Standard Cell Potential Calculator
Enter the standard reduction potentials of the cathode and anode to get the standard cell potential E°cell, the spontaneity verdict, and the free energy change ΔG° per mole of electrons.
Find the Cell Voltage from Two Half-Cell Potentials
Standard Cell Potential Formula
Where:
- \(E°_{cathode}\) — standard reduction potential of the half-reaction occurring at the cathode (V)
- \(E°_{anode}\) — standard reduction potential of the half-reaction occurring at the anode (V)
- \(n\) — number of electrons transferred in the balanced cell reaction
- \(F\) — Faraday constant, 96 485 C/mol
- \(\Delta G°\) — standard free energy change (J/mol, or kJ/mol when divided by 1000)
The first formula gives the voltage of the cell from two tabulated standard reduction potentials; the second links that voltage to the thermodynamic driving force of the cell reaction.
Standard Cell Potential – Calculation Example
Example: a Daniell cell with a copper electrode in Cu²⁺ solution as the cathode (E° = +0.34 V) and a zinc electrode in Zn²⁺ solution as the anode (E° = −0.76 V). These are the calculator's default values.
\[ E°_{cell} = E°_{cathode} - E°_{anode} = 0.34\ \text{V} - (-0.76\ \text{V}) = 1.10\ \text{V} \]
The reaction transfers \(n = 2\) electrons (Cu²⁺ + 2e⁻ → Cu and Zn → Zn²⁺ + 2e⁻), so the free energy is:
\[ \Delta G° = -nFE°_{cell} = -2 \times 96\,485 \times 1.10 = -212\,267\ \text{J} \approx -212\ \text{kJ/mol} \]
Result: E°cell = 1.10 V; ΔG° ≈ −106.13 kJ per mole of electrons (−212.27 kJ per mole of reaction). The positive cell potential and negative free energy confirm the reaction is spontaneous under standard conditions.
How to Use the Calculator
- Enter the Cathode Potential (E°cat) — the standard reduction potential of the half-reaction where reduction occurs. For the default example this is +0.34 V (Cu²⁺ + 2e⁻ → Cu).
- Enter the Anode Potential (E°an) — the standard reduction potential of the half-reaction where oxidation occurs, entered with the sign found in tables. Do not flip the sign yourself; the subtraction in the formula handles that. For the default example this is −0.76 V (Zn²⁺ + 2e⁻ → Zn).
- Read the Standard Cell Potential in volts, along with the spontaneity verdict, the free energy per mole of electrons, and a step-by-step breakdown of the calculation.
- If the result is negative, the calculator tells you what the potential would be with the electrodes swapped — the half-cell with the higher reduction potential is always the cathode of a spontaneous cell.
What Is Standard Cell Potential?
The standard cell potential \(E°_{cell}\) is the maximum voltage an electrochemical cell can deliver when both half-cells operate under standard conditions: 25 °C (298.15 K), 1 M concentrations of all dissolved species, and 1 atm (1 bar) gas pressures. It measures the tendency of the overall cell reaction to move electrons from the anode to the cathode, and every standard potential is measured relative to the standard hydrogen electrode (SHE), which is assigned exactly 0 V by convention.
The key rule is simple: both potentials must be entered as standard reduction potentials taken directly from a table, with their original signs. The cell potential is then just the cathode value minus the anode value. Many students make the mistake of manually reversing the sign of the anode potential before using it — this double-counts the reversal, because subtracting a reduction potential is mathematically identical to adding its oxidation potential.
The sign of \(E°_{cell}\) tells you the direction of the reaction. A positive value means the reaction is spontaneous as written: the cell is galvanic (voltaic) and produces electrical work, with \(\Delta G° < 0\). A negative value means the reaction is non-spontaneous and would require an external power source — an electrolytic cell. If the value is exactly zero, the two half-cells are identical and the cell is at equilibrium.
Two properties are worth remembering. First, \(E°_{cell}\) is independent of stoichiometry: multiplying a half-reaction by any factor does not change its potential, because a potential is an intensive quantity (energy per charge), not an extensive one. Second, \(E°_{cell}\) relates to the equilibrium constant through \(\log K = nE°_{cell}/0.0592\) at 25 °C, so even modest voltages correspond to very large equilibrium constants — the Daniell cell's 1.10 V corresponds to \(K \approx 10^{37}\) for n = 2.
Common Standard Reduction Potentials (25 °C)
| Half-reaction | E° (V) |
|---|---|
| F₂ + 2e⁻ → 2F⁻ | +2.87 |
| MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O | +1.51 |
| Cl₂ + 2e⁻ → 2Cl⁻ | +1.36 |
| Ag⁺ + e⁻ → Ag | +0.80 |
| Fe³⁺ + e⁻ → Fe²⁺ | +0.77 |
| Cu²⁺ + 2e⁻ → Cu | +0.34 |
| 2H⁺ + 2e⁻ → H₂ (SHE) | 0.00 |
| Pb²⁺ + 2e⁻ → Pb | −0.13 |
| Ni²⁺ + 2e⁻ → Ni | −0.26 |
| Fe²⁺ + 2e⁻ → Fe | −0.44 |
| Zn²⁺ + 2e⁻ → Zn | −0.76 |
| 2H₂O + 2e⁻ → H₂ + 2OH⁻ | −0.83 |
| Al³⁺ + 3e⁻ → Al | −1.66 |
| Mg²⁺ + 2e⁻ → Mg | −2.37 |
| Li⁺ + e⁻ → Li | −3.04 |
Assumptions, Limitations and Tips
- Standard conditions only. The result applies at 25 °C, 1 M activities and 1 atm gas pressures. At other concentrations, the actual voltage follows the Nernst equation and can differ substantially from \(E°_{cell}\).
- Activities, not concentrations. In concentrated or ionic solutions, effective activities deviate from molar concentrations; the tabulated potentials assume ideal behaviour.
- Junction potentials neglected. Real cells include small liquid-junction voltages at the salt bridge (typically a few millivolts) that this calculation ignores.
- The free energy is given per mole of electrons. To obtain \(\Delta G°\) for the full balanced reaction, multiply the displayed value by the number of electrons transferred, \(n\).
- Signs matter. Always enter the tabulated signs. A common error is entering +0.76 V for the zinc anode, which would give −0.42 V instead of +1.10 V for the Daniell cell.
- Electrode roles are determined by the potentials. In any galvanic cell the half-reaction with the more positive (higher) reduction potential runs as reduction at the cathode; the other runs as oxidation at the anode.
Related Calculators
Electrochemistry pairs naturally with other thermodynamic tools: to combine reaction energies from tabulated values, see the Hess's Law calculator; for solubility-driven potentials of sparingly soluble salts, try the Solubility Product (Ksp) calculator; and to study temperature effects on equilibrium constants, use the Van 't Hoff Equation calculator.
Frequently Asked Questions
How do I calculate the standard cell potential by hand?
Look up both half-reactions in a table of standard reduction potentials and copy their signs exactly as printed. Then subtract: \(E°_{cell} = E°_{cathode} - E°_{anode}\). The cathode is whichever half-reaction has the more positive potential. Do not flip the anode's sign yourself — the subtraction already does that. For example, with Cu²⁺/Cu (+0.34 V) and Zn²⁺/Zn (−0.76 V): 0.34 − (−0.76) = 1.10 V.
What is a good or typical value for a standard cell potential?
Most aqueous cells fall between about 0.5 V and 2.5 V. A single Daniell cell gives 1.10 V, an alkaline cell about 1.5 V, and a lead–acid cell 2.05 V per cell. Values above roughly 2 V usually involve non-aqueous or molten-salt chemistry, because water itself gets oxidized or reduced above about ±1.2 V per electrode. Anything negative means the reaction is not spontaneous as written.
Can I enter potentials in millivolts instead of volts?
No — the calculator expects volts, the unit used in standard tables. To convert, divide millivolts by 1000: 770 mV = 0.770 V. A quick sanity check: standard reduction potentials in water rarely exceed ±3 V, so if your entered value is larger than 6 in magnitude, you probably typed millivolts by mistake.
Why does my measured battery voltage differ from the calculated E°cell?
Standard potentials assume 25 °C, exactly 1 M solutions and 1 atm gases. Real batteries run at different concentrations and temperatures, so the actual voltage follows the Nernst equation: \(E = E° - (0.0592/n)\log Q\) at 25 °C. As a battery discharges, Q grows and the voltage drops below E°. Internal resistance and junction potentials also lower the voltage you measure under load.
How do I find n, the number of electrons transferred?
Balance both half-reactions so that they exchange the same number of electrons; that number is n. For Zn + Cu²⁺ → Zn²⁺ + Cu, zinc gives up 2 electrons and copper accepts 2, so n = 2. The calculator reports ΔG° per mole of electrons; multiply that value by n to get the free energy per mole of reaction. Note that n never affects E°cell itself.
Can a cell be spontaneous if both half-reactions have negative reduction potentials?
Yes. What matters is only the difference. The half-reaction with the higher (less negative) potential becomes the cathode. For example, coupling Fe²⁺/Fe (−0.44 V) as cathode with Zn²⁺/Zn (−0.76 V) as anode gives E°cell = −0.44 − (−0.76) = +0.32 V, a spontaneous galvanic cell even though both tabulated values are negative.
How do I convert the result to an equilibrium constant or to kcal?
At 25 °C use \(\log K = nE°_{cell}/0.0592\), so even 1 V corresponds to a huge K. For energy units, 1 V per electron equals 96.485 kJ/mol; divide by 4.184 to get 23.06 kcal/mol. The Daniell cell's 1.10 V therefore corresponds to about 25.4 kcal per mole of electrons, or roughly 50.7 kcal for the full n = 2 reaction.