Equilibrium constant from E° - ln K = nFE° - RT

What This Relationship Tells Us

The equilibrium constant K measures the ratio of products to reactants when a reaction stops changing. The standard cell potential E° measures the voltage when all species are at 1 M, 1 bar, 25°C. These seem like different worlds, but they're connected through the Gibbs free energy ΔG°:

$\Delta G° = -nFE° = -RT \ln K$

where:

n = number of electrons transferred per reaction unit (moles of e⁻)
F = Faraday constant ≈ 96,485 C/mol
R = gas constant = 8.314 J/(mol·K)
T = absolute temperature (Kelvin)

Derivation from First Principles

Why do voltage and equilibrium connect? Both describe the same energy landscape—just in different units.

Step 1: Gibbs Free Energy as the Master Function

For any process, the change in Gibbs free energy ΔG determines spontaneity:

ΔG < 0 → spontaneous (forward)
ΔG = 0 → equilibrium (no net change)
ΔG > 0 → non-spontaneous (reverse favored)

At equilibrium, ΔG = 0, which defines the equilibrium constant:

$\Delta G = \Delta G° + RT \ln Q$

At equilibrium, Q = K and ΔG = 0:

$0 = \Delta G° + RT \ln K$

$\Delta G° = -RT \ln K \quad \text{..(1)}$

Why this step? We're expressing equilibrium in thermodynamic language. ΔG° is the "standard energy price" of the reaction.

Step 2: Connecting Electrical Work to Free Energy

In an electrochemical cell, chemical energy converts to electrical work. The maximum electrical work for n moles of electrons moving through potential difference E° is:

$w_{\text{elec}} = -nFE°$

Why negative? Work done by the system on surroundings is negative in thermodynamics. A spontaneous cell does work, so w< 0.

At constant T and P, maximum work equals the free energy change:

$\Delta G° = -nFE° \quad \text{...(2)}$

Why this step? This is the electrochemistry-thermodynamics dictionary: voltage (E°) translates to energy per coulomb, and nF gives total charge moved.

Step 3: Equating and Solving

From equations (1) and (2):

$-RT \ln K = -nFE°$

$RT \ln K = nFE°$

$\boxed{\ln K = \frac{nFE°}{RT}}$

Why this matters: Now we can predict K from a measurable voltage, or calculate E° from equilibrium data.

Figure — Equilibrium constant from E° - ln K = nFE° - RT

Worked Examples

Worked example Example 1: Silver-Zinc Cell

Given: $\text{Zn}(s) + 2\text{Ag}^+(aq) \to \text{Zn}^{2+}(aq) + 2\text{Ag}(s)$ $E°_{\text{cell}} = E°_{\text{Ag}^+/\text{Ag}} - E°_{\text{Zn}^{2+}/\text{Zn}} = 0.80 - (-0.76) = 1.56 \text{ V}$

Find: Equilibrium constant K at 25°C.

Solution:

Step 1: Identify n (electrons transferred)

Zn → Zn²⁺ + 2e⁻
2Ag⁺ + 2e⁻ → 2Ag
==n = 2==

Why this step? The number of electrons must balance in the overall reaction. Each Ag⁺ takes one electron, two Ag⁺ need two electrons, which one Zn atom provides.

Step 2: Apply ln K = nFE°/RT at 298 K $\ln K = \frac{2 \times 96485 \times 1.56}{8.314 \times 298.15}$

$\ln K = \frac{301,033.2}{2,478.9} = 121.4$

Why this step? Direct substitution into our derived formula. The huge numerator (charge × voltage) vs. small denominator (thermal energy) signals a massive K.

Step 3: Solve for K

$K = e^{121.4} \approx 10^{52.7}$

Interpretation: K is astronomically large—this reaction goes essentially to completion. At equilibrium, [Zn²⁺]/[Ag⁺]² ≈ 10⁵³. The positive E° (1.56 V) means the reaction releases energy, so products dominate.

Worked example Example 2: Reverse Calculation (K → E°)

Given: $\text{Cu}^{2+} + \text{Cd} \rightleftharpoons \text{Cu} + \text{Cd}^{2+}$ has K = 1.2 × 10⁷ at 25°C.

Find: E°cell

Solution: Step 1: Identify n

Cu²⁺ + 2e⁻ → Cu
Cd → Cd²⁺ + 2e⁻
==n = 2== Step 2: Rearrange ln K = nFE°/RT

$E° = \frac{RT \ln K}{nF}$

Why this step? We're solving for the unknown E° instead of K. Algebra reversal.

Step 3: Substitute values

$\ln K = \ln(1.2 \times 10^7) = 16.30$

$E° = \frac{8.314 \times 298.15 \times 16.30}{2 \times 96485}$

$E° = \frac{40,422}{192,970} = 0.209 \text{ V}$

Interpretation: K > 1 means products favored, so E° must be positive. 0.21 V corresponds to a moderately favorable reaction.

Worked example Example 3: Unfavorable Reaction

Given: $\text{Fe}^{3+} + \text{Ag} \rightleftharpoons \text{Fe}^{2+} + \text{Ag}^+$

$E°_{\text{cell}} = E°_{\text{Fe}^{3+}/\text{Fe}^{2+}} - E°_{\text{Ag}^+/\text{Ag}} = 0.77 - 0.80 = -0.03 \text{ V}$

Find: K at 25°C Solution:

Step 1: n = 1 (one electron transferred) Step 2: Use the shortcut at 298 K

$\log K = \frac{nE°}{0.0592} = \frac{1 \times (-0.03)}{0.0592} = -0.507$

Why this step? The shortcut saves calculation time for standard conditions.

Step 3: Solve for K $K = 10^{-0.507} \approx 0.31$

Interpretation: K < 1 means reactants are favored at equilibrium. The negative E° tells us this reaction is "uphill"—you'd need to input energy to drive it forward. At equilibrium, [Fe²⁺][Ag⁺]/[Fe³⁺] ≈ 0.31, so mostly unreacted Fe³⁺ remains.

Common mistake Common Error: Forgetting n in Multi-Electron Transfers

Wrong thinking: "The formula is ln K = FE°/RT, just plug in E°."

Why it feels right: Students memorize formulas without tracking units. F has units C/mol, E° is J/C, so FE° is J/mol—looks like energy.

The steel-man: Actually, FE° is energy per mole of electrons. But the reaction might transfer 2 or 3 electrons per formula unit. If you forget n, you calculate K for a different reaction stoichiometry.

The fix: Always count electrons in the balanced half-reactions. For $\text{Zn} + \text{Cu}^{2+} \to \text{Zn}^{2+} + \text{Cu}$ , n = 2 (not 1). If you use n = 1, you get K¹/² instead of K.

Check yourself: If E° = 1.1 V and you calculate K, then double n—does K roughly square? It should. ln K scales linearly with n.

Common mistake Common Error: Using Non-Standard Concentrations for E°

Wrong thinking: "My cell has [Cu²⁺] = 0.5 M, so I'll use that in ln K = nFE/RT."

Why it feels right: The Nernst equation uses actual concentrations, so students mix it up with the equilibrium formula.

The steel-man: You are connecting concentrations and voltage, but the equilibrium formula specifically relates the standard potential E° (all 1 M) to K. If you use a non-standard E measured at arbitrary concentrations, the formula breaks.

The fix: ln K = nFE°/RT requires E° (standard conditions). If you have a measured Ecell at non-standard conditions, first use Nernst to find E°:

$E_{\text{cell}} = E° - \frac{RT}{nF} \ln Q$ Solve for E°, then calculate K.

Analogy: E° is like a "sea level" reference. You can't measure K by standing on a mountain (non-standard conditions) and using that altitude directly.

Physical Intuition: Why Voltage Predicts Equilibrium

Think of E° as a measure of the reaction's driving force. Positive E° means electrons "want" to flow from anode to cathode—energy releases. At equilibrium, the forward and reverse reactions balance, but the ratio of products to reactants reflects how far the reaction traveled before balancing.

Large positive E° → huge energy release → reaction runs almost to completion → K ≫ 1
E° ≈ 0 → barely any driving force → products and reactants similar at equilibrium → K ≈ 1
Negative E° → energy required to push forward → reverse direction favored → K < 1

The formula ln K = nFE°/RT is just an equation, but physically: the bigger the voltage, the exponentially larger K becomes. Each 0.059 V increase at 25°C (for n = 1) multiplies K by 10.

Recall Explain to a 12-Year-Old

Imagine a water slide. The height of the slide (like E°) tells you how much energy you get from sliding down. If the slide is super tall, you zoom down really fast and end up far from where you started—that's like a big K (lots of products). If the slide is flat, you barely move—small K (mostly reactants left). And if the slide goes uphill, you can't slide at all without someone pushing you—that's negative E°, K less than 1.

The formula ln K = nFE°/RT is like a calculator that takes the slide's height (E°) and tells you exactly where you'll end up (K). The n is how many friends ride down together (more friends = more total energy), and RT is like friction—higher temperature = more random jiggling = slightly changes where you stop.

Connections

Nernst Equation – describes E at non-standard conditions; at equilibrium, E = 0 and Q = K
Gibbs Free Energy and Spontaneity – ΔG° = -nFE° = -RT ln K unifies all three
Standard Reduction Potentials – E°cell comes from E°cathode - E°anode tables
Le Chatelier's Principle – large K means adding reactants barely shifts equilibrium
Thermodynamics of Electrochemical Cells – relates heat, work, and voltage
Faraday's Laws of Electrolysis – connects charge (nF) to moles of substance
Arrhenius Equation – similar exponential form: k = Ae^(-Ea/RT)

#flashcards/chemistry

What does a positive E° value tell you about K?

K > 1 (products favored at equilibrium). Positive E° means ΔG° < 0, so ln K > 0, thus K > 1.

Derive the relationship between ln K and E° from ΔG°.

ΔG° = -RT ln K (equilibrium condition) and ΔG° = -nFE° (electrochemical work). Equating: -RT ln K = -nFE°, so ln K = nFE°/RT.

For the reaction 2Fe³⁺ + Cu → 2Fe²⁺ + Cu²⁺ with E° = -0.43 V and n = 2, estimate K at 25°C.

log K = nE°/0.0592 = (2)(-0.43)/0.0592 ≈ -14.5, so K ≈ 10⁻¹⁴·⁵ ≈ 3 × 10⁻¹⁵ (reactants strongly favored).

Why does the formula include n (moles of electrons)?

Because nF gives the total charge transferred per mole of reaction. More electrons transferred = more electrical work = larger impact on ΔG° and thus K.

At 25°C, increasing E° by 0.059 V (with n=1) multiplies K by what factor?

10×. Because log K = E°/0.0592, adding 0.059 V to E° adds ~1 to log K, which multiplies K by 10.

If K = 1, what is E°?

E° = 0 V. When K = 1, ln K = 0, so nFE°/RT = 0, thus E° = 0.

What does the T in the denominator (RT) represent physically?

Thermal energy. Higher T means more random molecular motion, which "smears out" the equilibrium distribution, reducing the ln K impact of a given E°.

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Socho ek electrochemical cell hai jaise battery. Standard cell potential E° bata hai ki electrons kitne josh se flow karna chahte hain—positive E° matlab downhill flow, energy release hota hai. Lekin jab reaction equilibrium pe pahunchti hai (yani forward aur backward rate equal ho jate hain), tab voltage zero ho jaata hai, koi net current nahi. Par equilibrium pe products aur reactants ka ratio fixed hota hai—wo ratio hi equilibrium constant K hai.

Formula ln K = nFE°/RT ye bridge banata hai: E° (voltage) seedhe K (equilibrium ka number) nikaal sakte ho. Agar E° bada positive hai, matlab reaction bahut zy

Equilibrium constant from E° - ln K = nFE° - RT

What This Relationship Tells Us

Derivation from First Principles

Step 1: Gibbs Free Energy as the Master Function

Step 2: Connecting Electrical Work to Free Energy

Step 3: Equating and Solving

Worked Examples

Physical Intuition: Why Voltage Predicts Equilibrium

Connections

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — Redox & Electrochemistry (Intro)

Connections