Lesson 3.1.3
3.1.3 Ionic half-equations Quiz: Pearson Edexcel Chemistry, Unit 3
20 questions
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Lesson 3.1.3, Ionic half-equations: 20 multiple choice questions for the Pearson Edexcel Chemistry (9CH0), Unit 3: Redox I, written with Revision Ninja.
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The 20 questions
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What is an ionic half-equation?
- An equation showing the heat change of the reaction, written with the enthalpy change for each species
- An equation showing only the species that are oxidised or reduced, with electrons included
- An equation showing the molecular formulae of all the reactants and products in the reaction mixture
- An equation showing only the spectator ions that remain unchanged in the solution throughout the whole reaction
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Which is the correct half-equation for oxidation of Fe2+ to Fe3+?
- Fe2+ -> Fe3+ + e-
- Fe2+ + e- -> Fe+
- Fe2+ + 2e- -> Fe
- Fe3+ + e- -> Fe2+
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Which half-equation shows reduction of chlorine?
- 2Cl- -> Cl2 + 2e-
- Cl2 -> 2Cl- + 2e-
- Cl2 + 2e- -> 2Cl-
- Cl- + e- -> Cl2
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Which is the correct half-equation for the reduction of MnO4- in acid?
- MnO4- + 8H+ -> Mn2+ + 4H2O
- MnO4- + 8H+ + 5e- -> Mn2+ + 4H2O
- MnO4- + 8H+ + 3e- -> Mn2+ + 4H2O
- MnO4- + 5e- -> Mn2+
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Combining Fe2+ -> Fe3+ + e- with Cl2 + 2e- -> 2Cl- gives which full ionic equation?
- 2Fe2+ + 2Cl2 -> 2Fe3+ + Cl-
- 2Fe2+ + Cl2 -> 2Fe3+ + 2Cl-
- Fe2+ + 2Cl- -> Fe3+ + Cl2
- Fe2+ + Cl2 -> Fe3+ + Cl-
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Why must electrons balance when combining two half-equations?
- So that electrons lost in oxidation equal electrons gained in reduction
- So that the total mass of the products is larger than the mass of the reactants in the final equation
- So that the reaction releases a measurable amount of heat that can be detected by the apparatus
- So that the solution remains neutral in pH throughout the whole reaction as it proceeds to completion
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Which half-equation shows the oxidation of bromide ions?
- Br2 + 2e- -> 2Br-
- Br2 -> 2Br+ + 2e-
- 2Br- -> Br2 + 2e-
- Br- + e- -> Br
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What is the full ionic equation when Cl2 oxidises Br- to Br2?
- Cl2 + 2Br- -> Cl2 + Br2, which keeps the chlorine on both sides of the equation unchanged
- Cl + Br -> ClBr, which shows the two halogen atoms combining directly into a single molecule
- Cl2 + Br- -> Cl- + Br2, which is balanced for charge and atoms in the simplest way possible
- Cl2 + 2Br- -> 2Cl- + Br2
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Which half-equation is balanced correctly for the reduction of Cu2+?
- Cu2+ + 2e- -> Cu
- Cu2+ -> Cu + 2e-
- Cu2+ + 2e- -> Cu2
- Cu2+ + e- -> Cu
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What is the full ionic equation for zinc reacting with copper(II) ions?
- Zn + 2Cu2+ -> Zn2+ + 2Cu
- Zn2+ + Cu -> Zn + Cu2+
- Zn + Cu2+ -> Zn2+ + Cu
- Zn + Cu -> ZnCu
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In an ionic half-equation, which charges must balance?
- The number of protons on each side of the equation, which must be identical in every balanced half-equation
- Only the number of atoms of each element on the two sides of the equation, not the charges at all
- The mass of each species on each side of the equation, which must be identical for it to be balanced
- The total charge on each side, including electrons
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Which is the correct half-equation for the reduction of hydrogen ions to hydrogen gas?
- 2H+ -> H2 + 2e-
- 2H+ + 2e- -> H2
- H+ + e- -> H2
- H+ + 2e- -> H2
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Write the half-equation for the oxidation of sulfite ions, SO3 2-, to sulfate ions, SO4 2-, in neutral conditions.
- SO3 2- + H2O -> SO4 2- + 2H+ + 2e-
- SO3 2- -> SO4 2- + 2e-, which balances the sulfur atoms and the electrons but leaves the oxygen unbalanced
- SO3 2- + 2e- -> SO4 2-, which shows the sulfite gaining electrons and so being reduced in the reaction
- SO3 2- + H2O -> SO4 2- + 2e-, which balances the oxygen but omits the hydrogen ions needed for charge
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Why are ionic half-equations often preferred over molecular equations for redox reactions?
- They include all the spectator ions in the solution, so they show the full composition of the reaction mixture
- They omit the electrons from the equation, which makes them simpler to write and to balance for beginners
- They always give the mass of each product formed in the reaction, which is useful for calculating yields
- They show exactly which species gain or lose electrons
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Combine the half-equations Mg -> Mg2+ + 2e- and Ag+ + e- -> Ag. What is the balanced equation?
- Mg2+ + 2Ag -> Mg + 2Ag+
- Mg + 2Ag+ -> Mg + 2Ag
- Mg + 2Ag+ -> Mg2+ + 2Ag
- Mg + Ag+ -> Mg2+ + Ag
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Which equation correctly describes the disproportionation of chlorine in cold dilute NaOH?
- Cl2 + 2OH- -> 2Cl- + H2O, which forms only chloride and water and so does not show any oxidation of chlorine
- Cl2 + 2OH- -> Cl- + ClO- + H2O
- Cl2 + 2OH- -> ClO3- + H2O, which forms chlorate ions as the only chlorine-containing product of the reaction
- Cl2 + OH- -> Cl- + ClO- + H2, which gives hydrogen gas rather than water as the second product of the reaction
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A half-equation has 4 electrons on the reactant side and none on the product side. What does this represent?
- A neutralisation reaction, because hydrogen ions and hydroxide ions combine to form water in the reaction
- An oxidation, because electrons are being lost from the species and so its oxidation number is increasing
- A reduction, because electrons are being gained
- A precipitation reaction, because an insoluble solid forms from the dissolved ions in the solution
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Explain why the half-equation for oxidation of Fe2+ must be written with one electron on the product side.
- Fe2+ is neutral, so no electrons are involved, and the half-equation can be written without any electrons
- The electron is a spectator and is written on both sides of the half-equation so that the charges balance exactly
- Fe2+ gains one electron, which is written on the product side of the half-equation to show the gain clearly
- Each Fe2+ loses one electron to form Fe3+, and the charge must balance: +2 on the left and +3 + (-1) on the right
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Which set of half-equations, when combined, gives 2I- + Cl2 -> I2 + 2Cl-?
- I- -> I2 + e- and Cl2 + e- -> Cl-
- I2 -> 2I- + 2e- and 2Cl- -> Cl2 + 2e-
- 2I- -> I2 + 2e- and Cl2 + 2e- -> 2Cl-
- 2I- + 2e- -> I2 and Cl2 -> 2Cl- + 2e-
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Which half-equation is balanced for the oxidation of iodide to iodine?
- 2I- -> I2 + 2e-
- I- -> I2 + e-
- I2 -> 2I- + e-
- 2I- + 2e- -> I2
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