Question 18.PS.2: Determining Half-Reactions from Net Redox Reactions Aluminum......
Determining Half-Reactions from Net Redox Reactions
Aluminum will undergo a redox reaction with an acid such as HCl to produce Al^{3+}(aq) and H_2 (g).
(unbalanced equation) Al(s) + H^+(aq) → Al^{3+}(aq) + H_2(g)
Write the oxidation half-reaction and the reduction half-reaction equations, and combine them to give the balanced equation for the net reaction.
Oxidation half-reaction: Al (s) → Al^{3+}(aq) + 3 e^-
Reduction half-reaction: 2 H^+(aq) + 2 e^- → H_2(g)
Net reaction: 2 Al (s) + 6 H^+(aq) → 2 Al^{3+}(aq) + 3 H_2(g)
Strategy and Explanation To identify the half-reactions, we must first identify the species whose oxidation number increases (it is oxidized) and the species whose oxidation number decreases (it is reduced). Aluminum’s oxidation number increases from 0 to +3, so it is oxidized.
Al (s) → Al^{3+}(aq) + 3 e^-
This half-reaction must have three electrons on the right side to balance the 3+ charge of the aluminum ion and yield equal charges on the right side and the left side of the half-reaction. Hydrogen’s oxidation number decreases from +1 to 0 so it is reduced.
2 H^+(aq) + 2 e^- → H_2(g)
The half-reaction must have two electrons on the left side to balance the charge of the two hydrogen ions.
Notice that these two half-reactions contain different numbers of electrons. The two half-reactions are multiplied by 2 and 3, respectively, so that six e^- appear in each halfreaction.
2 [Al(s) → Al^{3+}(aq) + 3 e^-] gives 2 Al( s ) → 2 Al^{3+}(aq) + 6 e^-
\underline{3 [2 H^+(aq) + 2 e^- → H_2(g)] gives 6 H^+(aq) + 6 e^- → 3 H_2(g)}
Net reaction: 2 Al (s) + 6 H^+(aq) → 2 Al^{3+}(aq) + 3 H_2(g)
The net reaction is the sum of these two half-reactions multiplied by the proper whole numbers to make the number of electrons produced by the oxidation half-reaction equal to the number of electrons gained in the reduction half-reaction.