Worked examples — Nernst equation E = E° − (RT - nF) ln Q

This page is a firing range. The parent Nernst topic note built the equation; here we shoot at every kind of target it can throw at you. Before the examples, we lay out a scenario matrix so you can see there are no hidden cases.

We only ever use two forms of the equation, both already earned in the parent:

The scenario matrix

Every Nernst problem lives in exactly one of these boxes. The whole point of $Q$ is: is it bigger than 1, smaller than 1, or exactly 1? That single sign decides whether $E$ sits above, below, or equal to $E°$.

The rule "$Q>1\Rightarrow \log Q>0$, $Q<1\Rightarrow \log Q<0$, $Q=1\Rightarrow \log Q=0$" never changes — Cases D, E, H just require you to build $Q$ first (from the spontaneous direction, from $K$, or from pH) before you can read its sign, which is why the table flags them rather than pre-filling one symbol.

Notice what is missing on purpose: there is no "negative $n$" box (see the parent's Mistake 3 — $n$ is a count, always positive), and no "solid concentration" box (activities of pure solids/liquids are $1$, so they never sit inside $Q$).

Two quick anchors we'll lean on all page:

Look at the curve above: $E$ versus $\log Q$ is a straight line. Its height at the middle ($\log Q=0$) is $E°$, and it slopes downward with slope $-0.0592/n$. Every example below is just "find your $\log Q$ on the x-axis, read off $E$."

[!example] Ex 1 — Case A: products piled up ($Q>1$)

Statement. Zinc–copper cell $\text{Zn}(s)+{\text{Cu}}^{2+}(aq)\to {\text{Zn}}^{2+}(aq)+\text{Cu}(s)$, with $E°=1.10\text{V}$, $[{\text{Cu}}^{2+}]=0.01\text{M}$, $[{\text{Zn}}^{2+}]=1.0\text{M}$, at $25°\text{C}$. Find $E$.

Forecast: products (${\text{Zn}}^{2+}$) are 100× the reactant (${\text{Cu}}^{2+}$), so $Q>1$ — do you expect $E$ above or below $1.10$?

1. Write $Q$. $Q=\frac{[{\text{Zn}}^{2+}]}{[{\text{Cu}}^{2+}]}=\frac{1.0}{0.01}=100.$ Why this step? Solids Zn and Cu have activity 1, so only the two aqueous ions appear.
2. Find $n$. Each half-reaction moves 2 electrons, so $n=2$. Why this step? The $\frac{0.0592}{n}$ slope needs the electron count.
3. Read $\log Q$. ${\log }_{10}(100)=2$. Why this step? $Q=100=10^2$, two tens above one — this is Case A ($+$).
4. Apply Nernst. $E=1.10-\frac{0.0592}{2}(2)=1.10-0.0592=1.04\text{V}.$ Why this step? Plug the positive $\log Q$ into the drop term.

Verify: $E=1.04\text{V}<E°=1.10\text{V}$ — matches the forecast (products up ⇒ voltage down). Units: $\frac{\text{V}}{\text{(dimensionless }n)}\times (\text{dimensionless }\log Q)=\text{V}$. ✓

[!example] Ex 2 — Case B: reactants dominate ($Q<1$)

Statement. Same Zn–Cu cell, $E°=1.10\text{V}$, but now flip the concentrations: $[{\text{Cu}}^{2+}]=1.0\text{M}$, $[{\text{Zn}}^{2+}]=0.001\text{M}$, at $25°\text{C}$. Find $E$.

Forecast: now the reactant ion is 1000× the product ion — $Q<1$. Above or below $1.10$?

1. Write $Q$. $Q=\frac{[{\text{Zn}}^{2+}]}{[{\text{Cu}}^{2+}]}=\frac{0.001}{1.0}=\mathrm{0.001.}$ Why this step? Same reaction, same $Q$ recipe — only the numbers changed.
2. $n=2$ (unchanged; still 2 electrons).
3. Read $\log Q$. ${\log }_{10}(0.001)=-3$. Why this step? $0.001=10^{-3}$, three tens below one — Case B ($-$).
4. Apply Nernst. $E=1.10-\frac{0.0592}{2}(-3)=1.10+0.0888=1.1888\approx 1.19\text{V}.$ Why this step? Subtracting a negative adds — voltage rises above $E°$.

Verify: $E\approx 1.19\text{V}>E°=1.10\text{V}$ ✓. Sanity: fewer products, more reactants — reaction is "hungrier," pushes harder, exactly what Le Chatelier's Principle predicts.

[!example] Ex 3 — Case C: exactly standard ($Q=1$, degenerate)

Statement. Zn–Cu cell with $[{\text{Cu}}^{2+}]=0.5\text{M}$ and $[{\text{Zn}}^{2+}]=0.5\text{M}$, $25°\text{C}$, $E°=1.10\text{V}$. Find $E$.

Forecast: the two ions are equal. Does the correction term do anything?

1. Write $Q$. $Q=\frac{0.5}{0.5}=1.$ Why this step? Equal concentrations cancel — this is the degenerate boundary between Case A and B.
2. Find $n$. Still the Zn/Cu couple, still 2 electrons, so $n=2$. Why this step? Even when the ratio is 1, we must confirm $n$ before touching the slope — keep the routine identical every time.
3. Read $\log Q$. ${\log }_{10}(1)=0.$ Why this step? $Q=1=10^0$, exactly on the line at zero tens.
4. Apply Nernst. $E=1.10-\frac{0.0592}{2}(0)=1.10\text{V}.$

Verify: $E=E°=1.10\text{V}$ ✓. Key lesson: $Q=1$ does not require 1 M each — it only requires the ratio to be 1. The concentrations were $0.5$, not standard, yet $E=E°$ because the ratio landed on 1.

[!example] Ex 4 — Case D: concentration cell ($E°=0$)

Statement. Two silver half-cells, same electrode Ag/Ag⁺. Left $[{\text{Ag}}^{+}]=0.001\text{M}$ (anode), Right $[{\text{Ag}}^{+}]=1.0\text{M}$ (cathode), $25°\text{C}$. Find $E$.

Forecast: the electrodes are identical, so $E°=0$. Can a cell with zero standard potential still push a current?

1. Set $E°=0$. Identical half-cells ⇒ no intrinsic height difference. Why this step? This is what makes it Case D — voltage comes purely from the concentration gradient.
2. Find the spontaneous direction. Nature dilutes: ${\text{Ag}}^{+}(1.0)\to {\text{Ag}}^{+}(0.001)$. So the concentrated side is the cathode (reduction), dilute side the anode. Why this step? We need which side is products to build $Q$.
3. Write $Q$. $Q=\frac{[{\text{Ag}}^{+}{]}_{\text{anode}}}{[{\text{Ag}}^{+}{]}_{\text{cathode}}}=\frac{0.001}{1.0}=\mathrm{0.001.}$ Why this step? For ${\text{Ag}}_{\text{cathode}}^{+}\to {\text{Ag}}_{\text{anode}}^{+}$, product is the dilute side. This gives $Q<1$, the $\log Q<0$ branch of Case D.
4. Apply Nernst, $n=1$. $E=0-\frac{0.0592}{1}{\log }_{10}(0.001)=-0.0592(-3)=0.1776\approx 0.178\text{V}.$

Verify: $E\approx 0.178\text{V}>0$ ✓ — a real, positive voltage from concentration alone. This is exactly Battery Discharge Curves behaviour at the microscopic level: as the gradient shrinks, so does $E$.

[!example] Ex 5 — Case E: equilibrium, and the link to $K$

Statement. For the Zn–Cu cell ($E°=1.10\text{V}$, $n=2$, $25°\text{C}$), what is $E$ at equilibrium, and what is ${\log }_{10}K$?

Forecast: at equilibrium the reaction has "run out of downhill." Guess $E$ before reading.

1. State the equilibrium facts. At equilibrium $Q=K$ and (from Gibbs Free Energy and Spontaneity) $\Delta G=0$. Since $\Delta G=-nFE$, we get $E=0$. Why this step? Zero driving force means zero voltage — the dead battery.
2. Set Nernst to zero. $0=E°-\frac{0.0592}{n}{\log }_{10}K.$ Why this step? Substituting $E=0$ and $Q=K$ turns the equation into a bridge to $K$.
3. Solve for $\log K$. ${\log }_{10}K=\frac{nE°}{0.0592}=\frac{2\times 1.10}{0.0592}=\mathrm{37.16.}$ Why this step? Rearrange — this connects Reaction Quotient Q vs Equilibrium Constant K to voltage. Since $E°>0$ here, $\log K>0$ ($Q=K>1$), the positive branch flagged in the matrix.

Verify: $E=0$ at equilibrium ✓ (dead battery). ${\log }_{10}K\approx 37.2$, i.e. $K\approx 10^{37}$ — enormous, meaning Zn+Cu²⁺ goes essentially to completion, which fits its large positive $E°$. ✓

[!example] Ex 6 — Case F: not at $25°\text{C}$ (must use $RT/nF$)

Statement. The Ex 1 Zn–Cu cell ($E°=1.10\text{V}$, $Q=100$, $n=2$) is run hot at $T=350\text{K}$ instead of 298 K. Find $E$. Use $R=8.314{\text{J mol}}^{-1}{\text{K}}^{-1}$, $F=96485{\text{C mol}}^{-1}$.

Forecast: the $0.0592$ shortcut is locked to 298 K. Does the drop get bigger or smaller when we heat up?

1. Reject the shortcut. $0.0592$ hides $T=298$; here $T=350$, so we must use the raw $\frac{RT}{nF}\ln Q$ form. Why this step? Using $0.0592$ now would silently assume the wrong temperature — that's the Case F trap.
2. Compute the slope. $\frac{RT}{nF}=\frac{8.314\times 350}{2\times 96485}=0.01508\text{V}.$ Why this step? This is the per-$\ln Q$ voltage at 350 K.
3. Use $\ln Q$, not $\log Q$. Here $\ln$ is the natural logarithm (base $e$), which is the partner of the $RT/nF$ form (parent Mistake 2). $\ln (100)=\mathrm{4.6052.}$ Why this step? The $RT/nF$ form pairs with the natural log; using ${\log }_{10}$ here would be off by the factor $2.303$.
4. Apply Nernst. $E=1.10-0.01508\times 4.6052=1.10-0.0694=1.0306\approx 1.031\text{V}.$

Verify: At 298 K the drop was $0.0592$ V (Ex 1). At 350 K the drop is larger ($0.0694$ V), so $E=1.031\text{V}<1.04\text{V}$ ✓ — higher $T$ multiplies the correction, pulling $E$ further from $E°$. Units: $\frac{{\text{J mol}}^{-1}{\text{K}}^{-1}\cdot \text{K}}{{\text{C mol}}^{-1}}=\frac{\text{J}}{\text{C}}=\text{V}$ ✓.

[!example] Ex 7 — Case G: a gas enters $Q$ (partial pressure)

Statement. Hydrogen electrode reduction: $2{\text{H}}^{+}(aq)+2e^{-}\to {\text{H}}_2(g)$, paired so that $E°=0.00\text{V}$ for this reaction as written, $n=2$, $25°\text{C}$. Conditions: $[{\text{H}}^{+}]=1.0\text{M}$ (standard) but $P_{{\text{H}}_2}=4.0\text{atm}$. Find $E$.

Forecast: the product is a gas at 4 atm — more product than standard. Which case (A or B) does that make it?

1. Build $Q$ with a partial pressure. $Q=\frac{P_{{\text{H}}_2}}{[{\text{H}}^{+}{]}^2}=\frac{4.0}{(1.0{)}^2}=\mathrm{4.0.}$ Why this step? Gases enter $Q$ as their partial pressure in atm, not concentration; the $[{\text{H}}^{+}]$ is squared because of the 2 in the balanced equation.
2. Classify. $Q=4>1$ ⇒ Case A ($+$), so expect $E<E°$. Why this step? Piling up product H₂ should lower the drive.
3. Read $\log Q$. ${\log }_{10}(4)=\mathrm{0.60206.}$
4. Apply Nernst. $E=0.00-\frac{0.0592}{2}(0.60206)=-0.0296\times 0.60206=-0.01782\approx -0.0178\text{V}.$

Verify: $E\approx -0.018\text{V}<E°=0$ ✓, negative as Case A demands. Sanity: raising $P_{{\text{H}}_2}$ shoves the reaction backwards (Le Chatelier), so the forward drive dips just below zero. ✓

[!example] Ex 8 — Case H: pH-driven half-cell (word/exam twist)

Statement. For the same hydrogen electrode $2{\text{H}}^{+}+2e^{-}\to {\text{H}}_2$, $E°=0.00\text{V}$, $n=2$, keep $P_{{\text{H}}_2}=1.0\text{atm}$ but change the acid to $\text{pH}=3.0$. Find $E$. (See pH and Half-Cell Potentials.)

Forecast: pH 3 means the reactant H⁺ is scarce ($10^{-3}$ M). Reactant-poor — which case?

1. Convert pH to $[{\text{H}}^{+}]$. $\text{pH}=3\Rightarrow [{\text{H}}^{+}]=10^{-3}=0.001\text{M}.$ Why this step? Nernst needs a concentration; pH is just $-{\log }_{10}[{\text{H}}^{+}]$.
2. Build $Q$. $Q=\frac{P_{{\text{H}}_2}}{[{\text{H}}^{+}{]}^2}=\frac{1.0}{(0.001{)}^2}=\frac{1}{10^{-6}}=10^6.$ Why this step? The squared, tiny denominator makes $Q$ huge — this is why pH swings potentials so strongly. Here $Q>1$, so $\log Q>0$ — the positive branch flagged for Case H.
3. Read $\log Q$. ${\log }_{10}(10^6)=6.$
4. Apply Nernst. $E=0.00-\frac{0.0592}{2}(6)=-0.0296\times 6=-0.1776\approx -0.178\text{V}.$

Verify: $E\approx -0.178\text{V}$ ✓. Shortcut check: each pH unit shifts this electrode by $-0.0592$ V (since $\log Q$ gains 2 per pH unit but $n=2$ halves it → net $-0.0592$ V per pH unit). Three pH units ⇒ $-0.178$ V ✓. Reactant-scarce = Case B feel (product side heavy ⇒ negative) — consistent.

[!recall]- Quick self-test

Connections

• Standard Reduction Potentials — where every $E°$ used above comes from
• Gibbs Free Energy and Spontaneity — the $\Delta G=-nFE$ that makes $E=0$ at equilibrium
• Reaction Quotient Q vs Equilibrium Constant K — Ex 5's bridge to $K$
• Electrochemical Cells (Galvanic vs Electrolytic) — anode/cathode roles used in Ex 4
• Le Chatelier's Principle — the "push back" intuition behind Cases A/B/G
• pH and Half-Cell Potentials — Ex 8's pH twist
• Battery Discharge Curves — Case D and the real-world voltage sag