A Copper–Silver Cell That Reads 0.28 V Instead of 0.34 V
A student assembles the cell Cu(s) | Cu2+(aq, 0.010 mol L^-1) || Ag+(aq, 0.0010 mol L^-1) | Ag(s) at 298 K, using E°(Ag+/Ag) = +0.80 V and E°(Cu2+/Cu) = +0.34 V. A high-impedance digital voltmeter reads 0.28 V, reproducibly, on three freshly made cells. Electrodes are polished and the salt bridge is fresh. A demonstrator offers two explanations: the distilled water used to make the silver solution had been stored in glass and had picked up traces of chloride, or the Ag+ solution had been made up a few days earlier in a beaker that had previously held ammonia. The student then adds concentrated ammonia dropwise to the silver compartment: the solution clears, and the reading falls steadily. In a separate trial the copper compartment is made 1.0 mol L^-1 in Cu2+ while the silver compartment is left at 1.0 × 10^-8 mol L^-1 Ag+. Assume activity coefficients of unity and a negligible liquid-junction potential.
1. Calculate the cell emf expected from the stated concentrations and compare it with the 0.28 V reading. Is the discrepancy more likely to be meter error or a concentration problem? Work out what [Ag+] would actually explain 0.28 V.
Read the worked answer
Expected emf. The spontaneous reaction is Cu + 2Ag+ → Cu2+ + 2Ag, with n = 2 and E°cell = 0.80 − 0.34 = +0.46 V. The Nernst expression is E = 0.46 − (0.0592/2)·log([Cu2+]/[Ag+]^2). Inserting the numbers:
log(0.010/(1.0 × 10^-3)^2) = log(1.0 × 10^4) = 4 E = 0.46 − 0.0296 × 4 = 0.46 − 0.118 = 0.342 V
Diagnosis. The shortfall is about 0.06 V, far too large for a junction potential (typically a few mV) and not credible as a high-impedance meter fault when three cells agree. So the real concentrations differ from the nominal ones. Solving for the silver ion concentration that gives 0.28 V: log([Cu2+]/[Ag+]^2) = (0.46 − 0.28)/0.0296 = 6.08, so [Ag+]^2 = 0.010/1.2 × 10^6 = 8.3 × 10^-9 and [Ag+] ≈ 9 × 10^-5 mol L^-1 - roughly ten times lower than labelled.
Why the demonstrator is right to blame the silver side. Because [Ag+] enters squared, the silver half-cell is far more sensitive to loss of Ag+ than the copper side. With [Cu2+] ten times too high (0.10 mol L^-1) the emf would only fall to 0.31 V. But AgCl begins to precipitate as soon as [Cl^-] exceeds Ksp/[Ag+] = 1.8 × 10^-10/1.0 × 10^-3 ≈ 1.8 × 10^-7 mol L^-1, and ammonia would convert Ag+ into [Ag(NH3)2]+, so either contamination removes Ag+ quantitatively. Both stories fit a measured emf of 0.28 V.
2. Ammonia is added to the silver compartment and the emf falls. Explain the fall in terms of the Nernst equation and estimate the emf when the free [Ag+] has been driven to 1.0 × 10^-8 mol L^-1 and [Cu2+] is 1.0 mol L^-1.
Read the worked answer
Why it falls. Ammonia binds Ag+ as the diammine complex, [Ag(NH3)2]+ (log β2 ≈ 7.2). The silver electrode potential is E = E° + 0.0592 log[Ag+], so sequestering Ag+ lowers the potential of the cathode. The copper potential is unaffected, so the cell emf falls. The electrode reaction is unchanged; only the activity of the electroactive species has changed, which is exactly what the Nernst equation describes.
Calculation. E = 0.46 − 0.0296·log([Cu2+]/[Ag+]^2) = 0.46 − 0.0296·log(1.0/1.0 × 10^-16) = 0.46 − 0.0296 × 16 = 0.46 − 0.474 = −0.014 V.
Interpretation. The sign has flipped: the cell as written is now non-spontaneous. Silver metal is oxidised to Ag+ (immediately complexed) and Cu2+ is reduced to copper. The polarity of the voltmeter reverses, and the silver electrode becomes the anode. A useful judgement point is that an emf passing through zero does not mean 'the cell is used up' in the sense of charge exhaustion - it means the two half-cell potentials are equal, i.e. the system is at equilibrium for the cell reaction.
3. At [Cu2+] = 1.0 mol L^-1, find the free [Ag+] at which the cell delivers zero emf. A classmate claims that a cell reading a small positive voltage is 'nearly dead'. Assess that claim.
Read the worked answer
Zero-emf condition. Set E = 0 in the Nernst equation: 0.46 = 0.0296·log([Cu2+]/[Ag+]^2), so log([Cu2+]/[Ag+]^2) = 15.54. With [Cu2+] = 1.0 mol L^-1, [Ag+]^2 = 10^-15.54 = 2.9 × 10^-16, giving [Ag+] ≈ 1.7 × 10^-8 mol L^-1. Above this silver ion concentration the forward reaction is spontaneous; below it the direction reverses.
Assessing the claim. 'Nearly dead' is vague and partly wrong. A cell at open circuit with E = 0 must have Q = K for the cell reaction - the chemical driving force has vanished, not the charge. In a real discharging cell the emf sags for two different reasons: the Nernst term as products accumulate and reactants are consumed, and polarisation (kinetic overpotential, ohmic drop, concentration gradients near the electrodes). A small positive open-circuit emf is thermodynamic, whereas a small emf under load may be mostly kinetic. The correct distinction is: zero emf at open circuit means equilibrium (Q = K); a small emf under load means either near-equilibrium or a badly polarised cell.
Bonus inference. At an emf of order tens of millivolts, a saturated calomel or silver/silver chloride reference measurement of each half-cell separately is the only way to tell which electrode has drifted - a single cell measurement cannot separate anode drift from cathode drift.