22 4.6 Oxidation-Reduction Reactions
Practice questions
-
- Is this statement true? The following reaction is a redox reaction: [latex]\ce{2K{(s)} + Br_{2}{(l)} {\rightarrow} 2KBr {(s)}}[/latex]
- In the reaction [latex]\ce{Ca{(s)} + O_{2}{(g)} {\rightarrow} 2CaO}[/latex], indicate which element/compound has lost electrons and what element/compound has gained electrons.
- In the reaction [latex]\ce{2Li{(s)} + O_{2} {(g)} {\rightarrow} Li_{2}O_{2}{(s)}}[/latex], indicate which element/compound has been oxidised and which element/compound has been reduced.
- What is oxidation? Select all applicable answers.
-
- The loss of electrons
- The gaining of electrons
- An increase in the oxidation number
- A decrease in the oxidation number
-
- What is reduction? Select all applicable answers.
-
- Increase in oxidation number
- Decrease in the oxidation number
- Gain of electrons
- Loss of electrons
-
- Assign oxidation numbers to each atom in each substance:
-
- [latex]\ce{P}[/latex] in [latex]\ce{P_{4}}[/latex]
- [latex]\ce{S}[/latex] and [latex]\ce{O}[/latex] in [latex]\ce{SO_{2}}[/latex]
- [latex]\ce{S}[/latex] and [latex]\ce{O}[/latex] in [latex]\ce{SO_{2}^{2-}}[/latex]
- [latex]\ce{Ca}[/latex], [latex]\ce{N}[/latex] and [latex]\ce{O}[/latex] in [latex]\ce{Ca{(NO_{3})}_{2}}[/latex]
-
- Assign oxidation numbers to each atom in each substance:
-
- [latex]\ce{C}[/latex] and [latex]\ce{O}[/latex] in [latex]\ce{CO}[/latex]
- [latex]\ce{C}[/latex] and [latex]\ce{O}[/latex] in [latex]\ce{CO_{2}}[/latex]
- [latex]\ce{Ni}[/latex] and [latex]\ce{Cl}[/latex] in [latex]\ce{NiCl_{2}}[/latex]
- [latex]\ce{Ni}[/latex] and [latex]\ce{Cl}[/latex] in [latex]\ce{NiCl_{3}}[/latex]
-
- Identify what is being oxidised and reduced in the redox equation [latex]\ce{2NO + Cl_{2} {\rightarrow} 2NOCl}[/latex] by assigning oxidation numbers to the atoms. [latex]\ce{N}[/latex] is being ……………, and [latex]\ce{Cl}[/latex] is being ………… .
- Identify what is being oxidised and reduced in the redox equation [latex]\ce{2KrF_{2} + 2H_{2}O {\rightarrow} 2Kr + 4HF + O_{2}}[/latex] by assigning oxidation numbers to the atoms. [latex]\ce{O}[/latex] is being ……………, and [latex]\ce{Kr}[/latex] is being ………… .
- Identify the oxidising and reducing agents in the following reactions:
-
- [latex]\ce{Cu{(s)} + Pt^{2+}{(aq)} {\rightarrow} Cu^{2+}{(aq)} + Pt{(s)}}[/latex]
- [latex]\ce{2Mg{(s)} + CO_{2}{(g)} {\rightarrow} 2MgO{(s)} + C{(s)}}[/latex]
-
- Which of the following is the correct balanced net ionic equation for the reaction shown below? [latex]\ce{Cr_{2}O_{7}^{2-} + Fe^{2+} {\rightarrow} Cr^{3+} + Fe^{3+}}[/latex]
-
- [latex]\ce{14H^{+} + Cr_{2}O_{7}^{2-} + Fe^{2+} {\rightarrow} 2Cr^{3+} + 7H_{2}O + Fe^{3+}}[/latex]
- [latex]\ce{14H^{+} + Cr_{2}O_{7}^{2-} + 6Fe^{2+} {\rightarrow} 2Cr^{3+} + 14H_{2}O + 6Fe^{3+}}[/latex]
- [latex]\ce{14H^{+} + Cr_{2}O_{7}^{2-} + 6Fe^{2+} {\rightarrow} 2Cr^{3+} + 7H_{2}O + 6Fe^{3+}}[/latex]
- [latex]\ce{14H^{+} + Cr_{2}O_{7}^{2-} + 6Fe^{2+} {\rightarrow} 2Cr^{3+} + 7H_{2}O + 6Fe^{3+}}[/latex]
-
- Which of the following is the correct balanced net ionic equation for the reaction shown below? [latex]\ce{Pb{(NO_{3})}_{2}{(aq)} + 2KI{(aq)} {\rightarrow} 2KNO_{3}{(aq)} + PbI_{2}{(s)}}[/latex]
-
- [latex]\ce{Pb^{2+}{(aq)} + I^{-}{(aq)} {\rightarrow} PbI_{2}{(s)}}[/latex]
- [latex]\ce{Pb^{2+}{(aq)} + 2NO_{3}^{-}{(aq)} + 2K^{+} + 2I^{-}{(aq)} {\rightarrow} 2K^{+} {(aq)} + 2NO_{3}^{-} {(aq)} + PbI_{2}{(s)}}[/latex]
- [latex]\ce{Pb^{2+}{(aq)} + 2I^{-}{(aq)} {\rightarrow} PbI_{2}{(s)}}[/latex]
- [latex]\ce{Pb{(NO_{3})}_{2} {(aq)} + 2KI{(aq)} {\rightarrow} 2KNO_{3}{(aq)} + PbI_{2}}[/latex]
-
Answers
-
-
-
- True
- Lost electrons –[latex]\ce{Ca}[/latex] and gained electrons – [latex]\ce{O_{2}}[/latex]
- Oxidised – [latex]\ce{Li}[/latex], Reduced – [latex]\ce{O_{2}}[/latex]
- a and c
- b and c
- Oxidation numbers are as follows:
-
- [latex]\ce{P}[/latex] = 0
- [latex]\ce{S}[/latex] = +4 and [latex]\ce{O}[/latex] = -2
- [latex]\ce{S}[/latex] = +2 and [latex]\ce{O}[/latex] = -2
- [latex]\ce{Ca}[/latex] =+2, [latex]\ce{N}[/latex] = +5, and [latex]\ce{O}[/latex] = -2
-
- Oxidation numbers are as follows:
-
- [latex]\ce{C}[/latex] = +2 and [latex]\ce{O}[/latex] = -2
- [latex]\ce{C}[/latex] = +4 and [latex]\ce{O}[/latex] = -2
- [latex]\ce{Ni}[/latex] = +2 and [latex]\ce{Cl}[/latex] = -1
- [latex]\ce{Ni}[/latex] = +3 and [latex]\ce{Cl}[/latex] = -1
-
- Oxidised, reduced
- Oxidised, reduced
- Reducing and oxidising agents are as follows:
-
- Reducing agent [latex]\ce{Cu}[/latex] and oxidising agent [latex]\ce{Pt^{2+}}[/latex]
- Reducing agent [latex]\ce{Mg}[/latex] and oxidising agent [latex]\ce{CO_{2}}[/latex]
-
- d
- c
-
-