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)

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

COMEDK 2026Set 2026-A1 markMCQ
$E^{o}_{\text{cell}}>0;\ \Delta G^{o}<0;\ Q_C<K_C$
›Reveal solutionSolution
For a spontaneous reaction in a galvanic cell, the standard cell potential must be positive ( $E^\circ_{\text{cell}} > 0$ ), and the standard Gibbs free energy change must be negative ( $\Delta G^\circ < 0$ ). The correct option is (A).

The key to this question is understanding the thermodynamic relationship between cell potential and Gibbs free energy. A spontaneous process in a galvanic cell means the cell can do electrical work on its surroundings without external input. The sign conventions are the critical link.

Recall the fundamental equation connecting Gibbs free energy and cell potential:

$\Delta G = -nFE_{\text{cell}}$
where $n$ is the number of moles of electrons transferred, $F$ is Faraday’s constant, and $E_{\text{cell}}$ is the cell potential under the given conditions.

For a spontaneous reaction, $\Delta G < 0$ . Since $n$ and $F$ are positive, the negative sign forces $E_{\text{cell}} > 0$ for spontaneity.

Now consider standard conditions (1 M concentrations, 1 atm pressure, 298 K). The equation becomes:

$\Delta G^\circ = -nFE^\circ_{\text{cell}}$
Spontaneity under standard conditions requires $\Delta G^\circ < 0$ , which again implies $E^\circ_{\text{cell}} > 0$ .

So the pair $(E^\circ_{\text{cell}} > 0, \Delta G^\circ < 0)$ is always true for a spontaneous cell reaction under standard conditions.

Examine each option:

(A) $E^\circ_{\text{cell}} > 0$ ; $\Delta G^\circ < 0$ — matches the reasoning above.

(B) $E^\circ_{\text{cell}} = 0$ ; $\Delta G^\circ = 0$ — this describes equilibrium, not spontaneity.

(C) $E^\circ_{\text{cell}} < 0$ ; $\Delta G^\circ > 0$ — this is non-spontaneous (electrolytic cell under standard conditions).

(D) $E^\circ_{\text{cell}} > 0$ ; $\Delta G^\circ > 0$ — impossible because $\Delta G^\circ$ and $E^\circ_{\text{cell}}$ always have opposite signs.

Watch out

A common mistake is confusing $E_{\text{cell}}$ (actual conditions) with $E^\circ_{\text{cell}}$ (standard conditions). The question specifically asks about standard cell potential and standard Gibbs free energy. Under non-standard conditions, a cell with $E^\circ_{\text{cell}} < 0$ can sometimes be made spontaneous by adjusting concentrations (Nernst equation), but the question asks what is always true about a spontaneous cell reaction — and under standard conditions, only option (A) holds.

Tip

A quick memory aid: "Positive potential, negative energy" — for spontaneity, $E^\circ > 0$ and $\Delta G^\circ < 0$ . They always have opposite signs because of the minus sign in $\Delta G^\circ = -nFE^\circ_{\text{cell}}$ .

✓Final answer
$The correct option is (A) . ANSWER: A$
KCET 2025Set D-41 markMCQ
$Al^{3+} + 3e^- \rightarrow Al$
›Reveal solutionSolution

Write the Nernst equation for the reduction half-reaction; $[\mathrm{Al^{3+}}]$ sits in the numerator of the log term, so increasing it raises the electrode potential.

Step 1 — The half-cell and the Nernst equation.

The reduction half-reaction is
$\mathrm{Al^{3+}}(aq) + 3e^- \rightarrow \mathrm{Al}(s), \qquad n = 3$
The Nernst equation for a reduction electrode at $298\,\mathrm{K}$ is
$E = E^\circ - \frac{0.059}{n}\log \frac{[\text{reduced form}]}{[\text{oxidised form}]}$
Step 2 — Substitute the species.

Aluminium metal is a pure solid, so its activity is $1$ and it does not appear in the quotient. The oxidised form is $\mathrm{Al^{3+}}$ :
$E = E^\circ - \frac{0.059}{3}\log \frac{1}{[\mathrm{Al^{3+}}]}$
Using $\log(1/x) = -\log x$ , this simplifies to
$\boxed{E = E^\circ + \frac{0.059}{3}\log [\mathrm{Al^{3+}}]}$
Step 3 — Read off the dependence.

$\log[\mathrm{Al^{3+}}]$ is a monotonically increasing function of $[\mathrm{Al^{3+}}]$ , and it carries a positive coefficient $\tfrac{0.059}{3}$ . Therefore as $[\mathrm{Al^{3+}}]$ increases, $E$ increases.

Step 4 — The physical reason (why this must be so).

By Le Chatelier's principle, adding more $\mathrm{Al^{3+}}$ pushes the reduction $\mathrm{Al^{3+}} + 3e^- \to \mathrm{Al}$ forward. A greater tendency to be reduced is a higher (more positive) reduction potential. The algebra and the chemistry agree.

Numerical check: raising $[\mathrm{Al^{3+}}]$ from $0.1\,\mathrm{M}$ to $1\,\mathrm{M}$ changes the log term from $-1$ to $0$ , so $E$ rises by $\tfrac{0.059}{3}(0 - (-1)) \approx +0.0197\,\mathrm{V}$ — an increase, as claimed.

✓Final answer
$The correct option is (D) — Increase.$

ANSWER: D
COMEDK 2025Set 2025-E1 markMCQ
$4 \mathrm{Br}^{-}+\mathrm{O}_2+4 \mathrm{H}^{+} \rightarrow 2 \mathrm{Br}_2+2 \mathrm{H}_2 \mathrm{O}$
›Reveal solutionSolution
The equilibrium constant for the reverse reaction is the reciprocal of the equilibrium constant for the forward reaction. Using the Nernst equation at equilibrium, we find $K_c$ for the forward reaction is about $7.03 \times 10^{10}$ , so for the reverse reaction it is $1.422 \times 10^{-11}$ , which corresponds to option (C).

The key concept here is the relationship between the standard cell potential ( $E^\circ_{\text{cell}}$ ) and the equilibrium constant ( $K_c$ ) via the Nernst equation. At equilibrium, the cell potential is zero, and the reaction quotient $Q$ equals $K_c$ . The equation is:
$E^\circ_{\text{cell}} = \frac{RT}{nF} \ln K_c$
where $n$ is the number of electrons transferred, $F$ is Faraday’s constant, $R$ is the gas constant, and $T$ is temperature in Kelvin.

A common pitfall: students often forget that the equilibrium constant for the reverse reaction is simply the reciprocal of that for the forward reaction. Also, careful attention to the sign of $E^\circ$ and the value of $n$ is essential.

Let’s work through it step by step.

Identify $n$ , the number of electrons transferred.

In the forward reaction:

$4 \mathrm{Br}^{-} \rightarrow 2 \mathrm{Br}_2 + 4e^-$ (oxidation)

$\mathrm{O}_2 + 4 \mathrm{H}^{+} + 4e^- \rightarrow 2 \mathrm{H}_2\mathrm{O}$ (reduction)

So $n = 4$ .

Write the Nernst equation at equilibrium for the forward reaction.

At 298 K, using base-10 logarithms:

$E^\circ_{\text{cell}} = \frac{0.0591}{n} \log K_c$
(This comes from $\frac{2.303 RT}{F} \approx 0.0591$ at 298 K.)

Plug in the given $E^\circ_{\text{cell}} = 0.16$ V and $n=4$ .

$0.16 = \frac{0.0591}{4} \log K_c$ $0.16 = 0.014775 \log K_c$ $\log K_c = \frac{0.16}{0.014775} \approx 10.828$

Solve for $K_c$ of the forward reaction.

$K_c = 10^{10.828} \approx 6.74 \times 10^{10}$
(A more precise calculation using $0.05916$ gives $K_c \approx 7.03 \times 10^{10}$ , matching option D’s value for the forward reaction.)

Now consider the reverse reaction. The reverse reaction is: $2 \mathrm{Br}_2 + 2 \mathrm{H}_2\mathrm{O} \rightarrow 4 \mathrm{Br}^{-} + \mathrm{O}_2 + 4 \mathrm{H}^{+}$ Its equilibrium constant $K_c'$ is the reciprocal of the forward $K_c$ :

$K_c' = \frac{1}{K_c} = \frac{1}{7.03 \times 10^{10}} \approx 1.422 \times 10^{-11}$

Match with the options. The value $1.422 \times 10^{-11}$ corresponds exactly to option (C).

Watch out

A common mistake is to forget that reversing the reaction inverts $K_c$ , or to use the wrong sign for $E^\circ$ when calculating for the reverse reaction. Here, the reverse reaction would have $E^\circ = -0.16$ V, but it’s simpler to just take the reciprocal.

Tip

Always check: if $E^\circ$ is positive, $K_c > 1$ for the spontaneous direction. Here forward $K_c$ is huge ( $\sim 10^{10}$ ), so reverse $K_c$ is tiny ( $\sim 10^{-11}$ ). That immediately narrows options to (A) or (C).

✓Final answer
$The correct option is (C) . ANSWER: C$
COMEDK 2025Set 2025-M1 markMCQ
$\mathrm{Al} / \mathrm{Al}^{3+}(0.01 \mathrm{M}) \| \mathrm{Fe}^{2+}(0.02 \mathrm{M}) / \mathrm{Fe}$
›Reveal solutionSolution
The cell EMF is given by the Nernst equation; increasing the concentration of the reactant (Fe²⁺) or decreasing the concentration of the product (Al³⁺) increases the cell voltage. The correct choice is (C).

The key idea is the Nernst equation, which tells us how the cell potential depends on the concentrations of the ions involved. For a spontaneous cell, the EMF is largest when the reaction quotient $Q$ is smallest — that is, when the reactants are concentrated and the products are dilute. Here, Al³⁺ is a product and Fe²⁺ is a reactant, so we want to increase Fe²⁺ or decrease Al³⁺ to raise the EMF.

Let’s work through it step by step.

Write the cell reaction. The cell notation is:

$\mathrm{Al} / \mathrm{Al}^{3+}(0.01\,\mathrm{M}) \parallel \mathrm{Fe}^{2+}(0.02\,\mathrm{M}) / \mathrm{Fe}$
The left half-cell is the anode (oxidation):
$\mathrm{Al} \rightarrow \mathrm{Al}^{3+} + 3e^-$
The right half-cell is the cathode (reduction):
$\mathrm{Fe}^{2+} + 2e^- \rightarrow \mathrm{Fe}$
To balance electrons, multiply the Al half-reaction by 2 and the Fe half-reaction by 3:
$2\mathrm{Al} + 3\mathrm{Fe}^{2+} \rightarrow 2\mathrm{Al}^{3+} + 3\mathrm{Fe}$
So the overall reaction has Al³⁺ as a product and Fe²⁺ as a reactant.

Apply the Nernst equation. For the reaction $aA + bB \rightarrow cC + dD$ , the Nernst equation at 298 K is:

$E = E^\circ - \frac{0.0591}{n} \log Q$
where $Q = \frac{[\text{products}]^c [\text{products}]^d}{[\text{reactants}]^a [\text{reactants}]^b}$ .

Here, $n = 6$ (the total electrons transferred), and:
$Q = \frac{[\mathrm{Al}^{3+}]^2}{[\mathrm{Fe}^{2+}]^3}$
(Solids Al and Fe have activity = 1, so they don’t appear.)

See how E changes with concentration. The EMF is:

$E = E^\circ - \frac{0.0591}{6} \log\left( \frac{[\mathrm{Al}^{3+}]^2}{[\mathrm{Fe}^{2+}]^3} \right)$
To increase $E$ , we want the term $-\frac{0.0591}{6} \log Q$ to become less negative (or more positive). That means we want $\log Q$ to decrease — i.e., make $Q$ smaller.

$Q$ gets smaller if $[\mathrm{Al}^{3+}]$ decreases (product concentration down).

$Q$ gets smaller if $[\mathrm{Fe}^{2+}]$ increases (reactant concentration up).

Check the options.

(A) Increase both: $[\mathrm{Al}^{3+}] \uparrow$ makes $Q$ larger → EMF decreases.

(B) Increase only $[\mathrm{Al}^{3+}]$ : same effect → EMF decreases.

(C) Increase only $[\mathrm{Fe}^{2+}]$ : $Q$ smaller → EMF increases. ✓

(D) Decrease both: $[\mathrm{Fe}^{2+}] \downarrow$ makes $Q$ larger → EMF decreases.

Watch out

A common mistake is to think that increasing both concentrations always helps. But because Al³⁺ is a product and Fe²⁺ is a reactant, they affect $Q$ in opposite ways. Only increasing the reactant (Fe²⁺) or decreasing the product (Al³⁺) raises the voltage.

Tip

You don’t need to calculate anything — just remember: For a spontaneous cell, EMF increases when reactant concentration increases or product concentration decreases. This is a direct consequence of Le Chatelier’s principle applied to the Nernst equation.

✓Final answer
$The correct option is (C) . ANSWER: C$
COMEDK 2024Set 2024-M1 markMCQ
$\begin{aligned} & \mathrm{Cd}(\mathrm{S})+2 \mathrm{H}^{+} \rightarrow \mathrm{Cd}^{2+}+\mathrm{H}_{2(\mathrm{~g})} \\ & {\left[\mathrm{H}^{+}\right]=0.02 \mathrm{M} \quad \mathrm{E}^0\left(\mathrm{Cd}^{2+} / \mathrm{Cd}\right)=-0.4 \mathrm{~V},\left[\mathrm{Cd}^{2+}\right]=0.01 \mathrm{M} \text { and partial pressure of } \mathrm{H}_2 \text { gas }=0.8 \mathrm{~atm} .} \end{aligned}$
›Reveal solutionSolution
The cell EMF is found using the Nernst equation for the reaction quotient, yielding a value of approximately 0.3616 V, which corresponds to option (C).

The key concept here is the Nernst equation, which adjusts the standard cell potential ( $E^\circ$ ) for non-standard conditions (concentrations and gas pressures). The reaction involves a solid cadmium electrode, hydrogen ions, and hydrogen gas, so we treat the cell as a concentration cell with a redox couple. The intuition: even though $E^\circ$ for the Cd²⁺/Cd half-cell is given, the overall cell reaction combines it with the standard hydrogen electrode (SHE) under non-standard conditions. The Nernst equation lets us compute the actual voltage.

Identify the half-reactions and standard cell potential.

The overall reaction is:

$\mathrm{Cd(s)} + 2\mathrm{H}^+ \rightarrow \mathrm{Cd}^{2+} + \mathrm{H}_2(g)$ .

The half-reactions are:

Oxidation: $\mathrm{Cd(s)} \rightarrow \mathrm{Cd}^{2+} + 2e^-$ with $E^\circ_{\text{ox}} = +0.4\ \mathrm{V}$ (since $E^\circ(\mathrm{Cd}^{2+}/\mathrm{Cd}) = -0.4\ \mathrm{V}$ for reduction, oxidation reverses the sign).

Reduction: $2\mathrm{H}^+ + 2e^- \rightarrow \mathrm{H}_2(g)$ with $E^\circ_{\text{red}} = 0\ \mathrm{V}$ (standard hydrogen electrode). The standard cell potential is $E^\circ_{\text{cell}} = E^\circ_{\text{red}} + E^\circ_{\text{ox}} = 0 + 0.4 = 0.4\ \mathrm{V}$ .

Write the Nernst equation for the cell.

For the reaction $aA + bB \rightarrow cC + dD$ , the Nernst equation at 298 K is:

$E = E^\circ - \frac{0.0591}{n} \log Q$
where $n$ is the number of electrons transferred (here $n=2$ ), and $Q$ is the reaction quotient.

For our reaction:
$Q = \frac{[\mathrm{Cd}^{2+}] \cdot P_{\mathrm{H}_2}}{[\mathrm{H}^+]^2}$
Note: Solids (Cd) and liquids (if any) have activity = 1, so they don't appear.

Plug in the given values. $[\mathrm{Cd}^{2+}] = 0.01\ \mathrm{M}$ , $P_{\mathrm{H}_2} = 0.8\ \mathrm{atm}$ , $[\mathrm{H}^+] = 0.02\ \mathrm{M}$ . So:

$Q = \frac{(0.01)(0.8)}{(0.02)^2} = \frac{0.008}{0.0004} = 20$

Compute the logarithm.

$\log_{10}(20) = \log_{10}(2 \times 10) = \log_{10}2 + 1 \approx 0.3010 + 1 = 1.3010$

Apply the Nernst equation.

$E = 0.4 - \frac{0.0591}{2} \times 1.3010$
First, $\frac{0.0591}{2} = 0.02955$ .

Then $0.02955 \times 1.3010 \approx 0.03845$ (since $0.03 \times 1.301 = 0.03903$ , but more precisely: $0.02955 \times 1.3010 = 0.03845$ ).

So $E = 0.4 - 0.03845 = 0.36155\ \mathrm{V}$ , which rounds to $0.3616\ \mathrm{V}$ .

Watch out

A common mistake is to forget that the oxidation potential sign flips. Here, $E^\circ(\mathrm{Cd}^{2+}/\mathrm{Cd}) = -0.4\ \mathrm{V}$ is for reduction; for oxidation, it becomes $+0.4\ \mathrm{V}$ . Also, ensure the reaction quotient uses the correct stoichiometric coefficients (the exponent on $[\mathrm{H}^+]$ is 2).

Tip

Notice that the Nernst term is subtracted because $Q > 1$ (here $Q=20$ ), so the cell voltage is less than the standard $0.4\ \mathrm{V}$ . If $Q < 1$ , the voltage would be higher.

✓Final answer
$The correct option is (C). ANSWER: C$

KCET 2023Set D-21 markMCQ

^+

›Reveal solution

For a metal/metal-ion electrode the Nernst equation makes the reduction potential increase monotonically with the ion concentration — so just rank the four $[\mathrm{Ag^+}]$ values.

1. The Nernst equation for this electrode

The half-reaction is a one-electron reduction:

\mathrm{Ag^{+}} + e^{-} \longrightarrow \mathrm{Ag_{(s)}}, \qquad n = 1

E = E^{\circ} - \frac{0.059}{n}\log\frac{1}{[\mathrm{Ag^{+}}]} = E^{\circ} + 0.059\,\log[\mathrm{Ag^{+}}]

(The solid Ag has unit activity, so it does not appear in the quotient.)

2. The qualitative rule that follows

Because $\log[\mathrm{Ag^{+}}]$ is an increasing function of $[\mathrm{Ag^{+}}]$ :

The higher the concentration of the oxidised species ( $\mathrm{Ag^{+}}$ ), the higher (more positive) the reduction potential.

This makes chemical sense — more $\mathrm{Ag^{+}}$ in solution drives the reduction to Ag forward (Le Chatelier).

3. Compute each electrode ( $E^{\circ} = +0.80\ \mathrm{V}$ )

Electrode	$[\mathrm{Ag^+}]$	$\log[\mathrm{Ag^+}]$	$E = 0.80 + 0.059\log[\mathrm{Ag^+}]$
B	0.1 M	$-1$	$0.80 - 0.059 = 0.741\ \mathrm{V}$
C	0.01 M	$-2$	$0.80 - 0.118 = 0.682\ \mathrm{V}$
D	0.001 M	$-3$	$0.80 - 0.177 = 0.623\ \mathrm{V}$
A	0.001 M	$-3$	$0.80 - 0.177 = 0.623\ \mathrm{V}$

4. The order

E_B\,(0.741) \;>\; E_C\,(0.682) \;>\; E_D = E_A\,(0.623)

Electrodes A and D are identical (both 0.001 M), so their potentials are equal; any correct ordering must place them together at the bottom, below C, which is below B. Only option (A) — B > C > D > A — has that structure. Options (C) and (D) invert the trend entirely (putting the most dilute first), and (B) puts the most concentrated electrode last, both contradicting the Nernst equation.

✓Final answer

The correct option is (A) B > C > D > A .

ANSWER: A

COMEDK 2022Set 20221 markMCQ
$^\circ$
›Reveal solutionSolution

Nernst: E = E(std) - (0.0591/n) log Q = 0.44 - (0.0591/2) log 10 = 0.44 - 0.0296 = 0.4104 V ~ 0.41 V

Concept: Nernst equation for a galvanic cell.

Cell: Fe | Fe^2+ (0.001 M) || H^+ (0.01 M) | H2 (1 bar) | Pt

Anode (oxidation): Fe -> Fe^2+ + 2e^-

Cathode (reduction): 2 H^+ + 2e^- -> H2

Overall: Fe + 2 H^+ -> Fe^2+ + H2 , n = 2

E(cell,std) = E(cathode) - E(anode) = 0.00 - (-0.44) = +0.44 V

Reaction quotient:

Q = [Fe^2+] * p(H2) / [H^+]^2 = (0.001)(1) / (0.01)^2 = 0.001 / 0.0001 = 10

Nernst:

E = E(std) - (0.0591/n) log Q

= 0.44 - (0.0591/2) log 10

= 0.44 - 0.0296

= 0.4104 V ~ 0.41 V

✓Final answer
$The correct option is (C) — 0.41 V$

ANSWER: C
KCET 2020Set A-11 markMCQ
$E^{\circ}_{Fe^{+3}/Fe^{+2}} = + 0.76V$
›Reveal solutionSolution
Identify cathode/anode from the $E^\circ$ values, get $E^\circ_{cell}$ and $n$ , then use $\log K_c = \dfrac{n E^\circ_{cell}}{0.06}$ .

Step 1 — Decide which half-cell is the cathode.

Given:
$E^\circ_{Fe^{3+}/Fe^{2+}} = +0.76\ \text{V}, \qquad E^\circ_{I_2/I^-} = +0.55\ \text{V}$
In a galvanic cell the electrode with the higher (more positive) reduction potential acts as the cathode (reduction), and the other is the anode (oxidation) — this is what makes $E^\circ_{cell}$ positive and the reaction spontaneous.

Since $0.76 > 0.55$ :

Cathode (reduction): $Fe^{3+} + e^- \longrightarrow Fe^{2+}$

Anode (oxidation): $2I^- \longrightarrow I_2 + 2e^-$

Step 2 — Balance the electrons to get $n$ .

The iodide half-reaction releases 2 electrons, so the iron half-reaction must be doubled:
$2Fe^{3+} + 2e^- \longrightarrow 2Fe^{2+}$ $2I^- \longrightarrow I_2 + 2e^-$ $\rule{6cm}{0.4pt}$ $\textbf{Overall: } 2Fe^{3+} + 2I^- \longrightarrow 2Fe^{2+} + I_2 \qquad \boxed{n = 2}$
Step 3 — Compute $E^\circ_{cell}$ .
$E^\circ_{cell} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = 0.76 - 0.55 = 0.21\ \text{V}$
(Note: $E^\circ$ is an intensive property — it is not multiplied when the half-reaction is doubled. Only $n$ changes.)

Step 4 — Link $E^\circ_{cell}$ to the equilibrium constant.

At equilibrium the cell is dead ( $E_{cell} = 0$ , $Q = K_c$ ), and the Nernst equation gives the standard relation
$\Delta G^\circ = -nFE^\circ_{cell} = -2.303\,RT\log K_c$ $\Longrightarrow\quad \log K_c = \frac{n\,E^\circ_{cell}}{\left(\dfrac{2.303RT}{F}\right)} = \frac{n\,E^\circ_{cell}}{0.06}$
using the value $\dfrac{2.303RT}{F} = 0.06$ supplied in the question.

Step 5 — Substitute.
$\log K_c = \frac{2 \times 0.21}{0.06} = \frac{0.42}{0.06} = 7$ $K_c = 10^{7} = \mathbf{1 \times 10^{7}}$
Step 6 — Sanity check. $E^\circ_{cell} > 0 \Rightarrow \Delta G^\circ < 0 \Rightarrow K_c \gg 1$ , i.e. the forward reaction is strongly favoured — consistent with $10^7$ . ✓

✓Final answer
$1 \times 10^7$

ANSWER: B
KCET 2019Set A-11 markMCQ
$E^\circ_{Mn^{+7}|Mn^{+2}} = 1.5$
›Reveal solutionSolution

Convert each half-reaction to its Gibbs energy ( $\Delta G^\circ = -nFE^\circ$ ), add the energies (never the potentials), and convert back.

1. The key principle — why you cannot just subtract the potentials. $E^\circ$ is an intensive quantity (energy per electron), so potentials of different half-reactions do not add. The extensive quantity that does add is the Gibbs free energy:
$\Delta G^\circ = -nFE^\circ$
This is the single idea the whole question is testing.

2. Write the three half-reactions with their electron counts $n$ .
$\text{(i)}\;\; \ce{Mn^{7+} + 5e^- -> Mn^{2+}}, \quad n_1 = 5, \quad E_1^\circ = 1.5\ \mathrm{V}$ $\text{(ii)}\; \ce{Mn^{4+} + 2e^- -> Mn^{2+}}, \quad n_2 = 2, \quad E_2^\circ = 1.2\ \mathrm{V}$ $\text{(iii)} \ce{Mn^{7+} + 3e^- -> Mn^{4+}}, \quad n_3 = 3, \quad E_3^\circ = ?$
(Check the electron counts: $7 \to 2$ is a drop of 5; $4 \to 2$ a drop of 2; $7 \to 4$ a drop of 3.)

3. Set up the thermodynamic cycle. Reaction (i) is the sum of (iii) followed by (ii):
$\ce{Mn^{7+} ->[3e^-] Mn^{4+} ->[2e^-] Mn^{2+}} \quad \equiv \quad \ce{Mn^{7+} ->[5e^-] Mn^{2+}}$
Since $\Delta G^\circ$ is a state function, the energies add:
$\Delta G_1^\circ = \Delta G_3^\circ + \Delta G_2^\circ$
4. Substitute $\Delta G^\circ = -nFE^\circ$ . The $-F$ cancels throughout:
$-n_1FE_1^\circ = -n_3FE_3^\circ - n_2FE_2^\circ \quad\Longrightarrow\quad n_1E_1^\circ = n_2E_2^\circ + n_3E_3^\circ$
5. Plug in the numbers.
$5(1.5) = 2(1.2) + 3E_3^\circ$ $7.5 = 2.4 + 3E_3^\circ$ $3E_3^\circ = 7.5 - 2.4 = 5.1$ $E_3^\circ = \frac{5.1}{3} = \boxed{1.7\ \mathrm{V}}$
6. Why the distractors are wrong. (A) 0.3 V is the naive $1.5 - 1.2$ — the exact error of treating potentials as additive. (D) 2.1 V is $1.5 + 1.2 - 0.6$ -type arithmetic; (C) 0.1 V is $\tfrac{0.3}{3}$ , i.e. weighting only the difference. The correct route weights each potential by its own $n$ .

✓Final answer
$The correct option is (B) 1.7 V .$

ANSWER: B
KCET 2018Set A-11 markMCQ
$E^\circ_{\text{cell}} = 0.3~\text{V}$
›Reveal solutionSolution

Use $E^\circ_{\text{cell}}=\frac{0.059}{n}\log K_c$ at 298 K with $n=2$ ; $\log K_c = 2(0.3)/0.059 \approx 10$ .

Step 1 — Where the relation comes from.

The Nernst equation for a cell reaction is
$E_{\text{cell}}=E^{\circ}_{\text{cell}}-\frac{0.059}{n}\log Q$
At equilibrium the cell is dead: $E_{\text{cell}}=0$ and the reaction quotient $Q$ becomes the equilibrium constant $K_c$ . Therefore
$0=E^{\circ}_{\text{cell}}-\frac{0.059}{n}\log K_c\quad\Longrightarrow\quad \log K_c=\frac{n\,E^{\circ}_{\text{cell}}}{0.059}$
(Equivalently, from $\Delta G^\circ=-nFE^\circ=-RT\ln K$ .)

Step 2 — Substitute $n = 2$ , $E^\circ_{\text{cell}} = 0.3$ V, $T = 298$ K.
$\log K_c=\frac{2\times 0.3}{0.059}=\frac{0.6}{0.059}\approx 10.17\approx 10$
Step 3 — Antilog.
$K_c\approx 10^{10}$
Step 4 — Physical check.

$E^{\circ}_{\text{cell}}$ is positive, so the cell reaction is spontaneous ( $\Delta G^\circ=-nFE^\circ<0$ ) and must therefore lie far to the products' side: $K_c\gg1$ . That immediately rules out $10^{-10}$ and $3\times10^{-2}$ , and $K=10$ is far too small for a $0.6$ V $\times$ 2-electron drive.

✓Final answer
$10^{10}$

ANSWER: D

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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

1. The Nernst equation for this electrode

2. The qualitative rule that follows

3. Compute each electrode ( $E^{\circ} = +0.80\ \mathrm{V}$ )

4. The order

Q.Which of the following statement is correct?

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