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.

Showing the 12 most recent of 14 on this concept.

CBSE 2026Set ANNUAL1 mark
$Q. What is the potential difference between the two electrodes of the galvanic cell called?$
›Reveal solutionSolution

The potential difference between the two electrodes of a galvanic cell (measured when no current is drawn) is called the electromotive force (EMF) or cell potential, $E_{\text{cell}}$ .

Concept. In a galvanic (voltaic) cell, the two half-cells are at different electrode potentials. The difference between the cathode and anode potentials is what pushes electrons through the external circuit:

$E_{\text{cell}} = E_{\text{cathode}} - E_{\text{anode}}$

When this potential difference is measured under zero-current conditions (using a potentiometer, so the cell reaction is effectively at equilibrium and no IR drop occurs), it is the maximum potential difference the cell can deliver and is termed the electromotive force (EMF) of the cell.

✓Final answer
$E_{\text{cell}}$
CBSE 2026Set ANNUAL1 markMCQ
$Q. Consider the following statements about a reaction at equilibrium: A(g) + B(g) \leftrightarrow C(g). Statement I: Adding an inert gas at constant volume will shift the equilibrium to the right. Statement II: A catalyst changes the position of equilibrium.$
(a) i) Both statement I and II are correct
(b) ii) Both statement I and II are incorrect
(c) iii) Statement I is correct and statement II is incorrect
(d) iv) Statement I is incorrect and statement II is correct
›Reveal solutionSolution
[!TLDR]

ii) Both statement I and II are incorrect

Why

Adding an inert gas at constant volume does not change partial pressures/concentrations of reacting species, so it does not shift equilibrium (Statement I false). A catalyst speeds up attainment of equilibrium equally in both directions and never shifts its position (Statement II false).

[!ANSWER]

ii) Both statement I and II are incorrect
CBSE 2025Set ANNUAL1 mark
$Zn(s)+Cu^{2+}(aq) \rightarrow Zn^{2+}(aq)+Cu(s)$
›Reveal solutionSolution
An external potential exactly equal and opposite to the cell's own EMF brings the system to balance, so no net current flows in either direction — this is the basis of potentiometric EMF measurement.

The Daniell-type cell $Zn(s)|Zn^{2+}(aq)\|Cu^{2+}(aq)|Cu(s)$ spontaneously drives current in the galvanic direction (electrons flow from Zn anode to Cu cathode through the external circuit) with an EMF of 1.1 V under standard conditions.

If an external opposing potential is applied, it works against this spontaneous cell reaction:

If the external potential is less than 1.1 V, the cell's own EMF still dominates, and current continues to flow in the original (galvanic) direction, though at a reduced magnitude.

If the external potential is greater than 1.1 V, it overpowers the cell's own EMF, and current is forced to flow in the reverse direction (the cell now behaves as an electrolytic cell, being charged/driven backward).

If the external potential exactly equals 1.1 V (as given here), the two opposing potentials exactly cancel, and no net current flows — the system is in electrochemical balance (equilibrium).

(This exact-balance condition is the working principle behind the potentiometric method of accurately measuring a cell's true EMF.)

✓Final answer
$Since the applied external potential (1.1 V) exactly equals and opposes the cell's own EMF, the two cancel and no net current flows in either direction.$
CBSE 2025Set ANNUAL1 markMCQ
$Q. The correct statement in a cell of zinc and copper is$
(a) zinc acts as cathode and copper as anode
(b) zinc acts as anode and copper as cathode
(c) the standard reduction potential of zinc is more than that of copper
(d) the flow of electrons is from copper to zinc
›Reveal solutionSolution

Zinc has a lower (more negative) standard reduction potential than copper, so it is oxidized (anode) while copper is reduced (cathode).

In a Daniell-type zinc–copper cell, E°(Zn²⁺/Zn) = −0.76 V is lower than E°(Cu²⁺/Cu) = +0.34 V. The electrode with the lower (more negative) reduction potential is oxidized — zinc loses electrons and acts as the anode (Zn → Zn²⁺ + 2e⁻) — while the electrode with the higher reduction potential is reduced — copper gains electrons and acts as the cathode (Cu²⁺ + 2e⁻ → Cu). Electrons flow through the external circuit from zinc to copper (not the reverse), and since Zn has the more negative (smaller) standard reduction potential, statement (c) is false too.

✓Final answer
$(b) zinc acts as anode and copper as cathode.$
CBSE 2024Set D1 markMCQ
$Q. The electromotive force of the cell Zn | ZnSO4 || CuSO4 | Cu is 1.1 volt. Its cathode is$
(a) Zn
(b) Cu
(c) ZnSO4
(d) CuSO4
›Reveal solutionSolution
Reduction happens at the cathode; Cu2+ is reduced to Cu, so Cu is the cathode.

In the Daniell cell Zn | ZnSO4 || CuSO4 | Cu:

Anode (oxidation, left): Zn -> Zn2+ + 2e-

Cathode (reduction, right): Cu2+ + 2e- -> Cu

By convention the electrode written on the right of a cell notation is the cathode where reduction occurs. The standard EMF = E(cathode) - E(anode) = 0.34 - (-0.76) = +1.10 V, matching the given 1.1 V. Because copper has the higher (more positive) reduction potential, Cu2+ is reduced and copper is the cathode.

✓Final answer
$(b) Cu — the copper electrode is the cathode (reduction of Cu2+).$
CBSE 2024Set ANNUAL1 markMCQ
$Q. An electrochemical cell can behave like an electrolytic cell when _______.$
(a) Ecell = 0
(b) Ecell > Eext
(c) Eext > Ecell
(d) Ecell = Eext
›Reveal solutionSolution
A galvanic (electrochemical) cell starts behaving like an electrolytic cell when an external potential greater than the cell's own emf is applied against it, reversing the direction of current flow.

Consider a Daniell cell: Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu(s), which normally works as a galvanic cell producing a cell potential Ecell, with electrons flowing from Zn (anode) to Cu (cathode) through the external circuit.

If an external opposing emf (Eext) is applied to this cell:

When Eext < Ecell, the cell continues to work as a galvanic cell, but the current decreases.

When Eext = Ecell, no current flows through the cell (this is used to measure the cell's emf accurately, e.g. using a potentiometer).

When Eext > Ecell, the direction of current flow reverses. Electrons are now forced to flow from Cu to Zn, so Cu is oxidised and Zn2+ is reduced — exactly the reverse of the spontaneous cell reaction. The cell now behaves as an electrolytic cell, consuming electrical energy to drive a non-spontaneous reaction (this is the working principle of charging a rechargeable/secondary cell such as a lead storage battery).

✓Final answer
$The cell behaves as an electrolytic cell when Eext > Ecell (option c).$
CBSE 2024Set ANNUAL1 mark
$Q. Write True/False: A hydrogen bridge is used to maintain continuity of ion flow in a Daniell cell.$
›Reveal solutionSolution

This statement is FALSE. A Daniell cell uses a salt bridge (e.g. containing KCl or KNO3 in agar-agar gel), not any "hydrogen bridge", to complete the internal circuit.

A Daniell cell consists of a Zn electrode dipped in ZnSO4 solution (anode) and a Cu electrode dipped in CuSO4 solution (cathode), connected externally by a wire and internally by a salt bridge. The salt bridge allows ions to migrate between the two half-cells, maintaining electrical neutrality in each compartment as the cell reaction proceeds, and completes the internal circuit without letting the two solutions mix directly. There is no such thing as a "hydrogen bridge" performing this role.

✓Final answer
$False -- it is a salt bridge, not a hydrogen bridge, that maintains continuity of ion flow in a Daniell cell.$
CBSE 2023Set 56/1/11 markMCQ
$Zn + 2Ag^+ \rightarrow Zn^{2+} + 2Ag$
›Reveal solutionSolution
By convention the anode (oxidation) is written on the left and the cathode (reduction) on the right. Zinc is oxidised and silver ions are reduced, so the cell is $Zn \mid Zn^{2+} \parallel Ag^+ \mid Ag$ — option (D).

A cell diagram is written anode (left) $\parallel$ cathode (right), with each half-cell running from the electrode metal outward and the double bar $\parallel$ marking the salt bridge.

For the reaction
$Zn + 2Ag^+ \rightarrow Zn^{2+} + 2Ag$

Zinc loses electrons: $Zn \rightarrow Zn^{2+} + 2e^-$ (oxidation, anode, left).

Silver ions gain electrons: $Ag^+ + e^- \rightarrow Ag$ (reduction, cathode, right).

Writing the anode as metal $\mid$ ion and the cathode as ion $\mid$ metal gives
$Zn \mid Zn^{2+} \parallel Ag^+ \mid Ag$
Checking the options:

(A) $2Ag \mid Ag^+ \parallel Zn \mid Zn^{2+}$ — electrodes reversed and coefficients are never used.

(B) $Ag^+ \mid Ag \parallel Zn^{2+} \mid Zn$ — anode/cathode reversed.

(C) $Ag \mid Ag^+ \parallel Zn \mid Zn^{2+}$ — anode/cathode reversed.

(D) $Zn \mid Zn^{2+} \parallel Ag^+ \mid Ag$ — Zn (anode) on the left, Ag (cathode) on the right. Correct.

✓Final answer
$Zn \mid Zn^{2+} \parallel Ag^+ \mid Ag$
CBSE 2022Set E1 markMCQ
$Q. The standard electrode potentials for the following reactions are given ( At 25°C ): Ag+(aq) + e- -> Ag(s), E° Ag+/Ag = +0.80 V ; Sn2+(aq) + 2e -> Sn(s), E° Sn2+/Sn = -0.14 V. The electromotive force (EMF) of the given cell Sn | Sn2+ (1M) || Ag+ (1M) | Ag is$
(a) 0.66 V
(b) 0.80 V
(c) 1.08 V
(d) 0.94 V
›Reveal solutionSolution
For Sn | Sn2+ || Ag+ | Ag, EMF = E°(Ag+/Ag) - E°(Sn2+/Sn) = 0.80 - (-0.14) = 0.94 V.

In the cell notation the left electrode is the anode (oxidation) and the right is the cathode (reduction):

Cathode (reduction): Ag+ + e- -> Ag, E° = +0.80 V

Anode (oxidation): Sn -> Sn2+ + 2e-, E°(Sn2+/Sn) = -0.14 V

E°cell = E°cathode - E°anode = (+0.80) - (-0.14) = +0.94 V.

(The positive value confirms the reaction Sn + 2Ag+ -> Sn2+ + 2Ag is spontaneous. Standard potentials are intensive, so they are not multiplied by the number of electrons.)

✓Final answer
$(d) 0.94 V.$
CBSE 2022Set ANNUAL1 markMCQ
$Q. Which one of the following statements is incorrect for a voltaic cell ?$
(a) It converts chemical energy to electrical energy.
(b) It uses electrical energy to carry out chemical changes.
(c) It is based on a redox reaction.
(d) It has $-\Delta G$ .
›Reveal solutionSolution
A voltaic cell produces electricity from a spontaneous redox reaction — it does not consume electrical energy, so statement (b) describes an electrolytic cell instead.

Checking each option against what a voltaic (galvanic) cell actually does:

(a) True — a voltaic cell converts chemical energy into electrical energy.

(b) False — this describes an electrolytic cell, which uses externally supplied electrical energy to force a non-spontaneous chemical change. A voltaic cell does the opposite.

(c) True — it is based on a spontaneous redox (oxidation–reduction) reaction.

(d) True — a spontaneous cell reaction has $\Delta G < 0$ , i.e. $-\Delta G$ .

✓Final answer
$(b) — "It uses electrical energy to carry out chemical changes" is the incorrect statement for a voltaic cell.$
CBSE 2022Set ANNUAL1 markMCQ
$Mg\,|\,Mg^{2+}\,||\,Cu^{2+}\,|\,Cu$
(a) $Mg$ as cathode
(b) $Cu$ as cathode
(c) $Cu$ is oxidizing agent
(d) None of the above
›Reveal solutionSolution
By IUPAC convention the electrode written on the LEFT of a cell is the anode and the one on the RIGHT is the cathode. Here Mg is the anode and Cu is the cathode. Option (B).

The cell is written as $Mg\,|\,Mg^{2+}\,||\,Cu^{2+}\,|\,Cu$ .

Convention: anode (negative, oxidation) is written on the left; cathode (positive, reduction) is written on the right.

The electrode reactions are:

Anode (Mg, oxidation): $Mg \rightarrow Mg^{2+} + 2e^-$

Cathode (Cu, reduction): $Cu^{2+} + 2e^- \rightarrow Cu$

Since $E^\circ_{Mg^{2+}/Mg} = -2.37\ V$ and $E^\circ_{Cu^{2+}/Cu} = +0.34\ V$ , Mg (more negative) is oxidised and acts as the anode; $Cu^{2+}$ is reduced, so Cu is the cathode. ( $Cu^{2+}$ is the oxidising agent here, not Cu metal, so option C is wrong.)

✓Final answer
$Cu$
CBSE 2021Set A1 markMCQ
$Q. Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu(s) is$
(a) Weston cell
(b) Daniel cell
(c) Calomel cell
(d) None of these
›Reveal solutionSolution
A zinc-copper galvanic cell with this notation is the Daniell cell.

The cell Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu(s) is the Daniell cell, a galvanic (voltaic) cell.

At the anode (LHS): Zn(s) → Zn2+ + 2e- (oxidation).

At the cathode (RHS): Cu2+ + 2e- → Cu(s) (reduction).

Standard EMF = E°(cathode) - E°(anode) = 0.34 - (-0.76) = +1.10 V.

The Weston cell (standard cell) uses Cd/Hg, and the calomel electrode uses Hg/Hg2Cl2 — neither matches this notation.

✓Final answer
$(B) Daniel cell.$

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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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