JEE Mains Chemistry · Electrochemistry

Gibbs Energy, Equilibrium Constant and Combining Potentials

Turning a cell potential into Gibbs energy, an equilibrium constant, entropy or useful work, and finding a new E° from two known ones by adding Gibbs energies.

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Why this matters

Sixteen PYQs, eleven of them numerical, and four from 2026. Nine convert E° into ΔG°, K, ΔS° or the work a cell can do; seven combine two or three known potentials into a new one, often from a Latimer diagram. Two ideas cover the page.

Concept 1 of 2: Gibbs energy, K and work from E°

A cell's voltage is its Gibbs energy per unit of charge. Multiply by the charge that flows,

nF

, and you have

\Delta G

. From

\Delta G^\circ

the usual thermodynamics gives

K

, and the slope of

E^\circ

with temperature gives

\Delta S^\circ

Definition

$\Delta G^\circ=-nFE^\circ_{cell}$ and $\Delta G=-nFE_{cell}$ . $F=96500$ C mol⁻¹.
$\log K=\frac{nE^\circ_{cell}}{0.059}$ at 298 K.
The most negative $\Delta G^\circ$ belongs to the largest $nE^\circ$ , not the largest $E^\circ$ .
$\Delta S^\circ=nF\left(\frac{\partial E^\circ}{\partial T}\right)_P$ , and $\Delta H^\circ=\Delta G^\circ+T\Delta S^\circ$ .
The maximum electrical work is $-\Delta G$ : charge times potential, $nFE$ , never charge divided by potential.
A cell working at efficiency $\eta$ delivers $\eta\,nFE$ . Work against a constant pressure is $P_{ext}\Delta V$ .
$\Delta_fG^\circ$ is zero for elements and for $\mathrm{H^+(aq)}$ .

Gibbs energy and equilibrium constant

\Delta G^\circ=-nFE^\circ_{cell},\qquad \log K=\frac{nE^\circ_{cell}}{0.059}

Worked example

For the Daniell cell,

E^\circ_{cell}=1.10

V. Find

\Delta G^\circ

and

K

at 298 K.

Practice this conceptself-check · 5 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 2 Apr 2026 Shift 2 · Q48Moderate

Example 1 · Electrochemistry · Gibbs Energy, Equilibrium Constant and Combining Potentials

Consider the following two half-cell reactions along with the standard reduction potential given :

{CO}_{2} + 6H^{+} + 6e^{-} \rightarrow {CH}_{3}OH + H_{2}OE_{\text{red~}}^{o} = 0.02\text{ }V

\frac{1}{2}O_{2} + 2H^{+} + 2e^{-} \rightarrow H_{2}OE_{\text{red~}}^{o} = 1.23\text{ }V

The fuel cell was set up using the above two reactions such that the cell operates under the standard condition of 1 bar pressure and 298 K temperature. The fuel cell works with

80\%

efficiency. If the work derived from the cell using 1 mol of

{CH}_{3}OH

is used to compress an ideal gas isothermally against a constant pressure of 1 kPa , then the change in the volume of the gas,

\Delta V =

\_\_\_\_

m^{3}

. (Nearest integer) Given :

F = 96500C{mol}^{- 1}

Work as charge divided by potential

Electrical work is

Q\times E

, in joules. A statement that puts

E

in the denominator is the incorrect one.

Joules against kilojoules

nFE

comes out in joules. A blank asking for kJ mol⁻¹ needs a division by 1000 first.

Drill 8 more on gibbs energy, k and work from e°

Concept 2 of 2: Combining electrode potentials

Potentials do not add, but Gibbs energies do. To get a new

E^\circ

, turn each known step into

\Delta G^\circ=-nFE^\circ

, add or subtract the steps, and divide by the new

n

Definition

If step 3 = step 1 − step 2: $n_3E^\circ_3=n_1E^\circ_1-n_2E^\circ_2$ .
Example: $E^\circ(\mathrm{Fe^{3+}/Fe^{2+}})=3E^\circ(\mathrm{Fe^{3+}/Fe})-2E^\circ(\mathrm{Fe^{2+}/Fe})$ .
Latimer diagram (steps in series): $E^\circ=\frac{\sum n_iE^\circ_i}{\sum n_i}$ .
Insoluble-salt electrode: $\mathrm{MX(s)+e^-\to M(s)+X^-}$ is $\mathrm{M^++e^-\to M}$ plus $\mathrm{MX\to M^++X^-}$ . So $E^\circ(\mathrm{X^-/MX/M})=E^\circ(\mathrm{M^+/M})+0.059\log K_{sp}$ .
Two half-reactions combined into a FULL cell reaction are the one exception: there $E^\circ_{cell}=E^\circ_{cathode}-E^\circ_{anode}$ , because the electrons cancel.

Combining two steps

n_3E^\circ_3=n_1E^\circ_1\pm n_2E^\circ_2

Worked example

E^\circ(\mathrm{Cu^{2+}/Cu})=0.34

V and

E^\circ(\mathrm{Cu^+/Cu})=0.52

V. Find

E^\circ(\mathrm{Cu^{2+}/Cu^+})

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 8 Apr 2026 Shift 2 · Q31Moderate

Example 2 · Electrochemistry · Gibbs Energy, Equilibrium Constant and Combining Potentials

Given at

298\text{ }K:E_{Fe^{2 +}/Fe}^{\ominus}= X

Volt

E_{Fe^{3 +}/Fe}^{\ominus}= Y\text{~Volt~}

The

E_{Fe^{3 +}/Fe^{2 +}}^{\ominus}

in Volt at 298 K is given by :

Subtracting potentials directly

E^\circ(\mathrm{Fe^{3+}/Fe})-E^\circ(\mathrm{Fe^{2+}/Fe})

is not

E^\circ(\mathrm{Fe^{3+}/Fe^{2+}})

. Weight each potential by its electrons first.

Averaging a Latimer diagram

Two steps of 1 and 2 electrons are not averaged 50:50. Divide the electron-weighted sum by the TOTAL electrons.

Drill 6 more on combining electrode potentials

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

Gibbs energy, K and work from E°
Gibbs energy and equilibrium constant
$\Delta G^\circ=-nFE^\circ_{cell},\qquad \log K=\frac{nE^\circ_{cell}}{0.059}$
Combining electrode potentials
Combining two steps
$n_3E^\circ_3=n_1E^\circ_1\pm n_2E^\circ_2$

Watch out for (4)

Work as charge divided by potential
→ Gibbs energy, K and work from E°
Joules against kilojoules
→ Gibbs energy, K and work from E°
Subtracting potentials directly
→ Combining electrode potentials
Averaging a Latimer diagram
→ Combining electrode potentials

Test yourself on Electrochemistry

20 past JEE Mains questions from this chapter, timed at 48 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

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