Question 19.7: Predicting Whether a Redox Reaction Is Spontaneous Predict f...
Predicting Whether a Redox Reaction Is Spontaneous
Predict from Table 19.1 whether Pb^{2+}(aq) can oxidize Al(s) or Cu(s) under standard state conditions. Calculate E° for each reaction at 25 °C.
STRATEGY
To predict whether a redox reaction is spontaneous, remember that an oxidizing agent can oxidize any reducing agent that lies below it in the table but can’t oxidize one that lies above it. To calculate E° for a redox reaction, sum the E° values for the reduction and oxidation half-reactions. If the value of E° is positive then the reaction is spontaneous.
TABLE 19.1 Standard Reduction Potentials at 25 °C | ||||
Reduction Half-Reaction | E° (V) | |||
|
F_{2}(g) + 2 e^{-} → 2F^{-}(aq) | 2.87 |
|
|
H_{2}O_{2}(aq) + 2 H^{+}(aq) + 2 e^{-} → 2H_{2}O(l) | 1.78 | |||
MnO^{-}_{4}(aq) + 8 H^{+}(aq) + 5 e^{-}→ Mn^{2+}(aq) + 4 H_{2}O(l) | 1.51 | |||
Cl_{2}(g) + 2 e^{-} → 2Cl^{-}(aq) | 1.36 | |||
Cr_{2}O^{2-}_{7}(aq) + 14 H^{+}(aq) + 6 e^{-} → 2 Cr^{3+}(aq) + 7 H_{2}O(l) | 1.36 | |||
O_{2}(g) + 4 H^{+}(aq) + 4 e^{-} → 2 H_{2}O(l) | 1.23 | |||
Br_{2}(aq) + 2 e^{-} → 2 Br^{-}(aq) | 1.09 | |||
Ag^{+}(aq) + e^{-} → Ag(s) | 0.80 | |||
Fe^{3+}(aq) + e^{-} → Fe^{2+}(aq) | 0.77 | |||
O_{2}(g) + 2 H^{+}(aq) + 2 e^{-} → H_{2}O_{2}(aq) | 0.70 | |||
I_{2}(s) + 2 e^{-} → 2 I^{-}(aq) | 0.54 | |||
O_{2}(g) + 2 H_{2}O(l) + 4 e^{-} → 4 OH^{-}(aq) | 0.40 | |||
Cu^{2+}(aq) + 2 e^{-} → Cu(s) | 0.34 | |||
Sn^{4+}(aq) + 2 e^{-} → Sn^{2+}(aq) | 0.15 | |||
2 H^{+}(aq) + 2 e^{-} → H_{2}(g) | 0 | |||
Pb^{2+}(aq) + 2 e^{-} → Pb(s) | -0.13 | |||
Ni^{2+}(aq) + 2 e^{-} → Ni(s) | -0.26 | |||
Cd^{2+}(aq) + 2 e^{-} → Cd(s) | -0.40 | |||
Fe^{2+}(aq) + 2 e^{-} → Fe(s) | -0.45 | |||
Zn^{2+}(aq) + 2 e^{-} → Zn(s) | -0.76 | |||
2 H_{2}O(l) + 2 e^{-} → H_{2}(g) + 2 OH^{-}(aq) | -0.83 | |||
Al^{3+}(aq) + 3 e^{-} → Al(s) | -1.66 | |||
Mg^{2+}(aq) + 2 e^{-} → Mg(s) | -2.37 | |||
Na^{+}(aq) + e^{-} → Na(s) | -2.71 | |||
Li^{+}(aq) + e^{-} → Li(s) | -3.04 |
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