Question 18.PS.7: Predicting Redox Reactions (a) Will zinc metal react with a ......
Predicting Redox Reactions
(a) Will zinc metal react with a 1 \text{M} \text{Ag}^+(aq) solution? If so, what is E° for the reaction?
(b) Will a 1 \text{M} \text{Fe}^{2+}(aq) solution react with metallic tin? If so, what is E° for the reaction?
(a) Yes. E^\circ_{cell} = + 1.56 V
(b) No. E^\circ_{cell} is negative.
Strategy and Explanation We will answer the questions by referring to Table 18.1 and comparing the positions of the reactants there.
(a) Ag^+(aq) is above metallic zinc in Table 18.1, so it is a better oxidizing agent, and we predict that it can oxidize zinc, causing metallic zinc atoms to form Zn^{2+}(aq) ions. To be certain, we combine the half-cell reactions to give the net equation. We subtract the half-cell potentials, which yields a positive E^\circ_{cell}, so this reaction is product-favored, as we predicted from Table 18.1.
Zn( s ) → Zn^{2+} (aq, 1 M) + 2 e^- E^\circ_{anode} = – 0.763 V
\underline{2 [Ag^+ (aq, 1 M) + e^- → Ag(s)] E^\circ_{cathode} = + 0.80 V}
Zn( s ) + 2 Ag ^+ (aq, 1 M) → Zn^{2+} (aq, 1 M) + 2 Ag( s ) E^\circ_{cell} = ?
E^\circ_{cell} = E^\circ_{cathode} – E^\circ_{anode} = + 0.80 V – (- 0.763 V) = + 1.56 V
The positive value for E^\circ_{cell} shows that this is a product-favored reaction.
(b) We evaluate the reaction between Fe^{2+}(aq) and metallic Sn the same way. Fe^{2+}(aq) is on the left in Table 18.1 below Sn(s), which is on the right. Therefore, Fe^{2+}(aq) is not a strong enough oxidizing agent to oxidize Sn(s), and we predict that this reaction will not occur. The combined half-reactions are
Sn( s ) → Sn^{2+} (aq) + 2 e^- E^\circ_{anode} = – 0.14 V
\underline{Fe^{2+}(aq) + 2 e^- → Fe(s) E^\circ_{cathode} = – 0.44 V}
Sn( s ) + Fe^{2+} (aq) → Sn^{2+} (aq) + Fe(s) E^\circ_{cell} = ?
E^\circ_{cell} = E^\circ_{cathode} – E^\circ_{anode} = – 0.14 V – (- 0.14 V) = – 0.30 V
The negative value for E^\circ_{cell} shows that this process is reactant-favored, and it will not form appreciable quantities of products under standard conditions. In fact, iron metal will reduce Sn^{2+} (E^\circ_{cell} = + 0.30 V), the reverse of the net reaction shown above.
| Table 18.1 Standard Reduction Potentials in Aqueous Solution at 25 °C* | ||
| Reduction Half-Reaction | E° (V) | |
| F_2(g) + 2 e^- | →2 F^-(aq) | + 2.87 |
| H_2O_2(aq) + 2 H_3O^+(aq) + 2 e^- | →4 H2O(\ell) | + 1.77 |
| PbO_2(s) + SO_4^{2-} (aq) + 4 H_3O^+(aq) +2 e^- | →PbSO_4 (s) +6 H_2O(\ell) | + 1.685 |
| MnO_4^-(aq) + 8 H_3O^+(aq) + 5 e^- | →Mn^{2+}(aq) +12 H_2O(\ell) | + 1.51 |
| Au^{3+}(aq) + 3 e^- | →Au(s) | + 1.50 |
| Cl_2(g) + 2 e^- | →2 Cl^-(aq) | + 1.358 |
| Cr_2O_7^{2-}(aq) + 14 H_3O^+(aq) + 6 e^- | →2 Cr^{3+}(aq) + 21 H_2O(\ell) | + 1.33 |
| O_2(g) + 4 H_3O^+(aq) + 4 e^- | →6 H_2O(\ell) | + 1.229 |
| Br_2 (\ell) + 2 e^- | →2 Br^-(aq) | + 1.066 |
| NO_3^-(aq) + 4 H_3O^+(aq) + 3 e^- | →NO(g) + 6 H_2O(\ell) | + 0.96 |
| Ocl^-(aq) + H_2O(\ell) + 2 e^- | →Cl^-(aq) + 2 OH^-(aq) | + 0.89 |
| Hg^{2+}(aq) + 2 e^- | →Hg(\ell) | + 0.855 |
| Ag^+(aq) + e^- | →Ag(s) | + 0.7994 |
| Hg_2^{2+}(aq) + 2 e^- | →2 Hg(\ell) | + 0.789 |
| Fe^{3+}(aq) + e^- | →Fe^{2+}(aq) | + 0.771 |
| I_2 (s) + 2 e^- | →2 I^-(aq) | + 0.535 |
| O_2(g) + 2 H_2O(\ell) + 4 e^- | →4 OH^-(aq) | + 0.403 |
| Cu^{2+}(aq) + 2 e^- | →Cu(s) | + 0.337 |
| Sn^{4+}(aq) + 2 e^- | →Sn^{2+}(aq) | + 0.15 |
| 2 H_3O^+(aq) + 2 e^- | →H_2(g) + 2 H_2O(\ell) | 0.00 |
| Sn^{2+}(aq) + 2 e^- | →Sn(s) | – 0.14 |
| Ni^{2+}(aq) + 2 e^- | →Ni(s) | – 0.25 |
| PbSO_4 (s) + 2 e^- | →Pb(s) + SO_4^{2-}(aq) | – 0.356 |
| Cd^{2+}(aq) + 2 e^- | →Cd(s) | – 0.403 |
| Fe^{2+}(aq) + 2 e^- | →Fe(s) | – 0.44 |
| Zn^{2+}(aq) + 2 e^- | →Zn(s) | – 0.763 |
| 2 H_2O(\ell) + 2 e^- | →H_2(g) + 2 OH^-(aq) | – 0.8277 |
| Al^{3+}(aq) + 3 e^- | →Al(s) | – 1.66 |
| Mg^{2+}(aq) + 2 e^- | →Mg(s) | – 2.37 |
| Na^+(aq) + e^- | →Na(s) | – 2.714 |
| K^+(aq) + e^- | →K(s) | – 2.925 |
| Li^+(aq) + e^- | →Li (s) | – 3.045 |
| *In volts (V) versus the standard hydrogen electrode. | ||