Question 11.3: Practicing Spontaneity Using the data from Table 11.1, predi...
Practicing Spontaneity
Using the data from Table 11.1, predict whether 1 M HN O_{3} will dissolve gold metal to form a 1 M Au^{3+} solution.
Table 11.1
Standard Reduction Potentials at 25°C (298 K) for Many Common Half-reactions
| Half-reaction | \xi° ( V ) | Half-reaction | \xi° ( V ) |
| F_{2} + 2 e^{-} → 2 F^{-} | 2.87 | O_{2} + 2 H_{2}O + 4 e^{-} → 4 OH^{-} | 0.40 |
| Ag^{2+} + e^{-} → Ag^{+} | 1.99 | Cu^{2+} + 2 e^{-} → Cu | 0.34 |
| Co^{3+} + e^{-} → Co^{2+} | 1.82 | Hg_{2}Cl_{2} + 2 e^{-} → 2 Hg + 2 Cl^{-} | 0.27 |
| H_{2}O_{2} + 2 H^{+} + 2 e^{-} → 2 H_{2}O | 1.78 | AgCl + e^{-} → Ag + Cl^{-} | 0.22 |
| Ce^{4+} + e^{-} → Ce^{3+} | 1.70 | SO_{4}^{2-} + 4 H^{+} + 2 e^{-} → H_{2}SO_{3} + H_{2}O | 0.20 |
| PbO_{2} + 4 H^{+} + SO_{4}^{2-} + 2 e^{-} → Pb SO_{4} + 2 H_{2}O | 1.69 | Cu^{2+} + e^{-} → Cu^{+} | 0.16 |
| MnO_{4}^{-} + 4 H^{+} + 3 e^{-} → Mno_{2} + 2 H_{2}O | 1.68 | 2 H^{+} + 2 e^{-} → H _{2} | 0.00 |
| IO_{4}^{-} + 2 H^{+} + 2 e^{-} → IO_{3}^{-} + H_{2}O | 1.60 | Fe^{3+} + 3 e^{-} → Fe | – 0.036 |
| MnO_{4}^{-} + 8 H^{+} + 5 e^{-} → Mn^{2+} + 4 H_{2}O | 1.51 | Pb^{2+} + 2 e^{-} → Pb | – 0.13 |
| Au^{3+} + 3 e^{-} → Au | 1.50 | Sn^{2+} + 2 e^{-} → Sn | – 0.14 |
| PbO_{2} + 4 H^{+} + 2 e^{-} → Pb^{2+} + 2 H_{2}O | 1.46 | Ni^{2+} + 2 e^{-} → Ni | – 0.23 |
| Cl_{2} + 2 e^{-} → 2Cl ^{-} | 1.36 | Pb SO_{4} + 2 e^{-} → Pb + SO_{4} ^{2-} | – 0.35 |
| Cr_{2}O_{7} ^{2-} + 14 H^{+} + 6 e^{-} →2 Cr ^{3+}+ 7 H_{2}O | 1.33 | Cd ^{2+} + 2 e^{-} → Cd | – 0.40 |
| O_{2} + 4 H^{+} + 4 e^{-} → 2 H_{2}O | 1.23 | Fe^{2+} + 2 e^{-} → Fe | – 0.44 |
| MnO_{2} + 4 H^{+} + 2 e^{-} → Mn ^{2+} + 2 H_{2}O | 1.21 | Cr ^{3+} + e^{-} → Cr^{2+} | – 0.50 |
| IO_{3}^{-} + 6 H^{+} + 5 e^{-} → \frac{1}{2}I _{2} + 3 H_{2}O | 1.20 | Cr ^{3+} + 3 e^{-} → Cr | – 0.73 |
| Br _{2} + 2 e^{-} → 2 Br^{-} | 1.09 | Zn^{2+} + 2 e^{-} → Zn | – 0.76 |
| VO_{2}^{+} + 2 H^{+} + e^{-} → VO^{2+} + H_{2}O | 1.00 | 2 H_{2}O + 2 e^{-} → H_{2} + 2 OH^{-} | – 0.83 |
| AuCl_{4}^{-} + 3 e^{-} → Au + 4 Cl^{-} | 0.99 | Mn^{2+} + 2 e^{-} → Mn | – 1.18 |
| NO_{3}^{-} + 4 H^{+} + 3 e^{-} → NO + 2 H_{2}O | 0.96 | Al^{3+} + 3 e^{-} → Al | – 1.66 |
| ClO_{2} + e^{-} → ClO_{2}^{-} | 0.954 | H _{2} + 2 e^{-} → 2 H ^{-} | – 2.23 |
| 2 Hg ^{2+} + 2 e^{-} → Hg_{2} ^{2+} | 0.91 | Mg^{2+} + 2 e^{-} → Mg | – 2.37 |
| Ag^{+} + e^{-} →Ag | 0.80 | La^{3+} + 3 e^{-} → La | – 2.37 |
| Hg_{2} ^{2+} + 2 e^{-} → 2 Hg | 0.80 | Na^{+} + e^{-} → Na | – 2.71 |
| Fe^{3+} + e^{-} → Fe^{2+} | 0.77 | Ca^{2+} + 2 e^{-} → Ca | – 2.76 |
| O_{2} + 2 H^{+} + 2 e^{-} → H_{2} O_{2} | 0.68 | Ba^{2+} + 2 e^{-} → Ba | – 2.90 |
| MnO_{4}^{-} + e^{-} → MnO_{4}^{2-} | 0.56 | K^{+} + e^{-} → K | – 2.92 |
| I_{2} + 2 e^{-} → 2 I^{-} | 0.54 | Li^{+} + e^{-} → Li | – 3.05 |
| Cu ^{+} + e^{-} → Cu | 0.52 |
What are we trying to solve?
We want to determine if gold will dissolve in 1 M nitric acid (HNO_{3} ) to form a 1 M Au^{3+} solution. To make this determination, we need to see if the reaction between Au and HN O_{3} to form Au^{3+} is spontaneous or has a standard potential greater than zero. We must use the appropriate half-reactions. For this process to occur, gold would be oxidized, so the HNO_{3} would act as an oxidizing agent.
The half-reaction for HN O_{3} acting as an oxidizing agent is
NO_{3} ^{-} + 4 H ^{+} + 3 e ^{-} → NO + 2H_{2}O ξ° (cathode)= 0.96 V
The reaction for the oxidation of solid gold to Au^{3+} ions is
Au → Au^{3+} + 3 e ^{-} – ξ° (anode)= – 1.50 V
The sum of these half-reactions gives the required reaction:
Au (s) + NO_{3} ^{-} (aq) + 4 H ^{+} (aq) → Au^{3+} (aq) + NO (g) + 2 H_{2}O (l)
and
ξ° _{cell} = 0.96 V – 1.50 V = – 0.54 V
Since the ξ° value is negative, the process will not occur under standard conditions. That is, gold will not dissolve in 1 M HN O_{3} to give 1 M Au^{3+}. In fact, a mixture (1 : 3 by volume) of concentrated nitric and hydrochloric acid, called aqua regia, is required to dissolve gold.