JEE Mains Chemistry · Electrochemistry

Nernst Equation and Concentration Effects

How a cell's voltage moves away from E° when the concentrations, gas pressures or pH are not standard, and how to run the Nernst equation backwards to find an unknown.

All Electrochemistry notes JEE Mains Chemistry strategy

Why this matters

Thirty PYQs, the largest page in the chapter, and nine of them from 2026. Twenty-three are numerical. Eleven find a cell emf from given concentrations, nine run the equation backwards for a concentration, a ratio, n or E°, and ten put H⁺ or OH⁻ into the log term through pH. Three ideas cover the page.

Concept 1 of 3: Cell emf from the Nernst equation

E^\circ

is the voltage when every ion is at 1 M and every gas at 1 bar. Away from that, the voltage shifts by a small log term. More product pushes the voltage down; more reactant pushes it up.

Definition

$E_{cell}=E^\circ_{cell}-\frac{0.059}{n}\log Q$ at 298 K. Use 0.06 if the question gives 0.06.
$Q$ is products over reactants, each raised to its coefficient in the BALANCED reaction. Solids and pure liquids count as 1.
$n$ is the number of electrons in the balanced full reaction. When the half-reactions differ, take their LCM.
Concentration cell: the same couple on both sides, so $E^\circ=0$ and $E=\frac{0.059}{n}\log\frac{c_{cathode}}{c_{anode}}$ . It gives a positive voltage only when the cathode side is more concentrated.
As a cell runs, $Q$ rises and $E_{cell}$ falls, reaching 0 at equilibrium. $E^\circ_{cell}$ does not change.

Nernst equation at 298 K

E_{cell}=E^\circ_{cell}-\frac{0.059}{n}\log Q

Qreaction quotient: products over reactants, powers from the balanced equation
nelectrons transferred in the balanced reaction

Worked example

Find the emf at 298 K of

\mathrm{Mg|Mg^{2+}(0.001\,M)\|Cu^{2+}(0.0001\,M)|Cu}

E^\circ(\mathrm{Mg^{2+}/Mg})=-2.37

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

Practice this conceptself-check · 5 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 22 Jan 2026 Shift 1 · Q47Moderate

Example 1 · Electrochemistry · Nernst Equation and Concentration Effects

Consider the following electrochemical cell at 298 K

Pt\left| {HSnO}_{2}^{-}(aq) \right|Sn(OH)_{6}\ ^{2 -}(aq)\left| {Bi}_{2}O_{3}(s) \right|Bi(s)

. If the reaction quotient at a given time is

10^{6}

, then the cell EMF (

E_{\text{cell~}}

) is

\_\_\_\_

\times 10^{- 1}\text{ }V

(Nearest integer). Given the standard half-cell reduction potential as

E_{{Bi}_{2}O_{3}/{Bi}_{,{OH}^{-}}}^{0} = - 0.44\text{ }V

and

E_{Sn(OH)_{6}^{2 -}/{HSnO}_{2}^{-},OH}^{0} = - 0.90\text{ }V

Dropping the powers in Q

\mathrm{Zn+2Ag^+}

, the silver ion is squared in

Q

. The same balancing that fixes

n

fixes the powers, so do both from one balanced equation.

Q upside down

Q

is products over reactants. Writing it the other way flips the sign of the log term and moves the answer by twice the correction.

Drill 10 more on cell emf from the nernst equation

Concept 2 of 3: Solving the Nernst equation for an unknown

The same equation has four quantities that can be missing: a concentration, a ratio of concentrations,

n

, or

E^\circ

. Put in everything you know, isolate the log, and undo it.

Definition

Rearranged: $\log Q=\frac{n\,(E^\circ-E)}{0.059}$ .
If $E>E^\circ$ , then $\log Q<0$ , so $Q<1$ : reactants are in excess.
For a redox couple on Pt ( $\mathrm{Fe^{3+},Fe^{2+}}$ ), the unknown is usually the ratio $[\text{reduced}]/[\text{oxidised}]$ .
To find $n$ , use two readings or one reading with a known $Q$ : $n=\frac{0.059\log Q}{E^\circ-E}$ .
To find $E^\circ$ of one electrode, first find $E^\circ_{cell}$ from the measured $E$ , then subtract the known electrode.

Nernst equation, rearranged

\log Q=\frac{n\left(E^\circ_{cell}-E_{cell}\right)}{0.059}

Worked example

\mathrm{Cu|Cu^{2+}(1\,M)\|Ag^+(x\,M)|Ag}

reads 0.342 V at 298 K.

E^\circ_{cell}=0.46

V. Find

x

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 5 Apr 2026 Shift 2 · Q32Moderate

Example 2 · Electrochemistry · Nernst Equation and Concentration Effects

One half cell in a voltaic cell is constructed by dipping silver rod in

AgNO_{3}

solution of unknown concentration, other half cell is Zn rod dipped in 1 molar solution of

ZnSO_{4}

. A voltage of 1.60 V is measured at 298 K for this cell. What is the concentration of

Ag^{+}

ions used in terms of

\log x\left( x =\left\lbrack Ag^{+} \right\rbrack \right)

E_{Zn^{2 +}/Zn}^{\Theta}= - 0.76\text{ }V,E_{Ag^{+}/Ag}^{\Theta}= + 0.80\text{ }V

\frac{2.303RT}{F}= 0.059\text{ }V

Losing the sign of the log

If the measured

E

is ABOVE

E^\circ

\log Q

must be negative. Check this before you take the antilog; a sign slip turns 0.01 M into 100 M.

Square root forgotten

When

Q=1/x^2

, the log gives

x^2

. Take the square root at the end.

Drill 8 more on solving the nernst equation for an unknown

Concept 3 of 3: Electrodes that depend on pH

When

\mathrm{H^+}

\mathrm{OH^-}

sits in a half-reaction, its concentration goes into the log term like any other ion. So pH moves the potential: by 0.059 V per pH unit for every electron that carries one proton.

Definition

Hydrogen electrode, $\mathrm{2H^++2e^-\to H_2}$ : $E=-\frac{0.059}{2}\log\frac{p_{H_2}}{[\mathrm{H^+}]^2}=-0.059\,\mathrm{pH}-0.0295\log p_{H_2}$ .
For $E=0$ you need $p_{H_2}=[\mathrm{H^+}]^2$ . In pure water that is $10^{-14}$ bar.
Oxygen electrode: $E=1.23-0.059\,\mathrm{pH}$ (at 1 bar $\mathrm{O_2}$ ).
Quinhydrone electrode: $E=0.70-0.059\,\mathrm{pH}$ .
$\mathrm{MnO_4^-+8H^++5e^-\to Mn^{2+}+4H_2O}$ : $[\mathrm{H^+}]^8$ in the log. $\mathrm{Cr_2O_7^{2-}+14H^++6e^-\to 2Cr^{3+}+7H_2O}$ : $[\mathrm{H^+}]^{14}$ .
In a basic solution a metal ion is fixed by $K_{sp}$ : $[\mathrm{Cu^{2+}}]=K_{sp}/[\mathrm{OH^-}]^2$ .
For a buffer, find the pH first: $\mathrm{pH}=\mathrm{p}K_a+\log\frac{[\text{salt}]}{[\text{acid}]}$ .

Hydrogen electrode

E_{H^+/H_2}=-0.059\,\mathrm{pH}-\frac{0.059}{2}\log p_{H_2}

Worked example

A hydrogen electrode at 1 bar dips into a solution of pH 5 at 298 K. Find its potential. What

\mathrm{H_2}

pressure would make it zero in the same solution?

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 22 Jan 2026 Shift 2 · Q46Moderate

Example 3 · Electrochemistry · Nernst Equation and Concentration Effects

Consider the following electrochemical cell :

Pt\left| O_{2}(g)(lbar) \right|HCl(aq)\| M^{2 +}(aq,1.0M) \mid M(s)

The pH above which, oxygen gas would start to evolve at anode is

\_\_\_\_

(nearest integer).

\begin{bmatrix} \text{~Given :~} & \left. \ \begin{matrix} E_{M^{2 +}/M}^{0} = 0.994\text{ }V \\ E_{O_{2}/H_{2}O}^{0} = 1.23\text{ }V \end{matrix} \right\}\text{~standard reduction potential~} \\ & \text{~and~}\frac{RT}{F}(2.303) = 0.059\text{ }V\text{~at the given condition~} \end{bmatrix}

Electrode potential is not the cell emf

"Potential of the hydrogen electrode" means the single electrode's reduction potential,

-0.059\,\mathrm{pH}

. It is not an

E_{cell}

, and no second electrode is subtracted.

Forgetting the pressure term

A hydrogen electrode at 2 atm or 0.1 bar needs

-0.0295\log p_{H_2}

as well as the pH term.

Drill 9 more on electrodes that depend on ph

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 (3)

Cell emf from the Nernst equation
Nernst equation at 298 K
$E_{cell}=E^\circ_{cell}-\frac{0.059}{n}\log Q$
Solving the Nernst equation for an unknown
Nernst equation, rearranged
$\log Q=\frac{n\left(E^\circ_{cell}-E_{cell}\right)}{0.059}$
Electrodes that depend on pH
Hydrogen electrode
$E_{H^+/H_2}=-0.059\,\mathrm{pH}-\frac{0.059}{2}\log p_{H_2}$

Watch out for (6)

Dropping the powers in Q
→ Cell emf from the Nernst equation
Q upside down
→ Cell emf from the Nernst equation
Losing the sign of the log
→ Solving the Nernst equation for an unknown
Square root forgotten
→ Solving the Nernst equation for an unknown
Electrode potential is not the cell emf
→ Electrodes that depend on pH
Forgetting the pressure term
→ Electrodes that depend on pH

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