✔ 最佳答案
The prediction in Part (b) depends on what the electrodes are used.
2008-05-18 11:44:22 補充：
(a)
In the solution, Pb(NO3)2 is dissociated to give Pb2+(aq) and NO3-(aq) ions, and H2O is slightly ionized to give H+(aq) and OH-(aq) ions.
Anode: Oxygen gas is evolved.
The Pt electrode is inert. NO3-(aq) and OH-(aq) anions migrate to the anode. OH-(aq) ion is preferentially discharged to give O2(g) gas, because OH-(aq) ion is oxidized more readily than NO3-(aq).
4OH-(aq) → 2H2O(l) + O2(g) + 4e-
or: 2H2O(aq) → 4H+(aq) + O2(g) + 4e-
Cathode: Lead metal is deposited.
The cathode is unchanged in electrolysis. Pb2+(aq) and H+(aq) cations migrate to the cathode. Pb2+(aq) ion is preferentially discharged to give Pb(s), because Pb2+(aq) ion is more concentrated than H+(aq) ion.
Pb2+(aq) + 2e- → Pb(s)
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(b)
Assume that inert electrodes are used.
In the solution, NaBr is dissociated to give Na+(aq) and Br-(aq) ions, and H2O is slightly ionized to give H+(aq) and OH-(aq) ions.
Anode: Bromine solution is formed.
Assume that the anode is inert. Br-(aq) and OH-(aq) anions migrate to the anode. Br-(aq) ion is preferentially discharged to give Br2(aq) solution, because Br-(aq) ion is more concentrated than OH-(aq) ion.
2Br-(aq) → Br2(aq) + 2e-
Cathode: Hydrogen gas is evolved.
Na+(aq) and H+(aq) cations migrate to the cathode. H+(aq) ion is preferentially discharged to give H2(g) gas, because H+(aq) ion is reduced more readily than Na+(aq) ion.
2H+(aq) + 2e- → H2(g)
or 2H2O(aq) + 2e- → H2(g) + 2OH-(aq)
