During the electrolysis of aqueous NaBr, what is produced at the cathode and at the anode?
A
Sodium metal at the cathode and bromine gas at the anode
B
Sodium metal at the cathode and oxygen gas at the anode
C
Hydrogen gas at the cathode and oxygen gas at the anode
D
Hydrogen gas at the cathode and bromine gas at the anode
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1
Identify the species present in the aqueous NaBr solution: Na\(\textsuperscript{+}\), Br\(\textsuperscript{-}\), H\(\textsubscript{2}\)O molecules, and possibly H\(\textsuperscript{+}\) and OH\(\textsuperscript{-}\) from water dissociation.
At the cathode (reduction site), compare the reduction potentials of Na\(\textsuperscript{+}\) and H\(\textsuperscript{+}\) (from water). Since water is more easily reduced than Na\(\textsuperscript{+}\), hydrogen gas (H\(\textsubscript{2}\)) is produced instead of sodium metal.
At the anode (oxidation site), compare the oxidation potentials of Br\(\textsuperscript{-}\) and water. Bromide ions are more easily oxidized than water, so bromine gas (Br\(\textsubscript{2}\)) is produced at the anode.
Write the half-reactions for each electrode:
Cathode (reduction): \(2\mathrm{H^{+}} + 2e^{-} \rightarrow \mathrm{H_2(g)}\)
Anode (oxidation): \(2\mathrm{Br^{-}} \rightarrow \mathrm{Br_2(g)} + 2e^{-}\)
Combine the half-reactions to understand the overall electrolysis process, confirming that hydrogen gas forms at the cathode and bromine gas forms at the anode during the electrolysis of aqueous NaBr.