Question 4.19: An aluminum wire is dipped into a solution containing sodium...
An aluminum wire is dipped into a solution containing sodium, tin, and copper ions.
Which ions will be present in the solution after the reaction is complete?
Collect and Organize We are asked to predict whether the metal ions in solution will oxidize the aluminum wire. We use the activity series in Table 4.6
| Table 4.6 A n Activity Series for Metals in Aqueous Solution | |
| Metal | Oxidation Reaction |
| Lithium | Li(s) → Li^{+}(aq) + e^{-} |
| Potassium | K(s) → K^{+}(aq)+ e^{-} |
| Barium | Ba(s) → Ba^{2+}(aq) + 2 e^{-} |
| Calcium | Ca(s) → Ca^{2+}(aq) + 2 e^{-} |
| Sodium | Na(s) → Na^{+}(aq) + e^{-} |
| Magnesium | Mg(s) → Mg^{2+}(aq) + 2 e^{-} |
| Aluminum | Al(s) → Al^{3+}(aq) + 3 e^{-} |
| Manganese | Mn(s) → Mn^{2+}(aq) + 2 e^{-} |
| Zinc | Zn(s) → Zn^{2+}(aq) + 2 e^{-} |
| Chromium | Cr(s) → Cr^{3+}(aq) + 3 e^{-} |
| Iron | Fe(s) → Fe^{2+}(aq) + 2 e^{-} |
| Cobalt | Co(s) → Co^{2+}(aq) + 2 e^{-} |
| Nickel | Ni(s) → Ni^{2+}(aq) + 2 e^{-} |
| Tin | Sn(s) → Sn^{2+}(aq) + 2 e^{-} |
| Lead | Pb(s) → Pb^{2+}(aq)+ 2 e^{-} |
| Hydrogen | H_{2}(g) → 2 H^{+}(aq) + 2 e^{-} |
| Copper | Cu(s) → Cu^{2+}(aq) + 2 e^{-} |
| Silver | Ag(s) → Ag^{+}(aq) + e^{-} |
| Mercury | Hg(\ell) → Hg^{2+}(aq) + 2 e^{-} |
| Platinum | Pt(s) → Pt^{2+}(aq)+ 2 e^{-} |
| Gold | Au(s) → Au^{3+}(aq) + 3 e^{-} |
to guide our answer.
Analyze The activity series is arranged in terms of increasing oxidizing ability of cations as we proceed down Table 4.6. Alternatively, the ease of oxidation of the corresponding metal decreases down the list. Any metal in Table 4.6 is oxidized by a cation below it in the activity series.
Solve Cu^{2+} and Sn^{2+} are both listed below Al metal, so they will oxidize the aluminum wire. Sodium ion is above Al in the activity series, so it will not react. Thus both Cu^{2+} and Sn^{2+} ions are removed from the solution, but Na^{+} ions remain. Because electron transfer from aluminum to both Cu^{2+} and Sn^{2+} occurs, Al^{3+} ions are also present in the solution at the end of the experiment.
Think About It The activity series allows us to predict which metals are oxidized by other members of the series. Table 4.6 does not explain why a particular metal is a better oxidizing agent than another.