(a) (i) Calculate Ecell of a galvanic cell in which the following reaction takes place at 25 °C : Zn(s) + Pb²⁺

Question

Q. (a)

(i) Calculate

E_{cell}

of a galvanic cell in which the following reaction takes place at 25 °C :

Zn(s) + Pb^{2+}(0·02\,M) \longrightarrow Zn^{2+}(0·1\,M) + Pb(s)

[Given :

E°_{Zn^{2+}/Zn} = -0·76\,V

E°_{Pb^{2+}/Pb} = -0·13\,V

; log 2 = 0·3010, log 4 = 0·6021, log 5 = 0·6990].

(ii) State Faraday's first law of electrolysis. How much electricity, in terms of Faraday, is required to reduce one mol of

MnO_4^-

Mn^{2+}

ion ?

(OR)

(b)

(i) The resistance of a conductivity cell containing 0·001 M KCl solution at 298 K is 1000 ohm. What is the cell constant if conductivity of 0·001 M KCl solution at 298 K is

0·125 \times 10^{-3}\,S\,cm^{-1}

(ii) Calculate the

E_{Mg^{2+}/Mg}

potential for the following half cell at 25 °C :

Mg/Mg^{2+}(1 \times 10^{-4}\,M)

;

E°_{Mg^{2+}/Mg} = +2·36\,V

[Given : log 10 = 1]

(iii) What is the effect of temperature on the electrical conductance of metallic conductor ?

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Part (a)Concept understanding — Nernst Equation

The Nernst Equation: Why Batteries Don't Always Give Their Rated Voltage

Imagine you have a fresh AA battery. It says 1.5 V on the side. But if you measure it with a voltmeter, you might get 1.58 V when it's new, and 1.2 V when it's almost dead. Why does the voltage change? The Nernst equation is the tool that tells you exactly why.

The Core Idea: Concentration Drives Voltage

Every electrochemical cell works because of a chemical reaction that wants to happen. But here's the key: how badly the reaction wants to happen depends on how much of each chemical is present.

Think of it like a slope. A steep hill gives you more energy when you roll down. A shallow hill gives you less. In a battery, the "hill" is the difference in concentration (or more precisely, activity) of ions between the two electrodes. When the battery is fresh, the hill is steep — lots of reactants, few products. As the battery runs, reactants get used up, products build up, the hill flattens, and the voltage drops.

The Nernst equation is the mathematical formula that calculates the exact voltage for any given set of concentrations.

The Precise Statement

For a general electrochemical reaction:

aA + bB \rightarrow cC + dD

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

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

Where:

$E$ = cell potential under the given conditions (what you actually measure)
$E^\circ$ = standard cell potential (the voltage when all reactants and products are at 1 M concentration, 1 atm pressure, 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 reaction
$F$ = Faraday constant (96,485 C/mol)
$Q$ = reaction quotient = $\frac{[C]^c [D]^d}{[A]^a [B]^b}$ (using concentrations for now)

At 25°C (298 K), the constants combine into a simpler form:

E = E^\circ - \frac{0.0592}{n} \log_{10} Q

Note

The 0.0592 comes from $\frac{2.303RT}{F}$ at 298 K. The 2.303 converts natural log to base-10 log, which is more convenient for calculations.

What It Actually Means

The equation has three parts:

$E^\circ$ — the "ideal" voltage when everything is at standard conditions. This is what you'd get in a textbook table.
$\frac{RT}{nF}$ — a scaling factor. It tells you how sensitive the voltage is to concentration changes. More electrons transferred ( $n$ ) means less sensitivity.
$\ln Q$ — the "concentration penalty". When $Q$ is small (lots of reactants, few products), $\ln Q$ is negative, so $E$ is higher than $E^\circ$ . When $Q$ is large (products building up), $\ln Q$ is positive, so $E$ drops below $E^\circ$ .

A Concrete Example

Consider the Daniell cell: $\text{Zn} | \text{Zn}^{2+} || \text{Cu}^{2+} | \text{Cu}$

The reaction is: $\text{Zn} + \text{Cu}^{2+} \rightarrow \text{Zn}^{2+} + \text{Cu}$

$E^\circ = 1.10 \text{ V}$ , $n = 2$

If $[\text{Cu}^{2+}] = 0.1 \text{ M}$ and $[\text{Zn}^{2+}] = 1.0 \text{ M}$ :

Q = \frac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]} = \frac{1.0}{0.1} = 10

$E = 1.10 - \frac{0.0592}{2} \log_{10}(10) = 1.10 - 0.0296 \times 1 = 1.07 \text{ V}$ …

Part (b)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 …

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Part (a): $E_{cell} = 0.63 - \tfrac{0.059}{2}\log 5 = 0.6094\ \text{V}$ ; reducing 1 mol $\text{MnO}_4^-$ to $\text{Mn}^{2+}$ needs 5 F. Part (b): cell constant $= 0.125\ \text{cm}^{-1}$ ; $E_{\text{Mg}^{2+}/\text{Mg}} = 2.242\ \text{V}$ ; a metal's conductance decreases as temperature rises.

Part (a)

(i) Nernst equation.

Half‑cells: Zn $\rightarrow$ Zn $^{2+}$ +2e $^-$ (anode, $E^\circ=-0.76$ V); Pb $^{2+}$ +2e $^-\rightarrow$ Pb (cathode, $E^\circ=-0.13$ V).
$E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} = (-0.13) - (-0.76) = +0.63\ \text{V}$ (spontaneous).
With $n=2$ and $Q = \dfrac{[\text{Zn}^{2+}]}{[\text{Pb}^{2+}]} = \dfrac{0.1}{0.02} = 5$ :

E_{cell} = 0.63 - \frac{0.059}{2}\log 5 = 0.63 - 0.0295(0.6990) = 0.63 - 0.0206 \approx \mathbf{0.609\ V}

Watch out

Remember $n=2$ (two electrons transferred) and exclude the solids from $Q$ .

(ii) Faraday's first law. The mass of a substance deposited or liberated at an electrode is directly proportional to the quantity of electricity passed through the electrolyte, $m = ZQ$ ( $Z$ = electrochemical equivalent). …

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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 36 on this concept.

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 2025Set 56/4/11 markMCQ
$2Cu^{+} (aq) + Zn (s) \rightarrow 2Cu (s) + Zn^{2+} (aq)$
›Reveal solutionSolution
The cell voltage depends on the reaction quotient via the Nernst equation. As the reaction proceeds, $[Zn^{2+}]$ increases and $[Cu^+]$ decreases, so the voltage decreases. The correct option is (B).

The Nernst equation tells us that the actual voltage of an electrochemical cell under non-standard conditions is:
$E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.059}{n} \log Q$
where $Q$ is the reaction quotient. For the given reaction:
$2Cu^+ (aq) + Zn (s) \rightarrow 2Cu (s) + Zn^{2+} (aq)$
the reaction quotient is:
$Q = \frac{[Zn^{2+}]}{[Cu^+]^2}$
(Remember: pure solids like Zn and Cu have activity = 1, so they don’t appear in $Q$ .)

The number of electrons transferred, $n$ , is 2 (each Cu⁺ gains one electron, and two Cu⁺ ions are reduced; Zn loses two electrons).

So the Nernst equation becomes:
$E_{\text{cell}} = 1.28 - \frac{0.059}{2} \log \frac{[Zn^{2+}]}{[Cu^+]^2}$
Now, as the reaction progresses:

$[Zn^{2+}]$ increases — Zn metal is oxidised to Zn²⁺, so its concentration in solution rises.

$[Cu^+]$ decreases — Cu⁺ ions are reduced to Cu metal, so their concentration falls.

Both changes make the fraction $\frac{[Zn^{2+}]}{[Cu^+]^2}$ larger.

A larger $Q$ means $\log Q$ is larger (more positive).

Since we subtract this term, $E_{\text{cell}}$ decreases.

Watch out

A common mistake is to think that because $[Zn^{2+}]$ appears in the numerator, the voltage might increase. But the Nernst equation has a minus sign in front of the log term — so anything that increases $Q$ actually lowers the voltage. …
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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 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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CBSE 2024Set D1 markMCQ
$Q. Which of the following has the highest molar electrical conductance in aqueous solution?$
(a) [Pt(NH3)6]Cl4
(b) [Pt(NH3)5Cl]Cl3
(c) [Pt(NH3)4Cl2]Cl2
(d) [Pt(NH3)3Cl3]Cl
›Reveal solutionSolution
Molar conductance rises with the number of ions produced on dissociation. [Pt(NH3)6]Cl4 gives 5 ions, the most of the options, so it conducts best.

Count the ions each complex furnishes in water (only the counter-ions outside the coordination sphere ionise):

[Pt(NH3)6]Cl4 -> [Pt(NH3)6]4+ + 4 Cl- => 5 ions

[Pt(NH3)5Cl]Cl3 -> [Pt(NH3)5Cl]3+ + 3 Cl- => 4 ions

[Pt(NH3)4Cl2]Cl2 -> [Pt(NH3)4Cl2]2+ + 2 Cl- => 3 ions …
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CBSE 2024Set D1 markMCQ
$Q. The cell constant of a conductivity cell is$
(a) l/A
(b) A/l
(c) l.A
(d) R/A
›Reveal solutionSolution

Cell constant = l/A (distance between electrodes ÷ electrode area), unit cm^-1.

Conductance G of a solution in a conductivity cell is G = kappa (A/l), where kappa is conductivity, A is the electrode area and l is the distance between the electrodes.

Rearranging, kappa = G (l/A). The geometric factor (l/A) is called the CELL CONSTANT because it depends only on the fixed geometry of the cell.

…

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CBSE 2024Set B1 markMCQ
$Q. The unit of cell constant is$
(a) ohm cm
(b) cm^-1
(c) cm
(d) ohm^-1 cm^-1
›Reveal solutionSolution

Cell constant G* = l/A (distance between electrodes divided by their area), so its unit works out to cm^-1.

For a conductivity cell, the cell constant is defined as:

$G^{*} = \frac{l}{A}$

where l is the distance between the two electrodes (in cm) and A is the area of cross-section of the electrodes (in cm^2).

Units: cm / cm^2 = cm^-1.

…

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Q.(a)

Part (a)

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