Unit 9 · Topic 9.10 Beta

Cell Potential Under Nonstandard Conditions

Away from standard conditions, a cell’s potential depends on Q: it is above E° when Q < 1, equal to E° at Q = 1, below E° when Q > 1, and zero at equilibrium (Q = K).

Practice 4: Model AnalysisPractice 6: Argumentation

Question set for this topic

Part 1 · Hook

Why this matters

A fresh flashlight battery is bright; an old one is dim, then dead, even though it still holds plenty of zinc. The voltage of a cell is not fixed: it depends on how much reactant and product are present, and it falls to zero when the reaction reaches equilibrium.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. How is Q compared with K to predict the direction of a reaction?

  1. Q < K: net forward; Q > K: net reverse
  2. Q < K: net reverse; Q > K: net forward
  3. Q and K are always equal
  4. Only K matters
Show the answer

A reaction proceeds in the direction that moves Q toward K.

  • Correct: Q < K: net forward; Q > K: net reverse:
  • Q < K: net reverse; Q > K: net forward:
  • Q and K are always equal:
  • Only K matters:

2. For a galvanic cell under standard conditions, what is the sign of E°cell?

  1. Positive
  2. Negative
  3. Zero
  4. It depends on the salt bridge
Show the answer

A galvanic cell runs a favored reaction, so E°cell > 0.

  • Correct: Positive:
  • Negative:
  • Zero:
  • It depends on the salt bridge:

Part 4 · See it

See it first

A graph of cell potential against Q on a log scale: a straight line falling from left to right. Where Q is below 1, E is above E°; at Q equal to 1, E equals E°; above 1, E is below E°; the line reaches zero at Q equal to K, equilibrium.
Cell potential falls as Q rises: above E° when Q < 1, equal to E° at Q = 1, below E° when Q > 1, and zero at Q = K. LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

  1. A cell away from standard conditions has Q ≠ 1its actual drive, ΔG, differs from ΔG°, so E differs from E°
  2. Less product or more reactant makes Q < 1the reaction is further from equilibrium and E > E°
  3. As the cell runs, reactants are used and products formQ rises and E falls
  4. When Q reaches K, there is no net driveE = 0 and the battery is dead

Part 6 · Key ideas

Key ideas

  • Q < 1 → E > E°. Q = 1 → E = E°. Q > 1 → E < E°. Q = K → E = 0 (equilibrium, a dead battery).
  • The Nernst equation, E = E° − (RT/nF) ln Q, describes this; the exam focuses on the direction of the change.
  • Q uses only dissolved species and gases: solid electrodes and spectator ions do not affect E.
  • A concentration cell has the same half-reaction on both sides (E° = 0) and runs until the concentrations are equal.

Part 7 · Misconception

A common mistake

The wrong idea: A cell’s voltage is always E°, and a dead battery has run out of electrons.

What actually happens: E equals E° only when Q = 1. As the cell runs, Q rises toward K and E falls toward zero; a dead battery is a cell at equilibrium, with reactants often still present.

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

Data table

A zinc-copper cell at different concentrations

A student measures the voltage of the cell Zn(s) | Zn²⁺(aq) || Cu²⁺(aq) | Cu(s), which runs Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s) with E° = 1.10 V, at 25 °C with different starting concentrations.

Measured cell potential for five concentration pairs
Trial[Zn²⁺] (M)[Cu²⁺] (M)E (V)
11.01.01.100
20.101.01.130
30.0101.01.159
41.00.101.070
51.00.0101.041

1. In which trial is the cell at standard conditions, so that E = E°?

  1. Trial 3, where the voltage is highest
  2. Trial 1, where both ions are 1.0 M and Q = 1
  3. Trial 5, where the voltage is lowest
  4. Each of the five trials, because each is at 25 °C
Show the answer

Q = [Zn²⁺]/[Cu²⁺] = 1 only in Trial 1, so only there does E equal E°.

  • Trial 3, where the voltage is highest: The highest voltage comes from Q < 1, not from standard conditions.
  • Correct: Trial 1, where both ions are 1.0 M and Q = 1: Right: standard conditions mean 1 M for every dissolved species, so Q = 1 and E = E° = 1.100 V.
  • Trial 5, where the voltage is lowest: The lowest voltage comes from Q > 1, far from standard conditions.
  • Each of the five trials, because each is at 25 °C: Standard conditions also require 1 M concentrations, which only Trial 1 has.

2. Which statement explains why E is greater than E° in Trials 2 and 3?

  1. Less Zn²⁺ means fewer ions in solution, so the electrons meet less resistance in the wire.
  2. E° itself increases when [Zn²⁺] decreases.
  3. The reaction reaches equilibrium faster when [Zn²⁺] is low.
  4. Q < 1, so the cell is further from equilibrium and its drive is greater.
Show the answer

When Q < 1 the reaction has a larger drive to proceed (ΔG < ΔG°), so E > E°.

  • Less Zn²⁺ means fewer ions in solution, so the electrons meet less resistance in the wire.: Electrons in the wire do not depend on the ion concentration. The change in E comes from Q.
  • E° itself increases when [Zn²⁺] decreases.: E° is fixed (1.10 V); E changes with concentration through Q.
  • The reaction reaches equilibrium faster when [Zn²⁺] is low.: At equilibrium E would be zero. The higher E shows the cell is further from equilibrium.
  • Correct: Q < 1, so the cell is further from equilibrium and its drive is greater.: Right: a lower product concentration (Zn²⁺) or a higher reactant concentration makes Q < 1, raising E above E°.

3. A sixth trial uses [Zn²⁺] = 0.50 M and [Cu²⁺] = 0.050 M. How does its voltage compare with E°?

  1. Less than 1.10 V, because Q = 10 > 1
  2. Greater than 1.10 V, because both concentrations are below 1 M
  3. Equal to 1.10 V, because neither concentration is zero
  4. Zero, because the concentrations are not equal
Show the answer

Q > 1 means E < E°. The ratio, not the individual concentrations, decides it.

  • Correct: Less than 1.10 V, because Q = 10 > 1: Right: Q = 0.50/0.050 = 10, the same as Trial 4, so E ≈ 1.07 V.
  • Greater than 1.10 V, because both concentrations are below 1 M: What matters is the ratio Q = [Zn²⁺]/[Cu²⁺], here 10, not whether each is below 1 M.
  • Equal to 1.10 V, because neither concentration is zero: E equals E° only when Q = 1.
  • Zero, because the concentrations are not equal: E = 0 only at equilibrium, where Q = K (an enormous number here).

4. The cell from Trial 1 is left running until it no longer lights a bulb. What are E and Q at that point?

  1. E = E° and Q = 1, because the reaction has stopped
  2. E = 0 and Q = 0, because one reactant is used up
  3. E = 0 and Q = K: the reaction is at equilibrium
  4. E is negative and Q > K, because the reaction runs backward
Show the answer

As the cell runs, Q rises toward K and E falls toward 0. At equilibrium, Q = K and E = 0.

  • E = E° and Q = 1, because the reaction has stopped: E = E° only at Q = 1; as the cell runs, [Zn²⁺] rises and [Cu²⁺] falls, so Q grows.
  • E = 0 and Q = 0, because one reactant is used up: Q = [Zn²⁺]/[Cu²⁺] grows, not shrinks, as the cell runs.
  • Correct: E = 0 and Q = K: the reaction is at equilibrium: Right: a "dead" battery is at equilibrium: no net drive, so no voltage.
  • E is negative and Q > K, because the reaction runs backward: Left alone, a galvanic cell runs down to equilibrium; it does not overshoot.

5. For a galvanic cell, how does E compare with E° when Q > 1?

  1. E is smaller than E°
  2. E is larger than E°
  3. E equals E°
  4. E is zero
Show the answer

Q < 1: E > E°. Q = 1: E = E°. Q > 1: E < E°. Q = K: E = 0.

  • Correct: E is smaller than E°: Right: more products (or fewer reactants) than standard lowers the drive, so E < E°.
  • E is larger than E°: That is the case for Q < 1.
  • E equals E°: That is the case only for Q = 1.
  • E is zero: That is the case only at equilibrium, Q = K.

6. A galvanic cell runs Ni(s) + 2 Ag⁺(aq) → Ni²⁺(aq) + 2 Ag(s), starting at standard conditions. Predict the effect of each change on the measured cell potential, E.

VariableChange
Adding solid AgNO₃ to the silver half-cell, raising [Ag⁺]—
Adding solid Ni(NO₃)₂ to the nickel half-cell, raising [Ni²⁺]—
Using a larger nickel strip of the same metal—
Letting the cell run for an hour—
Replacing the KNO₃ salt bridge with a NaNO₃ salt bridge—
Show the answer

E depends on Q = [Ni²⁺]/[Ag⁺]². Raising reactants lowers Q and raises E; raising products does the opposite. Solids and spectator ions do not appear in Q.

  • Adding solid AgNO₃ to the silver half-cell, raising [Ag⁺]: increases. More reactant lowers Q, so E rises above E°.
  • Adding solid Ni(NO₃)₂ to the nickel half-cell, raising [Ni²⁺]: decreases. More product raises Q, so E falls below E°.
  • Using a larger nickel strip of the same metal: no change. Solids do not appear in Q, so the size of the strip does not change E.
  • Letting the cell run for an hour: decreases. As the cell runs, Ag⁺ is used and Ni²⁺ forms, so Q rises and E falls.
  • Replacing the KNO₃ salt bridge with a NaNO₃ salt bridge: no change. The salt bridge ions are spectators; they do not appear in Q.

7. For Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s), which change raises the cell potential the most?

  1. Lowering [Cu²⁺] to 0.010 M while keeping [Zn²⁺] at 1.0 M
  2. Lowering both concentrations to 0.010 M
  3. Using a zinc strip twice as large
  4. Lowering [Zn²⁺] to 0.010 M, with [Cu²⁺] at 1.0 M
Show the answer

E is largest when Q = [Zn²⁺]/[Cu²⁺] is smallest.

  • Lowering [Cu²⁺] to 0.010 M while keeping [Zn²⁺] at 1.0 M: Q = 100: E falls below E° to about 1.04 V.
  • Lowering both concentrations to 0.010 M: Q = 1, so E = E° = 1.10 V: no change.
  • Using a zinc strip twice as large: Solids do not appear in Q, so E is unchanged.
  • Correct: Lowering [Zn²⁺] to 0.010 M, with [Cu²⁺] at 1.0 M: Right: Q = 0.010, the smallest Q of the choices, so E is the largest (about 1.16 V).

Part 9 · Summary

Summary

Away from standard conditions, a cell’s potential depends on Q: it is above E° when Q < 1, equal to E° at Q = 1, below E° when Q > 1, and zero at equilibrium (Q = K). The Nernst equation expresses this. A concentration cell has E° = 0 and runs only on a concentration difference.

Part 10 · Up next

What comes next

Part 11 · Connections

Connections