Question 17.4: ARRANGING OXIDIZING AND REDUCING AGENTS IN ORDER OF INCREASI...
ARRANGING OXIDIZING AND REDUCING AGENTS IN ORDER OF INCREASING STRENGTH
(a) Arrange the following oxidizing agents in order of increasing strength under standard-state conditions: Br_{2}(aq), Fe^{3+}(aq), Cr_{2}O_{7}^{2-}(aq).
(b) Arrange the following reducing agents in order of increasing strength under standard-state conditions: Al(s), Na(s), Zn(s).
STRATEGY
Pick out the half-reactions in Table 17.1 that involve the given oxidizing or reducing agents, and list them, along with E° their values, in the order in which they occur in the table. The strength of an oxidizing agent increases as the E° value increases, and the strength of a reducing agent increases as the E° value decreases.
TABLE 17.1 Standard Reduction Potentials at 25 °C
| Reduction Half-Reaction | E° (V) | |||
| Stronger oxidizing agent Weaker |
F_{2}(g) + 2 e^{–} H_{2}O_{2}(aq) + 2 H^{+}(aq) + 2 e^{–} MnO_{4}^{–}(aq) + 8 H^{+}(aq) + 5 e^{–} Cl_{2}(g) + 2 e^{–} Cr_{2}O_{7}^{2–}(aq) + 14 H^{+}(aq) + 6 e^{–} O_{2}(g) + 4 H^{+}(aq) + 4 e^{–} Br_{2}(aq) + 2 e^{–} Ag^{+}(aq) + e^{–} Fe^{3+}(aq) + e^{–} O_{2}(g) + 2 H^{+}(aq) + 2 e^{–} I_{2}(s) + 2 e^{–} O_{2}(g) + 2 H_{2}O(l) + 4 e^{–} Cu^{2+}(aq) + 2 e^{–} Sn^{4+}(aq) + 2 e^{–} 2 H^{+}(aq) + 2 e^{–} Pb^{2+}(aq) + 2 e^{–} Ni^{2+}(aq) + 2 e^{–} Cd^{2+}(aq) + 2 e^{–} Fe^{2+}(aq) + 2 e^{–} Zn^{2+}(aq) + 2 e^{–} 2 H_{2}O(l) + 2 e^{–} Al^{3+}(aq) + 3 e^{–} Mg^{2+}(aq) + 2 e^{–} Na^{+}(aq) + e^{–} Li^{+}(aq) + e^{-} |
→ 2 F^{–}(aq)
→ 2 H_{2}O(l) → Mn^{2+}(aq) + 4 H_{2}O(l) → 2 Cl^{–}(aq) → 2 Cr^{3+}(aq) + 7 H_{2}O(l) → 2 H_{2}O(l) → 2 Br^{–}(aq) → Ag(s) → Fe^{2+}(aq) → H_{2}O_{2}(aq) → 2 I^{–}(aq) → 4 OH^{–}(aq) → Cu(s) → Sn^{2+}(aq) → H_{2}(g) →Pb(s) →Ni(s) →Cd(s) →Fe(s) →Zn(s) → H_{2} (g) + 2 OH^{-}(aq) → Al(s) → Mg(s) → Na(s) → Li(s) |
2.87
1.78 1.51 1.36 1.36 1.23 1.09 0.80 0.77 0.70 0.54 0.40 0.34 0.15 0 – 0.13 – 0.26 – 0.40 – 0.45 – 0.76 – 0.83 – 1.66 – 2.37 – 2.71 – 3.04 |
Weaker oxidizing agent Stronger |
(a) List the half-reactions that involve Br_{2}, Fe^{3+}, and Cr_{2}O_{7}^{2-} in the order in which they occur in Table 17.1:
Cr_{2}O_{7}^{2-}(aq) + 14 H^{+}(aq) + 6 e^{-} → 2 Cr^{3+}(aq) + 7 H_{2}O(l) E° = 1.36 V
Br_{2}(aq) + 2 e^{-} → 2 Br^{-}(aq) E° = 1.09 V
Fe^{3+}(aq) + e^{-} → Fe^{2+}(aq) E° = 0.77 V
We can see that Cr_{2}O_{7}^{2-} has the greatest tendency to be reduced (largest E°) and Fe^{3+} has the least tendency to be reduced (smallest E°). The species that has the greatest tendency to be reduced is the strongest oxidizing agent, so oxidizing strength increases in the order Fe^{3+} < Br_{2} < Cr_{2}O_{7}^{2-}. As a shortcut, simply note that the strength of the oxidizing agents, listed on the left side of Table 17.1, increases on moving up in the table.
(b) List the half-reactions that involve Al(s), Na(s), and Zn(s) in the order in which they occur in Table 17.1:
Zn^{2+}(aq) + 2 e^{-} → Zn(s) E° = -0.76 V
Al^{3+}(aq) + 3 e^{-} → Al(s) E° = -1.66 V
Na^{+}(aq) + e^{-} → Na(s) E° = -2.71 V
The last half-reaction has the least tendency to occur in the forward direction (most negative E°) and the greatest tendency to occur in the reverse direction. Therefore, Na is the strongest reducing agent, and reducing strength increases in the order Zn < Al < Na. As a shortcut, note that the strength of the reducing agents, listed on the right side of Table 17.1, increases on moving down in the table.