AP® Chemistry review sheet from Aim for Five (aimforfive.com/chem/units/9/9-10)
Unit 9 · Topic 9.10
9.10 Cell Potential Under Nonstandard Conditions
E° assumes standard conditions, where Q = 1. As a cell runs, Q moves toward K and the voltage falls, reaching zero at equilibrium (a dead battery). The Nernst equation, E = E° − (RT/nF) ln Q, shows that raising Q lowers the voltage; on the exam you use it to reason about direction, not to grind through calculations.
Key terms
- nonstandard conditions
- Nernst equation
- reaction quotient (Q)
- concentration cell
- E = 0 at equilibrium
Cell potential depends on concentrations
Standard conditions mean 1 M solutions and 1 atm gases, which makes Q = 1. Real cells rarely stay there. Think of E as the cell's push toward equilibrium. A cell far from equilibrium pushes hard (a large E); a cell close to equilibrium barely pushes (a small E).
As a galvanic cell runs, reactants are used up and products build up, so Q increases toward K. The voltage drops. When Q = K, the reaction is at equilibrium and E = 0. That's a dead battery: no more net reaction, no more current.
Reading the Nernst equation
E = E° − (RT/nF) ln Q. You don't need to calculate with it, but you should be able to read it (here for a galvanic cell, where E° > 0 and K > 1):
- If Q < 1, ln Q is negative, so E > E°. The cell is farther from equilibrium than standard conditions.
- If Q = 1, ln Q = 0, so E = E°.
- If Q > 1, ln Q is positive, so E < E°. The cell is closer to equilibrium.
- If Q = K, E = 0.
- The effect is modest. For a reaction with n = 2 at 25 °C, a tenfold change in Q changes E by only about 0.03 V.
Reasoning about changes
To predict how a concentration change affects the voltage, ask how it changes Q. For Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s), Q = [Zn²⁺]/[Cu²⁺]. Increasing [Cu²⁺] or decreasing [Zn²⁺] lowers Q, so E increases. Increasing [Zn²⁺] or decreasing [Cu²⁺] raises Q, so E decreases.
Justify these predictions with Q and the Nernst equation, not with Le Châtelier's principle. A running cell isn't at equilibrium, so equilibrium-shift arguments don't apply to it.
Concentration cells
A concentration cell has the same half-reaction on both sides but different concentrations, for example two copper electrodes, one in 0.010 M Cu²⁺ and one in 1.0 M Cu²⁺. Since both half-cells are identical, E° = 0, but E isn't zero because the concentrations differ.
The cell runs in the direction that makes the concentrations equal (its equilibrium). In the dilute half-cell, Cu is oxidized to Cu²⁺, raising that concentration, so it's the anode. In the concentrated half-cell, Cu²⁺ is reduced to Cu, lowering that concentration, so it's the cathode. Electrons flow from the dilute side to the concentrated side, and the voltage drops to zero when the concentrations become equal.
Worked examples
Try each one yourself first, then open the solution.
- Example 1
Predicting a voltage change
A Zn/Cu galvanic cell (E° = 1.10 V) is set up with 0.10 M Zn²⁺ and 1.0 M Cu²⁺. Is E greater than, less than or equal to 1.10 V? What happens to E as the cell runs?
Show the solutionHide the solution
- Step 1: Overall reaction: Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s). Q = [Zn²⁺]/[Cu²⁺] = 0.10 ÷ 1.0 = 0.10.
- Step 2: Q < 1, so ln Q is negative and E = E° − (RT/nF) ln Q is greater than E°. E > 1.10 V.
- Step 3: As the cell runs, [Zn²⁺] increases and [Cu²⁺] decreases, so Q increases. E decreases steadily.
- Step 4: When Q reaches K (which is enormous for this reaction), E = 0 and the cell is dead.
Answer: E is greater than 1.10 V at first, then decreases as Q rises, reaching 0 at equilibrium.
- Example 2
Trap: a concentration cell's direction
Two silver electrodes are placed in 0.010 M AgNO₃ and 1.0 M AgNO₃, connected by a wire and a salt bridge. A student says no current flows because E° = 0. Is the student right? If not, which electrode is the anode, and which way do electrons flow?
Show the solutionHide the solution
- Step 1: E° = 0 only because both half-reactions are the same. The concentrations are different, so the system isn't at equilibrium, and E ≠ 0. Current flows.
- Step 2: The cell runs toward equal concentrations. The dilute side's [Ag⁺] must rise, so there Ag(s) → Ag⁺(aq) + e⁻: oxidation, so that electrode is the anode.
- Step 3: On the concentrated side, Ag⁺ + e⁻ → Ag(s) lowers [Ag⁺]: reduction, the cathode.
- Step 4: Electrons flow through the wire from the electrode in 0.010 M AgNO₃ to the electrode in 1.0 M AgNO₃, until the two concentrations are equal and E = 0.
Answer: The student is wrong. The electrode in 0.010 M AgNO₃ is the anode; electrons flow from the dilute side to the concentrated side until the concentrations match.
Common mistakes
- Thinking E° = 0 means no current in a concentration cell. E depends on actual concentrations.
- Getting the Q direction backward: increasing a product concentration raises Q and lowers E.
- Using Le Châtelier's principle to justify a voltage change instead of Q and the Nernst equation.
- Thinking a dead battery has run out of all its chemicals. It has reached equilibrium (Q = K), where E = 0.
On the exam
- Expect 'will the voltage increase, decrease or stay the same?' questions. Write Q, say how it changes, and connect it to E through the Nernst equation.
- For concentration cells, identify the anode as the electrode in the more dilute solution, and explain using the drive toward equal concentrations.
Connected topics
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Check yourself
4 questions on 9.10 Cell Potential Under Nonstandard Conditions. Pick an answer to see if you got it, and why.
A galvanic cell is allowed to run until its voltage reads 0 V. Which of the following is true at that point?
A cell is built from two copper electrodes. One sits in 0.010 M Cu²⁺(aq) and the other in 1.0 M Cu²⁺(aq), with a salt bridge between them. Which of the following describes what happens as the cell operates?
A galvanic cell runs on Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s). Some Na₂S is added to the copper half-cell, and black CuS precipitates. How does the cell voltage change, and why?
A galvanic cell based on Fe(s) + Cu²⁺(aq) → Fe²⁺(aq) + Cu(s) starts with 1.0 M Fe²⁺ and 1.0 M Cu²⁺. As the cell operates, how does its voltage change, and why?
0 of 4 answered