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Metal displacement reactions

A metal displacement reaction is one in which a less reactive metal in a compound is replaced by a more reactive metal.

The displacement occurs because the overall energy of the system is lowered, making the forward reaction energetically favourable. Essentially, the more reactive metal has a greater ability to donate electrons, which drives the displacement reaction. For example, the thermite reaction (also known as the Goldschmidt process) is used to prepare small quantities of metallic iron:

2Al (s) + Fe2O3 (s) → Al2O3 (s) + 2Fe (s)

 

Question

What do you observe when you place a strip of magnesium in a beaker containing dilute aqueous copper sulphate?

Answer

The blue solution turns colourless with pink deposits at the bottom of the beaker.

Mg (s) + CuSO4 (aq) → MgSO4 (aq) + Cu (s)

 

 

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Metal reactions with water and acid

Metals have the ability to react with substances such as water and acids.

These reactions provide valuable insights into the reactivity of different metals and have numerous practical applications in industry, energy production, and everyday life.

 

Metal Reactions with Water

When metals come into contact with water, they may react to form metal hydroxides or metal oxides, along with the release of hydrogen gas. The vigor of the reaction depends on the metal’s position in the reactivity series.

Highly reactive metals such as potassium, sodium and calcium react readily with cold water. For example, sodium reacts vigorously with water to produce sodium hydroxide and hydrogen gas:

2Na + 2H₂O → 2NaOH + H₂

Less reactive metals, such as magnesium, react slowly with cold water but more rapidly with steam. Metals like copper, silver and gold show little or no reaction with water because of their low reactivity.

These reactions demonstrate how the reactivity of metals influences their behaviour in different environments and help scientists classify metals according to their chemical activity.

 

Metal Reactions with Acids

Many metals react with dilute acids to produce a salt and hydrogen gas. This occurs because metals can displace hydrogen ions from the acid solution. The general equation for such reactions is:

Metal + Acid → Salt + Hydrogen

For example, zinc reacts with dilute hydrochloric acid to form zinc chloride and hydrogen gas:

Zn + 2HCl → ZnCl₂ + H₂

Similarly, magnesium reacts vigorously with dilute sulfuric acid:

Mg + H₂SO₄ → MgSO₄ + H₂

The rate of reaction depends on the metal’s reactivity. Highly reactive metals react quickly, while less reactive metals react more slowly. Metals below hydrogen in the reactivity series, such as copper, generally do not react with dilute acids because they cannot displace hydrogen from the acid.

 

Summary

Understanding how metals react with water and acids is essential in chemistry. These reactions help explain corrosion, metal extraction, hydrogen production, and industrial manufacturing processes. They also provide a practical way to compare the reactivity of different metals and predict their behaviour in various chemical environments.

In conclusion, metal reactions with water and acids are fundamental chemical processes that illustrate the diverse reactivity of metals. By studying these reactions, scientists and students gain a deeper understanding of chemical principles and the practical uses of metals in modern society.

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Reactivity series of metals

The reactivity of metals broadly refers to the relative extent of change that metals undergo with respect to certain reactions. Some of these reactions are:

    1. Metal reaction with water
    2. Metal reaction with acid
    3. Metal displacement reactions
    4. Extraction of metals from metal oxides (or other ores)
    5. Heating of metal hydroxides, metal carbonates and metal nitrates

In general, the greater the extent to which a metal changes from its elemental form to a compound, the more reactive it is. For example, sodium is more reactive than copper, as elemental sodium reacts violently with cold water, while elemental copper does not react with water and steam. The reactivities of metals for reactions 1 to 5 are summarised in the following table, which is known as the metal reactivity series:

The two non-metals, C and H are included for reference.

 

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Effects of pH on redox reactions

What are the effects of pH on redox reactions?

pH affects the strength of an oxidising agent. For example, the oxidising strength of potassium manganate (VII), as measured by its standard electrode potential, is as follows:

Acidic medium

MnO4– + 8H++ 5e– → Mn2+ + 4H2O               Eored  = +1.51 V

Neutral/slightly basic medium

MnO4– + 2H2O + 3e– → MnO2 + 4OH–              Eored  = +1.23 V

Strongly basic medium

MnO4– + e– → MnO42-               Eored  = +0.56 V

Hence, potassium manganate (VII) is most strongly oxidising in an acidic medium and least oxidising in a strongly basic medium.

The behaviour of MnO4– in different pH conditions is best explained using the the reaction MnO4– + 2H2O + 3e– → MnO2 + 4OH– as a reference. If the pH of this reaction is lowered, excess H+ allows manganese dioxide to be further reduced to the more stable +2 oxidation state ion:

MnO2 + 4H++ 2e– → Mn2+ + 2H2O

Combining the half reactions MnO4– + 2H2O + 3e– → MnO2 + 4OH– and MnO2 + 4H++ 2e– → Mn2+ + 2H2O gives the overall reduction reaction of MnO4– to Mn2+ in acidic medium.

On the other hand, if the pH of the same reaction MnO4– + 2H2O + 3e– → MnO2 + 4OH– is increased, excess OH– will shift the equilibrium towards the left. As the solution becomes more strongly basic, the manganate (VII) ion is instead reduced to the manganate (VI) ion. This implies that the manganate (VI) ion is stable only in very basic conditions. It disproportionates in acidic, neutral and slightly basic conditions as follows:-

3MnO42- + 4H+ → 2MnO4– + MnO2 + 2H2O

Incidentally,

MnO4– purple
MnO42- green
MnO2 brown solid
Mn2+ pale pink

 

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How to construct and balance redox equations?

How do we construct and balance redox equations?

The way to construct a redox equation is to write two balanced ionic half-reaction equations, one for the oxidising agent and the other one for the reducing agent, and combine them. The method to balance each half-reaction equation is slightly different from that of a normal stoichiometric equation and involves the following steps:-

    1. Balance elements other than O and H
    2. Balance O by adding H2O
    3. Balance H by adding H+
    4. Balance excess positive charges by adding electrons, e–
    5. (For reaction in a basic solution) Balance H+ by adding OH– to both sides, then combine H+ and OH– to form H2O and cancel out any common terms

For example, the redox equation for the oxidation of acidified iron (II) sulphate by potassium dichromate (VI) is constructed as follows:

1st half-reaction equation

Fe2+ → Fe3+

Balancing the equation

Fe2+ → Fe3+ + e–            eq1

(note that steps 1 through 3 and step 5 do not apply)

2nd half-reaction equation

Cr2O72- → Cr3+

Balancing the equation

Cr2O72- → 2Cr3+

Cr2O72- → 2Cr3+ + 7H2O

Cr2O72-  + 14H+ → 2Cr3+ + 7H2O

Cr2O72-  + 14H+ + 6e–→ 2Cr3+ + 7H2O            eq2

(note that step 5 does not apply)

Next, combine the balanced half-reaction equations, cancel any common terms and ensure that all electrons are eliminated in the final equation. For this example, we multiply eq1 by 6 throughout before adding to eq2, giving:

6Fe2+(aq) + Cr2O72- (aq) + 14H+ (aq) → 6Fe3+(aq) + 2Cr3+ (aq) + 7H2O (l)

 

Question

Write the redox equation for the oxidation of sodium formate by potassium manganite (VII) to form sodium carbonate and manganese dioxide in slightly basic conditions.

Answer

1st half-reaction equation

HCO2– → CO32-

Balancing the equation

HCO2–   + H2O → CO32-

HCO2–   + H2O → CO32- + 3H+

HCO2–   + H2O → CO32- + 3H+ + 2e–

HCO2–   + H2O + 3OH– → CO32- + 3H+ + 2e–+ 3OH–

HCO2–   + H2O + 3OH– → CO32- + 3H2O + 2e–

HCO2– + 3OH– → CO32- + 2H2O + 2e–

2nd half-reaction equation

MnO4– → MnO2

Balancing the equation

MnO4– → MnO2 + 2H2O

MnO4– + 4H+ → MnO2 + 2H2O

MnO4– + 4H+ + 3e– → MnO2 + 2H2O

MnO4– + 4H+ + 3e– + 4OH– → MnO2 + 2H2O + 4OH–

MnO4– + 4H2O + 3e– → MnO2 + 2H2O + 4OH–

MnO4– + 2H2O + 3e– → MnO2 + 4OH–

Combining the balanced half-reaction equations, we have,

3HCO2– (aq) + OH– (aq) + 2MnO4– (aq) → 3CO32- (aq) + 2MnO2 (s) + 2H2O (l)

 

 

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Disproportionation reaction

A disproportionation reaction is one in which an atom in a reactant undergoes both oxidation and reduction. For example,

The O in H2O2 (oxidation state of O = -1) is oxidised to the O in O2 (oxidation state of O = 0) and reduced to the O in H2O (oxidation state of O = -2).

A disproportionation reaction happens when the oxidation state of an element in a compound is unstable and can achieve greater stability by splitting into two species — one at a higher oxidation state and one at a lower. This is often driven by the products having a lower total Gibbs energy than the reactant. It may also occur due to favourable changes in enthalpy and entropy.

 

 

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Redox reactions

A redox reaction is one in which oxidation and reduction occur concurrently. When two different reactants are involved, an atom, a molecule or an ion in one reactant undergoes oxidation, while an atom, a molecule or an ion in the other undergoes reduction. For example,

 

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Oxidising agent and reducing agent

An oxidising agent is a species that oxidises another species and a reducing agent is a species that reduces another species. For example,

Cu2+ in the above reaction is the oxidising agent as it oxidises Zn to Zn2+ while Zn is the reducing agent as it reduces Cu2+ to Cu.

 

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Oxidation state or oxidation number

The oxidation state/number of an atom or a monatomic ion is a number assigned to the atom or monatomic ion that corresponds to its degree of oxidation.

The rules for assigning an oxidation state/number to an atom or monatomic ion is as follows:-

    • The oxidation state/number of an uncombined atom or an atom in a neutral homonuclear molecule is zero, e.g.
 

Atom

Oxidation state/number

Cu

 Cu

0

Cl2

 Cl

0

C60

C

0
    • The oxidation state/number of a monatomic ion is equal to its charge, e.g.

Ion

Oxidation state/number

Fe3+

+3

Cl–

-1

Na+

+1
    • The oxidation state/number of an atom in a compound corresponds to the number of electrons the atom gains or loses when it hypothetically forms ionic bonds with its neighbours, e.g.

Compound

 Atom Assumed role

Oxidation state/number

MgCl2

 Mg Electron donor

+2

CO2

 C Electron donor

+4

KOH

 O Electron acceptor

-2

NH3

 N Electron acceptor

-3
    • The total oxidation state/number of all atoms in a species is equivalent to the overall charge of the species, e.g.

Species

 Overall charge of species

Oxidation state/number

NaCl

 0

+1 + (-1) = 0

NH4+

 +1

-3 + (+1 x 4) = +1

PO43-

 -3

+5 + (-2 x 4) = -3

 

Question

Determine the oxidation state/number of the underlined elements in AlH3, CH4 and Cr2O72-.

Answer

For AlH3, we assume that Al is the electron donor as it is more electropositive than H, and so, +3 + (H x 3) = 0. Therefore, H = -1.

For CH4, we assume that H is the electron donor as it is more electropositive than C, and so, C + (+1 x 4) = 0. Hence, C = -4.

For Cr2O72-, we assume that Cr is the electron donor as it is more electropositive than O, and so, (Cr x2) + (-2 x 7) = -2. Therefore, Cr = +6.

 

 

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Oxidation and reduction: introduction

Oxidation is the gain in oxygen, loss of hydrogen, loss of electron or increase in oxidation number, while reduction is the loss of oxygen, gain in hydrogen, gain in electron or decrease in oxidation number.

    • Gain/loss of oxygen, e.g.

The reactant H2 is oxidised as it gains an oxygen atom to form H2O, while O2 is reduced as it loses an oxygen to form H2O.

    • Gain/loss of hydrogen, e.g.

The reactant H2S is oxidised as it loses hydrogen atoms to form S, while Cl2 is reduced as it gains hydrogen atoms to form HCl.

    • Gain/loss of electron, e.g.

The reactant Zn is oxidised as it loses two electrons to form Zn2+, while Cu2+ is reduced as it gains two electrons to form Cu.

    • Increase in oxidation state or oxidation number (see next article for details), e.g.

The reactant Zn is oxidised as its oxidation state/number increases from 0 in Zn to +2 in Zn2+, while Cu2+ is reduced as its oxidation state/number decreases from +2 in Cu2+ to 0 in Cu.

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