Why the Daniell Cell Reads Low
In a school laboratory, a student assembles a Daniell cell at 298 K: Zn(s) | ZnSO4(aq, 0.010 M) || CuSO4(aq, 0.10 M) | Cu(s). The standard electrode potentials are E°(Zn²⁺/Zn) = −0.76 V and E°(Cu²⁺/Cu) = +0.34 V. The digital voltmeter, which has a high input impedance, reads 1.07 V. The copper solution is still pale blue, but the zinc strip has developed a dark, spongy coating after the cell has stood for 15 minutes. The student has fresh zinc strips, 0.10 M CuSO4, 0.010 M ZnSO4, and new KCl-agar salt bridges available. The teacher asks whether the low reading is simply a concentration effect or whether the cell itself has been compromised, and what should be done before recording a value for the practical record.
1. Calculate the cell potential expected from the stated concentrations using the Nernst equation. Does this explain the 1.07 V reading? What effective copper(II) concentration would be needed to give 1.07 V if the zinc concentration were unchanged?
Read the worked answer
For the cell reaction Zn + Cu²⁺ → Zn²⁺ + Cu, n = 2 and
E = E° − (0.0591/2) log([Zn²⁺]/[Cu²⁺])
E° = 0.34 − (−0.76) = 1.10 V.
E = 1.10 − 0.02955 log(0.010/0.10) = 1.10 − 0.02955 × (−1) = 1.13 V.
The theoretical value is about 1.13 V, not 1.07 V. To see what copper concentration would give 1.07 V with [Zn²⁺] = 0.010 M:
1.07 = 1.10 − 0.02955 log(0.010/[Cu²⁺]) log(0.010/[Cu²⁺]) = 1.015 0.010/[Cu²⁺] ≈ 10.4 [Cu²⁺] ≈ 9.6 × 10⁻⁴ M.
That is far below the bulk 0.10 M CuSO4. The low reading is therefore not explained by the bulk concentration alone; the copper concentration near the electrode has been lowered, or the electrode is contaminated.
2. The zinc strip has a dark spongy coating. Explain its origin and its effect on the measured potential. Give two practical corrections for the next trial.
Read the worked answer
The coating is copper deposited by direct displacement: Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s). This happens because copper(II) ions migrate through the salt bridge or diffuse into the zinc compartment and contact the zinc surface. The spongy copper forms tiny local galvanic cells on the zinc, so the zinc is partly short-circuited and the measured cell potential falls. It also increases the local Zn²⁺ concentration and can block part of the zinc surface, making the electrode behave less reversibly.
Two corrections are: (i) clean the zinc with emery paper and use a fresh zinc strip; (ii) use fresh CuSO4 solution and assemble/read the cell quickly, keeping the two compartments separated by a fresh agar-KCl salt bridge to reduce mixing. Avoiding prolonged standing before measurement is important.
3. If the cell is allowed to run for several hours, predict the direction of change in EMF and explain using the Nernst equation. What observation would indicate the cell is nearly exhausted?
Read the worked answer
As the cell discharges, Zn²⁺ increases and Cu²⁺ decreases. Therefore the reaction quotient Q = [Zn²⁺]/[Cu²⁺] increases. Since E = E° − (0.0591/2) log Q, an increase in Q makes log Q more positive and E decreases. The EMF therefore falls steadily from its initial value.
Near exhaustion, Cu²⁺ is almost consumed, so the ratio [Zn²⁺]/[Cu²⁺] becomes very large and E falls sharply toward zero. Visually, the blue colour of Cu²⁺ fades, copper deposition slows or stops, and the zinc may be heavily coated with displaced copper. These changes show that the cell is no longer able to deliver a useful potential.