Match the terms given in Column I with the units given in Column II. Column I: (i) Lambdam (ii) ECell (iii) ka

NCERT Exemplar · Q53

Q. Match the terms given in Column I with the units given in Column II. Column I:

(i)

\Lambda_m

(ii)

E_{Cell}

(iii)

\kappa

(iv)

G^*

Column II:

(a)

S\ cm^{-1}

(b)

m^{-1}

(c)

S\ cm^2\ mol^{-1}

(d)

V

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Concept understanding — Molar 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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This is a matching problem linking four electrochemical quantities to their correct SI or practical units. The key is to recall the definitions: molar conductivity $\Lambda_m$ is conductivity per mole, so its unit is $S\ cm^2\ mol^{-1}$ ; cell potential $E_{Cell}$ is voltage, so its unit is $V$ ; conductivity $\kappa$ is the reciprocal of resistivity, so its unit is $S\ cm^{-1}$ ; and the cell constant $G^*$ is the ratio of length to area, so its unit is $m^{-1}$ . The correct matches are: (i)→(c), (ii)→(d), (iii)→(a), (iv)→(b).

The question asks you to match four fundamental electrochemical quantities with their units. This is not about memorising a table — it is about understanding what each quantity physically represents. Once you know the definition, the unit follows naturally.

Let us go through each one.

1. Molar conductivity $\Lambda_m$

Molar conductivity is defined as the conductivity of a solution divided by the molar concentration of the electrolyte:

\Lambda_m = \frac{\kappa}{c}

Here $\kappa$ has units of $S\ cm^{-1}$ (siemens per centimetre) and $c$ has units of $mol\ cm^{-3}$ (or more commonly $mol\ L^{-1}$ , but we must be careful with volume units). In practice, $\kappa$ is measured in $S\ cm^{-1}$ and concentration in $mol\ cm^{-3}$ , so:

\text{Unit of } \Lambda_m = \frac{S\ cm^{-1}}{mol\ cm^{-3}} = S\ cm^2\ mol^{-1}

That matches option (c).

Tip

A common shortcut: molar conductivity always has "per mole" in the denominator and "area" in the numerator — think of it as conductivity per unit concentration, which gives $S\ cm^2\ mol^{-1}$ .

2. Cell potential $E_{Cell}$

The electromotive force (emf) of a cell is simply the potential difference between its two electrodes. The SI unit of electric potential is the volt ( $V$ ). So $E_{Cell}$ is measured in volts. That matches option (d).

3. Conductivity $\kappa$

Conductivity is the reciprocal of resistivity ( $\rho$ ). Resistivity has units of $\Omega\ cm$ (ohm-centimetre), so:

\kappa = \frac{1}{\rho} \quad \Rightarrow \quad \text{Unit of } \kappa = \frac{1}{\Omega\ cm} = S\ cm^{-1}

Here $S$ (siemens) is the unit of conductance, equal to $\Omega^{-1}$ . So $\kappa$ has units $S\ cm^{-1}$ , matching option (a). …

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Method: Unit Analysis by Dimensional Matching

This method works by recalling the defining formula for each term, then deriving its unit from the SI/base units of the quantities in the formula.

Step 1: Recall the definition of each term

Term	Definition / Formula	Key quantities involved
$\Lambda_m$ (Molar conductivity)	$\Lambda_m = \frac{\kappa}{c}$	$\kappa$ = conductivity, $c$ = concentration (mol/cm³)
$E_{Cell}$ (Cell potential)	$E = -\frac{\Delta G}{nF}$	Voltage = J/C = V
$\kappa$ (Conductivity)	$\kappa = \frac{1}{\rho}$ (reciprocal of resistivity)	Unit: S/cm or S/m
$G^*$ (Cell constant)	$G^* = \frac{l}{A}$	$l$ = length, $A$ = area → unit: cm⁻¹ or m⁻¹

Step 2: Derive the unit for each

$\Lambda_m$ :

$\kappa$ has unit $S\,cm^{-1}$ , $c$ has unit $mol\,cm^{-3}$

$\Rightarrow \Lambda_m = \frac{S\,cm^{-1}}{mol\,cm^{-3}} = S\,cm^2\,mol^{-1}$

→ matches (c)
$E_{Cell}$ :

Directly measured in volts

→ matches (d)
$\kappa$ :

Conductivity unit is siemens per centimetre …

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Here are the common mistakes students make when matching these electrochemical quantities with their units, and how to avoid each.

Mistake 1: Confusing $\Lambda_m$ (Molar Conductivity) with $\kappa$ (Conductivity)

The error:

Students often swap the units of molar conductivity ( $\Lambda_m$ ) and specific conductivity ( $\kappa$ ). They match $\Lambda_m$ with $S\ cm^{-1}$ (conductivity) and $\kappa$ with $S\ cm^2\ mol^{-1}$ (molar conductivity).

Why it happens:

Both terms sound similar and both involve "conductivity." The formula $\Lambda_m = \frac{\kappa}{C}$ makes it easy to mix them up.

How to avoid:

Remember the definition:

Molar conductivity ( $\Lambda_m$ ) is the conductivity of a solution containing 1 mole of electrolyte. Its unit must involve per mole → $S\ cm^2\ mol^{-1}$ .
Conductivity ( $\kappa$ ) is the conductance of a 1 cm cube of solution → unit is $S\ cm^{-1}$ .

Trick: The "molar" in $\Lambda_m$ means "per mole" → look for $mol^{-1}$ in the unit.

Mistake 2: Forgetting that $E_{Cell}$ is measured in Volts

The error:

Some students match $E_{Cell}$ with $S\ cm^{-1}$ or $m^{-1}$ , thinking it is a conductivity-related term.

Why it happens:

$E_{Cell}$ appears alongside conductivity terms, so students assume it must also be a conductivity unit.

How to avoid:

Recall that $E$ stands for electromotive force (emf) or cell potential. The unit of potential difference is always the volt (V). No other unit in Column II has $V$ .

Key fact: $E_{Cell} = E^\circ_{cathode} - E^\circ_{anode}$ → result is in volts.

Mistake 3: Confusing $G^*$ (Cell Constant) with Conductivity

The error:

Students match $G^*$ with $S\ cm^{-1}$ or $S\ cm^2\ mol^{-1}$ , thinking it is a type of conductance.

Why it happens:

The symbol $G$ is often used for conductance, so $G^*$ looks like it should have a conductance unit.

How to avoid:

Know the definition:

$G^*$ (cell constant) = $\frac{l}{A}$ , where $l$ is length and $A$ is area. Its unit is $cm^{-1}$ or $m^{-1}$ . It is not a conductance — it is a geometric factor.

Trick: The star (*) means it is a constant for the cell, not a measured quantity. Units are inverse length.

Mistake 4: Mixing up $cm^{-1}$ and $m^{-1}$

The error:

Students match $G^*$ with $S\ cm^{-1}$ (conductivity) instead of $m^{-1}$ . …

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

AHSEC Higher Secondary (HS) Final Examination 2026Set ANNUAL2 marks
$Q. State the variation of conductivity and molar conductivity of acetic acid with dilution.$
›Reveal solutionSolution

Diluting acetic acid lowers its conductivity but raises its molar conductivity (increasing dissociation).

Conductivity (κ) measures the conductance of ions present in unit volume of solution. On dilution the number of ions per unit volume decreases, so conductivity always decreases with dilution.

Molar conductivity (Λ_m = κ × 1000 / c) measures the conducting power of all the ions produced by one mole of electrolyte. On dilution the volume containing one mole increases, and for a weak electrolyte like acetic acid the degree of dissociation (α) rises steeply, releasing many more ions. Therefore molar conductivity increases with dilution, and the increase is very sharp near infini …

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AHSEC Higher Secondary (HS) Final Examination 2024Set ANNUAL2 marks
$Q. Define conductivity and molar conductivity for the solution of an electrolyte. Discuss their variation with concentration.$
›Reveal solutionSolution
Conductivity falls as a solution is diluted (fewer ions per unit volume), while molar conductivity rises on dilution (each mole's ions become more independent / more of a weak electrolyte dissociates).

Conductivity (κ, specific conductance):

Conductivity is defined as the conductance of a solution of unit length (1 cm) and unit cross-sectional area (1 cm²) — i.e. it is the reciprocal of resistivity (ρ):

κ = 1/ρ

Its SI unit is S m⁻¹ (commonly expressed as S cm⁻¹). It measures how well the solution as a whole (per unit volume) conducts electricity.

Molar conductivity (Λm):

Molar conductivity is the conducting power of all the ions produced by dissolving 1 mole of an electrolyte in solution, and is related to conductivity by:

Λm = κ × 1000 / C (C = molar concentration in mol L⁻¹, κ in S cm⁻¹)

Its unit is S cm² mol⁻¹.

Variation with concentration:

Conductivity (κ) always decreases as the solution is diluted, because the number of ions per unit volume of solution decreases with dilution, even though the degree of dissociation (for weak electrolytes) increases.

Molar conductivity (Λm) always increases as concentration decreases (i.e. on dilution), because it is normalised per mole of electrolyte: …
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AHSEC Higher Secondary (HS) Final Examination 2023Set ANNUAL2 marks
$Q. The following limiting molar conductivities are given as: lambda-m-zero(H2SO4) = x S cm^2 mol^-1, lambda-m-zero(K2SO4) = y S cm^2 mol^-1, lambda-m-zero(CH3COOK) = z S cm^2 mol^-1. Calculate limiting molar conductivity of acetic acid. OR The cell potential for the following cell is 0.576V at 298K. Calculate the pH of the solution: Pt | H2(g) | H+(aq) || Cu2+(0.01M) | Cu(s). Given, E-zero(Cu2+/Cu) = 0.34V.$
›Reveal solutionSolution

Option 1 combines the three given conductivities via Kohlrausch's law to cancel out K⁺ and SO₄²⁻; Option 2 uses the Nernst equation for the H₂|H⁺ vs Cu²⁺|Cu cell to back-calculate [H⁺].

Option 1 — Limiting molar conductivity of acetic acid via Kohlrausch's law:

By Kohlrausch's law of independent migration of ions, each limiting molar conductivity splits into ionic contributions:

$2\lambda^0(H^+) + \lambda^0(SO_4^{2-}) = x \quad \text{...(i), from } H_2SO_4$

$2\lambda^0(K^+) + \lambda^0(SO_4^{2-}) = y \quad \text{...(ii), from } K_2SO_4$

$\lambda^0(CH_3COO^-) + \lambda^0(K^+) = z \quad \text{...(iii), from } CH_3COOK$

Subtracting (ii) from (i): $2\lambda^0(H^+) - 2\lambda^0(K^+) = x - y \Rightarrow \lambda^0(H^+) = \lambda^0(K^+) + \dfrac{x-y}{2}$

We want $\lambda_m^0(CH_3COOH) = \lambda^0(CH_3COO^-) + \lambda^0(H^+)$ . From (iii), $\lambda^0(CH_3COO^-) = z - \lambda^0(K^+)$ , so:

$\lambda_m^0(CH_3COOH) = \left[z - \lambda^0(K^+)\right] + \left[\lambda^0(K^+) + \frac{x-y}{2}\right] = z + \frac{x-y}{2}$

(Equivalently, $\lambda_m^0(CH_3COOH) = \lambda_m^0(CH_3COOK) + \lambda_m^0(\tfrac12H_2SO_4) - \lambda_m^0(\tfrac12K_2SO_4)$ .)

Option 2 — pH from cell potential:

Cell: $Pt|H_2(g)|H^+(aq)\ ||\ Cu^{2+}(0.01M)|Cu(s)$ , with anode = H₂/H⁺ (oxidation, $E^0=0V$ ) and cathode = Cu²⁺/Cu (reduction, $E^0=0.34V$ ).

…

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AHSEC Higher Secondary (HS) Final Examination 2020Set ANNUAL2 marks
$Q. Define molar conductivity of a solution. Explain how molar conductivity changes with change in concentration of solution for a weak and a strong electrolyte. (1+1=2)$
›Reveal solutionSolution

Molar conductivity is conductivity per mole of dissolved electrolyte; it rises on dilution for both strong and weak electrolytes, but far more steeply for weak electrolytes because dilution increases their degree of dissociation.

Definition of molar conductivity

Molar conductivity (Λm) is the conducting power of all the ions produced by dissolving one mole of an electrolyte in solution. It is related to the specific conductivity (κ) and molar concentration (C) by:

Λm = κ × 1000 / C (with κ in S cm⁻¹ and C in mol L⁻¹, giving Λm in S cm² mol⁻¹)

Variation with concentration — strong electrolytes

For strong electrolytes (fully ionised at all concentrations, e.g. NaCl, KCl), Λm increases only slowly as concentration decreases (i.e. on dilution). This is because, even though the number of ions per unit volume stays proportional to concentration (the electrolyte is always ~100% ionised), interionic attractive forces between the oppositely charged ions reduce their mobility at higher concentrations; diluting the solution weakens these interionic forces, allowing ions to move a little more freely, so Λm rises gradually. Λm for a strong electrolyte varies linearly with √C (Debye-Hückel-Onsager equation), so Λm° (the limiting molar conductivity at infinite dilution) can be found by extrapolating the Λm vs √C plot to C = 0.

Variation with concentration — weak electrolytes …

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AHSEC Higher Secondary (HS) Final Examination 2020Set ANNUAL2 marks
$Q. (a) Following reactions occur at cathode during the electrolysis of aqueous silver chloride: Ag+(aq) + e- -> Ag(s), E° = +0.80V; H+(aq) + e- -> 1/2 H2(g), E° = 0.00V. On the basis of their standard reduction electrode potential (E°) values, which reaction is feasible at the cathode and why?$
(1)
(b) State Kohlrausch law of independent migration of ions. Write an expression for the molar conductivity of acetic acid at infinite dilution according to Kohlrausch law. (1/2+1/2=1)
›Reveal solutionSolution

(a) The half-reaction with the more positive standard reduction potential occurs preferentially at the cathode. (b) Kohlrausch's law lets us calculate the limiting molar conductivity of any electrolyte, including weak ones like acetic acid, by summing the independent ionic contributions.

(a) Which reaction occurs at the cathode?

Given: Ag+(aq) + e- → Ag(s), E° = +0.80 V and H+(aq) + e- → ½H2(g), E° = 0.00 V

At the cathode, reduction occurs, and between two competing reduction half-reactions, the one with the higher (more positive) standard reduction potential is thermodynamically more favourable and occurs preferentially, since it has the greater tendency to be reduced (to gain electrons). Since E°(Ag+/Ag) = +0.80 V is more positive than E°(H+/H2) = 0.00 V, silver ions are reduced in preference to hydrogen ions at the cathode:

Ag+(aq) + e- → Ag(s)

(b) Kohlrausch's law of independent migration of ions

At infinite dilution, when dissociation of an electrolyte is complete and interionic interactions vanish, each ion migrates independently of the other ion with which it is associated, and each ion makes its own definite contribution to the total molar conductivity of the electrolyte, regardless of the nature of the other ion present. Mathematically, for an electrolyte that dissociates into ν+ cations and ν- anions:

Λm° = ν+ λ°+ + ν- λ°-

where λ°+ and λ°- are the limiting (infinite-dilution) molar conductivities of the cation and anion respectively.

Molar conductivity of acetic acid at infinite dilution …

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