Nernst Equation Numericals
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Direct answer
Cell emf drifts below its standard value as products pile up, and the Nernst equation prices the drift: at 298 K, E(cell) = E°(cell) − (0.059 V/n) × log Q, with Q the reaction quotient of the cell reaction and n the electrons transferred. Standard potentials combine as E°(cell) = E°(cathode) − E°(anode) — for the Daniell cell, 0.34 − (−0.76) = 1.10 V. At equilibrium Q becomes Kc and E falls to zero, giving log Kc = nE°/0.059; and the electrical yield is ΔG = −nFE° with F = 96,500 coulomb per mol, so the Daniell cell's 1.10 V and n = 2 amount to about −212 kJ per mol of free energy. For concentration cells E° is zero and the entire emf comes from the log of the concentration ratio.
What you must remember
- Two Nernst forms: general E = E° − (2.303RT/nF) log Q, and the 298 K shortcut with 0.059 V replacing 2.303RT/F.
- Cell assembly: E°(cell) = E°(cathode) − E°(anode), using standard reduction potentials as tabulated; Daniell cell 1.10 V.
- Equilibrium link: at equilibrium E = 0 and Q = Kc, so log Kc = nE°/0.059 at 298 K.
- Energy link: ΔG° = −nFE°, F = 96500 C per mol; a spontaneous cell has positive E° and negative ΔG°.
- Concentration cells: E° = 0; E = (0.059/n) log(c2/c1), the more dilute side acting as anode.
- Q discipline: products over reactants, concentrations in mol per litre, gases as partial pressures in bar, powers equal to coefficients, pure solids and liquids excluded.
- SHE convention: the standard hydrogen electrode is assigned zero at every temperature, the reference all tabulated potentials hang from.
A Daniell cell taken through all three equations
Cell: Zn | Zn2+(0.1 M) || Cu2+(0.01 M) | Cu. Cell reaction Zn + Cu2+ → Zn2+ + Cu, so Q = [Zn2+]/[Cu2+] = 0.1/0.01 = 10, and n = 2. E = 1.10 − (0.059/2) × log 10 = 1.10 − 0.0295 ≈ 1.07 V. One line, one subtraction — but note how the more concentrated copper side was chosen cathode; reversing the ratio sign is the standard error.
Now the two conversions the same numbers unlock. Equilibrium: log Kc = 2 × 1.10/0.059 = 37.3, so Kc ≈ 2 × 10³⁷ — the reaction runs essentially to completion, which is precisely why the zinc-copper cell is the textbook workhorse. Energy: ΔG° = −nFE° = −2 × 96,500 × 1.10 ≈ −212,300 J ≈ −212 kJ per mol. Three exam questions, one data set; practise the chain until each step takes fifteen seconds.
Sign discipline in cell arithmetic
Sign discipline fails first: Q is always products over reactants of the written cell reaction, and electrons or electrode materials never appear in it. Second slip: using 0.059 when the question sets a temperature other than 298 K — the correct coefficient is 2.303RT/nF, which at 308 K is about 0.061. Third, n counted per atom instead of per equation: dichromate's reduction consumes 6 electrons per ion, not 3. In concentration cells, candidates reach for E° values that do not exist — both electrodes are the same couple, so E° = 0 and the entire emf is the 0.059 log ratio. Finally, the hydrogen electrode questions: pH enters through Q as [H+]², so E = −0.059 × pH for the SHE-coupled hydrogen electrode — the bridge between this chapter and acid-base calculations that NEET crosses regularly.
Frequently asked questions
What is the Nernst equation at 25 °C?
E(cell) = E°(cell) − (0.059/n) log Q, where n is electrons transferred and Q the reaction quotient of the cell reaction.
What is E of a Daniell cell with 0.1 M zinc and 0.01 M copper ion?
Q = 10 and n = 2, so E = 1.10 − 0.0295 × 1 ≈ 1.07 V.
How is the equilibrium constant extracted from E°?
At equilibrium E = 0, so log Kc = nE°/0.059 at 298 K; large positive E° means enormous Kc.
What is ΔG° for the Daniell cell?
ΔG° = −nFE° = −2 × 96,500 × 1.10 ≈ −212 kJ per mol — spontaneous by a wide margin.
Why does a concentration cell have E° = 0?
Both electrodes are the same couple, so their standard potentials cancel; the measurable emf comes only from the concentration difference, E = (0.059/n) log(c2/c1).