Cell EMF Calculator
Enter the cathode and anode reduction potentials to find the cell EMF instantly. Switch to Nernst mode to account for non-standard concentrations and temperature. A built-in electrode picker lets you choose from 17 common half-cells, and the show-your-work panel walks through every step of the math.
Formula
Worked example
Daniell cell: Zn anode (-0.76 V) and Cu cathode (+0.34 V). E-cell = 0.34 - (-0.76) = 1.10 V. DeltaG = -2 x 96485 x 1.10 = -212.3 kJ/mol. K = exp(2 x 96485 x 1.10 / (8.314 x 298.15)) = ~10^37. If [Zn2+] = 2 M and [Cu2+] = 0.5 M, Q = 2/0.5 = 4, Nernst gives E = 1.10 - (8.314 x 298.15)/(2 x 96485) x ln(4) = 1.10 - 0.0178 = 1.082 V.
What is electromotive force (EMF)?
Electromotive force, abbreviated EMF and measured in volts, is the maximum potential difference a cell can deliver when no current flows. It is not a force in the mechanical sense but a measure of the energy per unit charge that the chemical reaction can supply. In a galvanic (voltaic) cell, a spontaneous redox reaction drives electrons from the anode (where oxidation occurs) through the external circuit to the cathode (where reduction occurs). The greater the difference in reduction potential between the two half-cells, the larger the EMF and the more electrical work the cell can produce.
The cell EMF formula and how to use it
The standard cell EMF is calculated as E-cell = E-cathode minus E-anode, where both potentials are standard reduction potentials taken from a reference table at 25 degC, 1 mol/L, and 1 atm. The cathode is always the electrode with the higher reduction potential - it attracts electrons. The anode, with the lower reduction potential, releases electrons and is oxidised. For the classic Daniell cell, the copper cathode has E = +0.34 V and the zinc anode has E = -0.76 V, giving E-cell = 0.34 - (-0.76) = 1.10 V. A positive EMF means the forward reaction is spontaneous and the cell can do useful work.
The Nernst equation for non-standard conditions
Standard potentials apply at exactly 25 degC, 1 M concentrations, and 1 atm. Real cells operate at other conditions, so the actual EMF is corrected by the Nernst equation: E = E-cell - (RT / nF) x ln(Q), where R is the gas constant (8.314 J/mol/K), T is the absolute temperature in kelvin, n is the number of electrons transferred per formula unit, F is the Faraday constant (96485 C/mol), and Q is the reaction quotient. When Q equals 1 (standard conditions), ln(Q) = 0 and E collapses back to E-cell. As the reaction proceeds and products accumulate, Q rises, ln(Q) grows, and the Nernst correction reduces the voltage until E reaches zero at equilibrium.
Gibbs free energy and the equilibrium constant
The connection between electrochemistry and thermodynamics runs through two key equations. The Gibbs free energy change is DeltaG = -nFE-cell: negative DeltaG means the cell reaction is spontaneous at standard conditions, and the magnitude tells you how much energy per mole is released as electrical work. The equilibrium constant K is related by K = exp(nFE-cell / RT). For a Daniell cell with E-cell = 1.10 V and n = 2, K is approximately 10^37, meaning the reaction runs essentially to completion before equilibrium is reached. These relationships show that any cell with a positive EMF has K greater than one, confirming the forward reaction is thermodynamically favored.
Standard Reduction Potentials (25 degC, 1 M, 1 atm)
| Half-cell | Half-reaction | E (V vs SHE) | Typical role |
|---|---|---|---|
| Li/Li+ | Li+ + e- -> Li | -3.04 | Anode (strong reducer) |
| K/K+ | K+ + e- -> K | -2.93 | Anode |
| Ca/Ca2+ | Ca2+ + 2e- -> Ca | -2.87 | Anode |
| Na/Na+ | Na+ + e- -> Na | -2.71 | Anode |
| Mg/Mg2+ | Mg2+ + 2e- -> Mg | -2.37 | Anode |
| Al/Al3+ | Al3+ + 3e- -> Al | -1.66 | Anode |
| Zn/Zn2+ | Zn2+ + 2e- -> Zn | -0.76 | Anode (common) |
| Fe/Fe2+ | Fe2+ + 2e- -> Fe | -0.44 | Anode |
| Ni/Ni2+ | Ni2+ + 2e- -> Ni | -0.25 | Anode |
| Sn/Sn2+ | Sn2+ + 2e- -> Sn | -0.14 | Anode |
| Pb/Pb2+ | Pb2+ + 2e- -> Pb | -0.13 | Anode |
| H2/H+ | 2H+ + 2e- -> H2 | 0.00 | Reference (SHE) |
| Cu/Cu2+ | Cu2+ + 2e- -> Cu | +0.34 | Cathode (common) |
| Ag/Ag+ | Ag+ + e- -> Ag | +0.80 | Cathode |
| Hg/Hg2+ | Hg2+ + 2e- -> Hg | +0.85 | Cathode |
| Pt/Pt2+ | Pt2+ + 2e- -> Pt | +1.20 | Cathode |
| Au/Au3+ | Au3+ + 3e- -> Au | +1.52 | Cathode (weak reducer) |
Selected standard reduction potentials referenced to the Standard Hydrogen Electrode (SHE = 0.00 V). More negative values are stronger reducing agents (anodes); more positive values are stronger oxidizing agents (cathodes).
Frequently asked questions
What does a negative cell EMF mean?
A negative EMF means the chosen cathode actually has a lower reduction potential than the chosen anode. The reaction as written is non-spontaneous - it would require external electrical energy to proceed. To get a working galvanic cell, swap the roles of the two electrodes (the one with the higher reduction potential becomes the cathode).
Why is the standard hydrogen electrode assigned 0.00 V?
Absolute electrode potentials cannot be measured directly - only differences between two half-cells are measurable. By convention, the reduction of H+ to H2 gas at 1 atm pressure and 1 M concentration is assigned exactly 0.00 V, and all other half-cell potentials are measured relative to it. This is called the Standard Hydrogen Electrode (SHE).
How does temperature affect the cell EMF?
Temperature enters through the Nernst equation. A higher temperature increases the magnitude of the correction term (RT/nF), which amplifies the effect of any deviation of Q from 1. If Q is greater than 1 (products exceed reactants), the voltage drops faster with rising temperature. If Q is less than 1, the voltage rises slightly. At standard conditions (Q = 1), temperature has no first-order effect on the standard EMF, although the standard potential itself does vary slightly with temperature through thermodynamic relationships.
What is the difference between EMF and terminal voltage?
EMF is the open-circuit (no-current) potential of the cell - the maximum possible voltage. When current flows, the internal resistance of the electrodes and electrolyte causes a voltage drop, so the terminal voltage is always less than the EMF. This calculator computes EMF, not the loaded terminal voltage. For that you would also need the cell current and internal resistance: terminal voltage = EMF minus (current x internal resistance).
How do I find the number of electrons transferred (n)?
Write out the two balanced half-reactions and scale them so that the electrons gained at the cathode equal the electrons lost at the anode. The number n is that common electron count. For the Daniell cell: Zn -> Zn2+ + 2e- (anode) and Cu2+ + 2e- -> Cu (cathode), so n = 2. For a reaction such as Fe + Au3+ -> Fe3+ + Au, you would balance to 3 electrons each, giving n = 3.
Can I use this calculator for batteries and fuel cells?
Yes. Any electrochemical device that converts chemical energy to electrical energy follows the same formula. The standard potentials of common battery couples - such as Zn/MnO2 in alkaline cells (about 1.5 V) or Li/CoO2 in lithium-ion cells (about 3.6 V) - can be entered directly. Fuel cell operation is also described by the Nernst equation when partial pressures of hydrogen and oxygen are specified.