Two Cells, One Diluted Half-Cell: Judging Feasibility with the Nernst Equation
A student assembles two cells at 298 K and records the emf. Cell I is Zn(s) | Zn²⁺(0.10 M) || Cu²⁺(0.0010 M) | Cu(s), for which E°cell = 1.10 V; the measured value is close to 1.04 V. Cell II is used to test whether copper turnings will displace silver from a 0.0010 M AgNO3 solution in which the copper(II) concentration is already 0.10 M. The standard potentials available are E°(Ag⁺/Ag) = +0.80 V and E°(Cu²⁺/Cu) = +0.34 V. A classmate predicts 'no reaction, because the silver solution is far too dilute - the standard potential is only 0.46 V and dilution will kill it'. The teacher says the prediction of no reaction is wrong, although the driving force is certainly reduced.
1. Calculate the emf of Cell I and explain why the measured value falls below the standard value.
Read the worked answer
Reaction: Zn + Cu²⁺ → Zn²⁺ + Cu, with n = 2. E = E° − (0.0592/2) log([Zn²⁺]/[Cu²⁺]) = 1.10 − 0.0296 × log(0.10/0.0010) = 1.10 − 0.0296 × log 100 = 1.10 − 0.0296 × 2 = 1.10 − 0.059 = 1.041 V.
The measured 1.04 V agrees well. It is below E° because E° refers to unit activities of both ions; here the oxidant Cu²⁺ is much more dilute (0.0010 M) than the product Zn²⁺ (0.10 M), so Q = [Zn²⁺]/[Cu²⁺] = 100 and the log term is positive. In addition, small liquid-junction potentials, activity coefficients and temperature differences keep the measured value a little away from the ideal Nernst value.
2. Calculate the cell potential for the copper–silver displacement and judge whether the classmate is right.
Read the worked answer
Reaction: Cu(s) + 2Ag⁺ → Cu²⁺ + 2Ag(s), n = 2. E° = 0.80 − 0.34 = 0.46 V. E = 0.46 − (0.0592/2) log([Cu²⁺]/[Ag⁺]²) = 0.46 − 0.0296 × log(0.10/(1.0 × 10⁻⁶)) = 0.46 − 0.0296 × log(1.0 × 10⁵) = 0.46 − 0.0296 × 5 = 0.46 − 0.148 = 0.312 V.
E is positive, so the displacement is still feasible; dilution has reduced the driving force from 0.46 V to about 0.31 V but has not reversed it. The classmate's error is to treat E° as if it were a threshold: feasibility depends on the sign of E, i.e. of ΔG = −nFE, not on how large E° is. Reaction continues until E = 0, i.e. until [Cu²⁺]/[Ag⁺]² = 10^(0.46/0.0296) ≈ 10¹⁵·⁵, so silver is essentially completely displaced.
3. In Cell I the student replaces the copper half-cell with 0.0010 M Cu²⁺ that is also 1.0 M in NH3, forming [Cu(NH3)4]²⁺. Predict and justify the effect on the cell emf.
Read the worked answer
Ammonia binds Cu²⁺ strongly (Kf for [Cu(NH3)4]²⁺ is of the order of 10¹²–10¹³), so almost all the copper is converted to the complex and the free [Cu²⁺] falls from 0.0010 M to roughly 10⁻¹⁵–10⁻¹⁶ M.
In E = 1.10 − 0.0296 log([Zn²⁺]/[Cu²⁺]), a much smaller denominator makes the log term much larger, so E falls sharply; with [Cu²⁺] ≈ 10⁻¹⁵ M and [Zn²⁺] = 0.10 M, log(10¹⁴) = 14 and E ≈ 1.10 − 0.41 = 0.69 V. The direction can even reverse if the free copper ion concentration is driven low enough.
The judgement being tested is that complexation is a way of controlling the effective concentration of a metal ion, and therefore the electrode potential, without changing the standard potential at all.