Write the Nernst equation and calculate emf of the following cell at 298 K : $Cr\,|\,Cr^{3+}\ (0.1\ M)\,||\,Fe^{2+}\ (0.01\ M)\,|\,Fe$ Given : $E^\circ_{Cr^{3+}/Cr} = -0.75$ V, $E^\circ_{Fe^{2+}/Fe} = -0.45$ V (log 10 = 1)

> [!TLDR] > With $E^\circ_{\text{cell}}=0.30\ \text{V}$, $n=6$ and $Q=10^4$, the Nernst equation gives $E_{\text{cell}}\approx 0.26\ \text{V}$. **Nernst equation.** For a cell reaction transferring $n$ electrons at 298 K, $$E_{\text{cell}}=E^\circ_{\text{cell}}-\frac{0.0591}{n}\log Q.$$ **1. Half-reactions.** - Anode (oxidation): $\text{Cr}\rightarrow \text{Cr}^{3+}+3e^-$ - Cathode (reduction): $\text{Fe}^{2+}+2e^-\rightarrow \text{Fe}$ **2. Balanced cell reaction** (LCM of 3 and 2 is 6): $$2\,\text{Cr}+3\,\text{Fe}^{2+}\rightarrow 2\,\text{Cr}^{3+}+3\,\text{Fe},\qquad n=6.$$ **3. Standard cell potential.** $$E^\circ_{\text{cell}}=E^\circ_{\text{cathode}}-E^\circ_{\text{anode}}=(-0.45)-(-0.75)=+0.30\ \text{V}.$$ **4. Reaction quotient** (pure solids omitted): …

Write the Nernst equation and calculate emf of the following cell at 298 K : Cr|Cr³⁺(0.1M)||Fe²⁺(0.01M)|Fe Giv

Question

Cr\,|\,Cr^{3+}\ (0.1\ M)\,||\,Fe^{2+}\ (0.01\ M)\,|\,Fe

CBSECBSE Class XII Board 2022Subjective· 3mImportance★★★★★

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Concept understanding — Nernst Equation

The Nernst Equation: Why Batteries Don't Always Give Their Rated Voltage

Imagine you have a fresh AA battery. It says 1.5 V on the side. But if you measure it with a voltmeter, you might get 1.58 V when it's new, and 1.2 V when it's almost dead. Why does the voltage change? The Nernst equation is the tool that tells you exactly why.

The Core Idea: Concentration Drives Voltage

Every electrochemical cell works because of a chemical reaction that wants to happen. But here's the key: how badly the reaction wants to happen depends on how much of each chemical is present.

Think of it like a slope. A steep hill gives you more energy when you roll down. A shallow hill gives you less. In a battery, the "hill" is the difference in concentration (or more precisely, activity) of ions between the two electrodes. When the battery is fresh, the hill is steep — lots of reactants, few products. As the battery runs, reactants get used up, products build up, the hill flattens, and the voltage drops.

The Nernst equation is the mathematical formula that calculates the exact voltage for any given set of concentrations.

The Precise Statement

For a general electrochemical reaction:

aA + bB \rightarrow cC + dD

The cell potential $E$ under non-standard conditions is:

E = E^\circ - \frac{RT}{nF} \ln Q

Where:

$E$ = cell potential under the given conditions (what you actually measure)
$E^\circ$ = standard cell potential (the voltage when all reactants and products are at 1 M concentration, 1 atm pressure, 25°C)
$R$ = universal gas constant (8.314 J/mol·K)
$T$ = temperature in Kelvin
$n$ = number of moles of electrons transferred in the balanced reaction
$F$ = Faraday constant (96,485 C/mol)
$Q$ = reaction quotient = $\frac{[C]^c [D]^d}{[A]^a [B]^b}$ (using concentrations for now)

At 25°C (298 K), the constants combine into a simpler form:

E = E^\circ - \frac{0.0592}{n} \log_{10} Q

Note

The 0.0592 comes from $\frac{2.303RT}{F}$ at 298 K. The 2.303 converts natural log to base-10 log, which is more convenient for calculations.

What It Actually Means

The equation has three parts:

$E^\circ$ — the "ideal" voltage when everything is at standard conditions. This is what you'd get in a textbook table.
$\frac{RT}{nF}$ — a scaling factor. It tells you how sensitive the voltage is to concentration changes. More electrons transferred ( $n$ ) means less sensitivity.
$\ln Q$ — the "concentration penalty". When $Q$ is small (lots of reactants, few products), $\ln Q$ is negative, so $E$ is higher than $E^\circ$ . When $Q$ is large (products building up), $\ln Q$ is positive, so $E$ drops below $E^\circ$ .

A Concrete Example

Consider the Daniell cell: $\text{Zn} | \text{Zn}^{2+} || \text{Cu}^{2+} | \text{Cu}$

The reaction is: $\text{Zn} + \text{Cu}^{2+} \rightarrow \text{Zn}^{2+} + \text{Cu}$

$E^\circ = 1.10 \text{ V}$ , $n = 2$

If $[\text{Cu}^{2+}] = 0.1 \text{ M}$ and $[\text{Zn}^{2+}] = 1.0 \text{ M}$ :

Q = \frac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]} = \frac{1.0}{0.1} = 10

$E = 1.10 - \frac{0.0592}{2} \log_{10}(10) = 1.10 - 0.0296 \times 1 = 1.07 \text{ V}$ …

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With $E^\circ_{\text{cell}}=0.30\ \text{V}$ , $n=6$ and $Q=10^4$ , the Nernst equation gives $E_{\text{cell}}\approx 0.26\ \text{V}$ .

Nernst equation. For a cell reaction transferring $n$ electrons at 298 K,

E_{\text{cell}}=E^\circ_{\text{cell}}-\frac{0.0591}{n}\log Q.

1. Half-reactions.

Anode (oxidation): $\text{Cr}\rightarrow \text{Cr}^{3+}+3e^-$
Cathode (reduction): $\text{Fe}^{2+}+2e^-\rightarrow \text{Fe}$

2. Balanced cell reaction (LCM of 3 and 2 is 6):

2\,\text{Cr}+3\,\text{Fe}^{2+}\rightarrow 2\,\text{Cr}^{3+}+3\,\text{Fe},\qquad n=6.

3. Standard cell potential.

E^\circ_{\text{cell}}=E^\circ_{\text{cathode}}-E^\circ_{\text{anode}}=(-0.45)-(-0.75)=+0.30\ \text{V}.

4. Reaction quotient (pure solids omitted): …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2025Set 56/4/11 markMCQ
$2Cu^{+} (aq) + Zn (s) \rightarrow 2Cu (s) + Zn^{2+} (aq)$
›Reveal solutionSolution
The cell voltage depends on the reaction quotient via the Nernst equation. As the reaction proceeds, $[Zn^{2+}]$ increases and $[Cu^+]$ decreases, so the voltage decreases. The correct option is (B).

The Nernst equation tells us that the actual voltage of an electrochemical cell under non-standard conditions is:
$E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.059}{n} \log Q$
where $Q$ is the reaction quotient. For the given reaction:
$2Cu^+ (aq) + Zn (s) \rightarrow 2Cu (s) + Zn^{2+} (aq)$
the reaction quotient is:
$Q = \frac{[Zn^{2+}]}{[Cu^+]^2}$
(Remember: pure solids like Zn and Cu have activity = 1, so they don’t appear in $Q$ .)

The number of electrons transferred, $n$ , is 2 (each Cu⁺ gains one electron, and two Cu⁺ ions are reduced; Zn loses two electrons).

So the Nernst equation becomes:
$E_{\text{cell}} = 1.28 - \frac{0.059}{2} \log \frac{[Zn^{2+}]}{[Cu^+]^2}$
Now, as the reaction progresses:

$[Zn^{2+}]$ increases — Zn metal is oxidised to Zn²⁺, so its concentration in solution rises.

$[Cu^+]$ decreases — Cu⁺ ions are reduced to Cu metal, so their concentration falls.

Both changes make the fraction $\frac{[Zn^{2+}]}{[Cu^+]^2}$ larger.

A larger $Q$ means $\log Q$ is larger (more positive).

Since we subtract this term, $E_{\text{cell}}$ decreases.

Watch out

A common mistake is to think that because $[Zn^{2+}]$ appears in the numerator, the voltage might increase. But the Nernst equation has a minus sign in front of the log term — so anything that increases $Q$ actually lowers the voltage. …
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CBSE 2024Set ANNUAL1 mark
$Q. In an electrochemical cell the free energy change is related to EMF of the cell as ______.$
›Reveal solutionSolution

The free energy change of a cell reaction is related to its EMF by Delta G = -nFE, which is the thermodynamic basis for the Nernst equation.

The electrical work done by a galvanic cell is equal to the product of the total charge passed and the EMF of the cell. The total charge passed when n moles of electrons flow is nF (F = Faraday constant = 96500 C/mol).

Maximum electrical work obtainable = nFE (E = EMF of the cell)

This maximum work done by the system equals the decrease in Gibbs free energy of the system, so:

Delta G = -nFE

…

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CBSE 2021Set A1 markMCQ
$Q. The Electromotive force (EMF) of the cell for the cell reaction at equilibrium state is$
(a) positive
(b) zero
(c) negative
(d) none of these
›Reveal solutionSolution

At equilibrium a galvanic cell is fully discharged, so its EMF = 0.

As a galvanic cell operates, the concentrations change until the reaction reaches equilibrium (the cell is 'dead').

From the Nernst equation: Ecell = E°cell − (0.059/n) log Q.

At equilibrium Q = K and Ecell = 0, giving the relation E°cell = (0.059/n) log K.

…

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CBSE 2020Set ANNUAL1 mark
$Q. What is the relation between standard Gibbs' free energy and standard emf of the cell?$
›Reveal solutionSolution

The standard Gibbs free energy change of a cell reaction is related to the standard cell emf by $\Delta G^\circ = -nFE^\circ_{cell}$ .

The maximum electrical work obtainable from a galvanic cell equals the decrease in Gibbs free energy of the cell reaction. The electrical work done is the product of the total charge passed (nF, where n is the number of moles of electrons transferred in the balanced cell reaction and F is Faraday's constant, 96500 C/mol) and the cell's emf:

Electrical work = nFE_cell

Since this work is done at the expense of the free energy of the system, $\Delta G = -nFE_{cell}$ , and under standard conditions:

…

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CBSE 2020Set ANNUAL1 mark
$Q. How is equilibrium constant related to standard Gibb's energy?$
›Reveal solutionSolution

The standard Gibbs energy change of a reaction and its equilibrium constant are linked by $\Delta G^\circ = -RT\ln K$ .

For any reaction at equilibrium, thermodynamics gives the relation

$\Delta G^\circ = -RT\ln K = -2.303\,RT\log K$

where $R$ is the gas constant, $T$ the absolute temperature, and $K$ the equilibrium constant.

This connects directly to electrochemistry through the Nernst equation. Since $\Delta G^\circ = -nFE^\circ_{cell}$ (where $n$ = number of electrons transferred, $F$ = Faraday constant, $E^\circ_{cell}$ = standard cell potential), equating the two expressions for $\Delta G^\circ$ gives

$-nFE^\circ_{cell} = -RT\ln K \quad\Rightarrow\quad E^\circ_{cell} = \frac{RT}{nF}\ln K = \frac{0.0591}{n}\log K \text{ (at 298 K)}$

…

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CBSE 2019Set ANNUAL1 mark
$Q. Write the Nernst equation for following cell: Sn(s) | Sn²⁺ || H⁺ | H₂(g)(1bar) | Pt(s).$
›Reveal solutionSolution

For the cell Sn(s) | Sn²⁺ || H⁺ | H₂(g)(1 bar) | Pt(s), n = 2 and the Nernst equation is $E_{cell}=E^\circ_{cell}-\frac{0.0591}{2}\log\frac{[Sn^{2+}]p_{H_2}}{[H^+]^2}$ .

From the cell notation, the left electrode (Sn) is the anode (oxidation) and the right electrode (Pt, with H₂/H⁺) is the cathode (reduction):

Anode (oxidation): $Sn(s) \rightarrow Sn^{2+}(aq) + 2e^-$

Cathode (reduction): $2H^+(aq) + 2e^- \rightarrow H_2(g)$

Overall cell reaction: $Sn(s) + 2H^+(aq) \rightarrow Sn^{2+}(aq) + H_2(g)$

Here the number of electrons transferred, $n = 2$ .

The general Nernst equation is $E_{cell}=E^\circ_{cell}-\frac{0.0591}{n}\log Q$ , where $Q$ is the reaction quotient (products over reactants, each raised to its stoichiometric coefficient, gases as partial pressure, solids/pure liquids omitted):

$Q = \dfrac{[Sn^{2+}]\, p_{H_2}}{[H^+]^2}$

So:

…

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CBSE 2019Set ANNUAL1 markMCQ
$Q. The electrode potential of any electrode does not depend on$
(a) the nature of metal and its ions
(b) concentration of ions present in solution
(c) pressure
(d) temperature
›Reveal solutionSolution
By the Nernst equation, electrode potential depends on the nature of the metal/ions, their concentration and temperature, but not on pressure (for a metal electrode), so option (c).

The potential of an electrode is governed by the Nernst equation:

E = E° - (RT/nF) ln (1/[M^n+])

From this and the nature of the half-cell:

It depends on the nature of the metal and its ions (through E°).

It depends on the concentration of the ions in solution (the [M^n+] term). …
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CBSE 2018Set ANNUAL1 markMCQ
$Q. For HO-C6H4-OH ⇌ O=C6H4=O + 2H+ + 2e-, E° = 1.30 V. At pH = 2, Electrode potential is -$
(a) 1.36 V
(b) 1.30 V
(c) 1.42 V
(d) 1.20 V
›Reveal solutionSolution

E = E° + 0.059·pH for this couple → 1.30 + 0.118 = 1.42 V.

For the quinhydrone-type couple written as H₂Q ⇌ Q + 2H⁺ + 2e⁻ (n = 2), the Nernst equation for the reaction as written is

E = E° − (0.059/2)·log([Q][H⁺]²/[H₂Q]).

Taking activities of quinone and hydroquinone as unity: …

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CBSE 2016Set ANNUAL1 mark
$Q. Write an equation for the relation between standard free energy change and standard cell potential.$
›Reveal solutionSolution

Standard free energy change and standard cell potential are linked by $\Delta G^\circ = -nFE^\circ_{cell}$ .

The maximum electrical work obtainable from a galvanic cell equals the decrease in Gibbs free energy of the cell reaction. Electrical work done = charge $\times$ potential = $nFE^\circ_{cell}$ , and since this work is done by the system (free energy decreases), we get

$\Delta G^\circ = -nFE^\circ_{cell}$

…

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