Standard reduction potentials
Each half-reaction is written as a reduction, and its potential under standard conditions (1 M, 1 bar, usually 25 °C) is its standard reduction potential标准还原电势 E°, measured relative to 2H⁺ + 2e⁻ → H₂, which is set to zero. The larger E°, the more readily the species on the left gains electrons, so the stronger an oxidizing agent it is; the smaller E°, the more readily the species on the right loses electrons, so the stronger a reducing agent it is.
The standard potential of a galvanic cell is
Use the reduction potentials from the table for both. E° is an intensive property: multiplying an equation by 2 doesn’t change E° (each electron gets the same energy).
E°, ΔG° and K
When 1 mol of electrons moves through a potential difference of E volts, the electrical work is F·E, where F = 96 485 C/mol is the Faraday constant法拉第常数. So:
Here n is the number of moles of electrons transferred in the overall reaction. At 25 °C, combining the two gives log K = nE° ÷ 0.0592 V. The three are linked:
| E°cell | ΔG° | K | Under standard conditions |
|---|---|---|---|
| > 0 | < 0 | > 1 | spontaneous forward |
| = 0 | = 0 | = 1 | at equilibrium |
| < 0 | > 0 | < 1 | spontaneous in reverse |
In the table, choose a cathode (reduced) and then an anode (oxidized) to see the overall reaction, E°cell, ΔG° and K (data: OpenStax Appendix L):
Example 1: can Fe³⁺ oxidize I⁻?
E°: Fe³⁺ + e⁻ → Fe²⁺ +0.77 V; I₂ + 2e⁻ → 2I⁻ +0.54 V. Decide whether 2Fe³⁺(aq) + 2I⁻(aq) → 2Fe²⁺(aq) + I₂(s) is spontaneous under standard conditions, and find ΔG° and K.
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Cathode and anode. Fe³⁺ gains electrons and is reduced (cathode); I⁻ loses electrons and is oxidized (anode).
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E°cell. E° of the cathode minus E° of the anode (in V):
E°cell > 0, so the reaction is spontaneous under standard conditions. The Fe³⁺ half-reaction must be doubled to balance electrons with I⁻, but E° is intensive and is not doubled.
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Electrons transferred, n. Two Fe³⁺ each gain 1 electron and two I⁻ lose 2 electrons in all: n = 2.
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ΔG°. Use ΔG° = −nFE° (in J):
That is , negative, consistent with E°cell > 0.
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K. At 25 °C, log K = nE°/0.0592:
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Check. E°cell > 0, ΔG° < 0, and K > 1 all agree.
Example 2: which halogen displaces which?
E°: Cl₂ + 2e⁻ → 2Cl⁻ +1.36 V; Br₂ + 2e⁻ → 2Br⁻ +1.09 V. If bromine water is added to NaCl solution, does Cl₂ form? What if chlorine water is added to NaBr solution?
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Bromine water + Cl⁻: reaction and E°cell. Suppose Br₂ oxidizes Cl⁻: Br₂ + 2Cl⁻ → 2Br⁻ + Cl₂, with Br₂ at the cathode (in V):
E°cell < 0, not spontaneous: no Cl₂ forms.
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Chlorine water + Br⁻: reaction and E°cell. Cl₂ + 2Br⁻ → 2Cl⁻ + Br₂, with Cl₂ at the cathode:
E°cell > 0, spontaneous: Br₂ forms and the solution turns orange-yellow.
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ΔG° for the second reaction. n = 2, in J:
That is .
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Check. The two reactions are reverses of each other, and their E°cell values are exact opposites (−0.27 and +0.27). Consistent.
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Connect to the pattern. Cl₂, with the larger E°, is the stronger oxidizing agent and can oxidize Br⁻ from the couple with the smaller E°; the reverse doesn’t happen.
- Zn + Ni²⁺ → Zn²⁺ + Ni (Zn²⁺/Zn −0.76 V, Ni²⁺/Ni −0.26 V)
E° = +0.50 V.
- Cu + 2H⁺ → Cu²⁺ + H₂ (Cu²⁺/Cu +0.34 V)
E° = −0.34 V: copper doesn’t dissolve in dilute hydrochloric acid.
- Mg + 2H⁺ → Mg²⁺ + H₂ (Mg²⁺/Mg −2.37 V)
E° = +2.37 V.
- 2Ag + Cu²⁺ → 2Ag⁺ + Cu
E° = 0.34 − 0.80 = −0.46 V.
- Cl₂ + 2I⁻ → 2Cl⁻ + I₂
E° = 1.36 − 0.54 = +0.82 V.
- I₂ + 2Br⁻ → 2I⁻ + Br₂
E° = 0.54 − 1.09 = −0.55 V.
- 2Fe³⁺ + Cu → 2Fe²⁺ + Cu²⁺
E° = 0.77 − 0.34 = +0.43 V: FeCl₃ solution can etch copper boards.
- Pb + Fe²⁺ → Pb²⁺ + Fe (Pb²⁺/Pb −0.13 V, Fe²⁺/Fe −0.45 V)
E° = −0.45 − (−0.13) = −0.32 V.
Common mistakes
Mistake 1: “Doubling a half-reaction doubles its E°.” E° is intensive and doesn’t depend on how the equation is written. The coefficients affect ΔG° (through n), not E°.
Mistake 2: “E°cell = E°(anode) − E°(cathode),” or “flip the sign of the anode’s E° and then subtract.” Use the table’s reduction potentials for both and subtract the anode from the cathode; changing the sign twice gives the wrong answer.
Mistake 3: “The larger E°cell, the faster the reaction.” Like ΔG°, E° says whether a reaction is favorable, not how fast it is.
Quick check
E°: Mg²⁺/Mg −2.37 V, Sn²⁺/Sn −0.14 V. A student calculates E°cell = −2.51 V for Mg + Sn²⁺ → Mg²⁺ + Sn. What is the correct value, and what was the student’s error?
Sn²⁺ is reduced (cathode), so E°cell = −0.14 − (−2.37) = +2.23 V; −2.51 V is (−0.14) + (−2.37). E° isn’t multiplied by coefficients.
A cell reaction has n = 2 and E°cell = +0.50 V. What is ΔG°?
ΔG° = −2 × 96 485 × 0.50 J = −9.65 × 10⁴ J = −96.5 kJ.
From E°: Cl₂/Cl⁻ +1.36 V, Ag⁺/Ag +0.80 V, Cu²⁺/Cu +0.34 V, Zn²⁺/Zn −0.76 V, the strongest oxidizing agent is:
The species on the left of the half-reaction with the largest E° gains electrons most readily.
A redox reaction has E°cell = −0.20 V. Which statement is correct?
E° < 0 corresponds to ΔG° > 0 and K < 1: the reverse reaction is spontaneous under standard conditions.
How many electrons, n, are transferred in 2Al + 3Cu²⁺ → 2Al³⁺ + 3Cu?
Each of the 2 Al atoms loses 3 electrons, 6 in all; each of the 3 Cu²⁺ ions gains 2, also 6.
At 25 °C a reaction has n = 1 and E°cell = +0.0592 V. What is K?
log K = 1 × 0.0592 ÷ 0.0592 = 1, so K = 10.
AP-style free response
When a car’s lead–acid battery discharges, the half-reactions at the two electrodes (written as reductions) are PbO₂(s) + SO₄²⁻(aq) + 4H⁺(aq) + 2e⁻ → PbSO₄(s) + 2H₂O(l), E° = +1.69 V, and PbSO₄(s) + 2e⁻ → Pb(s) + SO₄²⁻(aq), E° = −0.35 V (OpenStax Appendix L).
(a) Write the overall reaction during discharge, and identify the substances at the anode and the cathode.
(b) Calculate E°cell for one cell.
(c) Calculate ΔG° for the discharge reaction, and state whether it is thermodynamically favorable.
(d) A car battery is several cells connected in series, giving about 12 V in total. Estimate how many cells are needed.
(e) When the battery is recharged, is it acting as a galvanic or an electrolytic cell? During recharging, what does PbSO₄ become at each electrode?
Model answer and scoring
(a) Pb(s) + PbO₂(s) + 2SO₄²⁻(aq) + 4H⁺(aq) → 2PbSO₄(s) + 2H₂O(l); the anode is Pb and the cathode is PbO₂.
(b) E°cell = 1.69 − (−0.35) = 2.04 V.
(c) n = 2, so ΔG° = −2 × 96 485 × 2.04 J = −3.94 × 10⁵ J = −394 kJ < 0, which is thermodynamically favorable.
(d) 12 ÷ 2.04 ≈ 6 cells (6 × 2.04 = 12.2 V).
(e) An electrolytic cell: the external source drives the reverse of the discharge reaction (ΔG > 0). At the former cathode, PbSO₄ is oxidized to PbO₂; at the former anode, PbSO₄ is reduced to Pb.
Scoring: (a) overall reaction and electrodes; (b) E°cell; (c) ΔG° and judgment; (d) number of cells; (e) electrolytic cell and the product at each electrode.
Summary
- E°cell = E°(cathode) − E°(anode), both reduction potentials; E° doesn’t change with the coefficients.
- ΔG° = −nFE°; E° > 0 ↔ ΔG° < 0 ↔ K > 1. At 25 °C, log K = nE°/0.0592.
- The larger E° in the table, the stronger an oxidizing agent the species on the left; the smaller E°, the stronger a reducing agent the species on the right.
AP tip: In ΔG° = −nFE°, n is the number of electrons transferred in the overall reaction (the least common multiple of the two half-reactions’ electrons), F is 96 485 C/mol, and the result is in J; divide by 1000 to convert to kJ.