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Edexcel IGCSE Chemistry 复习笔记 1.9.3 Writing Half-Equations

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Edexcel IGCSE Chemistry 复习笔记 1.9.3 Writing Half-Equations

Writing Half-Equations

In electrochemistry we are mostly concerned with the transfer of electrons, hence the definitions of oxidation and reduction are applied in terms of electron loss or gain rather than the addition or removal of oxygen
Oxidation is when a substance loses electrons and reduction is when a substance gains electrons
As the ions come into contact with the electrode, electrons are either lost or gained and they form neutral substances
These are then discharged as products at the electrodes
At the anode, negatively charged ions lose electrons and are thus oxidised
At the cathode, the positively charged ions gain electrons and are thus reduced
This can be illustrated using half equations which describe the movement of electrons at each electrode

Electrolysis of molten lead(II) bromide

In the electrolysis of molten lead(II) bromide the half equation at the negative electrode (cathode) is:

Pb²⁺ + 2e^– ⟶ Pb Reduction

At the positive electrode (anode) bromine gas is produced by the discharge of bromide ions:

2Br^– – 2e^– ⟶ Br₂Oxidation

2Br^– ⟶ Br₂+ 2e^–

Exam Tip

At the anode, it doesn't matter whether you subtract the electrons on the left or add them on the right. Most chemists prefer to add them on the right, because chemical equations, by convention, generally involve the addition of materials rather than the subtraction.

Electrolysis of aqueous sodium chloride

In the electrolysis of aqueous sodium chloride the half equation at the negative electrode (cathode) is:

2H⁺ + 2e^– ⟶ H₂Reduction

At the positive electrode (anode) chlorine gas is produced by the discharge of chloride ions:

2Cl^– – 2e^– ⟶ Cl₂Oxidation

2Cl^– ⟶ Cl₂+ 2e^–

Electrolysis of dilute sulfuric acid

In the electrolysis of dilute sulfuric acid the half equation at the negative electrode (cathode) is:

2H⁺ + 2e^– ⟶ H₂Reduction

At the positive electrode (anode) oxygen gas is produced by the discharge of water molecules:

2H₂O – 4e^– ⟶ O₂ + 4H⁺Oxidation

2H₂O ⟶ O₂ + 4H⁺+ 4e^–

Electrolysis of aqueous copper(II) sulfate

In the electrolysis of aqueous copper(II)sulfate the half equation at the negative electrode (cathode) is:

Cu²⁺ + 2e^– ⟶ Cu Reduction

At the positive electrode (anode) oxygen gas is produced by the discharge of water molecules:

2H₂O – 4e^– ⟶ O₂ + 4H⁺Oxidation

2H₂O ⟶ O₂ + 4H⁺+ 4e^–

Exam Tip

In electrode half equations the charges on each side of the equation should always balance.It may seem odd that water molecules are discharged and not hydroxide ions, but remember that acidic solutions will not contain any hydroxide ions. Even copper(II)sulfate is slightly acidic in water, so will not contain hydroxide ions.

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