Which of the following statement is correct? (i) ECell and Δr G of cell reaction both are extensive properties

NCERT Exemplar · Q3

Q. Which of the following statement is correct?

(i)

E_{Cell}

and

\Delta_r G

of cell reaction both are extensive properties.

(ii)

E_{Cell}

and

\Delta_r G

of cell reaction both are intensive properties.

(iii)

E_{Cell}

is an intensive property while

\Delta_r G

of cell reaction is an extensive property.

(iv)

E_{Cell}

is an extensive property while

\Delta_r G

of cell reaction is an intensive property.

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Concept understanding — Cell Representation Nernst Equation

Cell Representation and the Nernst Equation: From Intuition to Precision

Imagine you have a Daniell cell — a zinc rod in zinc sulphate solution connected by a salt bridge to a copper rod in copper sulphate solution. You know it produces a voltage. But what happens if you dilute the copper sulphate solution? Or if you change the temperature? The voltage changes. The Nernst equation is the tool that tells you exactly how much it changes.

The Intuition First

A battery works because the two half-cells "want" to react — zinc wants to lose electrons, copper ions want to gain them. This "want" is measured as a tendency, or potential. But the strength of that tendency depends on how crowded the ions are.

Think of it like this: If you have a room full of people who all want to leave (like zinc ions wanting to form), the push to get out is stronger when the room is packed. If the room is nearly empty, the push is weaker. Similarly, for copper ions wanting to enter the metal (gain electrons), the pull is stronger when there are many copper ions around, and weaker when there are few.

The Nernst equation quantifies this: the actual cell potential depends on the concentrations (or activities) of the ions involved.

The Precise Statement

For a general cell reaction:

aA + bB \rightarrow cC + dD

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

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

Where:

$E$ = cell potential under the given conditions (in volts)
$E^\circ$ = standard cell potential (when all reactants/products are at 1 M, 1 atm, 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 half-reactions
$F$ = Faraday constant (96,485 C/mol)
$Q$ = reaction quotient = $\frac{[C]^c [D]^d}{[A]^a [B]^b}$ (using concentrations for dilute solutions)

At 25°C (298 K), the equation simplifies to a very practical form:

E = E^\circ - \frac{0.0591}{n} \log_{10} Q

The 0.0591 comes from $\frac{2.303RT}{F}$ at 298 K. Notice it uses $\log_{10}$ (common log), not natural log.

Cell Representation: How We Write It

In electrochemistry, we represent a cell with a shorthand notation. For the Daniell cell:

\text{Zn(s)} \mid \text{Zn}^{2+}(\text{aq}) \parallel \text{Cu}^{2+}(\text{aq}) \mid \text{Cu(s)}

The single vertical line $\mid$ represents a phase boundary (solid electrode | solution). The double line $\parallel$ represents the salt bridge.

The anode (oxidation) is written on the left, the cathode (reduction) on the right. Electrons flow from left to right in the external circuit.

Applying the Nernst Equation to a Cell Representation

For the Daniell cell, the half-reactions are:

Anode (oxidation): $\text{Zn(s)} \rightarrow \text{Zn}^{2+}(\text{aq}) + 2e^-$
Cathode (reduction): $\text{Cu}^{2+}(\text{aq}) + 2e^- \rightarrow \text{Cu(s)}$

Overall: $\text{Zn(s)} + \text{Cu}^{2+}(\text{aq}) \rightarrow \text{Zn}^{2+}(\text{aq}) + \text{Cu(s)}$

Here $n = 2$ (two electrons transferred). The reaction quotient is:

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

So the Nernst equation becomes:

E = E^\circ - \frac{0.0591}{2} \log_{10} \frac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]}

Note

Solids (Zn, Cu) do not appear in $Q$ because their concentrations are constant (activity = 1).

A Worked Example

Suppose you have a Daniell cell where $[\text{Zn}^{2+}] = 0.1$ M and $[\text{Cu}^{2+}] = 1.0$ M at 25°C. $E^\circ$ for the cell is 1.10 V.

E = 1.10 - \frac{0.0591}{2} \log_{10} \frac{0.1}{1.0}

E = 1.10 - \frac{0.0591}{2} \log_{10}(0.1)

\log_{10}(0.1) = -1

E = 1.10 - \frac{0.0591}{2}(-1) = 1.10 + 0.02955 = 1.1296 \text{ V}

The cell voltage is slightly higher than standard because the zinc ion concentration is lower (less "push" from the anode side, so the net driving force is larger).

Watch out

A common mistake is to forget that $n$ must match the balanced equation. If you write the half-reactions with different numbers of electrons, you'll get the wrong $n$ . Always check that the overall reaction is balanced.

The Key Insight

The Nernst equation is not just a formula — it's a statement that electrochemical potential is a logarithmic function of concentration. This means:

Diluting the reactant side (lowering $[ \text{Cu}^{2+} ]$ ) decreases $E$
Diluting the product side (lowering $[ \text{Zn}^{2+} ]$ ) increases $E$
At equilibrium, $E = 0$ and $Q = K$ (the equilibrium constant), giving $\ln K = \frac{nFE^\circ}{RT}$

This last point connects electrochemistry directly to thermodynamics — the Nernst equation is just the Gibbs free energy equation ( $\Delta G = -nFE$ ) written in terms of concentrations.

Cell representation and the Nernst equation together form a heavily tested pair within the NCERT/CBSE Class 12 Chemistry Electrochemistry chapter, and ‘how to write cell representation’ or ‘Nernst equation for cell reaction’ are common important-question searches for board exams, JEE Main and NEET. Being fluent in both the notation and the formula is essential for solving electrochemistry numericals quickly in competitive exams.

Why this formula?

Cell Representation & the Nernst Equation: Why It Works

The Core Question

Why does a cell's voltage change when concentrations change? The Nernst equation answers this — but the reason lies in the link between chemical free energy and electrical work.

1. The Fundamental Link: Gibbs Free Energy & Cell Potential

A galvanic cell does electrical work. The maximum useful work a cell can do equals the change in Gibbs free energy ( $\Delta G$ ):

\Delta G = -nFE_{\text{cell}}

Where:

$n$ = moles of electrons transferred
$F$ = Faraday constant ( $96,485 \, \text{C mol}^{-1}$ )
$E_{\text{cell}}$ = cell potential (volts)

Why negative? A spontaneous reaction has $\Delta G < 0$ and $E_{\text{cell}} > 0$ — the negative sign makes this consistent.

2. The Chemical Side: $\Delta G$ Depends on Concentration

For a general redox reaction:

aA + bB \rightarrow cC + dD

The Gibbs free energy under non-standard conditions is:

\Delta G = \Delta G^\circ + RT \ln Q

Where $Q$ is the reaction quotient:

Q = \frac{[C]^c [D]^d}{[A]^a [B]^b}

Why this form? It comes from the relationship between chemical potential and concentration — the entropy of mixing drives concentration dependence.

3. Combining Both Sides: The Derivation

Set the electrical work equal to the chemical free energy change:

-nFE_{\text{cell}} = -nFE^\circ_{\text{cell}} + RT \ln Q

Divide both sides by $-nF$ :

E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{RT}{nF} \ln Q

This is the Nernst equation.

4. The "Why" in Plain Terms

Concept	Physical Meaning
$E^\circ_{\text{cell}}$	Voltage when all species are at 1 M (standard state)
$-\frac{RT}{nF} \ln Q$	Correction factor — adjusts voltage for real concentrations
$Q$	Tells you how far the reaction is from equilibrium

Key insight: When $Q = K$ (equilibrium), $E_{\text{cell}} = 0$ — the battery is dead because no net reaction occurs.

5. The Common Form (log base 10)

At $25^\circ \text{C}$ ( $298 \, \text{K}$ ):

\frac{RT}{F} \ln 10 \approx 0.0592 \, \text{V}

So:

E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0592}{n} \log_{10} Q

Why convert to log? Exam convenience — most concentration values are powers of 10.

6. Cell Representation: How to Write $Q$

For a cell written as:

\text{Zn(s)} \mid \text{Zn}^{2+}(aq) \parallel \text{Cu}^{2+}(aq) \mid \text{Cu(s)}

The reaction is:

\text{Zn(s)} + \text{Cu}^{2+}(aq) \rightarrow \text{Zn}^{2+}(aq) + \text{Cu(s)}

Solids are omitted from $Q$ (activity = 1):

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

Why omit solids? Their concentration doesn't change — they're pure phases with fixed chemical potential.

7. Exam-Ready Summary

Step	What to Do	Why
1	Write balanced half-reactions	Identify $n$ (electrons transferred)
2	Write overall reaction	Determine $Q$ form
3	Plug into Nernst	Corrects $E^\circ$ for real conditions
4	Use $\log_{10}$ at $25^\circ \text{C}$	$0.0592/n$ is exam standard

Final takeaway: The Nernst equation is thermodynamics in disguise — it's the Gibbs free energy equation rewritten in electrical units. Every time you use it, you're balancing chemical potential against electrical potential.

The cell potential $E_{\text{Cell}}$ does not depend on the size of the cell or the amount of reactants — it is an intensive property. The Gibbs free energy change $\Delta_r G$ does depend on the amount of substance reacting — it is an extensive property. Hence option (iii) is correct.

The distinction between intensive and extensive properties is a fundamental idea in thermodynamics. An intensive property does not change when you scale the system up or down — temperature, pressure, and density are examples. An extensive property scales with the amount of matter — mass, volume, and total internal energy are examples.

Now, where does $E_{\text{Cell}}$ fit? Think about a Daniell cell. Whether you build a tiny cell with a few millilitres of solution or a giant industrial cell with litres of electrolyte, the voltage you measure between the two electrodes remains the same (assuming same concentrations, temperature, and pressure). That voltage is determined only by the nature of the half-reactions and the conditions — not by how much zinc or copper you have. So $E_{\text{Cell}}$ is intensive.

What about $\Delta_r G$ ? This is the Gibbs free energy change for the cell reaction as written. If you double the amount of reactants, you double the number of moles reacting, and therefore you double the free energy change. It scales with the extent of reaction. So $\Delta_r G$ is extensive.

Let's confirm this with the relation that connects them:

\Delta_r G = -n F E_{\text{Cell}}

Here $n$ is the number of moles of electrons transferred per mole of reaction as written. $F$ is Faraday's constant. Notice: $E_{\text{Cell}}$ is intensive, but when you multiply it by $n$ (which is a fixed stoichiometric number for the balanced equation) and $F$ , you get $\Delta_r G$ — which is extensive because it refers to the reaction as written. If you write the reaction for twice the amount, $n$ stays the same (it's per mole of reaction), but $\Delta_r G$ doubles because you have two moles of reaction. The equation is consistent.

Now let's walk through the options:

Option (i) says both are extensive. That is wrong because $E_{\text{Cell}}$ does not depend on the size of the system.
Option (ii) says both are intensive. That is wrong because $\Delta_r G$ scales with the amount of reaction.
Option (iii) says $E_{\text{Cell}}$ is intensive and $\Delta_r G$ is extensive. This matches our reasoning exactly.
Option (iv) reverses the two — incorrect.

Watch out

A common mistake is to think that because $\Delta_r G = -nFE_{\text{Cell}}$ , and $n$ and $F$ are constants, $\Delta_r G$ must also be intensive. But $n$ is a stoichiometric coefficient — it does not change with the amount of reaction. The extensivity of $\Delta_r G$ comes from the fact that it is defined for a specific amount of reaction (usually one mole of reaction as written). If you have two moles of reaction, $\Delta_r G$ doubles, while $E_{\text{Cell}}$ stays the same.

Tip

A quick way to test: ask yourself — "If I put two identical cells in parallel, does the voltage change?" No — voltage is like pressure, intensive. "If I double the amount of reactants, does the total free energy change double?" Yes — free energy is like mass, extensive.

✓Final answer

The correct option is (iii) .

Common Mistakes & How to Avoid Them

Mistake 1: Confusing Intensive vs. Extensive Properties

The Error: Students often think that because $E_{Cell}$ depends on the cell reaction (which involves amounts), it must be extensive. Similarly, they may assume $\Delta_r G$ is intensive because it's related to $E_{Cell}$ .

The Truth:

Intensive property: Independent of the amount of substance (e.g., temperature, density, $E_{Cell}$ ).
Extensive property: Depends on the amount of substance (e.g., mass, volume, $\Delta_r G$ ).

Why $E_{Cell}$ is intensive:

The cell potential is determined by the nature of the electrodes and the concentration of ions — not by how much electrode material or solution you use. A bigger cell gives the same voltage.

Why $\Delta_r G$ is extensive:

$\Delta_r G$ is the Gibbs energy change for the reaction as written. If you double the reaction (2 moles instead of 1), $\Delta_r G$ doubles. The relation is:

\Delta_r G = -n F E_{Cell}

Here, $n$ (number of electrons) scales with the reaction stoichiometry, making $\Delta_r G$ extensive.

How to Avoid:

Ask yourself: "If I double the size of the cell, does the voltage change?" (No → intensive). "Does the total energy change double?" (Yes → extensive).

Mistake 2: Misapplying the Nernst Equation

The Error: Students think the Nernst equation:

E_{Cell} = E^\circ_{Cell} - \frac{RT}{nF} \ln Q

implies $E_{Cell}$ depends on $n$ , so it must be extensive.

The Correction:

$n$ is a ratio (number of electrons per mole of reaction), not an amount. The term $\frac{RT}{nF}$ is a constant for a given reaction — it doesn't make $E_{Cell}$ scale with quantity.

How to Avoid:

Remember: $n$ cancels out when you write the full relation. The intensive nature of $E_{Cell}$ is preserved because $Q$ (reaction quotient) is also intensive.

Mistake 3: Forgetting the Sign Convention

The Error: Students pick option (iv) because they think "voltage is like energy, so it must be extensive."

The Correction:

Voltage (potential difference) is analogous to pressure or temperature — it's a driving force, not a quantity of energy. $\Delta_r G$ is the actual energy available, which scales with amount.

How to Avoid:

Think of a battery:

Voltage = how hard it pushes (intensive)
Total energy = voltage × charge (extensive, because charge depends on amount)

Mistake 4: Relying on Memory Instead of Logic

The Error: Students memorize "cell potential is intensive" without understanding why, then get confused when $\Delta_r G$ appears.

The Correction:

Use the test of doubling:

Double the cell size → $E_{Cell}$ stays same → intensive ✓
Double the reaction amount → $\Delta_r G$ doubles → extensive ✓

How to Avoid:

Always apply the "double the system" test before answering.

Final Answer

The correct option is (iii): $E_{Cell}$ is an intensive property while $\Delta_r G$ of cell reaction is an extensive property.

Quick Mnemonic:

E = Everywhere same (Intensive)
G = Grows with amount (Extensive)

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Q.Which of the following statement is correct?

Concept understanding — Cell Representation Nernst Equation

Cell Representation and the Nernst Equation: From Intuition to Precision

The Intuition First

The Precise Statement

Cell Representation: How We Write It

Applying the Nernst Equation to a Cell Representation

A Worked Example

The Key Insight

Why this formula?

Cell Representation & the Nernst Equation: Why It Works

The Core Question

1. The Fundamental Link: Gibbs Free Energy & Cell Potential

2. The Chemical Side: $\Delta G$ Depends on Concentration

3. Combining Both Sides: The Derivation

4. The "Why" in Plain Terms

5. The Common Form (log base 10)

6. Cell Representation: How to Write $Q$

7. Exam-Ready Summary

Method: Property Classification by Dependence on Amount of Substance

Step 1: Recall definitions

Step 2: Analyze $E_{\text{Cell}}$

Step 3: Analyze $\Delta_r G$ (Gibbs free energy change of reaction)

Step 4: Match with options

Common Mistakes & How to Avoid Them

Mistake 1: Confusing Intensive vs. Extensive Properties

Mistake 2: Misapplying the Nernst Equation

Mistake 3: Forgetting the Sign Convention

Mistake 4: Relying on Memory Instead of Logic

Final Answer

Q.Which of the following statement is correct?

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