(a) Following reaction takes place in the cell : $Zn\,(s) + Ag_2O\,(s) + H_2O\,(l) \rightarrow Zn^{2+}\,(aq) + 2Ag\,(s) + 2OH^-\,(aq)$ Calculate $\Delta_r G^{\circ}$ of the reaction. [Given : $E^{\circ}_{(Zn^{2+}/Zn)} = -0.76\ V$, $E^{\circ}_{(Ag^+/Ag)} = 0.80\ V$, $1\,F = 96{,}500\ C\ mol^{-1}$] (b) How can you determine limiting molar conductivity, ($\Lambda^{\circ}_m$) for strong electrolyte and weak electrolyte ?

(a) Following reaction takes place in the cell : Zn(s) + Ag₂O(s) + H₂O(l) arrow Zn²⁺(aq) + 2Ag(s) + 2OH^-(aq)

Question

Zn\,(s) + Ag_2O\,(s) + H_2O\,(l) \rightarrow Zn^{2+}\,(aq) + 2Ag\,(s) + 2OH^-\,(aq)

(b) How can you determine limiting molar conductivity, (

\Lambda^{\circ}_m

) for strong electrolyte and weak electrolyte ?

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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.

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

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.

--- …

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The cell reaction involves zinc oxidation and silver oxide reduction. From the half-cell potentials we construct $E^\circ_{\text{cell}} = 1.56\,\text{V}$ , giving $\Delta_r G^\circ = -301{,}080\,\text{J mol}^{-1}$ or $-301.08\,\text{kJ mol}^{-1}$ . Limiting molar conductivity for strong electrolytes comes from extrapolating a linear $\Lambda_m$ vs. $\sqrt{c}$ plot to zero concentration; for weak electrolytes we use Kohlrausch's law of independent migration of ions.

Part (a): Standard Gibbs energy from cell potential

The connection between electrochemistry and thermodynamics is direct: a spontaneous cell reaction releases free energy, and the electrical work the cell can perform equals $\Delta_r G$ . For standard conditions the relationship is

\Delta_r G^\circ = -nFE^\circ_{\text{cell}},

where $n$ is the number of electrons transferred per formula unit, $F$ is the Faraday constant, and $E^\circ_{\text{cell}}$ is the standard cell potential. The negative sign reflects the convention that a positive cell potential corresponds to a spontaneous (negative $\Delta G$ ) reaction.

1. Identify the half-reactions

The overall reaction is

\text{Zn}(s) + \text{Ag}_2\text{O}(s) + \text{H}_2\text{O}(l) \rightarrow \text{Zn}^{2+}(aq) + 2\,\text{Ag}(s) + 2\,\text{OH}^-(aq).

Zinc is oxidized:

\text{Zn}(s) \rightarrow \text{Zn}^{2+}(aq) + 2\,e^- \quad (\text{anode}).

Silver oxide is reduced. In alkaline medium $\text{Ag}_2\text{O}$ accepts electrons:

\text{Ag}_2\text{O}(s) + \text{H}_2\text{O}(l) + 2\,e^- \rightarrow 2\,\text{Ag}(s) + 2\,\text{OH}^-(aq) \quad (\text{cathode}).

Each half-reaction involves $n = 2$ electrons.

2. Find the standard reduction potential for the silver oxide half-cell

We are given $E^\circ(\text{Ag}^+/\text{Ag}) = +0.80\,\text{V}$ for the couple

\text{Ag}^+(aq) + e^- \rightarrow \text{Ag}(s).

The silver oxide reduction in alkaline solution can be related to the $\text{Ag}^+/\text{Ag}$ couple. The half-reaction

\text{Ag}_2\text{O}(s) + \text{H}_2\text{O}(l) + 2\,e^- \rightarrow 2\,\text{Ag}(s) + 2\,\text{OH}^-(aq)

is equivalent to the combination of $\text{Ag}^+$ reduction and the solubility equilibrium of $\text{Ag}_2\text{O}$ . For a silver–silver oxide alkaline cell the standard reduction potential is known to be $E^\circ = +0.34\,\text{V}$ (this is a standard value for the $\text{Ag}_2\text{O}/\text{Ag}$ couple in base). However, the problem expects us to work with the given data.

Tip

In many exam problems the $\text{Ag}_2\text{O}/\text{Ag}$ potential in alkaline medium is taken as $+0.34\,\text{V}$ , but if the question provides only $E^\circ(\text{Ag}^+/\text{Ag})$ and $E^\circ(\text{Zn}^{2+}/\text{Zn})$ , a common shortcut is to treat the silver oxide cathode potential as $+0.80\,\text{V}$ (the $\text{Ag}^+/\text{Ag}$ value) for calculation purposes, yielding $E^\circ_{\text{cell}} = 0.80 - (-0.76) = 1.56\,\text{V}$ . This is the intended approach here.

3. Calculate the standard cell potential

The cell potential is

E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = 0.80 - (-0.76) = 1.56\,\text{V}.

4. Compute $\Delta_r G^\circ$

Substitute into the Gibbs–cell potential relation:

\Delta_r G^\circ = -nFE^\circ_{\text{cell}} = -2 \times 96{,}500\,\text{C mol}^{-1} \times 1.56\,\text{V}.

Since $1\,\text{V} = 1\,\text{J C}^{-1}$ ,

\Delta_r G^\circ = -2 \times 96{,}500 \times 1.56 = -301{,}080\,\text{J mol}^{-1} = -301.08\,\text{kJ mol}^{-1}.

Part (b): Determining limiting molar conductivity

Limiting molar conductivity $\Lambda^\circ_m$ is the molar conductivity of an electrolyte at infinite dilution, where inter-ionic interactions vanish and each ion migrates independently.

For strong electrolytes

Strong electrolytes dissociate completely at all concentrations. Their molar conductivity $\Lambda_m$ decreases with increasing concentration $c$ because of inter-ionic attractions (the electrophoretic and relaxation effects). Kohlrausch discovered that for strong electrolytes

\Lambda_m = \Lambda^\circ_m - A\sqrt{c},

where $A$ is a constant. This is Kohlrausch's law (the square-root law).

Method: Measure $\Lambda_m$ at several low concentrations, plot $\Lambda_m$ versus $\sqrt{c}$ , and extrapolate the straight line to $c = 0$ . The intercept is $\Lambda^\circ_m$ .

Concentration $c$	$\sqrt{c}$	$\Lambda_m$
$c_1$	$\sqrt{c_1}$	$\Lambda_{m,1}$

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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:

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

…

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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 …
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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). …
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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 …

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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 …
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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). …
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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 …

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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: …
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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.

…
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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. …
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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): $Cu^{2+} + 2e^- \rightarrow Cu$ …
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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). …
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Part (a): Standard Gibbs energy from cell potential

1. Identify the half-reactions

2. Find the standard reduction potential for the silver oxide half-cell

3. Calculate the standard cell potential

4. Compute ΔrG∘\Delta_r G^\circΔr​G∘

Part (b): Determining limiting molar conductivity

For strong electrolytes

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4. Compute $\Delta_r G^\circ$

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