Cell Potential Under Nonstandard Conditions非标准条件下的电池电势

Big Idea · Energy (ENE)AP Science Practice 6 · ArgumentationAP link · Unit 9~40 min

Learning goals

  • Decide whether the cell potential E is larger or smaller than E° from whether Q is less than or greater than 1
  • Explain that Q increases and E decreases as a cell discharges, and that E = 0 when Q = K
  • Explain why a concentration cell produces a voltage although E° = 0, and do simple calculations with the Nernst equation

Before you start

  • Reaction quotient and equilibrium constant (U7.3)
  • Cell potential and free energy (U9.8)

Change the concentrations, change the potential

E° is the potential when every ion is at 1 M (and every gas at 1 bar). At other concentrations, the actual cell potential E depends on the reaction quotient Q (U7.3). Qualitatively, it follows Le Châtelier’s principle:

  • Q < 1 (more reactant ions, fewer product ions): the reaction is “pushed” forward more, so E > E°.
  • Q = 1: E = E°.
  • Q > 1 (more product ions): E < E°.
  • Q = K: the reaction is at equilibrium, E = 0, and the cell can no longer supply electrical energy.

The quantitative relationship is the Nernst equation能斯特方程 (at 25 °C, RT/F × ln 10 ≈ 0.0592 V):

E=E∘−RTnFln⁡QE=E∘−0.0592 Vnlog⁡Q(25 °C)\begin{gathered} E = E^\circ - \frac{RT}{nF}\ln Q \\ E = E^\circ - \frac{0.0592\ \text{V}}{n}\log Q \quad (25\,°\text{C}) \end{gathered}

It is the same relationship as ΔG = ΔG° + RT ln Q, divided through by −nF. Solids (the electrode metals) are left out of Q.

As a cell discharges, reactant ions are used up and product ions form, so Q increases and E gradually drops; when Q reaches K, E = 0 and the cell is “dead.” Drag the ion concentrations on each side, then let the cell discharge:

Try it: how concentration affects cell potential

Concentration cells

A concentration cell浓差电池 has two half-cells with identical electrodes and ions, differing only in concentration. Its E° = 0, but as long as the concentrations differ, Q ≠ 1 and E isn’t zero. The electrode on the dilute side is oxidized (producing ions) and the ions on the concentrated side are reduced, until the two concentrations are equal and E = 0.

Example 1: changing one concentration

For Ni(s) | Ni²⁺(aq) ‖ Ag⁺(aq) | Ag(s), the overall reaction is Ni + 2Ag⁺ → Ni²⁺ + 2Ag, with E° = 1.06 V and n = 2. (1) If only Ag⁺ is diluted to 0.010 M, how does E change? (2) What if only Ni²⁺ is diluted to 0.010 M? Estimate E in each case.

Step-by-step solution

  1. Write Q and the Nernst equation. Solid Ni and Ag are left out of Q; at 25 °C:

    Q=[NiX2+][AgX+]2E=E∘−0.0592nlog⁡Q\begin{aligned} Q &= \frac{[\ce{Ni^2+}]}{[\ce{Ag+}]^2} \\ E &= E^\circ - \frac{0.0592}{n}\log Q \end{aligned}

  2. (1) Dilute Ag⁺: find Q. [Ni²⁺] is still 1.0 M:

    Q=1.0(0.010)2=1.0×104Q = \frac{1.0}{(0.010)^2} = 1.0 \times 10^4

    Q > 1, so E < E°.

  3. (1) Find E. log Q = 4 and n = 2 (in V):

    E=1.06−0.05922×4=0.94 V\begin{aligned} E &= 1.06 - \frac{0.0592}{2} \times 4 \\ &= \boxed{0.94\ \text{V}} \end{aligned}

  4. (2) Dilute Ni²⁺: find Q and E. [Ag⁺] is still 1.0 M:

    Q=0.010(1.0)2=0.010E=1.06−0.05922×(−2)=1.12 V\begin{aligned} Q &= \frac{0.010}{(1.0)^2} = 0.010 \\ E &= 1.06 - \frac{0.0592}{2} \times (-2) \\ &= \boxed{1.12\ \text{V}} \end{aligned}

    Q < 1, so E > E°. Consistent.

  5. Explain qualitatively. Removing reactant ions (Ag⁺) weakens the push toward products, so E falls; removing product ions (Ni²⁺) strengthens it, so E rises. This matches the direction Le Chatelier’s principle predicts.

Example 2: a copper concentration cell

Both half-cells have a copper strip in CuSO₄ solution, one with [Cu²⁺] = 0.010 M and the other 1.0 M, joined by a salt bridge. Which electrode is the anode? About what is E? When does the cell stop working?

Step-by-step solution

  1. Find what differs. Both electrodes are Cu/Cu²⁺ with the same E°, so E°cell = 0; the only difference is concentration. The reaction runs in the direction that evens out the two concentrations.

  2. Anode. Copper on the dilute side is oxidized, making the dilute solution more concentrated:

    Cu→CuX2+(dil)+2e−\ce{Cu -> Cu^2+}(\text{dil}) + 2e^-

  3. Cathode. Cu²⁺ on the concentrated side is reduced, making that solution more dilute:

    CuX2+(conc)+2e−→Cu\ce{Cu^2+}(\text{conc}) + 2e^- \ce{-> Cu}

  4. Find E. The overall change is Cu²⁺(conc) → Cu²⁺(dil), with n = 2:

    Q=[dil][conc]=0.0101.0=0.010E=0−0.05922×(−2)=+0.059 V\begin{aligned} Q &= \frac{[\text{dil}]}{[\text{conc}]} = \frac{0.010}{1.0} = 0.010 \\ E &= 0 - \frac{0.0592}{2} \times (-2) \\ &= \boxed{+0.059\ \text{V}} \end{aligned}

  5. When it stops. The dilute side gets more concentrated and the concentrated side more dilute; when they are equal, Q = 1 = K, log Q = 0, and E = 0, so the cell stops working.

Practice: for Zn(s) | Zn²⁺ ‖ Pb²⁺ | Pb(s), with Q = [Zn²⁺] ÷ [Pb²⁺], how does E compare with E°?
  1. [Zn²⁺] = 0.10 M, [Pb²⁺] = 1.0 M
  2. [Zn²⁺] = 1.0 M, [Pb²⁺] = 0.010 M
  3. Both ions at 0.50 M
  4. Starting from standard conditions, [Pb²⁺] is increased
  5. Starting from standard conditions, [Zn²⁺] is increased
  6. Starting from standard conditions, the cell discharges for a while
  7. Both solutions (originally 1 M each) are diluted tenfold
  8. Starting from standard conditions, Na₂SO₄ is added on the lead side, precipitating some Pb²⁺ as PbSO₄

Common mistakes

Mistake 1: “E° changes with concentration.” E° is a fixed value for standard conditions; what changes with concentration is the actual potential E.

Mistake 2: “A dead battery has used up all its reactants.” E = 0 means the reaction has reached equilibrium (Q = K); the reactants aren’t necessarily used up, though when K is large very few reactant ions remain.

Mistake 3: “Two identical half-cells can’t produce a voltage.” A concentration cell has E° = 0, but when the concentrations differ, Q ≠ 1 and E ≠ 0.

Quick check

A cell reaction has n = 2, and at some set of concentrations Q = 1.0 × 10⁻³. How does E compare with E°?

When a cell reaction reaches equilibrium, which statement is correct?

In a copper concentration cell, one side has [Cu²⁺] = 0.10 M and the other 1.0 M. Where is the cathode?

A flashlight battery's voltage slowly drops with use. The most reasonable explanation in terms of Q is:

For Zn | Zn²⁺(1.0 M) ‖ Pb²⁺(0.0010 M) | Pb, E° = 0.63 V and n = 2. E is about:

At some set of concentrations a cell has E = +0.40 V (n = 2). For the reaction under these conditions, ΔG is:

AP-style free response

AP-style free response

A concentration cell can be used to measure [Ag⁺] in a solution: one half-cell is a silver wire in 1.0 M AgNO₃, the other is a silver wire in the test solution, and they are joined by a KNO₃ salt bridge. At 25 °C, the measured voltage is 0.177 V.

(a) What is E° for this cell? Explain.

(b) The test solution is known to be more dilute than 1.0 M. Which side is the anode? Explain.

(c) Use the Nernst equation (n = 1) to calculate [Ag⁺] in the test solution.

(d) Which way do electrons flow in the wire?

(e) If the cell keeps operating, how does the voltage change? What eventually happens?

Model answer and scoring

(a) E° = 0: both half-cells have the same electrode reaction (Ag⁺ + e⁻ → Ag), so E°(cathode) − E°(anode) = 0.

(b) The test-solution side. Silver on the dilute side is oxidized to Ag⁺, and Ag⁺ on the concentrated side is reduced, bringing the two concentrations closer together.

(c) Q = [Ag⁺, test] ÷ 1.0; 0.177 = 0 − 0.0592 × log Q, so log Q = −2.99 and [Ag⁺] ≈ 1.0 × 10⁻³ M.

(d) From the silver wire in the test solution (anode) to the silver wire in the 1.0 M solution (cathode).

(e) [Ag⁺] rises on the anode side and falls on the cathode side, so Q approaches 1 and the voltage gradually decreases; when the two concentrations are equal, the voltage is zero.

Scoring: (a) E° = 0 with reason; (b) anode with reason; (c) Nernst calculation; (d) electron direction; (e) voltage falls to zero.

Summary

  • When Q < 1, E > E°; when Q > 1, E < E°; when Q = K, E = 0.
  • E = E° − (0.0592/n) log Q (25 °C); as a cell discharges, Q increases and E falls.
  • A concentration cell has E° = 0 and produces a voltage from the difference in concentration; the dilute side is the anode.

AP tip: For qualitative questions, write: which ion’s concentration changed → does Q get larger or smaller → is E larger or smaller than E°. Don’t say “E° changed.”

Practice this chapter: Unit 9 · Thermodynamics and Electrochemistry · 12 questionsPractice this topic: U9.9 · Cell Potential Under Nonstandard Conditions · 6 questions

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