(a) Give reasons : (2 × 1) (i) Mercury cell delivers a constant potential during its life time. (ii) In the ex

Question

Q. (a) Give reasons : (2 \times 1)

(i) Mercury cell delivers a constant potential during its life time.

(ii) In the experimental determination of electrolytic conductance, Direct Current (DC) is not used.

(OR)

(b) Define fuel cell with an example. What advantages do the fuel cells have over primary and secondary batteries ?

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Part (a)Concept understanding — Conductance And Conductivity

From Resistance to Conductance: Flipping the Idea

You already know resistance ( $R$ ) — it tells you how much a material opposes the flow of current. A high resistance means the wire fights the current; a low resistance means it lets current through easily.

Now flip that thought. Instead of asking "how much does it resist?", ask "how easily does it let current flow?" That's exactly what conductance measures.

Note

Conductance ( $G$ ) is the reciprocal of resistance:

$G = \frac{1}{R}$

Unit: siemens (S) — named after Werner von Siemens. 1 S = 1 A/V (ampere per volt).

If a wire has $R = 10\ \Omega$ , its conductance is $G = 0.1\ \text{S}$ . If $R = 0.5\ \Omega$ , $G = 2\ \text{S}$ — it conducts twice as well.

Ohm's Law in Conductance Form

You know $V = IR$ . Rearranging:

I = \frac{V}{R} = G V

So current = conductance × voltage. A high-conductance material draws a large current for the same voltage — it's a "good conductor."

Now, Conductivity: The Material's Intrinsic Property

Resistance depends on two things: the material itself (its "resistivity" $\rho$ ) and the geometry (length $L$ , cross-sectional area $A$ ):

R = \rho \frac{L}{A}

Conductance also depends on geometry. A thicker wire (larger $A$ ) or a shorter wire (smaller $L$ ) has higher conductance. To isolate the material's inherent ability to conduct, we define conductivity ( $\sigma$ ):

\sigma = \frac{1}{\rho}

Conductivity is the reciprocal of resistivity. It tells you how well the material itself conducts, independent of shape and size.

Unit: siemens per metre (S/m).
High $\sigma$ → good conductor (copper: $\approx 5.8 \times 10^7\ \text{S/m}$ ).
Low $\sigma$ → poor conductor / insulator (glass: $\approx 10^{-12}\ \text{S/m}$ ).

Watch out

Don't confuse conductance (property of a specific object, depends on geometry) with conductivity (property of the material, independent of geometry). A short thick copper wire has high conductance; a long thin copper wire has lower conductance — but both have the same conductivity.

The Big Picture in One Table

Quantity	Symbol	Definition	Depends on	Unit
Resistance	$R$	$V/I$	Material + geometry	$\Omega$
Resistivity	$\rho$	$R A / L$	Material only	$\Omega \cdot \text{m}$
Conductance	$G$	$1/R$	Material + geometry	S
Conductivity	$\sigma$	$1/\rho$	Material only	S/m

Intuitive Analogy

Think of a water pipe:

Resistance = how hard it is to push water through (narrow, long pipe).
Conductance = how easily water flows (wide, short pipe). …

Why this formula?

Conductance and Conductivity: Why the Formulas Hold

Let's build this from first principles — understanding the why before the what.

1. The Core Idea: How Easily Does Current Flow?

Think of a conductor (like a copper wire). When you apply a voltage across it, electrons drift through the material. Two questions arise:

How much current flows for a given voltage? → This is conductance ( $G$ ).
How well does the material itself allow current? → This is conductivity ( $\sigma$ ).

The key distinction: Conductance depends on the size and shape of the object. Conductivity is an intrinsic property of the material.

2. Ohm's Law in Terms of Conductance

You know Ohm's law:

V = IR

But we can rewrite it as:

I = \frac{V}{R}

Define conductance $G$ as the reciprocal of resistance:

G = \frac{1}{R}

So:

I = G V

Why this makes sense:

A larger $G$ means more current for the same voltage — the conductor "conducts" better.
$G$ has units of siemens (S) = $\text{A/V}$ .

3. From Resistance to Conductivity: The Geometry Factor

Resistance of a uniform conductor depends on:

Length $L$ (longer → more resistance)
Cross-sectional area $A$ (thicker → less resistance)
Material property $\rho$ (resistivity)

The formula:

R = \rho \frac{L}{A}

Now, conductivity $\sigma$ is the reciprocal of resistivity:

\sigma = \frac{1}{\rho}

So:

R = \frac{1}{\sigma} \cdot \frac{L}{A}

Why this form?

If you double the length, electrons have to travel twice as far, colliding more → resistance doubles.
If you double the area, there's twice as many "lanes" for electrons → resistance halves.

4. The Key Formula: Conductance in Terms of Conductivity

Since $G = 1/R$ , we get:

G = \sigma \frac{A}{L}

This is the central relationship. Let's see why it holds:

$\sigma$ tells you how well the material conducts (intrinsic).
$A/L$ tells you how the geometry amplifies or reduces that.

Intuition:

A fat, short wire ( $A$ large, $L$ small) has high conductance.
A thin, long wire ( $A$ small, $L$ large) has low conductance.
A material with high $\sigma$ (like copper) gives higher $G$ than one with low $\sigma$ (like iron), for the same shape.

5. Microscopic Derivation (Why $\sigma$ Exists)

At the microscopic level, conductivity arises from electron motion:

\sigma = n e \mu

Where:

$n$ = number of free electrons per unit volume
$e$ = electron charge …

Part (b)Concept understanding — Fuel Cell Fuels

Fuel Cell Fuels — From Intuition to Precision

Imagine you have a battery that never runs out as long as you keep feeding it a specific fuel. That's the core idea of a fuel cell. But unlike a car engine that burns petrol in a violent explosion, a fuel cell quietly combines a fuel with oxygen to produce electricity, water, and heat — no flames, no moving parts.

The question is: what can you feed it? That's where fuel cell fuels come in.

The Intuition: What Makes a Good Fuel?

A fuel cell works by stripping electrons from a fuel molecule at one electrode (the anode) and passing those electrons through an external circuit to do work. At the other electrode (the cathode), oxygen grabs those electrons and combines with the fuel's leftover ions.

For this to work efficiently, the fuel must:

Be easy to ionise (give up electrons readily)
React quickly at the anode (good kinetics)
Not poison the catalyst (usually platinum or nickel)
Be storable and transportable in practice

The simplest fuel that satisfies all these? Hydrogen gas ( $H_2$ ). It's the gold standard.

The Precise Statement

Fuel cell fuels are substances that can be oxidised at the anode of a fuel cell to release electrons, producing an electric current. The most common fuel is hydrogen, but other fuels like methanol, ethanol, natural gas (methane), and even ammonia or formic acid are used depending on the application.

The general anode reaction for any fuel is:

\text{Fuel} \rightarrow \text{oxidised products} + \text{electrons}

And the overall cell reaction combines this with oxygen reduction at the cathode:

\text{Fuel} + O_2 \rightarrow \text{oxidation products} + \text{water} + \text{electricity}

The Main Fuel Types (Exam-Ready)

Fuel	Anode Reaction (simplified)	Key Advantage	Key Challenge
Hydrogen ( $H_2$ )	$H_2 \rightarrow 2H^+ + 2e^-$	Highest energy density by mass; zero carbon emissions	Storage (low density); production cost
Methanol ( $CH_3OH$ )	$CH_3OH + H_2O \rightarrow CO_2 + 6H^+ + 6e^-$	Liquid at room temp; easy to transport	Lower efficiency; $CO_2$ produced; catalyst poisoning
Ethanol ( $C_2H_5OH$ )	$C_2H_5OH + 3H_2O \rightarrow 2CO_2 + 12H^+ + 12e^-$	Renewable (bioethanol); less toxic than methanol	Slower reaction kinetics; C–C bond breaking is hard
Methane ( $CH_4$ )	$CH_4 + 2H_2O \rightarrow CO_2 + 8H^+ + 8e^-$	Abundant (natural gas); existing infrastructure	Requires high temperature (solid oxide fuel cells)
Ammonia ( $NH_3$ )	$2NH_3 \rightarrow N_2 + 6H^+ + 6e^-$	Carbon-free; easier to store than $H_2$	Toxic; requires cracking to $H_2$ first

Watch out

A common mistake: thinking any fuel can be used in any fuel cell. Each fuel cell type is designed for a specific fuel. For example, a Proton Exchange Membrane Fuel Cell (PEMFC) runs on pure hydrogen — putting methanol in it would destroy the membrane. Direct methanol fuel cells (DMFCs) are built differently.

Why Not Just Burn the Fuel?

This is the key conceptual leap. In a fuel cell, the fuel's chemical energy is converted directly to electricity, bypassing the heat → mechanical work → generator steps of a thermal power plant. This means:

Higher efficiency (40–60% vs 30–35% for combustion engines)
No moving parts (silent, low maintenance)
No pollutants (if using hydrogen, the only byproduct is water)

Note

For hydrogen fuel cells: $2H_2 + O_2 \rightarrow 2H_2O$ — the only exhaust is pure water. This is why hydrogen fuel cells are considered zero-emission.

The Real-World Catch …

Why this formula?

Fuel Cell Fuels: Why the Key Formulas Hold

Fuel cells convert chemical energy directly into electrical energy. The core idea is electrochemical combustion — instead of burning a fuel to make heat, then turning a turbine, you let the fuel react electrochemically with oxygen to produce electricity directly.

The Fundamental Reaction

For a hydrogen–oxygen fuel cell (the simplest and most common):

2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O}

But this happens in two half-reactions:

Anode (oxidation):

2\text{H}_2 \rightarrow 4\text{H}^+ + 4e^-

Cathode (reduction):

\text{O}_2 + 4\text{H}^+ + 4e^- \rightarrow 2\text{H}_2\text{O}

The electrons flow through an external circuit — that’s your electricity.

Key Formula 1: Cell Potential from Gibbs Free Energy

The maximum electrical work a fuel cell can do equals the change in Gibbs free energy ( $\Delta G$ ) of the overall reaction:

W_{\text{max}} = -\Delta G

For an electrochemical cell, electrical work is:

W_{\text{electrical}} = n F E

where:

$n$ = number of electrons transferred per mole of fuel (here $n=4$ for $\text{H}_2$ )
$F$ = Faraday constant ( $96\,485\ \text{C/mol}$ )
$E$ = cell potential (volts)

Why this holds:

The Gibbs free energy is the “useful” energy available at constant temperature and pressure. In a fuel cell, that energy is converted into the work of moving electrons through a potential difference. Equating the two gives:

\Delta G = -n F E

So the reversible cell voltage is:

E^\circ = -\frac{\Delta G^\circ}{nF}

For the $\text{H}_2/\text{O}_2$ fuel cell at standard conditions:

$\Delta G^\circ = -237\ \text{kJ/mol}$ (for liquid water product)
$n = 4$
$F = 96\,485\ \text{C/mol}$

E^\circ = -\frac{(-237\,000)}{4 \times 96\,485} \approx 1.23\ \text{V}

This is the theoretical maximum voltage — no fuel cell can exceed this because it’s set by thermodynamics.

Key Formula 2: Efficiency Limit

The thermodynamic efficiency of a fuel cell is:

\eta = \frac{\Delta G}{\Delta H}

where $\Delta H$ is the enthalpy change (the total chemical energy released).

Why this holds:

$\Delta H$ is the total energy released when the fuel burns (the “heat of combustion”).
$\Delta G$ is the portion of that energy that can be converted to electrical work.
The rest ( $T\Delta S$ ) is lost as heat due to entropy change.

For hydrogen:

$\Delta H^\circ = -286\ \text{kJ/mol}$ (higher heating value)
$\Delta G^\circ = -237\ \text{kJ/mol}$

\eta_{\text{max}} = \frac{237}{286} \approx 0.83\ (83\%)

Why this is higher than a heat engine:

A Carnot engine is limited by temperature difference. A fuel cell is not a heat engine — it converts chemical energy directly, so its efficiency limit is set by $\Delta G/\Delta H$ , not by Carnot. This is why fuel cells can theoretically exceed 50% efficiency while internal combustion engines are stuck below ~40%.

Key Formula 3: Nernst Equation for Real Conditions

In practice, the cell voltage depends on concentrations (or partial pressures) of reactants and products:

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

where $Q$ is the reaction quotient.

For the hydrogen fuel cell:

Q = \frac{P_{\text{H}_2\text{O}}}{P_{\text{H}_2}^2 \cdot P_{\text{O}_2}}

So: …

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The mercury cell holds a constant ~1.35 V because its reaction changes no ionic concentration; DC is avoided in conductance work because it electrolyses/polarises the cell, so AC is used.
A fuel cell (e.g. $\text{H}_2$ – $\text{O}_2$ ) converts fuel energy directly and continuously into electricity, beating batteries on refuelling, efficiency and cleanliness.

Part (a)

(i) Mercury cell — constant potential

The mercury (button) cell has a zinc-amalgam anode and an $\text{HgO}$ cathode in a KOH/ZnO paste. During discharge:

\text{Zn(Hg)} + \text{HgO} \rightarrow \text{ZnO} + \text{Hg}

No dissolved ions are consumed or generated — the KOH concentration and the electrode potentials remain essentially unchanged. Its internal resistance therefore stays constant, and the voltage holds near $1.35\ \text{V}$ until the reactants run out. This steady voltage is why it is used in hearing aids, watches and pacemakers.

Watch out

The constant voltage is due to the unchanging electrolyte composition, not merely because the cell is non-rechargeable.

(ii) Why AC (not DC) is used for electrolytic conductance

If DC is passed:

the ions are discharged at the electrodes (electrolysis), altering the concentration of the solution near them;
products build up on the electrodes (polarisation), adding a back-emf and a false extra resistance. …

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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 38 on this concept.

CBSE 2026Set V11 markMCQ
$Q. Among the following cells, the cell used in the apollo space program for providing electric power is$
(a) SHE
(b) $\text{H}_2\text{-O}_2$ fuel cell
(c) Daniel cell
(d) Mercury cell
›Reveal solutionSolution
The Apollo space programme was powered by the H $_2$ –O $_2$ fuel cell.

A fuel cell converts the energy of combustion of a fuel directly into electrical energy. In the hydrogen–oxygen fuel cell, H $_2$ and O $_2$ are bubbled through porous carbon electrodes in concentrated aqueous NaOH/KOH:

Anode: $\;2\text{H}_2(g) + 4\text{OH}^-(aq) \rightarrow 4\text{H}_2\text{O}(l) + 4e^-$

Cathode: $\;\text{O}_2(g) + 2\text{H}_2\text{O}(l) + 4e^- \rightarrow 4\text{OH}^-(aq)$

Overall: $\;2\text{H}_2(g) + \text{O}_2(g) \rightarrow 2\text{H}_2\text{O}(l)$ …
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CBSE 2026Set A1 markMCQ
$Q. On increasing dilution, the specific conductance of an electrolyte$
(a) increases
(b) decreases
(c) remains constant
(d) none of these
›Reveal solutionSolution

Specific conductance (conductance per unit volume) falls on dilution because the number of current-carrying ions per unit volume decreases.

Specific conductance (κ) is the conductance of a solution held between electrodes 1 cm apart with 1 cm² area, i.e. conductance of unit volume. On dilution the number of ions per unit volume decreases, …

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CBSE 2026Set A1 markMCQ
$Q. The number of ions in aqueous solution of [Co(NH3)5Cl]Cl2 is$
(a) 3
(b) 4
(c) 2
(d) 6
›Reveal solutionSolution

Only the ions outside the coordination sphere are free; [Co(NH3)5Cl]Cl2 gives one complex cation plus two chloride ions = 3 ions.

In a coordination compound, only the counter ions outside the square brackets dissociate in water; the ligands inside the coordination sphere stay bound to the metal. Here one Cl and five NH3 are coordinated to cobalt, and two Cl are counter ions:

[Co(NH3)5Cl]Cl2 -> [Co(NH3)5Cl]2+ + 2 Cl-

…

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CBSE 2026Set ANNUAL1 markMCQ
$Q. The unit of cell constant is:$
(a) Ohm^-1 cm^2
(b) cm^-1
(c) Ohm^-1 cm^-1
(d) Ohm^-1 cm^2/ g eq
›Reveal solutionSolution

Cell constant $G^{*}=l/A$ has the unit of reciprocal length, i.e. cm^-1.

The cell constant of a conductivity cell is defined as the ratio of the distance between the two electrodes (l) to the area of cross-section of the electrodes (A): $G^{*} = \dfrac{l}{A}$ . Since l has units of cm and A has units of cm^2, the cell constant has units of $\dfrac{cm}{cm^2} = cm^{-1}$ .

…

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CBSE 2026Set ANNUAL1 markMCQ
$Q. The unit of specific conductivity is:$
(a) ohm⁻¹
(b) ohm⁻¹ cm⁻¹
(c) ohm cm
(d) ohm cm⁻¹
›Reveal solutionSolution

Specific conductance ( $\kappa$ ) is measured in ohm⁻¹ cm⁻¹ (S cm⁻¹).

Specific conductivity ( $\kappa$ ), also called conductivity, is the conductance of a 1 cm cube of a solution of an electrolyte. Conductance ( $G$ ) is the reciprocal of resistance and is measured in ohm⁻¹ (siemens, S). Since $\kappa = G \times (l/A)$ , where $l/A$ (t …

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CBSE 2026Set ANNUAL1 mark
$Q. Fill in the blank: Fuel cells produce electricity with an efficiency of about ______ percentage compared to thermal plants whose efficiency is about 40%.$
›Reveal solutionSolution

Fuel cells work at about 70% efficiency, versus about 40% for thermal plants.

A fuel cell (e.g. the H2-O2 fuel cell) converts the chemical energy of a fuel directly into electrical energy through controlled electrochemical reactions, avoiding the wasteful heat-to-work steps of a thermal plant. Its efficiency is therefore high, a …

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CBSE 2025Set 56/5/11 markMCQ
$Q. Two statements are given — one labelled as Assertion (A) and the other labelled as Reason (R). Select the correct answer from the codes (A), (B), (C) and (D) as given below. (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A). (B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A). (C) Assertion (A) is true, but Reason (R) is false. (D) Assertion (A) is false, but Reason (R) is true. Assertion (A) : For measuring resistance of an ionic solution an AC source is used. Reason (R) : Concentration of ionic solution will change if DC source is used.$
›Reveal solutionSolution

AC is used to measure ionic solution resistance because DC causes electrolysis, which changes the solution's composition and hence its conductance; both statements are true and the reason correctly explains the assertion.

When we measure the resistance (or conductance) of an ionic solution, we're essentially probing how easily ions can carry current through the liquid. The choice between AC and DC isn't arbitrary—it stems from what happens at the electrode-solution interface.

Why DC causes problems

In an ionic solution, current flows via the movement of ions: cations migrate toward the cathode, anions toward the anode. With a DC source, these ions don't just move—they undergo redox reactions at the electrodes. For instance, in a $\text{NaCl}$ solution, $\text{Cl}^-$ ions get oxidized at the anode ( $2\text{Cl}^- \to \text{Cl}_2 + 2e^-$ ) and $\text{H}^+$ from water gets reduced at the cathode ( $2\text{H}^+ + 2e^- \to \text{H}_2$ ). This is electrolysis.

The consequence? The concentration of ions in the solution changes continuously. As ions are consumed or new species are produced, the conductance of the solution drifts. You're no longer measuring the property of the original solution—you're measuring a changing system. The reading becomes unreliable and time-dependent.

Why AC solves this

An alternating current reverses direction many times per second (typically at 1000 Hz or so in conductivity bridges). In one half-cycle, a tiny bit of electrolysis might begin, but in the next half-cycle the current reverses and the reaction is essentially undone. The net chemical change over many cycles is negligible. The solution composition remains stable, and the resistance measurement reflects the true, steady-state property of the ionic solution. …
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CBSE 2025Set 56/6/11 markMCQ
$Q. Which of the following cell converts the energy of combustion of fuel into electrical energy ? (A) Mercury cell (B) Fuel cell (C) Dry cell (D) Lead storage cell$
›Reveal solutionSolution
A fuel cell directly converts the chemical energy from the combustion of a fuel (like hydrogen) into electrical energy, without burning the fuel in a flame. The correct answer is (B).

The key idea here is the direct conversion of chemical energy to electrical energy. In a fuel cell, the fuel (e.g., hydrogen) and an oxidant (e.g., oxygen) are supplied continuously. The reaction that happens inside the cell is essentially the same as combustion — hydrogen combining with oxygen to form water — but it occurs electrochemically, not as a flame. This means electrons are forced to travel through an external circuit, producing electricity.

Let’s look at each option to see why only one fits.

Fuel cell (Option B) — This is the only cell that continuously consumes a fuel and an oxidant from outside. The “combustion of fuel” happens at the electrodes: at the anode, fuel is oxidised (loses electrons), and at the cathode, oxygen is reduced (gains electrons). The overall reaction is the same as burning the fuel, but the energy is released as electrical work, not heat. This is exactly what the question describes.

Mercury cell (Option A) — This is a primary (non-rechargeable) battery. It uses a zinc anode and a mercuric oxide cathode. The reaction is not a combustion; it’s a redox reaction between solid chemicals inside the sealed cell. No fuel is burned, and no external fuel is supplied.

Dry cell (Option C) — The common Leclanché cell (zinc-carbon) or alkaline cell. Again, it’s a sealed primary battery. The reactants are inside the cell (zinc, manganese dioxide, etc.). There is no combustion of a fuel — the reaction is a one-time conversion of stored chemicals. …
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CBSE 2025Set ANNUAL1 mark
$Q. Define the following — Fuel Cells$
›Reveal solutionSolution

A fuel cell is a galvanic cell fed continuously with fuel and oxidant to give a steady electric current.

A fuel cell is a galvanic (voltaic) cell in which the chemical energy of combustion of a fuel (such as $H_2$ , $CH_4$ , $CH_3OH$ ) is converted directly into electrical energy. Fuel (at the anode) and oxygen/air (at the cathode) are fed continuously, and the products are removed continuously, so the cell keeps working as long as the reactants are supplied. Example: the $H_2$ – $O_2$ fuel cell, where

Anode: $2H_2 + 4OH^- \to 4H_2O + 4e^-$

Cathode: $O_2 + 2H_2O + 4e^- \to 4OH^-$

…

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CBSE 2025Set D1 markMCQ
$Q. The unit of specific conductance is$
(a) ohm cm^-1
(b) ohm cm^-2
(c) ohm^-1 cm^-1
(d) ohm^-1 cm^-2
›Reveal solutionSolution

Specific conductance = 1/(specific resistance), so its unit is ohm^-1 cm^-1 (S cm^-1).

Specific conductance (conductivity), kappa, is the reciprocal of specific resistance (resistivity), rho:

kappa = 1/rho

Specific resistance has the unit ohm cm, so its reciprocal has the unit:

…

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CBSE 2025Set A1 mark
$Q. Write the value of conductivity of superconductor.$
›Reveal solutionSolution

Since conductivity is the reciprocal of resistivity, and a superconductor's resistivity drops to exactly zero, its conductivity becomes infinite.

Certain materials, when cooled below a characteristic critical temperature, lose all electrical resistance completely — this state is called superconductivity, and such materials are superconductors. Electrical conductivity ( $\kappa$ ) and resistivity ( $\rho$ ) are reciprocals of each other: $\kappa = 1/\rho$ . Because a superconductor's res …

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CBSE 2025Set ANNUAL1 markMCQ
$Q. SI unit of resistivity (specific resistance) is -$
(a) Ω
(b) Ω^-1
(c) Ωm
(d) Ωm^-1
›Reveal solutionSolution

Resistivity (specific resistance) has SI unit ohm-metre (Ωm).

Resistance of a conductor is related to its resistivity by:

R = rho x (l/A)

where l is length (m) and A is cross-sectional area (m^2). Rearranging:

rho = R x A / l

Units: rho = (ohm) x (m^2) / (m) = ohm x m = Ωm

…

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Q.(a) Give reasons : (2 × 1)

Part (a)

(i) Mercury cell — constant potential

(ii) Why AC (not DC) is used for electrolytic conductance

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