Nernst Cell Voltage & Electrochemistry Lab

Build galvanic cells from standard reduction couples, set ion activities and temperature, and toggle between linear and log Q axes to see why cell voltage holds flat until depletion.

Worked example — Nernst Cell Voltage & Electrochemistry Lab

Activity a(Cu2+)	1.0000
Activity a(Zn2+)	1.0000
Cell Voltage E	1.1037 V
Standard Voltage E°	1.1037 V
Nernst Slope	29.58 mV/dec
Electrons n	2 e⁻
Reaction Quotient Q	1.0000
log₁₀ K	37.31

These are the values the calculator shows when the page opens. Change any input above to recompute them.

Loading interactive simulation...

Concentration barely moves a cell's voltage: 59.16 mV per decade, divided by n 🖖

The Nernst equation E = E° − (RT / nF) ln Q shows that voltage depends logarithmically on the concentration ratio Q. At 25 °C and n = 2, a 10-fold change in Q moves the potential by just 29.58 mV. A million-fold shift in the product-to-reactant ratio (Q = 10⁶) costs under 0.18 V. A battery reads near nominal voltage until it dies because voltage is set by the active chemistry, not by the amount of reactant left, until solid depletion or internal resistance terminates current flow.

The 59 mV rule is a temperature, not a constant 🖖

The factor 2.303 RT / F is 59.16 mV per decade at 25 °C, but expands to 66.10 mV at 60 °C — a 12% shift over ordinary laboratory temperature ranges. This thermal drift is why pH meters and analytical electrochemistry probes require automatic temperature compensation; calibrating at one temperature and measuring at another introduces immediate systematic errors. The temperature control moves the slope and nothing else: E° is held at its 25 °C value however you set T, which amounts to assuming the reaction's entropy change is zero, so ΔG° for the Daniell cell reads −213.0 kJ/mol at 25 °C and at 60 °C alike. It is also why the dead-cell preset is only dead at the temperature it was built for. Heat that same concentration ratio to 60 °C and the cell voltage card reads −0.1295 V, with ΔG back above zero at 25.0 kJ/mol.

A dead cell reached equilibrium: one relation, two readings 🖖

When a galvanic cell flattens to E = 0 V, reaction quotient Q equals equilibrium constant K. Setting E = 0 yields log₁₀ K = n E° / 0.05916 V at 25 °C. For a Daniell cell, log₁₀ K = 37.31 (K ≈ 2.05 × 10³⁷). Flattening the cell by concentration ratio alone would require a product-to-reactant ratio of 10³⁷ — which is why a cell with a large standard potential runs essentially to completion, though a pairing with a small one need not: Ag⁺ against Fe³⁺/Fe²⁺ is spontaneous at 0.0286 V and settles at K = 3.04, linked directly by ΔG° = −nFE° = −RT ln K.

Problems solved in full

Energy per kilogram of the metals a Daniell cell consumes 6 steps
A Daniell cell reads 1.1037 V. How much energy is that per kilogram of the metals it consumes, and which single substitution multiplies the figure by five?
1. The standard cell potential is the cathode reduction potential minus the anode one: copper sits at +0.3419 V, zinc at −0.7618 V, and the difference is 1.1037 V. Both ions are at unit activity here, so Q = 1 and log Q = 0, which kills the Nernst correction entirely. That is why the tool prints the same 1.1037 V in the Cell Voltage card as in the Standard Voltage card.
  
  $E_{cell}^{\circ} = E_{cat}^{\circ} - E_{an}^{\circ} = 0.3419 V - (- 0.7618 V) = 1.1037 V$
2. Zinc releases two electrons and copper(II) accepts two, so n = 2 and one mole of reaction moves 2F = 192,970.7 coulombs. The Electrons card reads 2.
  
  $n F = 2 \times 96485.33 C mo l^{- 1} = 192, 970.7 C mo l^{- 1}$
3. Charge times potential is work: ΔG° = −nFE° = −212,982 J per mole of reaction, which the ΔG card rounds to −213.0 kJ/mol. That is the entire electrical budget of one mole.
  
  $Δ G^{\circ} = - n F E^{\circ} = - 192, 970.7 \times 1.1037 = - 212, 982 J mo l^{- 1} = - 213.0 kJ mo l^{- 1}$
4. Now weigh it. One mole of reaction eats one mole of zinc metal, 65.38 g, and plates one mole of copper drawn from solution, 63.55 g. Call it 128.93 g of metal per 213.0 kJ, counting only the two metals: the sulfate that carries the copper into solution is real mass and is ignored here, so every energy figure below is a ceiling.
  
  $M = M_{Zn} + M_{Cu} = 65.38 + 63.55 = 128.93 g mo l^{- 1}$
5. 212,982 J divided by 0.12893 kg gives 1.652 MJ/kg, or 459 Wh/kg.
  
  $\frac{212 , 982 J}{0.12893 kg} = 1.652 MJ k g^{- 1} = 459 Wh k g^{- 1}$
6. Swap the zinc for lithium, which sits in the tool’s electrode list at −3.0401 V. E° becomes 3.3820 V, and n stays at 2 because each lithium atom gives up one electron and the copper needs two: Cu²⁺ + 2Li → Cu + 2Li⁺. ΔG° is −652.6 kJ/mol and the reactants now weigh 2 × 6.94 + 63.55 = 77.43 g, which is 2341 Wh/kg.
  
  $E^{\circ} = 0.3419 + 3.0401 = 3.3820 V, \frac{652 , 627 J}{0.07743 kg} = 2341 Wh k g^{- 1}$
Answer

Zinc and copper store 459 Wh in every kilogram they consume; lithium and copper store 2341, and the voltage is only part of the reason. The potential rises 3.064× and the reactants get 1.665× lighter, and the product of those two is the 5.10. Which means the voltage card ranks couples in the wrong order for anyone choosing a battery: pair the lithium with bromine instead and E° climbs again to 4.1061 V while the specific energy falls to 1267 Wh/kg, because Br₂ weighs 159.81 g where copper weighs 63.55. There is no molar mass anywhere on this page, so that comparison is one the tool cannot make however many couples you try. One warning before you build the lithium cell: dropped into aqueous copper sulfate, lithium metal attacks the water and not the copper. A standard potential says which reaction can run, never which one you get.

$\frac{2341}{459} = 5.10 = E^{\circ} 3.064 \times M 1.665$
The dead cell, warmed to sixty degrees 6 steps
Load Temperature shift (60 °C): the same Daniell cell, thirty-five degrees hotter. The Nernst Slope card moves and the Cell Voltage card does not. Work out what happened to K, then decide whether the exhausted cell from Dead cell at equilibrium (Q = K) is still exhausted at this temperature.
1. Both activities are one, so Q is one and its logarithm is zero. The Nernst correction vanishes and the cell voltage equals the standard voltage — at 25 °C, at 60, at any temperature the tool will accept. ΔG does not move either.
  
  $Q = 1 \Rightarrow lo g Q = 0 \Rightarrow E = E^{\circ} = 1.1037 V at every temperature; Δ G^{\circ} = - n F E^{\circ} = - 213.0 kJ mo l^{- 1}$
2. The slope does, because it is 2.303RT/F with the T still in it. Thirty-five kelvin is a 12% rise, and the two card values sit in exactly that ratio.
  
  $\frac{2.303 R T}{F} : 59.16 mV (298.15 K) \to 66.10 mV (333.15 K), \frac{66.10}{59.16} = 1.1174 = \frac{333.15}{298.15}$
3. The equilibrium constant is the same slope used differently. Set E to zero and log₁₀ K is the standard potential divided by the per-decade slope for two electrons.
  
  $lo g_{10} K = \frac{n F E ^{\circ}}{2.303 R T} = \frac{E ^{\circ}}{2.303 R T / n F} = \frac{1.1037}{0.029580} = 37.31 ⟶ \frac{1.1037}{0.033052} = 33.39$
4. Nearly four decades gone, from warming a cell by thirty-five degrees.
  
  $K : 2.05 \times 1 0^{37} ⟶ 2.47 \times 1 0^{33}, Δ lo g_{10} K = 3.92 decades = a factor of 8, 318$
5. Which settles the dead cell. Its activity ratio was chosen to equal K at 25 °C. At 60 °C that ratio sits 3.92 decades above K, and the voltage comes out negative.
  
  $Q = 2.05 \times 1 0^{37} at 60^{\circ} C : E = - 0.033052 \times 3.920 = - 0.1295 V, Δ G = + 25.0 kJ mo l^{- 1}$
6. One thing to be honest about before believing the size of it: the tool holds E° fixed as the temperature changes, and a standard free energy that does not vary with temperature has no entropy in it at all.
  
  $Δ S^{\circ} = - (\frac{\partial Δ G ^{\circ}}{\partial T})_{p} = 0 here, because E^{\circ} is held fixed ⟹ Δ H^{\circ} = Δ G^{\circ}$
Answer

No. At 60 °C that cell reads −129.5 mV, and ΔG is +25.0 kJ/mol. Zinc plates out and copper dissolves; the cell has become a load rather than a source, and connecting it now drives the reaction backwards until the ratio falls to the new, smaller K.

Real cells do have a temperature coefficient of E°, which this page does not model, so treat the four decades as a consequence of the tool’s own assumption rather than as a measurement. What survives the assumption is the shape: at fixed E°, log₁₀ K carries a 1/T, so a quoted equilibrium constant is a statement about a temperature as much as about a couple. The card reads 37.31 on one screen and 33.39 on another, for the same two metals in the same beaker.

$lo g_{10} K = \frac{n F E ^{\circ}}{2.303 R T} \propto \frac{1}{T}; \frac{K ( 298.15 )}{K ( 333.15 )} = 8, 318$

References (2)

Every standard reduction potential in the half-cell table Vanýsek, P. Electrochemical Series, in CRC Handbook of Chemistry and Physics (97th ed.), CRC Press, 2016.
R and F, which set the 59.16 mV per decade the tool is built around CODATA 2018 / SI: molar gas constant R = 8.314 462 618 J mol⁻¹ K⁻¹ (exact); Faraday constant F = 96 485.332 12 C mol⁻¹ (exact).

By the 42Math team · Updated 2026-08-23

Reviewed against published sources. Corrections welcome. Editorial policy · Corrections

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Example problems

Standard Daniell cell (Q = 1) - Standard Zn/Cu cell at Q = 1: E° = 1.1037 V, E = 1.1037 V, ΔG = -213.0 kJ/mol, log₁₀ K = 37.31.
Nearly flat cell (Q = 10⁶) - A million-fold concentration ratio shift (Q = 10⁶) costs only 0.1775 V, leaving voltage at E = 0.9262 V.
Temperature shift (60 °C) - Heating from 25 °C to 60 °C increases the Nernst slope from 59.16 mV to 66.10 mV per decade.
Dead cell at equilibrium (Q = K) - At equilibrium Q = K ≈ 2.05 × 10³⁷, cell voltage drops to E = 0.0000 V and ΔG = 0.0 kJ/mol.