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) Fe3++3e−⟶Fe
(b) Cr⟶Cr3++3e−
(c) MnO2−4⟶MnO−4+e−
(d) Li++e−⟶Li
- Given the following pairs of balanced half-reactions, determine the balanced reaction for each pair of half-reactions in an acidic solution.
(a) Ca⟶Ca2++2e−,F2+2e−⟶2F−
(b) Li⟶Li++e−,Cl2+2e−⟶2Cl−
(c) Fe⟶Fe3++3e−,Br2+2e−⟶2Br−
(d) Ag⟶Ag++e−,MnO−4+4H++3e−⟶MnO2+2H2O
- Identify the species that undergoes oxidation, the species that undergoes reduction, the oxidizing agent, and the reducing agent in each of the reactions.
(a) H2O2+Sn2+⟶H2O+Sn4+
(b) PbO2+Hg⟶Hg2+2+Pb2+
(c) Al+Cr2O2−7⟶Al3++Cr3+
- Identify the species that was oxidized, the species that was reduced, the oxidizing agent, and the reducing agent in each of the reactions.
(a) SO2−3(aq)+Cu(OH)2(s)⟶SO2−4(aq)+Cu(OH)(s)
(b) O2(g)+Mn(OH)2(s)⟶MnO2(s)
(c) NO−3(aq)+H2(g)⟶NO(g)
(d) Al(s)+CrO2−4(aq)⟶Al(OH)3(s)+Cr(OH)−4(aq)
- 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) F2+Ca⟶2F−+Ca2+; (b) Cl2+2Li⟶2Li++2Cl−; (c) 3Br2+2Fe⟶2Fe+3+6Br−; (d) MnO−4+4H++3Ag⟶3Ag++MnO2+2H2O
- 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.