Cell EMF Calculator

• Choose Standard to compute E°cell from cathode and anode reduction potentials.
• Choose Nernst to compute E at nonstandard conditions from E°, n, Q, and T.
• Pick what to solve for, fill the other inputs, then click Calculate.
• Turn on ΔG to compute ΔG = −nFE (you’ll need n).

• Standard mode: E°cell = E°cathode − E°anode.
• Nernst mode: E = E° − (RT/nF) ln Q with SI units (K, J, C).
• Spontaneity: E > 0 spontaneous; E < 0 nonspontaneous.
• Free energy: ΔG = −nFE (reported in kJ/mol too).

Example 1 — Daniell cell (Zn/Cu)

Given E°(Cu²⁺/Cu)=0.34 V and E°(Zn²⁺/Zn)=−0.76 V. Find E°cell.

1. Cathode is Cu (higher reduction potential): E°cathode = 0.34 V.
2. Anode is Zn: E°anode = −0.76 V.
3. Compute: E°cell = 0.34 − (−0.76) = 1.10 V.

Example 2 — Ag/Cu cell (standard conditions)

Given E°(Ag⁺/Ag)=0.80 V and E°(Cu²⁺/Cu)=0.34 V. Find E°cell and determine spontaneity.

1. Identify the cathode (higher reduction potential): Ag is the cathode since 0.80 > 0.34.
2. So E°cathode = 0.80 V and E°anode = 0.34 V (use reduction potentials).
3. Compute: E°cell = E°cathode − E°anode = 0.80 − 0.34 = 0.46 V.
4. Spontaneity: since E°cell > 0, the reaction is spontaneous as written.

Tip: Don’t flip signs just because the anode is “oxidation” — the formula already accounts for that by subtracting the anode reduction potential.

Example 3 — Daniell cell (Nernst, nonstandard conditions)

For the Daniell cell, assume E° = 1.10 V, n = 2, Q = 10, and T = 25°C. Find E.

1. Start with the Nernst equation: E = E° − (RT/nF)\,lnQ.
2. Convert temperature to Kelvin: T = 25 + 273.15 = 298.15\ \(\text{K}\).
3. Compute the correction term: (RT/nF)\,lnQ = \(\frac{(8.314)(298.15)}{(2)(96485)}\)\(\ln\)(10).
4. Numerically, \(\ln\)(10)\(\approx\) 2.303, so the correction is about 0.0296\ \(\text{V}\).
5. Compute: E \(\approx\) 1.10 − 0.0296 = 1.07\ \(\text{V}\).
6. Interpretation: since Q > 1, the reaction is “product-heavy,” so E drops below E°.

Quick check: if Q = 1, then lnQ = 0 and E = E°.

Example 4 — Free energy from cell potential (ΔG = −nFE)

Suppose a cell has E = 1.07\ \(\text{V}\) at the current conditions, and the balanced redox reaction transfers n = 2 electrons. Find ΔG (per mole of reaction as written).

1. Use the relationship: ΔG = −nFE.
2. Plug in values (F = 96485\ \(\text{C/mol e⁻}\)): ΔG = −(2)(96485)(1.07).
3. Compute: ΔG \(\approx\) −206{,}000\ \(\text{J/mol}\).
4. Convert to kJ/mol: ΔG \(\approx\) −206\ \(\text{kJ/mol}\).
5. Interpretation: negative ΔG means the reaction is thermodynamically favorable (spontaneous) under these conditions.

Note: E must be in volts (not mV) when using F in C/mol. Your calculator handles unit conversion automatically.