Daniell Cell Reading Low: Diagnosing a Diluted Half-Cell
A student sets up a Daniell cell: a zinc electrode in 1.0 M ZnSO₄ and a copper electrode in a blue solution labelled '1.0 M CuSO₄', joined by a KNO₃ salt bridge. A digital voltmeter reads 1.04 V at 21 °C. Standard electrode potentials are Zn²⁺/Zn = −0.76 V and Cu²⁺/Cu = +0.34 V. The copper solution is noticeably paler than a fresh 1.0 M CuSO₄ sample. The student first blames the salt bridge and adds more KNO₃, then blames the voltmeter and connects a second meter, which reads the same. They ask whether the cell is working, why the reading is low, and what would happen if they replaced the copper solution with fresh 1.0 M CuSO₄. The ZnSO₄ solution is confirmed 1.0 M.
1. Calculate the standard cell EMF and the expected EMF for the stated concentrations. Compare with 1.04 V and assess the salt-bridge and voltmeter explanations.
Read the worked answer
E°_cell = E°_cathode − E°_anode = 0.34 − (−0.76) = 1.10 V. For [Zn²⁺] = [Cu²⁺] = 1.0 M, the Nernst term is zero, so E = 1.10 V at 25 °C; at 21 °C the change is only about 0.0002 V. The measured 1.04 V is therefore 0.06 V low. Adding KNO₃ does not change either electrode potential; it only affects junction potentials and conductivity. A second high-impedance voltmeter giving the same reading argues against meter loading. The cell is working, but the copper half-cell is not at the stated concentration.
2. Use the Nernst equation to infer [Cu²⁺] from 1.04 V and decide whether the pale colour supports your inference.
Read the worked answer
Use E = 1.10 − (0.0592/2) log([Zn²⁺]/[Cu²⁺]). With [Zn²⁺] = 1.0 M and E = 1.04 V: 0.06 = 0.0296 log(1/[Cu²⁺]), so log(1/[Cu²⁺]) = 2.03 and [Cu²⁺] = 9.3 × 10⁻³ M ≈ 0.009 M. That is about one-hundredth of the labelled concentration. The pale blue colour is exactly what a roughly 0.01 M copper(II) solution looks like compared with 1.0 M. The most likely cause is that the '1.0 M CuSO₄' was made by a dilution error or was mislabelled; the salt bridge and voltmeter are not the cause.
3. What would happen to the voltage and the cell reaction if fresh 1.0 M CuSO₄ were used? Would the cell still work if the salt bridge were removed?
Read the worked answer
With fresh 1.0 M CuSO₄, [Cu²⁺] rises by about 100-fold, so the Nernst term returns to zero and the voltage rises to about 1.10 V. The cell reaction is Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s); higher [Cu²⁺] makes the forward reaction more favourable, so the copper electrode gains mass and the zinc electrode loses mass faster. Removing the salt bridge breaks the ionic circuit: no sustained current can flow, although a high-impedance voltmeter may show a transient or stray reading. The salt bridge is needed to maintain electroneutrality, not to set the cell EMF.