Unit 6 Electrochemistry
6.1 Exercises
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Section 6.1 Exercises
- For each of the following balanced half-reactions, determine whether an oxidation or reduction is occurring.
(a) [latex]\text{Fe}^{3+}\;+\;3\text{e}^{-}\;{\longrightarrow}\;\text{Fe}[/latex]
(b) [latex]\text{Cr}\;{\longrightarrow}\;\text{Cr}^{3+}\;+\;3\text{e}^{-}[/latex]
(c) [latex]\text{MnO}_4^{\;\;2-}\;{\longrightarrow}\;\text{MnO}_4^{\;\;-}\;+\;\text{e}^{-}[/latex]
(d) [latex]\text{Li}^{+}\;+\;\text{e}^{-}\;{\longrightarrow}\;\text{Li}[/latex]
- Given the following pairs of balanced half-reactions, determine the balanced reaction for each pair of half-reactions in an acidic solution.
(a) [latex]\text{Ca}\;{\longrightarrow}\;\text{Ca}^{2+}\;+\;2\text{e}^{-},\;\text{F}_2\;+\;2\text{e}^{-}\;{\longrightarrow}\;2\text{F}^{-}[/latex]
(b) [latex]\text{Li}\;{\longrightarrow}\;\text{Li}^{+}\;+\;\text{e}^{-},\;\text{Cl}_2\;+\;2\text{e}^{-}\;{\longrightarrow}\;2\text{Cl}^{-}[/latex]
(c) [latex]\text{Fe}\;{\longrightarrow}\;\text{Fe}^{3+}\;+\;3\text{e}^{-},\;\text{Br}_2\;+\;2\text{e}^{-}\;{\longrightarrow}\;2\text{Br}^{-}[/latex]
(d) [latex]\text{Ag}\;{\longrightarrow}\;\text{Ag}^{+}\;+\;\text{e}^{-},\;\text{MnO}_4^{\;\;-}\;+\;4\text{H}^{+}\;+\;3\text{e}^{-}\;{\longrightarrow}\;\text{MnO}_2\;+\;2\text{H}_2\text{O}[/latex]
- Identify the species that undergoes oxidation, the species that undergoes reduction, the oxidizing agent, and the reducing agent in each of the reactions.
(a) [latex]\text{H}_2\text{O}_2\;+\;\text{Sn}^{2+}\;{\longrightarrow}\;\text{H}_2\text{O}\;+\;\text{Sn}^{4+}[/latex]
(b) [latex]\text{PbO}_2\;+\;\text{Hg}\;{\longrightarrow}\;\text{Hg}_2^{\;\;2+}\;+\;\text{Pb}^{2+}[/latex]
(c) [latex]\text{Al}\;+\;\text{Cr}_2\text{O}_7^{\;\;2-}\;{\longrightarrow}\;\text{Al}^{3+}\;+\;\text{Cr}^{3+}[/latex]
- Identify the species that was oxidized, the species that was reduced, the oxidizing agent, and the reducing agent in each of the reactions.
(a) [latex]\text{SO}_3^{\;\;2-}(aq)\;+\;\text{Cu(OH)}_2(s)\;{\longrightarrow}\;\text{SO}_4^{\;\;2-}(aq)\;+\;\text{Cu(OH)}(s)[/latex]
(b) [latex]\text{O}_2(g)\;+\;\text{Mn(OH)}_2(s)\;{\longrightarrow}\;\text{MnO}_2(s)[/latex]
(c) [latex]\text{NO}_3^{\;\;-}(aq)\;+\;\text{H}_2(g)\;{\longrightarrow}\;\text{NO}(g)[/latex]
(d) [latex]\text{Al}(s)\;+\;\text{CrO}_4^{\;\;2-}(aq)\;{\longrightarrow}\;\text{Al(OH)}_3(s)\;+\;\text{Cr(OH)}_4^{\;\;-}(aq)[/latex]
- Why is it not possible for hydrogen ion (H+) to appear in either of the half-reactions or the overall equation when balancing oxidation-reduction reactions in basic solution?
Solutions
- (a) reduction; (b) oxidation; (c) oxidation; (d) reduction
- (a) [latex]\text{F}_2 + \text{Ca} \longrightarrow 2\text{F}^{-} + \text{Ca}^{2+}[/latex]; (b) [latex]\text{Cl}_{2} + 2\text{Li} \longrightarrow 2\text{Li}^{+} + 2\text{Cl}^{-}[/latex]; (c) [latex]3\text{Br}_2 + 2\text{Fe} \longrightarrow 2\text{Fe}^{+3} + 6 \text{Br}^{-}[/latex]; (d) [latex]\text{MnO}_4^- + 4\text{H}^{+} + 3 \text{Ag} \longrightarrow 3\text{Ag}^{+} + \text{MnO}_2 + 2\text{H}_2 \text{O}[/latex]
- Oxidized: (a) Sn2+; (b) Hg; (c) Al; reduced: (a) H2O2; (b) PbO2; (c) Cr2O72−; oxidizing agent: (a) H2O2; (b) PbO2; (c) Cr2O72−; reducing agent: (a) Sn2+; (b) Hg; (c) Al
- Oxidized = reducing agent: (a) SO32−; (b) Mn(OH)2; (c) H2; (d) Al; reduced = oxidizing agent: (a) Cu(OH)2; (b) O2; (c) NO32−; (d) CrO42−
- In basic solution at 298.15 K, [OH−] > 1.0 × 10−7M > [H+]. Hydrogen ion cannot appear as a reactant because its concentration is essentially zero. If it were produced, it would instantly react with the excess hydroxide ion to produce water. Thus, hydrogen ion should not appear as a reactant or product in basic solution.
