases it: Sn + 2HCl → SnCl₂ + H₂.
- Potassium in copper(II) sulfate solution
K reacts with water first, forming KOH and H₂. The KOH then forms a blue Cu(OH)₂ precipitate with the CuSO₄.
- Silver in copper(II) nitrate solution
Ag is below Cu: NR.
- Magnesium in silver nitrate solution
Mg + 2AgNO₃ → Mg(NO₃)₂ + 2Ag.
- Bromine water in potassium iodide solution
Br₂ is more active than I₂: Br₂ + 2KI → 2KBr + I₂, and the solution turns brown.
- Iodine in sodium chloride solution
I₂ is less active than Cl₂: NR.
Common mistakes
Mistake 1: “Iron and hydrochloric acid make FeCl₃.” In single replacement, iron forms the iron(II) compound: Fe + 2HCl → FeCl₂ + H₂, a pale green solution. FeCl₃ solution is yellow and is made a different way.
Mistake 2: “Sodium is above copper, so it replaces copper from copper(II) sulfate solution.” K, Ca, and Na react with the water in any solution first. With copper(II) sulfate you get hydrogen gas and a blue copper(II) hydroxide precipitate, not copper.
Mistake 3: “A metal above hydrogen releases hydrogen from any acid.” The rule works for acids like hydrochloric acid and dilute sulfuric acid. Nitric acid and concentrated sulfuric acid are strong oxidizers and usually do not release hydrogen from metals.
Quick check
Metals X and Y both release hydrogen from dilute hydrochloric acid, but Z does not. When X is placed in a solution of a Y salt, nothing happens. What is the order from most to least active?
X and Y are above hydrogen and Z is below it. X cannot replace Y, so Y is more active than X. The order is Y > X > Z.
Which reaction actually happens?
Halogen activity is Cl₂ > Br₂ > I₂, and only a more active halogen replaces a less active one. Only Cl₂ replacing I₂ works.
Summary
- In single replacement, a more active element replaces a less active one in a compound. Use the activity series for metals and F₂ > Cl₂ > Br₂ > I₂ for halogens.
- K, Ca, and Na react with cold water to form a hydroxide and H₂, and they react with the water in salt solutions first. Metals above H replace hydrogen in acids; a metal replaces any metal below it in a salt solution.
- Write products from ion charges: iron forms iron(II) compounds, aluminum forms Al³⁺ compounds. Write NR when there is no reaction.
On to AP (for later)
AP Chemistry treats single replacement as oxidation–reduction: the more active metal loses electrons (is oxidized) and the ion in solution gains them (is reduced). The table of standard reduction potentials in Unit 9 is a more precise activity series, and Zn + Cu²⁺ becomes a galvanic cell there. AP questions usually ask for the net ionic equation: Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s) (2.7.4).