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Standard Cell Potential Calculator

Chemistry

E°cell = E°(cathode) − E°(anode), with both written as standard reduction potentials. Pick two half-reactions to get the voltage, the balanced equation, ΔG°, K and the Nernst voltage; convert between E°, ΔG° and K; or combine half-reactions of one element through ΔG°.

Quick Examples:

Settings

The higher potential is the cathode (a galvanic cell).
Changes only the half-cell values shown. E°cell is a difference, so it never changes.

ΔS° and ΔH° from the cell's temperature coefficient. The field also appears when T ≠ 25 °C.

Half-cell 1

Cathode · reduction

Cu²⁺(aq) + 2e⁻ → Cu(s)

E° = +0.3419 V vs SHE

Half-cell 2

Anode · oxidation

Zn²⁺(aq) + 2e⁻ → Zn(s)

E° = −0.7618 V vs SHE

Reaction not in the table? Balance it with the Chemical Equation Balancer →

E° values: CRC Handbook of Chemistry and Physics, 'Electrochemical Series' (P. Vanýsek), 25 °C, 1 bar.

Result

Standard cell potential
E°cell = +1.1037 V
Spontaneous as written: a galvanic cell delivering up to 1.1037 V.

Cathode (reduction)

Cu²⁺(aq) + 2e⁻ → Cu(s)

E° = +0.3419 V

Anode (oxidation)

Zn(s) → Zn²⁺(aq) + 2e⁻

E° = −0.7618 V

Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s)

  • Cathode (reduction): Cu²⁺/Cu, E° = +0.3419 V
  • Anode (oxidation): Zn²⁺/Zn, E° = −0.7618 V
  • n = 2 electrons transferred
  • ΔG° = −212.98 kJ/mol (−50.904 kcal/mol), per mole of electrons: −F·E° = −106.49 kJ
  • K = 2.1 × 10³⁷ (log₁₀K = 37.31)
  • K for the overall equation as written above (activities relative to 1 M and 1 bar)
  • Each 1 mV in E° changes K by 8.1 % (n = 2)
  • dE°/dT = −0.109 mV/K
  • ΔS° = nF·(dE°/dT) = −21.0 J/(mol·K)
  • ΔH° = −219.2 kJ/mol
  • Thermodynamic efficiency limit ΔG°/ΔH° = 97.1 %
  • Temperature coefficients are derived from standard entropies (NBS tables) and assume ΔH° and ΔS° don't change with T.

Cell notation (anode on the left)

Zn(s) | Zn²⁺(aq) ‖ Cu²⁺(aq) | Cu(s)

Where the two half-reactions sit in the series

water is stable here (pH 0.00)−3.0−2.5−2.0−1.5−1.0−0.50.0+0.5+1.0+1.5+2.0+2.5+3.0SHE, 0 VAu³⁺/Au, +1.498 VHg²⁺/Hg, +0.851 VAg⁺/Ag, +0.7996 VHg₂²⁺/Hg, +0.7973 VCu⁺/Cu, +0.521 VCu²⁺/Cu, +0.3419 VFe³⁺/Fe, −0.037 VPb²⁺/Pb, −0.1262 VSn²⁺/Sn, −0.1375 VNi²⁺/Ni, −0.257 VCo²⁺/Co, −0.28 VCd²⁺/Cd, −0.4030 VFe²⁺/Fe, −0.447 VCr³⁺/Cr, −0.744 VZn²⁺/Zn, −0.7618 VMn²⁺/Mn, −1.185 VAl³⁺/Al, −1.662 VMg²⁺/Mg, −2.372 VNa⁺/Na, −2.71 VCa²⁺/Ca, −2.868 VBa²⁺/Ba, −2.912 VK⁺/K, −2.931 VLi⁺/Li, −3.0401 VF₂/F⁻, +2.866 VCl₂/Cl⁻, +1.35827 VBr₂/Br⁻, +1.066 VI₂/I⁻, +0.5355 VO₃/O₂, +2.076 VH₂O₂/H₂O, +1.776 VO₂/H₂O, +1.229 VO₂/H₂O₂, +0.695 VO₂/OH⁻, +0.401 VH⁺/H₂ (SHE), 0 V (exact)H₂O/H₂, −0.8277 VS₂O₈²⁻/SO₄²⁻, +2.010 VCe⁴⁺/Ce³⁺, +1.72 VHg²⁺/Hg₂²⁺, +0.920 VFe³⁺/Fe²⁺, +0.771 V[Fe(CN)₆]³⁻/[Fe(CN)₆]⁴⁻, +0.358 VCu²⁺/Cu⁺, +0.153 VSn⁴⁺/Sn²⁺, +0.151 VMnO₄⁻/Mn²⁺, +1.507 VCr₂O₇²⁻/Cr³⁺, +1.232 VMnO₂/Mn²⁺, +1.224 VNO₃⁻/NO, +0.957 VHg₂Cl₂/Hg (calomel, 1 M Cl⁻), +0.26808 VAgCl/Ag, +0.22233 VAgBr/Ag, +0.07133 VAgI/Ag, −0.15224 VPbO₂/PbSO₄, +1.6913 VAg₂O/Ag, +0.342 VPbSO₄/Pb, −0.3588 VCd(OH)₂/Cd, −0.809 VZnO/Zn, −1.260 VCu²⁺/Cu: E° = +0.3419 V (cathode)Cu²⁺/Cu · +0.3419 V · cathodeZn²⁺/Zn: E° = −0.7618 V (anode)Zn²⁺/Zn · −0.7618 V · anodeE°cell =1.1037 Ve⁻ flow in the wire↑ stronger oxidizing agents (left-hand species)V vs SHE↓ stronger reducing agents (right-hand species)

Grey ticks are every couple in the table (hover or tap for its E°). Filled markers are E°. The shaded band is where water itself is neither oxidized nor reduced, which is why Li, Na and F₂ can't work in water.

Cell diagram

V1.1037 Ve⁻e⁻salt bridge (KNO₃)NO₃⁻ → anodeK⁺ → cathodeZn(s)Zn²⁺ (1 M)Anode (−) · oxidationZn²⁺/ZnCu(s)Cu²⁺ (1 M)Cathode (+) · reductionCu²⁺/Cu

Zn(s) | Zn²⁺(aq) ‖ Cu²⁺(aq) | Cu(s)

E against log₁₀Q

E (V)

shaded above 0: forward spontaneous · below: reverse spontaneous

Q (log scale)

Each tenfold rise in Q lowers E by 29.6 mV (n = 2, 25.00 °C). E reaches 0 only at Q = K = 2.1 × 10³⁷, far beyond any real concentration, which is why the cell holds its voltage until a reactant runs out.

Step-by-Step Solution

Cathode (reduction): Cu²⁺(aq) + 2e⁻ → Cu(s) E° = +0.3419 V

Anode (oxidation): Zn(s) → Zn²⁺(aq) + 2e⁻ E° = −0.7618 V (its reduction potential)

E°cell = E°cathode − E°anode = 0.3419 − (−0.7618) = +1.1037 V

(equivalently E°red(cathode) + E°ox(anode) = 0.3419 + 0.7618)

n = lcm(2, 2) = 2, so neither half-reaction is scaled

Overall: Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s)

ΔG° = −nFE° = −2 × 96485.33212 × 1.1037 = −212 982 J/mol = −212.98 kJ/mol

log₁₀K = nE° / (2.302585·RT/F) = 2 × 1.1037 / 0.0591593 = 37.31 → K = 2.1 × 10³⁷

ΔS° = nF·(dE°/dT) = 2 × 96485.33212 × (−0.0001087) = −21.0 J/(mol·K)

ΔH° = −nF·[E°(25 °C) − 298.15·dE°/dT] = −2 × 96485.33212 × (1.1037 − 298.15 × (−0.0001087)) = −219.2 kJ/mol

About This Tool

Standard Cell Potential Calculator – From Half-Reactions to Volts, ΔG° and K

A galvanic cell pairs two half-reactions: reduction at the cathode and oxidation at the anode. This calculator takes any two couples from a table of 54 standard reduction potentials, or your own values. It works out which electrode is which, balances the electrons and writes the overall equation and cell notation. It then gives E°cell, ΔG°, the equilibrium constant K and, at real concentrations, the Nernst equation voltage.

Why every potential needs a reference

You can only measure a voltage between two electrodes. Chemists therefore define the standard hydrogen electrode (SHE), H⁺(aq, 1 M) | H₂(g, 1 bar) | Pt, as exactly 0 V and quote every other couple against it. Labs often use a calomel (SCE, +0.241 V) or Ag/AgCl (+0.197 V) reference instead. Switching the reference moves both electrodes by the same amount, so E°cell never changes. The "Potentials shown vs" setting lets you check this.

Reading the electrochemical series

In the series, strong oxidizing agents (F₂, MnO₄⁻) sit at the top and strong reducing agents (Li, K) at the bottom. In a galvanic cell the higher couple is the cathode. The ladder chart shows the gap between the two couples, which is the cell voltage, and the band of potentials in which water itself is stable.

The Daniell cell, step by step

For copper and zinc, E°cell = E°cathode − E°anode = 0.3419 − (−0.7618) = +1.1037 V. Both values are reduction potentials. The usual mistake is to flip the sign of the anode's potential and subtract it, which counts the reversal twice. The cell notation is Zn(s) | Zn²⁺(aq) ‖ Cu²⁺(aq) | Cu(s), with the anode on the left. To check oxidation states in a less familiar couple, use the Oxidation Number Calculator.

Why E° is never multiplied

A volt is a joule per coulomb, so E° is intensive. When Ag⁺ + e⁻ → Ag is doubled to match copper's two electrons, its E° stays +0.7996 V. What doubles is ΔG° = −nFE°, through n.

E°, ΔG° and K

Because log₁₀K = nE°/(0.05916 V) at 25 °C, a small voltage means a huge K. The Daniell cell's K is about 2 × 10³⁷, so the reaction goes to completion, and each millivolt of E° moves K by about 8 % when n = 2. The second tab converts directly between E°, ΔG° and K. For K from measured equilibrium amounts, use the Equilibrium Constant Calculator.

The Nernst equation in practice

Concentration enters only through a logarithm, so a battery's voltage barely sags until a reactant runs out. A concentration cell, the same couple at two concentrations, runs on the logarithm alone. pH can still flip the answer: Q for permanganate oxidizing chloride contains [H⁺]¹⁶. The reaction is spontaneous at pH 0, but by pH 3 E has fallen to −0.135 V.

Temperature coefficients and thermodynamics

The slope of E° against temperature gives the entropy change, ΔS° = nF·(dE°/dT), and ΔH° then follows. For the hydrogen fuel cell, ΔG°/ΔH° caps the efficiency at about 83 %. The table's coefficients are derived from standard entropies, like those in the Entropy Change Calculator.

Combining half-reactions

Potentials of consecutive steps don't add, but their ΔG° values do. The third tab weights each E° by its electrons, draws Latimer and Frost diagrams, and flags species that disproportionate, such as Cu⁺ and H₂O₂.

The formulas

E°cell = E°(cathode) − E°(anode)
ΔG° = −n·F·E°
log₁₀K = n·F·E° / (R·T·ln 10)
E = E° − (R·T / n·F)·ln Q
E°(combined) = Σ eᵢ·E°ᵢ / Σ eᵢ
Limits of the model
The Nernst equation here uses concentration/1 M as the activity, which drifts above about 0.01 M (see the Ionic Strength Calculator). E° is thermodynamic, and real electrolysis needs extra voltage (overpotential). Some couples, such as Ce⁴⁺/Ce³⁺, have formal potentials that depend on the acid. Lithium cells use non-aqueous electrolytes.

Frequently Asked Questions

Is the Standard Cell Potential Calculator free?

Yes, Standard Cell Potential Calculator is totally free :)

Can I use the Standard Cell Potential Calculator offline?

Yes, you can install the webapp as PWA.

Is it safe to use Standard Cell Potential Calculator?

Yes, any data related to Standard Cell Potential Calculator only stored in your browser (if storage required). You can simply clear browser cache to clear all the stored data. We do not store any data on server.

How do I calculate the standard cell potential from two half-reactions?

Write both as reductions and look up their standard reduction potentials. The one with the higher E° is the cathode, and E°cell = E°(cathode) − E°(anode). For the Daniell cell that is 0.3419 − (−0.7618) = +1.1037 V. Pick the two couples here and the tool also balances the electrons and writes the overall equation.

Why don't I multiply E° when I multiply a half-reaction?

A potential is energy per coulomb, so it is intensive: doubling the reaction doubles the charge and the energy together. ΔG° = −nFE° is what doubles, through n. That is why Ag⁺ + e⁻ → Ag keeps +0.7996 V when it is multiplied by 2 to pair with copper.

How are E°cell, ΔG° and the equilibrium constant related?

ΔG° = −nFE° and log₁₀K = nE°/(0.05916 V) at 25 °C, so a positive E° means a negative ΔG° and K above 1. Because K depends exponentially on E°, a 1.10 V cell with n = 2 has K ≈ 10³⁷, and each millivolt of E° changes K by about 8 %.

What does the Nernst equation tell me, and why does a battery's voltage drop?

E = E° − (RT/nF)·ln Q gives the voltage at real concentrations. Concentrations enter through a logarithm, so a Daniell cell only loses about 30 mV for every tenfold change in Q. The voltage stays nearly flat and then collapses as a reactant runs out and Q climbs steeply.

Which electrode is the anode, and is it positive or negative?

The anode is where oxidation happens and the cathode is where reduction happens, in every cell. In a galvanic cell, which produces current, the anode is negative. In an electrolytic cell, driven by an external source, the anode is positive. Electrons always flow through the wire from the anode to the cathode.

Why can't I add half-reaction potentials to get a new one?

When the electrons don't cancel, the potentials aren't additive; the ΔG° values are. Fe³⁺ → Fe²⁺ (+0.771 V, 1 e⁻) and Fe²⁺ → Fe (−0.447 V, 2 e⁻) combine to (0.771 − 0.894)/3 = −0.041 V for Fe³⁺ → Fe, not 0.324 V. The Combine half-reactions tab does this weighting for you.