Question 19.8: Calculating the Voltage of a Galvanic Cell under Standard-St...
Calculating the Voltage of a Galvanic Cell under Standard-State Conditions
What is the voltage of the galvanic cell shown in the illustration? What is the anode half-reaction, and what is the cathode half-reaction? Which direction do the electrons flow?
STRATEGY
In a galvanic cell, there must be a reduction half-reaction and an oxidation half reaction whose values of E° sum to give a positive value of E° for the cell. Any spontaneous redox reaction must have a positive value of E°. Use Table 19.1 to look up the values of E° for half-reactions and then determine which half-reaction is the reduction and which is the oxidation.
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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