19.6 Cell Potential, Delta G, and the Equilibrium Constant
PRE-HEALTH PREP
Calculating E°cell, ΔG° & Keq
Electrochemistry Calculations Solved
General Chemistry 19.6
The standard cell potential (E°cell), Gibbs free energy (ΔG°), and equilibrium constant (Keq) are mathematically interdependent quantities that each describe the spontaneity and extent of a redox reaction from a different perspective. Two key equations—ΔG° = −nFE°cell and E°cell = (RT/nF) ln Keq—allow calculation of any unknown quantity from a single known value, with careful attention to units and the number of electrons transferred. A complete worked example using the zinc–copper galvanic cell demonstrates each calculation step, from standard reduction potentials through to an equilibrium constant of 1.5 × 1037.
Table of Contents
How E°cell, ΔG°, and Keq Are Related in Electrochemistry
The standard cell potential (E°cell), standard Gibbs free energy (ΔG°), and the equilibrium constant (Keq) are all mathematically interdependent. If you know any one of these three quantities, you can calculate the other two. This is one of the most powerful relationships in electrochemistry, and you can absolutely expect to be tested on it.
Why this matters: These three quantities each describe whether a reaction favors products or reactants—just from different perspectives. E°cell measures the electrical driving force, ΔG° measures the thermodynamic favorability, and Keq describes the equilibrium position. Connecting them gives you a complete picture of any redox reaction.
How to Calculate ΔG° from E°cell (ΔG° = −nFE°cell)
The relationship between Gibbs free energy and the standard cell potential is given by:
ΔG° = −nFE°cell
Where:
- n = number of moles of electrons transferred in the balanced reaction
- F = Faraday's constant = 96,500 C/mol e−
- E°cell = standard cell potential (in volts)
Notice how the negative sign in this equation enforces the consistency between the two spontaneity criteria you already know. A positive E°cell means a spontaneous reaction. Because F is always positive and n is always a positive number of moles, the negative sign ensures that a positive E°cell produces a negative ΔG°—exactly what we expect for a spontaneous process.
Spontaneity check: If E°cell > 0, the reaction is spontaneous. If ΔG° < 0, the reaction is spontaneous. This equation guarantees these two criteria always agree.
| SPONTANEOUS REACTIONS — Three Equivalent Criteria | ||
|---|---|---|
| ΔG° | E°cell | Keq vs. Q |
| ΔG° < 0 | E°cell > 0 | Q < K (forward reaction favored) |
| ΔG° = 0 | E°cell = 0 | Q = K (at equilibrium) |
| ΔG° > 0 | E°cell < 0 | Q > K (reverse reaction favored) |
How to Calculate Keq from E°cell Using 3 Variables
Back in the thermodynamics chapter, you saw the relationship ΔG° = −RT ln Keq. This connects Gibbs free energy directly to the equilibrium constant. In a similar fashion, we can derive a direct link between E°cell and Keq.
Recall that the Nernst equation relates the non-standard cell potential to the standard value. At equilibrium, the cell potential under non-standard conditions drops to zero and the reaction quotient Q equals Keq. Applying these conditions yields:
E°cell = (RT / nF) × ln Keq
Where:
- R = gas constant = 8.314 J/(mol·K)
- T = temperature in kelvins
- n = moles of electrons transferred
- F = Faraday's constant = 96,500 C/mol e−
Now you have 2 bridge equations. The first relates ΔG° to Keq, and the second relates E°cell to Keq. Together with the ΔG°–E°cell relationship, any one of the three quantities unlocks the other two.
Temperature matters: Both the ΔG°–Keq and E°cell–Keq relationships depend on temperature. Most problems assume standard conditions at 298 K unless stated otherwise.
| Known Quantity | Find ΔG° | Find E°cell | Find Keq |
|---|---|---|---|
| E°cell | ΔG° = −nFE°cell | — | ln K = nFE°cell / RT |
| ΔG° | — | E°cell = −ΔG° / nF | ΔG° = −RT ln K |
| Keq | ΔG° = −RT ln K | E°cell = (RT/nF) ln K | — |
Worked Example: Solving E°cell, ΔG°, and Keq for the Zinc–Copper Cell
Let's put all three electrochemistry equations to work with a classic galvanic cell. We will start by calculating E°cell from a table of reduction potentials, then use that value to find ΔG° and Keq.
Step 1 — Calculate E°cell from Standard Reduction Potentials
Consider the zinc–copper galvanic cell. The relevant standard reduction potentials are:
| Half-Reaction (Reduction) | E° (V) |
|---|---|
| Cu2+ + 2e− → Cu | +0.34 |
| Zn2+ + 2e− → Zn | −0.76 |
In this cell, zinc is oxidized (it loses electrons), so its half-reaction runs in reverse. Copper(II) is reduced (it gains electrons), so its half-reaction proceeds as written. There are 2 equivalent ways to compute the standard cell potential:
Method 1 — Flip the anode sign:
Reverse the zinc reduction potential to get an oxidation potential of +0.76 V. Then add:
E°cell = +0.34 V + 0.76 V = +1.10 V
Method 2 — Cathode minus anode:
E°cell = E°cathode − E°anode = 0.34 − (−0.76) = +1.10 V
Both approaches give the same result: E°cell = +1.10 V. The positive value confirms that this reaction is spontaneous under standard conditions.
| Standard Reduction Potentials Reference Table | |
|---|---|
| Half-Reaction (Reduction) | E° (V) |
| Al3+ + 3e- → Al | −1.66 |
| Mn2+ + 2e- → Mn | −1.18 |
| Zn2+ + 2e- → Zn | −0.76 |
| Cr3+ + 3e- → Cr | −0.74 |
| Fe2+ + 2e- → Fe | −0.44 |
| Co2+ + 2e- → Co | −0.28 |
| Ni2+ + 2e- → Ni | −0.25 |
| 2H+ + 2e- → H2 | 0.00 |
| Cu2+ + 2e- → Cu | +0.34 |
| Ag+ + 1e- → Ag | +0.80 |
| Br2 + 2e- → 2Br- | +1.07 |
Step 2 — Calculate ΔG° from Standard Cell Potential
Now we apply the Gibbs free energy equation. First, identify n: whether you look at copper gaining two electrons or zinc losing two electrons, this is a two-electron transfer (n = 2).
Calculation:
ΔG° = −nFE°cell
ΔG° = −(2)(96,500 C/mol)(1.10 V)
Since 1 volt = 1 joule per coulomb, the units work out directly:
ΔG° = −212,300 J = −212.3 kJ
Watch your units: Using volts in this equation gives ΔG° in joules, because 1 V = 1 J/C. On multiple-choice exams, answer choices are often listed in kilojoules—don't forget to divide by 1,000.
The negative value of ΔG° confirms what the positive E°cell already told us: the reaction is spontaneous. The two criteria are in perfect agreement, just as the equation guarantees.
Step 3 — Calculate Keq from E°cell at 298 K
We could use ΔG° to find Keq through the thermodynamic relationship, but let's take the direct electrochemical route instead. At 298 K, the combination RT/F simplifies to 0.0257 V, giving us a convenient shortcut:
Simplified equation at 298 K:
E°cell = (0.0257 / n) × ln Keq
Rearranging to solve for Keq:
Calculation:
ln K = nE°cell / 0.0257
ln K = (2)(1.10) / 0.0257 = 85.6
To solve for K, take the exponential of both sides:
K = e85.6 = 1.5 × 1037
An equilibrium constant of 1.5 × 1037 is astronomically large, which means this reaction overwhelmingly favors the products at equilibrium. This makes intuitive sense—zinc readily displaces copper from solution, and the reverse reaction is essentially negligible.
Tip: Solving for a variable trapped inside a natural logarithm is the trickiest algebra in this calculation. Remember that the exponential function (ex) is the inverse of ln, so applying e to both sides eliminates the natural log.
Key Takeaways
- E°cell, ΔG°, and Keq are mathematically interdependent. Knowing any one of these three quantities allows you to calculate the other two.
- ΔG° = −nFE°cell connects Gibbs free energy to cell potential. A positive E°cell guarantees a negative ΔG°, confirming spontaneity from both perspectives.
- E°cell = (RT/nF) ln Keq links the standard cell potential directly to the equilibrium constant. At 298 K, the factor RT/F simplifies to 0.0257 V.
- n represents the moles of electrons transferred in the balanced redox reaction and must be identified from the half-reactions before plugging into any equation.
- Units require careful attention: using volts in the ΔG° equation yields joules (since 1 V = 1 J/C). Convert to kilojoules when comparing to tabulated thermodynamic data.
- A large positive E°cell corresponds to a very large Keq and a large negative ΔG°, all indicating a reaction that strongly favors product formation.