Cell Potential Under Nonstandard Conditions: Study Guide (Chemistry 9.10)

Standard conditions are just one point on the journey

In Topic 9.9 every calculation used standard conditions: 1 M solutions and 1 atm gases. Under those conditions the reaction quotient Q equals 1, and the cell potential is E°.

A real cell does not stay at standard conditions. As a galvanic cell runs, reactant ions are used up and product ions build up. So Q changes, and the voltage on the meter, E (no ° sign), changes with it. A torch battery does not give its full voltage for ever: it runs down.

The key idea in this topic is:

The cell potential is a driving force towards equilibrium. The further the reaction mixture is from equilibrium, the larger the size of E. At equilibrium there is no drive left, so E = 0.

Comparing Q with 1 and with K

Remember Q has the same form as K but uses the concentrations right now. Pure solids and liquids are left out. For Fe(s) + Cu²⁺(aq) → Fe²⁺(aq) + Cu(s):

Q = [Fe²⁺] ÷ [Cu²⁺]

A galvanic cell has E° > 0, so K is greater than 1 (usually much greater). As the cell runs, Q rises from its starting value towards K. Four cases cover everything:

Situation	Where the cell is	Effect on E
Q = 1	standard conditions	E = E°
Q < 1 (relatively more reactant, less product)	further from equilibrium than standard	E larger than E°
1 < Q < K (relatively more product, less reactant)	closer to equilibrium than standard	E smaller than E°, but still positive
Q = K	at equilibrium	E = 0; no current flows

The Nernst equation, read without a calculator

The Nernst equation puts Figure 1 into symbols:

E = E° − (RT / nF) ln Q

You can read every key fact straight off it:

• If Q = 1, ln Q = 0, so E = E°.
• If Q < 1, ln Q is negative, so you subtract a negative number: E > E°.
• If Q > 1, ln Q is positive: E < E°.
• At equilibrium E = 0 and Q = K, so E° = (RT/nF) ln K. A positive E° matches K > 1, as you saw in Topic 9.9.

The factor RT/nF is small (RT/F is only 0.0257 V at 298 K). So a tenfold change in Q usually shifts E by only a few hundredths of a volt. Changing concentrations moves E above or below E°, but for most cells it does not change the sign of E unless the mixture is pushed past equilibrium.

This course treats the Nernst equation as a reasoning tool. Plugging numbers in is not enough on its own: always say what happens to Q, and why that moves the cell closer to or further from equilibrium.

Why not Le Châtelier’s principle?

You may be tempted to say “adding reactant shifts the equilibrium to the right, so the voltage goes up”. The prediction happens to be right, but the reasoning is not. Le Châtelier’s principle describes how a system at equilibrium responds to a disturbance. A working cell is not at equilibrium: that is exactly why it has a voltage. A cell at equilibrium has E = 0 and does nothing.

So use this chain instead: change in concentration → change in Q → cell closer to or further from equilibrium (Q compared with K) → smaller or larger E.

Worked example 1: an iron–copper cell away from standard

Question. The cell Fe(s) | Fe²⁺(aq) || Cu²⁺(aq) | Cu(s) runs the reaction Fe(s) + Cu²⁺(aq) → Fe²⁺(aq) + Cu(s), with E° = +0.78 V. Predict whether E is greater than, less than or equal to 0.78 V in each case, and justify each answer.

(a) [Cu²⁺] = 1.0 M, [Fe²⁺] = 0.010 M (b) [Cu²⁺] = 0.010 M, [Fe²⁺] = 1.0 M (c) Starting from 1.0 M of each ion, both half-cells are diluted tenfold with water. (d) Starting from 1.0 M of each ion, some Na₂S is added to the copper half-cell, and CuS precipitates.

(a) Q = 0.010 ÷ 1.0 = 0.010, which is less than 1. There is more reactant and less product than at standard conditions, so the cell is further from equilibrium: E > 0.78 V.

(b) Q = 1.0 ÷ 0.010 = 100, which is greater than 1 (but still far below K). The cell is closer to equilibrium: E < 0.78 V.

(c) Both concentrations fall to 0.10 M, so Q = 0.10 ÷ 0.10 = 1. Q is unchanged, so E = 0.78 V. (This works because both ions have the same coefficient. It is not true for every cell: see Practice Q6.)

(d) Precipitating CuS removes Cu²⁺, the reactant ion. [Cu²⁺] falls, so Q rises above 1. The cell is closer to equilibrium: E < 0.78 V.

Numerical check. With n = 2, E = 0.78 − (0.0257/2) ln Q. For (a), E = 0.78 + 0.059 = 0.84 V; for (b), E = 0.78 − 0.059 = 0.72 V. The numbers agree with the reasoning, and show that a hundredfold change in Q shifts E by only about 0.06 V here.

Concentration cells

A concentration cell uses the same half-reaction in both half-cells, at different concentrations. Because the two half-reactions are identical, E° = 0. Even so, the cell gives a voltage, because Q ≠ 1. It is far from equilibrium, which for this cell means equal concentrations on both sides.

To find the direction, ask: which way does the system need to go to reach equilibrium? The concentrations must become equal. The dilute side can only gain Ag⁺ if silver metal is oxidised there, so the dilute side is the anode. The concentrated side loses Ag⁺ when the ions are reduced to silver, so it is the cathode.

Worked example 2: running a concentration cell

Question. For the cell in Figure 2 (equal volumes of solution in each beaker): (a) state the direction of electron flow; (b) describe how E changes as the cell runs; (c) find the Ag⁺ concentration in each beaker when the cell stops; (d) use the Nernst equation to estimate the starting potential.

(a) Electrons flow through the wire from the silver strip in 0.0010 M AgNO₃ to the silver strip in 0.50 M AgNO₃, for the reason given above.

(b) As the cell runs, the dilute side becomes more concentrated and the concentrated side becomes more dilute. The cell moves closer to equilibrium, so E falls, reaching zero when the concentrations are equal.

(c) One Ag⁺ is made at the anode for every Ag⁺ removed at the cathode, and the volumes are equal, so the final concentration is the average: (0.0010 + 0.50) ÷ 2 = 0.2505 M (about 0.25 M) in both beakers.

(d) Overall: Ag⁺(0.50 M) → Ag⁺(0.0010 M), so Q = 0.0010 ÷ 0.50 = 0.0020 and n = 1. E = 0 − (0.0257/1) ln(0.0020) = +0.160 V. The positive value confirms the direction in (a); it is small because E° = 0.

Worked example 3: a battery running down

Question. A zinc–copper cell (E° = +1.10 V) starts with 1.0 M Zn²⁺ and 1.0 M Cu²⁺ and is left connected to a bulb. Explain what happens to Q and E over time, and what a “dead” cell means.

1. The reaction Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s) makes Zn²⁺ and uses Cu²⁺, so Q = [Zn²⁺] ÷ [Cu²⁺] rises from 1.
2. As Q rises towards K, the cell gets closer to equilibrium, so E falls below 1.10 V.
3. When Q = K, E = 0. No more net reaction happens and no current flows: the cell is “dead”.

Interpretation. A dead cell is a cell at equilibrium, not necessarily one with no reactant left. For this cell K is about 10³⁷, so at equilibrium [Cu²⁺] is extremely small, but in principle it is not zero.

Common misconceptions

• “Adding reactant shifts the equilibrium right, so E goes up (Le Châtelier).” The cell is not at equilibrium. Reason with Q: less product or more reactant means Q falls, so the cell is further from equilibrium and E is larger.
• “Changing the concentration changes E°.” E° is fixed (at a given temperature). Only E changes.
• “E = 0 means all the reactants are used up.” E = 0 means Q = K: the cell is at equilibrium.
• “A concentration cell gives no voltage because E° = 0.” It gives a voltage because Q ≠ 1; it stops only when the concentrations are equal.
• “Diluting both half-cells never changes E.” Only if Q does not change. If the ions have different coefficients in the equation, dilution changes Q.
• “Electrons in a concentration cell flow towards the dilute side to ‘spread out’.” They flow from the dilute side (anode) to the concentrated side (cathode).
• “A Nernst calculation is a full answer.” Explain the direction with Q first; the number is a check.

Where this leads

Next, Topic 9.11 turns to electrolytic cells and connects the current and time to the amount of substance produced. Try the practice questions now, then use the revision notes and the checklist to consolidate.