Nernst Equation

Why E° Is Not Enough

A standard electrode potential, $E^\circ$, is defined only when the active species are at unit activity — aqueous ions at 1 mol L⁻¹ and gases at 1 bar, measured against the standard hydrogen electrode. The moment a galvanic cell begins to deliver current, those concentrations start to drift: the reactant ions are consumed and the product ions accumulate. The measured emf therefore changes continuously, and $E^\circ$ alone can no longer describe it.

Walther Nernst showed how the potential depends on concentration. NCERT states the result directly for the electrode reaction $\ce{M^{n+}(aq) + ne^- -> M(s)}$: the electrode potential at any concentration, measured with respect to the standard hydrogen electrode, is fixed once you know $E^\circ$ and the ion concentration. This single correction term is what the rest of this page unpacks.

Nernst Equation for a Single Electrode

For the reduction half-reaction $\ce{M^{n+}(aq) + ne^- -> M(s)}$, the activity of the pure solid metal is taken as unity, so only the ion concentration enters. The electrode potential is:

$$ E_{(\ce{M^{n+}/M})} = E^\circ_{(\ce{M^{n+}/M})} - \frac{RT}{nF}\,\ln \frac{1}{[\ce{M^{n+}}]} $$

Here $R = 8.314~\text{J K}^{-1}\text{mol}^{-1}$ is the gas constant, $F = 96487~\text{C mol}^{-1}$ is the Faraday constant, $T$ is the temperature in kelvin, $n$ is the number of electrons transferred, and $[\ce{M^{n+}}]$ is the molar concentration of the ion. Because the log carries $1/[\ce{M^{n+}}]$, raising the ion concentration raises the (reduction) electrode potential — a more concentrated solution is a stronger oxidising environment, exactly as chemical intuition demands.

Nernst Equation for a Complete Cell

A complete cell potential is the cathode (reduction) potential minus the anode (reduction) potential. Take the Daniell cell, $\ce{Zn(s) | Zn^{2+}(aq) || Cu^{2+}(aq) | Cu(s)}$, whose net reaction is $\ce{Zn(s) + Cu^{2+}(aq) -> Zn^{2+}(aq) + Cu(s)}$. Writing Nernst for each electrode and subtracting, NCERT obtains:

$$ E_{(\text{cell})} = E^\circ_{(\text{cell})} - \frac{RT}{2F}\,\ln \frac{[\ce{Zn^{2+}}]}{[\ce{Cu^{2+}}]} $$

The emf rises when $[\ce{Cu^{2+}}]$ (the reactant ion) increases and falls when $[\ce{Zn^{2+}}]$ (the product ion) increases — the same logic generalises to any cell. For the general reaction $\ce{aA + bB + ne^- -> cC + dD}$, the Nernst equation takes its master form:

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

$Q$ is the reaction quotient: products over reactants, each raised to its stoichiometric coefficient, with pure solids and pure liquids omitted (their activities are unity) and gases entered as partial pressures in bar.

Derivation Outline

The Nernst equation is not a separate postulate — it follows from joining two relations you already know. The first ties cell emf to Gibbs energy; the second ties Gibbs energy to the reaction quotient.

Step 1 reflects that the maximum work obtainable from a galvanic cell, drawn reversibly, equals the decrease in its Gibbs energy. NCERT stresses that $E_{(\text{cell})}$ is an intensive property (independent of how much material reacts) whereas $\Delta_r G$ is extensive and scales with $n$ — doubling the balanced equation doubles $\Delta_r G$ but leaves emf unchanged.

The 0.059/n log Q Form at 298 K

For numerical work it is far easier to convert the natural logarithm to base 10 (introducing the factor 2.303) and to lump the constants together at 298 K. The grouped term is:

$$ \frac{2.303\,RT}{F} = \frac{2.303 \times 8.314 \times 298}{96487} \approx 0.0592~\text{V} \approx 0.059~\text{V} $$

With this substitution the master equation collapses into the form that wins NEET marks:

$$ E_{(\text{cell})} = E^\circ_{(\text{cell})} - \frac{0.059}{n}\,\log Q $$

and, for the Daniell cell specifically, $$ E_{(\text{cell})} = E^\circ_{(\text{cell})} - \frac{0.059}{2}\,\log \frac{[\ce{Zn^{2+}}]}{[\ce{Cu^{2+}}]}. $$

The figure below shows how $E_{(\text{cell})}$ falls linearly as $\log Q$ rises — a straight line of slope $-0.059/n$ whose intercept on the vertical axis is $E^\circ_{(\text{cell})}$ (the point where $Q = 1$, i.e. standard conditions).

Worked Example: Concentration Effect on EMF

The cleanest test of the Nernst equation is to change concentrations and ask which way emf moves. This is exactly the form NEET 2017 took with the Daniell cell, and the same arithmetic recurs across years.

Concentration Cells

A concentration cell is the most elegant consequence of the Nernst equation. Both electrodes are the same metal dipping into the same ion, differing only in concentration — for example $\ce{Cu | Cu^{2+}(c_1) || Cu^{2+}(c_2) | Cu}$. Because the two half-cells are chemically identical, their standard potentials cancel exactly: $E^\circ_{(\text{cell})} = 0$. The entire emf therefore arises from the Nernst log term alone:

$$ E_{(\text{cell})} = -\frac{0.059}{n}\,\log \frac{c_{\text{dilute}}}{c_{\text{conc}}} = \frac{0.059}{n}\,\log \frac{c_{\text{conc}}}{c_{\text{dilute}}} $$

The dilute side acts as the anode (the metal dissolves there to raise its ion concentration) and the concentrated side as the cathode (ions deposit, lowering concentration). The cell runs spontaneously until the two concentrations equalise, at which point the emf vanishes. The schematic below traces this flow.

Link to the Equilibrium Constant

Run a galvanic cell long enough and it dies: the voltmeter reads zero. NCERT interprets this as the cell reaching equilibrium. When $E_{(\text{cell})} = 0$, the reaction quotient $Q$ has become the equilibrium constant $K_c$. Setting the Nernst equation to zero:

$$ 0 = E^\circ_{(\text{cell})} - \frac{0.059}{n}\log K_c \quad\Longrightarrow\quad E^\circ_{(\text{cell})} = \frac{0.059}{n}\log K_c $$

and in general $E^\circ_{(\text{cell})} = \dfrac{2.303\,RT}{nF}\log K_c$. For the Daniell cell ($E^\circ = 1.1~\text{V}$, $n = 2$), this gives $\log K_c = \dfrac{1.1 \times 2}{0.059} = 37.3$, so $K_c \approx 2 \times 10^{37}$ — an enormous constant that could never be measured by direct titration but falls straight out of one emf reading.

The same chain of relations also yields the standard Gibbs energy: $\Delta_r G^\circ = -nFE^\circ_{(\text{cell})}$. For the Daniell cell this is $-2 \times 1.1~\text{V} \times 96487~\text{C mol}^{-1} = -212.27~\text{kJ mol}^{-1}$, confirming a strongly spontaneous reaction.

Applications and Common Forms

The Nernst equation reappears in several standard guises NEET likes to test. The table collects the forms worth memorising.

The hydrogen-electrode form underlies the classic NEET 2016 question on the $\ce{H2}$ pressure that drives the electrode potential to zero. For $\ce{2H+ + 2e^- -> H2(g)}$ the potential is $E = E^\circ + \dfrac{0.059}{2}\log\dfrac{[\ce{H+}]^2}{p_{\ce{H2}}}$; with $E^\circ = 0$ and pure water ($[\ce{H+}] = 10^{-7}$ M), setting $E = 0$ forces $p_{\ce{H2}} = [\ce{H+}]^2 = 10^{-14}$ bar.