Nernst equation

A standard electrode potential is only true at exactly 1 M concentration, 298 K and 1 atm — real cells almost never sit there. Dilute the electrolyte, run the reaction, or just leave a battery on the shelf, and the concentrations drift away from standard, yet the cell still has a definite EMF at every instant. The Nernst equation is the bridge: it converts a fixed standard potential $E^\circ$ into the actual, moment-by-moment potential $E$ at any concentration and temperature. It is also the single formula that ties electrochemistry to the rest of thermodynamics — from $E_{cell}$ you get $\Delta G$ via $\Delta G = -nFE_{cell}$ , and by pushing the equation to equilibrium you get the equilibrium constant $K$ of a redox reaction. This section builds the equation from the free-energy relation, works the classic Zn–Cu numerical, and flags the sign and unit traps that decide marks in board exams.

The core insight

The Nernst equation, $E_{cell} = E^\circ_{cell} - \dfrac{0.0591}{n}\log_{10} Q$ at 298 K, rescales the standard cell potential by the actual reaction quotient $Q$ , so that at equilibrium ( $Q = K$ , $E_{cell} = 0$ ) it directly yields $\log K = nE^\circ_{cell}/0.0591$ .

The core idea

From a single electrode to a full cell. For a general reduction half-reaction $M^{n+} + ne^- \rightarrow M$ , the electrode potential away from standard conditions is

E = E^\circ - \frac{RT}{nF}\ln\frac{[\text{products}]}{[\text{reactants}]}

where: $E^\circ$ is the standard electrode potential, $R = 8.314\ \text{J K}^{-1}\text{mol}^{-1}$ , $T$ is the temperature in kelvin, $n$ is the number of electrons in the half-reaction, and $F = 96500\ \text{C mol}^{-1}$ is the Faraday constant. Combining the two half-reactions of a full cell $aA + bB \rightarrow cC + dD$ gives the cell form:

E_{cell} = E^\circ_{cell} - \frac{RT}{nF}\ln Q, \qquad Q = \frac{[C]^c[D]^d}{[A]^a[B]^b}

where $Q$ is the reaction quotient built from actual concentrations (pure solids and liquids are left out — their activity is taken as 1) and $n$ is the number of electrons transferred in the overall balanced cell reaction.

The 298 K working form. Converting the natural log to $\log_{10}$ (a factor of $2.303$ ) and substituting $R$ , $T = 298\ \text{K}$ and $F$ collapses the constants into a single number, $0.0591$ :

E_{cell} = E^\circ_{cell} - \frac{0.0591}{n}\log_{10} Q \quad \text{(at 298 K)}

This is the version used in nearly every board numerical, so it is worth remembering that the $0.0591$ already contains the $2.303$ conversion — applying $2.303$ a second time is the single commonest arithmetic slip on this topic.

Why concentration changes the potential at all, in one picture: a reduction half-reaction consumes the metal ion, so a higher ion concentration pushes the reaction further forward, raising the electrode's tendency to be reduced (higher $E$ ); a lower concentration does the opposite. The Nernst equation is simply Le Chatelier's principle written in the language of thermodynamic potential.

Figure

Three panels showing standard concentration gives standard potential, higher ion concentration raises the potential above standard, and lower ion concentration lowers it below standard

Derivation: from Nernst equation to log K

Derivation AReaching the equilibrium constant from the Nernst equation

The goal is to show that at chemical equilibrium the Nernst equation does not vanish uselessly to $0 = 0$ but instead collapses into a genuinely useful relation, $\log K = nE^\circ_{cell}/0.0591$ , that lets you compute a redox equilibrium constant directly from tabulated standard potentials without running any experiment at equilibrium at all.

Step 1. Start from the 298 K Nernst equation for a full cell, $E_{cell} = E^\circ_{cell} - \dfrac{0.0591}{n}\log_{10} Q$ , and recall the physical meaning of $Q$ : it is the actual reaction quotient at whatever concentrations the cell currently holds, built from products over reactants exactly as in ordinary equilibrium expressions.

Step 2. Apply the equilibrium condition. As the cell reaction proceeds, reactants are consumed and products form, so $Q$ changes and the driving EMF $E_{cell}$ shrinks. At equilibrium the cell can no longer do any net work: $E_{cell} = 0$ . At that same instant the reaction quotient has reached its equilibrium value, $Q = K$ , by definition of the equilibrium constant.

Step 3. Substitute both conditions into the Nernst equation: $0 = E^\circ_{cell} - \dfrac{0.0591}{n}\log_{10} K$ . Rearranging for $\log_{10} K$ isolates the equilibrium constant on one side:

\log_{10} K = \frac{nE^\circ_{cell}}{0.0591}

Step 4. Read the result. A larger positive $E^\circ_{cell}$ (a more spontaneous standard reaction) forces $\log K$ to be large and positive, so $K \gg 1$ — the reaction goes essentially to completion. This single line is also how $\Delta G^\circ = -nFE^\circ_{cell} = -RT\ln K$ connects three separately-taught quantities — standard EMF, standard free energy, and the equilibrium constant — into one consistent thermodynamic statement.

Worked example

Problem: Calculate the EMF of the cell $\text{Zn(s)} \mid \text{Zn}^{2+}(0.1\ M) \parallel \text{Cu}^{2+}(1.0\ M) \mid \text{Cu(s)}$ at 298 K given $E^\circ_{cell} = 1.10\ \text{V}$ , then find the equilibrium constant $K$ and $\Delta G^\circ$ (in kJ/mol) for the same reaction.

Step 1. Write the overall cell reaction and count electrons: $\text{Zn(s)} + \text{Cu}^{2+}(aq) \rightarrow \text{Zn}^{2+}(aq) + \text{Cu(s)}$ , with $n = 2$ electrons transferred.

Step 2. Build the reaction quotient, leaving out the pure solids Zn(s) and Cu(s): $Q = \dfrac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]} = \dfrac{0.1}{1.0} = 0.1$ .

Step 3. Apply the 298 K Nernst equation: $E_{cell} = E^\circ_{cell} - \dfrac{0.0591}{n}\log Q = 1.10 - \dfrac{0.0591}{2}\log(0.1)$ . Since $\log(0.1) = -1$ , this gives $E_{cell} = 1.10 + 0.02955 = 1.1296\ \text{V} \approx \mathbf{1.13\ V}$ — slightly higher than standard, because lowering the product-side $\text{Zn}^{2+}$ concentration favours the forward reaction, exactly as Le Chatelier's principle predicts.

Step 4. Find $K$ from $\log K = \dfrac{nE^\circ_{cell}}{0.0591} = \dfrac{2 \times 1.10}{0.0591} = 37.23$ , so $K = 10^{37.23} \approx \mathbf{1.7 \times 10^{37}}$ — an enormous value showing the reaction goes essentially to completion.

Step 5. Find $\Delta G^\circ = -nFE^\circ_{cell} = -2 \times 96500 \times 1.10 = -212{,}300\ \text{J mol}^{-1} = \mathbf{-212.3\ kJ\ mol^{-1}}$ . The large negative $\Delta G^\circ$ and the huge $K$ are two consistent expressions of the same strongly spontaneous forward reaction.

Competitive stretch

A concentration cell $\text{Cu(s)} \mid \text{Cu}^{2+}(c_1) \parallel \text{Cu}^{2+}(c_2) \mid \text{Cu(s)}$ has identical electrodes, so $E^\circ_{cell} = 0$ — yet the Nernst equation still gives a non-zero $E_{cell}$ if $c_1 \ne c_2$ , because each electrode individually sits at a different potential. The more dilute solution's electrode becomes the anode, and the EMF is driven purely by the concentration ratio, a useful reminder that "standard" driving force and actual driving force are not the same thing.

Common traps

Applying the $2.303$ conversion factor a second time on top of the already-converted $0.0591$ constant, effectively double-counting it.
Including pure solids or the pure solvent inside $Q$ — only ions and gases at their actual concentrations or partial pressures belong there.
Writing $Q$ upside down (reactants over products) or forgetting that it must match the reaction exactly as written, with the cathode as reduction and anode as oxidation combined.
Using the stoichiometric coefficient of one species as $n$ instead of the true number of electrons transferred in the balanced overall cell reaction (the LCM of the two half-reactions).

Ask the tutor: want to work through another Nernst numerical with a different concentration ratio, or see how a concentration cell's EMF is calculated step by step?

Why does the Nernst equation use log base 10 with 0.0591 instead of natural log with RT over nF?

Both forms are correct and mean the same physics — they differ only by the constant $2.303$ that converts $\ln x$ to $\log_{10}x$ . The $\log_{10}$ form is preferred for numericals because log tables and calculators traditionally worked in base 10, and because it is easier to spot integer changes like $Q = 0.1 \Rightarrow \log Q = -1$ . Just remember the $2.303$ is already baked into $0.0591$ , so never multiply by it again.

Why is the standard electrode potential defined at exactly 1 M concentration?

Because a cell's potential genuinely depends on concentration, a single "electrode potential" would be meaningless without fixing a reference state — otherwise the same electrode could report infinitely many values. Chemists agreed on 1 M ion concentration, 1 atm gas pressure and 298 K as the reference, giving $E^\circ$ . The Nernst equation is precisely the tool that converts this one fixed reference number back into the real potential under whatever conditions actually exist in the cell.

If a concentration cell has E standard cell equal to zero, how can it still produce current?

$E^\circ_{cell} = 0$ only means the two electrodes are chemically identical, so neither has an inherent thermodynamic advantage under standard conditions. But the Nernst equation's log term does not require $E^\circ_{cell}$ to be non-zero — it only needs $Q \ne 1$ . With different ion concentrations at the two electrodes, $Q \ne 1$ , so $-\dfrac{0.0591}{n}\log Q$ is non-zero on its own, and that term alone drives current until the concentrations equalise and $Q$ reaches 1.