When the Voltmeter Disagrees with the Table
A student assembles a cell at 298 K: a zinc electrode in 0.010 M ZnSO4 joined by a KNO3 salt bridge to a copper electrode in 0.50 M CuSO4. A high-impedance digital voltmeter reads 1.15 V. The student expected 1.10 V from the data sheet, since E°cell = E°(Cu²⁺/Cu) − E°(Zn²⁺/Zn) = 0.34 − (−0.76) = 1.10 V, and suspects the meter is faulty. The same student wants to light a red LED with a forward voltage of 1.8 V and asks whether 'tuning the concentrations' will do it. As an alternative they propose replacing the zinc half-cell with a silver electrode in 0.0010 M AgNO3 (E°(Ag⁺/Ag) = +0.80 V) because silver is 'more noble and will give more volts'. The salt bridge is freshly prepared and the electrodes are cleaned. Note: the LED draws several milliamps, which the student intends to ignore.
1. Account for the 1.15 V reading using the Nernst equation, and identify the correct direction in which concentrations must be changed to raise the cell potential.
Read the worked answer
The table value applies only at unit activity. For the reaction Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s), n = 2 and Q = [Zn²⁺]/[Cu²⁺].
E = E° − (0.0592/2) log Q = 1.10 − 0.0296 log(0.010/0.50) = 1.10 − 0.0296 log(0.020) = 1.10 − 0.0296(−1.699) = 1.10 + 0.050 = 1.15 V
The meter is correct; the student compared a non-standard cell with a standard potential. Since the zinc solution is tenfold more dilute than 1 M, the dissolution of zinc is favoured and the driving force rises.
To increase E further: lower [Zn²⁺] (products) and raise [Cu²⁺] (reactants). Both push Q down and log Q more negative. As a limit, with [Cu²⁺] = 1.0 M and [Zn²⁺] = 1 × 10⁻⁶ M, E = 1.10 + 0.0296 × 6 = 1.28 V. The cell is dead (E = 0) when Q = 10^(1.10/0.0296) ≈ 10³⁷, which is unreachable in practice.
2. Decide whether concentration tuning alone can light the 1.8 V LED, justify the answer with numbers, and give a practical recommendation.
Read the worked answer
Even the theoretical maximum for this chemistry, taking [Cu²⁺] = 1 M and a vanishingly small [Zn²⁺], is about 1.28 V (see previous part). Real zinc solutions below roughly 10⁻⁵ M are not usable, and as soon as the LED draws current the electrode potentials polarise and the terminal voltage sags below the open-circuit value. The gap to 1.8 V is about 0.5 V, far beyond anything concentration adjustment can supply.
The standard potentials fix the ceiling: 1.10 V is the thermodynamic limit for unit activities, and no arrangement of Zn/Cu concentrations exceeds roughly 1.3 V.
Recommendation: connect two such cells in series. Two Daniell cells give about 2.3 V open-circuit, comfortably above the 1.8 V forward voltage, with a series resistor (for example 100–220 Ω) to limit current; without a resistor the LED or the cells will be damaged. Alternatively use an LED whose forward voltage is below the available EMF - a red LED driven at low current may strike near 1.7 V, still too high, so the two-cell series route is the sound choice. The student should also check polarity: the copper electrode is the positive terminal.
3. The student proposes an Ag/Ag⁺ half-cell at 0.0010 M AgNO3 instead of Zn/Zn²⁺ against the same Cu/Cu²⁺ half-cell. Calculate the new cell potential and assess whether the student's reasoning ('more noble means more volts') is correct.
Read the worked answer
New cell: Cu(s) is now the anode? Compare E°: Ag⁺/Ag = +0.80 V, Cu²⁺/Cu = +0.34 V, so silver is the cathode and copper the anode. E°cell = 0.80 − 0.34 = 0.46 V.
The reaction is Cu(s) + 2Ag⁺(aq) → Cu²⁺(aq) + 2Ag(s), n = 2, Q = [Cu²⁺]/[Ag⁺]². E = 0.46 − 0.0296 log(0.50/(0.0010)²) = 0.46 − 0.0296 log(5.0 × 10⁵) = 0.46 − 0.0296(5.699) = 0.46 − 0.169 = 0.29 V
This is lower than the original 1.15 V, not higher. The student's mistake is to compare the silver half-cell with the zinc half-cell rather than with the copper half-cell actually fitted, and to assume that a more positive E° automatically means a bigger cell EMF. What matters is the difference between the two half-cells: 0.80 − 0.34 is small, whereas 0.34 − (−0.76) is large. The dilute AgNO3 also costs about 0.17 V through the Nernst term. The proposal should be rejected.