Question 4.20: Predict the products of the reaction between zinc metal and ...
Predict the products of the reaction between zinc metal and aqueous hydrochloric acid. Write balanced molecular and net ionic equations for the reaction. Identify the oxidation and reduction half-reactions.
Collect and Organize The reactants in this process are Zn(s) and HCl(aq). The products are determined by the activity series.
Analyze Zinc metal is listed above H^{+} 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^{-} |
thus it is oxidized by H^{+} from hydrochloric acid, producing Zn^{2+} and hydrogen gas. After writing a molecular equation for the reaction, we will convert it to a net ionic equation, as in Sample Exercise 4.7. After assigning oxidation numbers, we can identify the half reactions.
Solve The unbalanced molecular equation is
Zn(s) + HCl(aq) → ZnCl_{2}(aq) + H_{2}(g)
Adding a coefficient of 2 in front of the HCl balances all the elements:
Zn(s) + 2 HCl(aq) → ZnCl_{2}(aq) + H_{2}(g)
The overall ionic equation is:
Zn(s) + 2 H^{+}(aq) + 2 Cl^{-}(aq) → Zn^{2+}(aq) + 2 Cl^{-}(aq) + H_{2}(g)
The net ionic equation becomes:
Zn(s) + 2 H^{+}(aq) + \sout{2 Cl^{-}(aq)} → Zn^{2+}(aq) +\sout{2 Cl^{-}(aq)} + H_{2}(g)
Zn (s) + 2H^+ (aq) → Zn^{2+} (aq) +H_2(g)
The oxidation numbers of Zn, Zn^{2+}, and H^{+} are 0, +2, and +1, respectively. The oxidation number of H_{2} gas is zero because it is the elemental form of hydrogen. Zinc is oxidized and hydrogen ion is reduced. The two half-reactions are
Oxidation: Zn(s)→Zn^{2+}(aq)+2e^{-}
Reduction: 2e^{-}+2H^{+}(aq)→ H_{2}(g)
Think About It Including the spectator ions in a redox equation makes it more difficult to identify which atoms or ions are involved in the electron transfer. That is why most redox equations in this chapter and elsewhere in this book are written as net ionic equations.