Where a cell meets equilibrium
Consider the Daniell cell, in which the spontaneous reaction $\ce{Zn(s) + Cu^{2+}(aq) -> Zn^{2+}(aq) + Cu(s)}$ proceeds while the cell delivers current. As NCERT notes, once the circuit is closed the concentration of $\ce{Zn^{2+}}$ keeps rising and that of $\ce{Cu^{2+}}$ keeps falling. The voltmeter reading falls in step. After some time the concentrations stop changing and the voltmeter reads exactly zero. The cell has reached chemical equilibrium: there is no further net reaction and therefore no further driving force.
This is the conceptual heart of the subtopic. The working cell potential $E_{\text{cell}}$ is not a fixed number — it is the instantaneous driving force, and it bleeds away to zero as the reaction quotient $Q$ climbs toward the equilibrium constant $K$. The standard potential $E^{\circ}_{\text{cell}}$, by contrast, is a fixed thermodynamic constant defined at unit activities. The relation we derive below exploits the difference between these two quantities.
The cell potential decays along the Nernst curve as Q rises; it reaches zero precisely when Q = K. Setting $E_{\text{cell}}=0$ at that point is what unlocks the K–E°cell relation.
The equilibrium condition Ecell = 0
Start from the Nernst equation for the Daniell cell at 298 K (derived in the companion note on the Nernst equation):
$$E_{\text{cell}} = E^{\circ}_{\text{cell}} - \frac{0.059}{n}\,\log \frac{[\ce{Zn^{2+}}]}{[\ce{Cu^{2+}}]}$$
At equilibrium the cell does no work, so $E_{\text{cell}} = 0$, and the concentration ratio becomes the equilibrium constant, $\dfrac{[\ce{Zn^{2+}}]}{[\ce{Cu^{2+}}]} = K_C$. Substituting:
$$0 = E^{\circ}_{\text{cell}} - \frac{2.303RT}{nF}\,\log K_C \quad\Longrightarrow\quad E^{\circ}_{\text{cell}} = \frac{2.303RT}{nF}\,\log K_C$$
For the general balanced cell reaction $\ce{aA + bB <=> cC + dD}$ exchanging $n$ electrons, this generalises (NCERT Eq. 2.14) to the master relation of this subtopic:
$$\boxed{\,E^{\circ}_{\text{cell}} = \frac{2.303RT}{nF}\,\log K = \frac{0.059}{n}\,\log K \;\;(\text{at } 298\,\text{K})\,}$$
It is E°cell — not Ecell — that equals (0.059/n) log K
Students often write $E_{\text{cell}} = (0.059/n)\log K$. That is wrong. The substitution sets the working potential $E_{\text{cell}} = 0$; what survives on the left is the standard potential $E^{\circ}_{\text{cell}}$. The log argument is $K$ (equilibrium), not the running $Q$.
At equilibrium: $E_{\text{cell}} = 0$, $Q = K$, and $E^{\circ}_{\text{cell}} = \dfrac{0.059}{n}\log K$.
Deriving K from E°cell
The boxed relation is most often used in reverse: a standard cell potential is trivially measurable with a high-impedance voltmeter, whereas $K$ may be far too large or far too small to obtain by analysing equilibrium concentrations. Rearranging gives the form you will compute in the exam:
$$\log K = \frac{n\,E^{\circ}_{\text{cell}}}{0.059} \qquad\Longrightarrow\qquad K = 10^{\,n E^{\circ}_{\text{cell}}/0.059}$$
NCERT stresses the practical payoff: equilibrium constants of reactions that are "difficult to measure otherwise" can be calculated from the corresponding $E^{\circ}$ value of the cell. A modest potential difference produces an enormous $K$ because of the exponential. For the Daniell cell, $E^{\circ}_{\text{cell}} = 1.1\,\text{V}$ with $n = 2$ gives $\log K_C = (1.1 \times 2)/0.059 \approx 37.3$, i.e. $K_C \approx 2 \times 10^{37}$ at 298 K.
$E^{\circ}_{\text{cell}}$ is a straight-line function of $\log K$ through the origin, with slope $0.059/n$. A positive standard potential forces $K > 1$; the relation passes through $(0,0)$ since $\log 1 = 0$.
ΔG° = −nFE°cell and the K link
The maximum non-expansion (electrical) work a galvanic cell can deliver, obtained by passing the charge reversibly, equals the decrease in Gibbs energy. With $nF$ the charge passed and $E_{\text{cell}}$ the emf, NCERT Eq. 2.15 gives:
$$\Delta_r G = -nFE_{\text{cell}}$$
When every reacting species is at unit concentration, $E_{\text{cell}} = E^{\circ}_{\text{cell}}$ and the standard-state version (Eq. 2.16) follows:
$$\Delta_r G^{\circ} = -nFE^{\circ}_{\text{cell}}$$
Independently, thermodynamics gives $\Delta_r G^{\circ} = -RT\ln K$. Equating the two standard-state expressions reproduces the master relation from a second direction:
$$-nFE^{\circ}_{\text{cell}} = -RT\ln K \;\Longrightarrow\; E^{\circ}_{\text{cell}} = \frac{RT}{nF}\ln K = \frac{0.059}{n}\log K$$
This closes a tidy triangle: the standard cell potential, the standard Gibbs energy, and the equilibrium constant are three faces of the same thermodynamic information. Measuring any one of them fixes the other two.
Every relation here begins with the concentration form. Revisit the Nernst equation to see exactly how the 0.059/n slope and the reaction quotient Q arise.
Sign map: E°cell, ΔG° and K
Because the three quantities are locked together, a single sign determines all of them. This is the table NEET examiners build Assertion–Reason and "predict spontaneity" items from.
| E°cell | ΔG° = −nFE°cell | log K = nE°cell / 0.059 | K | Reaction |
|---|---|---|---|---|
| Positive | Negative | Positive | K > 1 | Spontaneous (forward favoured) |
| Zero | Zero | Zero | K = 1 | At equilibrium in standard state |
| Negative | Positive | Negative | K < 1 | Non-spontaneous (reverse favoured) |
The NEET 2022 spontaneity item used exactly this logic: a reaction "cannot occur" when $E^{\circ}_{\text{cell}} = E^{\circ}_{\text{cathode}} - E^{\circ}_{\text{anode}}$ comes out negative, which by the table forces $\Delta_r G^{\circ} > 0$ and $K < 1$. To compute electrode-potential differences correctly, keep the galvanic cells and electrode potential conventions firmly in mind.
Worked NEET numericals
Calculate the equilibrium constant of $\ce{Cu(s) + 2Ag^{+}(aq) -> Cu^{2+}(aq) + 2Ag(s)}$, given $E^{\circ}_{\text{cell}} = 0.46\ \text{V}$. (NCERT Example 2.2)
The reaction transfers $n = 2$ electrons. Apply $E^{\circ}_{\text{cell}} = \dfrac{0.059}{n}\log K_C$:
$$\log K_C = \frac{n\,E^{\circ}_{\text{cell}}}{0.059} = \frac{2 \times 0.46}{0.059} = 15.6$$
Therefore $K_C = 10^{15.6} \approx 3.92 \times 10^{15}$. The large value confirms a strongly forward-favoured reaction — consistent with the positive $E^{\circ}_{\text{cell}}$.
For the Daniell cell, $\ce{Zn(s) + Cu^{2+}(aq) -> Zn^{2+}(aq) + Cu(s)}$, $E^{\circ}_{\text{cell}} = 1.1\ \text{V}$. Find $\Delta_r G^{\circ}$. (NCERT Example 2.3 / NIOS Example 13.9)
Here $n = 2$ and $F = 96487\ \text{C mol}^{-1}$. Using $\Delta_r G^{\circ} = -nFE^{\circ}_{\text{cell}}$:
$$\Delta_r G^{\circ} = -2 \times 96487 \times 1.1 = -212270\ \text{J mol}^{-1} = -212.27\ \text{kJ mol}^{-1}$$
The strongly negative $\Delta_r G^{\circ}$ tells us the reaction is spontaneous in the standard state — the same conclusion the $K_C \approx 2 \times 10^{37}$ value delivers.
A cell reaction with $E^{\circ}_{\text{cell}} = 0.236\ \text{V}$ at 298 K transfers $n = 2$ electrons. Calculate $\Delta_r G^{\circ}$ and $K$. (NCERT Intext 2.6)
Gibbs energy: $\Delta_r G^{\circ} = -nFE^{\circ}_{\text{cell}} = -2 \times 96487 \times 0.236 \approx -45.5\ \text{kJ mol}^{-1}$.
Equilibrium constant: $\log K = \dfrac{2 \times 0.236}{0.059} = 8.0$, so $K = 10^{8.0} = 1.0 \times 10^{8}$. Both quantities flow from the single measured $E^{\circ}_{\text{cell}}$.
Intensive Ecell vs extensive ΔrG
A subtle but frequently examined distinction: $E_{\text{cell}}$ and $E^{\circ}_{\text{cell}}$ are intensive properties — they do not change when the balanced equation is multiplied by a factor. The Gibbs energy $\Delta_r G$ is extensive and scales with the stoichiometry. NCERT illustrates this with the Daniell reaction:
$$\ce{Zn(s) + Cu^{2+} -> Zn^{2+} + Cu(s)} \;\Rightarrow\; \Delta_r G = -2FE_{\text{cell}}$$ $$\ce{2Zn(s) + 2Cu^{2+} -> 2Zn^{2+} + 2Cu(s)} \;\Rightarrow\; \Delta_r G = -4FE_{\text{cell}}$$
The emf is identical for both because it is a per-electron driving force; only $n$ (and hence $\Delta_r G$) doubles. This is precisely the point a NEET Assertion–Reason question turned on — see the PYQ snapshot below.
Doubling the equation does not double K either
If you multiply the cell reaction by 2, both $n$ and the exponent in $\log K$ are affected: the new equilibrium constant is $K^2$, not $2K$, since $\log K' = \dfrac{(2n)E^{\circ}_{\text{cell}}}{0.059} = 2\log K$. $E^{\circ}_{\text{cell}}$ stays put because it is intensive.
Multiply reaction by $m$: $E^{\circ}_{\text{cell}}$ unchanged, $\Delta_r G^{\circ} \to m\Delta_r G^{\circ}$, $K \to K^{m}$.
Common errors and exam discipline
Most lost marks in this subtopic come from a handful of mechanical slips rather than conceptual gaps. The table collects the recurring ones.
| Slip | Correct discipline |
|---|---|
| Using $E_{\text{cell}}$ instead of $E^{\circ}_{\text{cell}}$ in the K relation | Only the standard potential equals $(0.059/n)\log K$; $E_{\text{cell}} = 0$ at equilibrium |
| Forgetting to multiply by $n$ inside $\log K$ | $\log K = nE^{\circ}_{\text{cell}}/0.059$ — the $n$ is in the numerator, not just the slope |
| Dropping the negative sign in $\Delta_r G^{\circ} = -nFE^{\circ}_{\text{cell}}$ | Positive $E^{\circ}_{\text{cell}}$ must give negative $\Delta_r G^{\circ}$ (spontaneous) |
| Mixing units — using $F$ in kC or $E^{\circ}$ in mV | Use $F = 96487\ \text{C mol}^{-1}$, $E^{\circ}$ in volts; answer in J mol⁻¹ |
| Reading the supplied factor as 0.059 when the paper gives 2.303RT/F | Confirm the given constant; NCERT uses 0.059 V at 298 K |
Equilibrium Constant from Nernst Equation
- At equilibrium a working cell reads $E_{\text{cell}} = 0$ and $Q = K$; substituting this into the Nernst equation gives $E^{\circ}_{\text{cell}} = \dfrac{0.059}{n}\log K$ at 298 K.
- Inverted for computation: $\log K = \dfrac{nE^{\circ}_{\text{cell}}}{0.059}$, so a small positive potential yields an enormous $K$.
- $\Delta_r G = -nFE_{\text{cell}}$ and $\Delta_r G^{\circ} = -nFE^{\circ}_{\text{cell}}$; combining with $\Delta_r G^{\circ} = -RT\ln K$ regenerates the K relation.
- Sign rule: $E^{\circ}_{\text{cell}} > 0 \Leftrightarrow \Delta_r G^{\circ} < 0 \Leftrightarrow K > 1$ (spontaneous).
- $E_{\text{cell}}$ is intensive; $\Delta_r G$ is extensive. Multiplying the reaction by $m$ leaves $E^{\circ}_{\text{cell}}$ unchanged but sends $\Delta_r G^{\circ} \to m\Delta_r G^{\circ}$ and $K \to K^{m}$.