Equilibrium constant from E° - ln K = nFE° - RT: Worked examples

Before anything, let's re-anchor the three quantities so we never use a symbol we haven't earned.

The scenario matrix

Every problem this topic can throw is one (or a blend) of these cells. Each example below is tagged with the cell it hits.

Cell	What's special about it	Example
A. Big positive $E °$ , find $K$	huge $K$ , reaction "goes to completion"	Ex 1
B. Small negative $E °$ , find $K$	$K < 1$ , reactants win, sign discipline	Ex 2
C. Multi-electron ( $n > 1$ ), find $K$	$n$ multiplies the exponent — easy to drop	Ex 3
D. Reverse direction: given $K$ , find $E °$	algebra flipped	Ex 4
E. Degenerate: $E ° = 0$	the "tie" case, $K = 1$ exactly	Ex 5
F. Limiting behaviour ( $E ° \to$ large, or $n \to$ large)	how fast $K$ blows up	Ex 6 (figure)
G. Real-world word problem	strip a story down to $E °$ and $n$	Ex 7
H. Exam twist: non-standard $E$ given	must find $E °$ first via Nernst	Ex 8

We'll cover A → H in order. Every numeric answer is machine-checked at the bottom.

The worked examples

Worked example Example 1 — Cell A: big positive

$E °$ , find $K$ Statement. $Zn (s) + 2 Ag^{+} \to Zn^{2 +} + 2 Ag (s)$ , with $E °_{cell} = 1.56 V$ at 25 °C. Find $K$ .

Forecast: $E °$ is strongly positive → do you expect $K ≫ 1$ , $\approx 1$ , or $≪ 1$ ? Guess the order of magnitude of the exponent before reading on.

Count $n$ . Zn loses 2 electrons; two Ag⁺ each grab one → 2 total. So $n = 2$ . Why this step? $n$ sits in the numerator; getting it wrong scales the whole exponent.
Use the 25 °C shortcut. $lo g_{10} K = \frac{n E °}{0.0592} = \frac{2 \times 1.56}{0.0592} = 52.7$ Why this step? The shortcut already bakes in $F, R, T$ at 298 K, so we only feed it $n$ and $E °$ .
Read the answer. $K = 1 0^{52.7} \approx 5 \times 1 0^{52}$ . Why this step? $lo g_{10} K$ is the power of ten — no more work needed.

Verify: Cross-check with the natural-log form: $ln K = \frac{2 ( 96485 ) ( 1.56 )}{8.314 ( 298.15 )} = 121.4$ , and $121.4/2.303 = 52.7$ ✓. A near-completion reaction should give an astronomically large $K$ — it does. Sign of $E °$ positive ⇒ $K > 1$ ✓.

Worked example Example 3 — Cell C: multi-electron (

$n = 3$ ), find $K$ Statement. $Al (s) + 3 Ag^{+} \to Al^{3 +} + 3 Ag (s)$ . Using $E °_{Al^{3 +} / Al} = - 1.66 V$ and $E °_{Ag^{+} / Ag} = + 0.80 V$ , find $K$ .

Forecast: With $n = 3$ and a big $E °$ , this $K$ will be even more extreme than Example 1 — guess whether the exponent beats 100.

Get $E °_{cell}$ . $E °_{cell} = E °_{cathode} - E °_{anode} = 0.80 - (- 1.66) = 2.46 V$ . Why this step? Cell voltage is always the reduction happening at the cathode minus the reduction potential of the species being oxidised.
Count $n$ . Al loses 3 electrons; three Ag⁺ each take one → $n = 3$ . Why this step? Here $n > 1$ genuinely triples the exponent — the exact trap the parent's "forgetting $n$ " mistake warns about.
Shortcut. $lo g_{10} K = \frac{3 \times 2.46}{0.0592} = 124.7$ So $K \approx 1 0^{124.7}$ .

Verify: Compare to $n = 1$ hypothetical: $lo g_{10} K$ would be $2.46/0.0592 = 41.6$ . Our real answer is $3 \times$ that ( $124.7 = 3 \times 41.6$ ) ✓ — exactly the "double/triple $n$ ⇒ multiply the exponent" scaling the parent note predicts.

Worked example Example 6 — Cell F: limiting behaviour (how fast does

$K$ explode?) Statement. For a fixed $n = 1$ at 25 °C, tabulate $lo g_{10} K$ as $E °$ sweeps from $- 0.3 V$ to $+ 0.3 V$ , and describe the trend.

Forecast: Straight line or curve? Steep or shallow?

Write the relation as a line. $lo g_{10} K = \frac{1}{0.0592} E ° = 16.9 E °$ . Why this step? With $n$ fixed, $lo g_{10} K$ is a linear function of $E °$ with slope $16.9$ per volt — a straight line on log-axes.
Read the slope's meaning. Every extra $0.0592 V$ multiplies $K$ by 10. Why this step? Slope $16.9$ means $+ 1$ in $lo g_{10} K$ (a factor of ten in $K$ ) for each $1/16.9 = 0.0592$ V — the origin of that magic number.
Look at the picture.

Look at the violet line: it is dead straight. The orange point at $E ° = 0$ sits on $lo g_{10} K = 0$ (the tie from Example 5). Crossing into positive $E °$ (magenta region) sends $lo g_{10} K$ soaring; negative $E °$ drives it deeply negative. Limiting fact: even a modest $E °$ of a few tenths of a volt gives astronomically large or small $K$ — voltage is an exquisitely sensitive thermometer for equilibrium.

Verify: At $E ° = + 0.3$ : $lo g_{10} K = 16.9 (0.3) = 5.07$ ; at $E ° = - 0.3$ : $- 5.07$ — symmetric about the origin, as a line through $(0, 0)$ must be ✓.

Worked example Example 7 — Cell G: real-world word problem

Statement. A chemist reads in a data table that the reaction $Cu^{2 +} + Zn (s) \to Cu (s) + Zn^{2 +}$ "goes essentially to completion." Their measured $E °_{cell} = 1.10 V$ . They want a number for the sales brochure: "products outnumber reactants by a factor of about ____." $n = 2$ .

Forecast: "Essentially complete" — should the brochure number have a few digits or dozens of digits?

Strip the story to $E °$ and $n$ . All the "goes to completion" language is flavour; only $E ° = 1.10 V$ and $n = 2$ matter. Why this step? Word problems bury two numbers in prose — find $E °$ and $n$ , ignore the rest.
Shortcut. $lo g_{10} K = \frac{2 \times 1.10}{0.0592} = 37.2$ Why this step? $lo g_{10} K$ answers "factor of how many powers of ten?" — exactly the brochure's question.
State it in words. $K \approx 1 0^{37}$ — products outnumber reactants by roughly $1 0^{37}$ to 1.

Verify: Sanity: this is the classic Daniell cell; textbooks quote $K \sim 1 0^{37}$ for it ✓. "Essentially to completion" (dozens of digits) confirmed. See Le Chatelier's Principle for why such a lopsided $K$ resists being pushed back.

Worked example Example 8 — Cell H: the exam twist (non-standard

$E$ given) Statement. An exam gives you a measured $E_{cell} = 1.00 V$ for $Zn + Cu^{2 +} \to Zn^{2 +} + Cu$ when $[Zn^{2 +}] = 1.0 M$ and $[Cu^{2 +}] = 0.010 M$ , $n = 2$ , 25 °C. Find $K$ .

Forecast: Trap alert — can you plug $1.00 V$ straight into $lo g K = n E /0.0592$ ? (No — and this example shows why.)

Spot the trap. $E_{cell} = 1.00 V$ is a non-standard voltage (concentrations aren't all 1 M). The $K$ -formula demands the standard $E °$ . Why this step? This is the parent note's "using non-standard concentrations" mistake — the whole point of the twist.
Recover $E °$ with the Nernst equation. $E_{cell} = E ° - \frac{0.0592}{n} lo g_{10} Q$ , with $Q = \frac{[ Zn ^{2 +} ]}{[ Cu ^{2 +} ]} = \frac{1.0}{0.010} = 100$ . $1.00 = E ° - \frac{0.0592}{2} lo g_{10} (100) = E ° - 0.0296 (2) = E ° - 0.0592$ $E ° = 1.00 + 0.0592 = 1.059 V$ Why this step? The Nernst Equation converts a real-world voltage back to the standard-conditions reference — the "sea-level" the parent note describes.
Now use the $K$ -formula with the true $E °$ . $lo g_{10} K = \frac{2 \times 1.059}{0.0592} = 35.8 \Rightarrow K \approx 1 0^{35.8}$

Verify: Had we (wrongly) used $1.00 V$ : $lo g_{10} K = 33.8$ — off by two full orders of magnitude, precisely the size of the $lo g Q$ correction we added ✓. So the Nernst step mattered.

Recall Quick self-test

A cell has $E ° = + 0.0592 V$ and $n = 1$ at 25 °C. What is $K$ ? ::: $lo g_{10} K = 0.0592/0.0592 = 1$ , so $K = 10$ . You compute $lo g_{10} K = - 0.51$ and write $K = + 0.51$ . What went wrong? ::: You forgot to raise 10 to the power; $K = 1 0^{- 0.51} = 0.31$ , not $- 0.51$ . A ratio $K$ can never be negative. You given a measured $E$ at 0.5 M concentrations. Can you use $lo g K = n E /0.0592$ directly? ::: No — first recover $E °$ via the Nernst equation, then use the $K$ -formula (Cell H). Doubling $n$ (same $E °$ ) does what to $lo g_{10} K$ ? ::: Doubles it — so $K$ gets squared.

See also: Thermodynamics of Electrochemical Cells · Standard Reduction Potentials · Nernst Equation · Gibbs Free Energy and Spontaneity