Daniell cell EMF that will not match the Nernst equation
A Class XII group is asked to verify the Nernst equation for the Daniell cell at 298 K. They prepare Zn | Zn²⁺(0.10 M) || Cu²⁺(0.10 M) | Cu and read 1.10 V on a digital voltmeter, exactly the standard value. They then replace the copper half-cell with Cu²⁺(0.0010 M), keeping [Zn²⁺] = 0.10 M, and read 0.98 V. Earlier, to clean the copper strip, one student had swirled it in dilute ammonia; a few drops of that ammonia were left in the beaker before the dilute CuSO₄ solution was poured in. Using the nominal concentrations, the group finds a mismatch of about 0.06 V. The teacher rules out meter error, because the same meter reproduced 1.10 V minutes earlier. The group must decide what the discrepancy means, whether the cell is close enough to equilibrium for a Nernst calculation, and how the value should be reported.
1. Calculate the EMF the group should have observed at 298 K and decide whether the 0.06 V gap is experimental error or a real chemical effect.
Read the worked answer
E = E° − (0.059 V / 2) log([Zn²⁺]/[Cu²⁺]) = 1.10 − 0.0295 log(0.10/0.0010) = 1.10 − 0.0295 × 2 = 1.041 V.
The measured 0.98 V is 0.061 V below this. Liquid-junction potentials and the small current drawn by a digital voltmeter are normally worth less than 0.01 V, so a gap six times larger needs a chemical explanation. The ammonia is the suspect: it complexes Cu²⁺ as [Cu(NH₃)₄]²⁺ with a very large formation constant, so the free Cu²⁺ concentration is far below the nominal 0.0010 M.
We can even size the effect. 0.061 V = 0.0295 log(ratio), so log(ratio) ≈ 2.07 and the free [Cu²⁺] is about 0.0010/120 ≈ 8 × 10⁻⁶ M. That is precisely the direction and magnitude expected when ammonia is present. Conclusion: the Nernst equation is not failing; the concentration substituted into it is the wrong quantity.
2. The group wants to recover an EMF close to the calculated value. State two changes they should make and justify each quantitatively.
Read the worked answer
First, discard that copper half-cell and rebuild it with fresh CuSO₄ solution and a strip that has not been in contact with ammonia. Removing the complexing agent restores free [Cu²⁺] to the nominal value, and the cell should read about 1.04 V again. The reason is quantitative: every tenfold increase in the free [Cu²⁺] raises E by 0.0295 V.
Second, improve the concentration ratio itself. Raising [Cu²⁺] to 1.0 M or lowering [Zn²⁺] from 0.10 M to 0.010 M changes [Zn²⁺]/[Cu²⁺] by a further factor of ten and adds roughly another 0.03 V, giving about 1.07 V. Any change that makes the oxidised form at the cathode more concentrated, or the oxidised form at the anode less concentrated, drives E up.
What the group should not do is 'correct' the Nernst calculation by adjusting the temperature term or adding a fudge factor for junction potential; neither can account for 0.06 V, and both would hide a genuine contamination problem.
3. A classmate argues: 'EMF is a thermodynamic quantity, so a measured value must equal the Nernst value exactly.' Evaluate this claim.
Read the worked answer
The claim is wrong in two different ways, and it is worth separating them.
First, a voltmeter reading is taken with a tiny current flowing, so the cell is not strictly at equilibrium; the reading approaches the reversible EMF only because the current is small. Standard potentials are also defined for unit activities, zero liquid-junction potential and 298 K. At ionic strengths of 0.1 M, activity coefficients are well below 1, so activities differ from concentrations by perhaps 20–40 %, which is worth only a few millivolts here.
Second, and decisively, the 0.06 V gap is far larger than any of these corrections and matches the ammonia complexation quantitatively. So the correct inference is not that the Nernst equation has broken down, but that the number placed into it was the total copper concentration rather than the free Cu²⁺ concentration. The examinable point is that the equation uses activities of the species actually present, and a masking or complexing agent silently removes the free ion.