Q.(a) The conductivity of 0.001 mol L−1 acetic acid is 4.95×10−5 S cm−1. Calculate the dissociation constant if Λmo for acetic acid is 390.5 S cm2 mol−1.
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🔒 Start your 14-day free trial to unlock the full solution →Part (a)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.
Conductance (G) is the reciprocal of resistance:
G=R1
Unit: siemens (S) — named after Werner von Siemens. 1 S = 1 A/V (ampere per volt).
If a wire has R=10 Ω, its conductance is G=0.1 S. If R=0.5 Ω, G=2 S — it conducts twice as well.
Ohm's Law in Conductance Form
You know V=IR. Rearranging:
I=RV=GV
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" ρ) and the geometry (length L, cross-sectional area A):
R=ρAL
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 (σ):
σ=ρ1
And for a uniform wire:
G=σLA
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 σ → good conductor (copper: ≈5.8×107 S/m).
- Low σ → poor conductor / insulator (glass: ≈10−12 S/m).
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 | Ω |
| Resistivity | ρ | RA/L | Material only | Ω⋅m |
| Conductance | G | 1/R | Material + geometry | S |
| Conductivity | σ | 1/ρ | 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 (σ).
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=RV
Define conductance G as the reciprocal of resistance:
G=R1
So:
I=GV
Why this makes sense:
- A larger G means more current for the same voltage — the conductor "conducts" better.
- G has units of siemens (S) = 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 ρ (resistivity)
The formula:
R=ρAL
Now, conductivity σ is the reciprocal of resistivity:
σ=ρ1
So:
R=σ1⋅AL
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=σLA
This is the central relationship. Let's see why it holds:
- σ 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 σ (like copper) gives higher G than one with low σ (like iron), for the same shape.
5. Microscopic Derivation (Why σ Exists)
At the microscopic level, conductivity arises from electron motion:
σ=neμ
Where:
- n = number of free electrons per unit volume
- e = electron charge …
Part (b)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→cC+dD
The cell potential E under non-standard conditions is given by:
E=E∘−nFRTlnQ
Where:
- E = cell potential under the given conditions (in volts)
- E∘ = 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 = [A]a[B]b[C]c[D]d (using concentrations for dilute solutions)
At 25°C (298 K), the equation simplifies to a very practical form:
E=E∘−n0.0591log10Q
The 0.0591 comes from F2.303RT at 298 K. Notice it uses log10 (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:
Zn(s)∣Zn2+(aq)∥Cu2+(aq)∣Cu(s)
The single vertical line ∣ represents a phase boundary (solid electrode | solution). The double line ∥ 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): Zn(s)→Zn2+(aq)+2e−
- Cathode (reduction): Cu2+(aq)+2e−→Cu(s)
Overall: Zn(s)+Cu2+(aq)→Zn2+(aq)+Cu(s)
Here n=2 (two electrons transferred). The reaction quotient is:
Q=[Cu2+][Zn2+]
So the Nernst equation becomes:
E=E∘−20.0591log10[Cu2+][Zn2+]
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 [Zn2+]=0.1 M and [Cu2+]=1.0 M at 25°C. E∘ for the cell is 1.10 V.
E=1.10−20.0591log101.00.1 …
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 (ΔG):
ΔG=−nFEcell
Where:
- n = moles of electrons transferred
- F = Faraday constant (96,485C mol−1)
- Ecell = cell potential (volts)
Why negative? A spontaneous reaction has ΔG<0 and Ecell>0 — the negative sign makes this consistent.
2. The Chemical Side: ΔG Depends on Concentration
For a general redox reaction:
aA+bB→cC+dD
The Gibbs free energy under non-standard conditions is:
ΔG=ΔG∘+RTlnQ
Where Q is the reaction quotient:
Q=[A]a[B]b[C]c[D]d
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:
−nFEcell=−nFEcell∘+RTlnQ
Divide both sides by −nF:
Ecell=Ecell∘−nFRTlnQ
This is the Nernst equation.
4. The "Why" in Plain Terms
| Concept | Physical Meaning |
|---|---|
| Ecell∘ | Voltage when all species are at 1 M (standard state) |
| −nFRTlnQ | Correction factor — adjusts voltage for real concentrations |
| Q | Tells you how far the reaction is from equilibrium |
Key insight: When Q=K (equilibrium), Ecell=0 — the battery is dead because no net reaction occurs.
5. The Common Form (log base 10)
At 25∘C (298K):
FRTln10≈0.0592V
So:
Ecell=Ecell∘−n0.0592log10Q
Why convert to log? Exam convenience — most concentration values are powers of 10.
--- …
Part (a)
- Dissociation constant of acetic acid.
Λm=cκ×1000=0.0014.95×10−5×1000=49.5 S cm2mol−1
α=Λm∘Λm=390.549.5=0.1268
Ka=1−αcα2=1−0.12680.001×(0.1268)2=0.87321.607×10−5=1.84×10−5
- Nernst equation for 2Al+3Cu2+→2Al3++3Cu (n=6): Ecell=Ecell∘−60.0591log[Cu2+]3[Al3+]2 …
Part (a): Λm=49.5 S cm2mol−1, α=0.127, so Ka≈1.84×10−5; the Nernst equation for the Al/Cu cell uses n=6; secondary batteries are rechargeable (lead–acid). Part (b): the Al/Ni cell is Al∣Al3+(0.01M)∥Ni2+(0.1M)∣Ni with Ecell≈1.42 V; Λm∘ of a weak electrolyte is obtained from Kohlrausch's law, not by extrapolation.
Part (a)
(a) Ka of acetic acid
First convert conductivity to molar conductivity (note the factor 1000 cm3L−1):
Λm=c (mol L−1)κ×1000=0.0014.95×10−5×1000=49.5 S cm2mol−1.
Degree of dissociation:
α=Λm∘Λm=390.549.5=0.1268.
Ostwald's dilution law:
Ka=1−αcα2=1−0.12680.001×(0.1268)2=0.87320.001×0.016078=0.87321.6078×10−5=1.84×10−5.
A common slip is to compute Λm=κ/c and forget the ×1000 unit conversion — that makes Λm (and hence Ka) wrong by a factor of 1000. The correct Ka≈1.8×10−5 matches the known value for acetic acid.
(b) Nernst equation for 2Al+3Cu2+→2Al3++3Cu
Al: 0→+3 (loses 3e− each, 2 atoms → 6e−); Cu: +2→0 (3 ions → 6e−). So n=6 and Q=[Cu2+]3[Al3+]2 (solids omitted):
Ecell=Ecell∘−60.0591log[Cu2+]3[Al3+]2(at 25∘C).
(c) Secondary batteries …
Showing the 12 most recent of 46 on this concept.
- CBSE 2026Set ANNUAL1 markQ.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, Ecell.
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:
Ecell=Ecathode−Eanode
…
- CBSE 2026Set ANNUAL1 markMCQQ.Consider the following statements about a reaction at equilibrium: A(g) + B(g) ↔ 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 …
- CBSE 2025Set ANNUAL1 markQ.What is the SI unit of molar conductivity?
›Reveal solutionSolution
Molar conductivity's SI unit is S m² mol⁻¹.
Molar conductivity Λm=Cκ, where κ (conductivity) has SI unit Sm−1 and concentration C has SI unit molm−3.
Λm=molm−3Sm−1=Sm2mol−1
…
- CBSE 2025Set ANNUAL1 markQ.Define the following — Limiting molar conductivity
›Reveal solutionSolution
Limiting molar conductivity is molar conductivity extrapolated to zero concentration.
Limiting molar conductivity (Λm0 or Λm∞) is the molar conductivity of an electrolyte solution when the concentration approaches zero (i.e. at infinite dilution). At infinite dilution, dissociation of the electrolyte is essentially complete and inter-ionic attractions vanish, so each ion conducts independently and to its maximum extent. For strong electrolytes, Λm0 is obtained by extrapolating the Λm …
- CBSE 2025Set ANNUAL1 markMCQQ.Equivalent conductances of sodium acetate, sodium chloride and hydrochloric acid at infinite dilution are 224, 38.2, 203 ohm^-1 cm^2 eqv^-1 respectively at 298K. So the (lambda)0 CH3COOH is:(a) 288.5 ohm^-1 cm^2 eqv.^-1(b) 288.8 ohm^-1 cm^2 eqv.^-1(c) 388.8 ohm^-1 cm^2 eqv.^-1(d) 59.2 ohm^-1 cm^2 eqv.^-1
›Reveal solutionSolution
λ0(CH3COOH) = λ0(CH3COONa) + λ0(HCl) − λ0(NaCl) = 224 + 203 − 38.2 = 388.8 ohm^-1 cm^2 eqv^-1.
CH3COOH is a weak electrolyte, so its limiting equivalent conductance cannot be found by direct extrapolation. Instead, Kohlrausch's law of independent migration of ions lets us combine the limiting conductances of related strong electrolytes.
We want λ0(CH3COO-) + λ0(H+). Note that:
λ0(CH3COONa) = λ0(CH3COO-) + λ0(Na+) = 224
λ0(HCl) = λ0(H+) + λ0(Cl-) = 203
λ0(NaCl) = λ0(Na+) + λ0(Cl-) = 38.2
…
- CBSE 2025Set ANNUAL1 markQ.Fill in the blank: Molar conductivity ________ with decrease in concentration.
›Reveal solutionSolution
Molar conductivity increases as concentration decreases (i.e., on dilution), reaching a maximum limiting value at infinite dilution.
Molar conductivity is given by:
Λm = κ x 1000 / M
As a solution is diluted (concentration M decreases):
- For weak electrolytes: the degree of dissociation increases sharply with dilution, so more ions are produced per mole, increasing Λm markedly. …
- CBSE 2025Set ANNUAL1 markMCQQ.The unit of molar conductivity is(a) S cm^-2 mol^-1(b) S cm^2 mol^-1(c) S^-1 cm^2 mol^-1(d) S cm^2 mol
›Reveal solutionSolution
Molar conductivity relates conductivity (S/cm) to concentration (mol/cm^3), giving the composite unit S cm^2 mol^-1.
Molar conductivity is defined as:
Λm=Cκ×1000
where κ (specific conductivity) has units S cm^-1 and C (concentration) has units mol L^-1 (mol per 1000 cm^3).
…
- CBSE 2025Set ANNUAL1 markQ.For the electrochemical cell Zn(s)+Cu2+(aq)→Zn2+(aq)+Cu(s) the cell produces an electrical potential of 1.1 volt, when [Zn2+] and [Cu2+] are unity. State the direction of flow of current on applying external potential of 1.1 volt.
›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)∣Zn2+(aq)∥Cu2+(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). …
- CBSE 2025Set ANNUAL1 markMCQQ.The molar conductivity of a 0.1mol L−1 solution of KCl with electrolytic conductivity 0.0129 S cm−1 at 298 K is –(a) 12.9 S cm2 mol−1(b) 1.29 S cm2 mol−1(c) 0.0129 S cm2 mol−1(d) 129 S cm2 mol−1
›Reveal solutionSolution
Converting conductivity (per cm) into molar conductivity requires dividing by the molar concentration expressed per cm³, giving a factor of 1000 in the standard formula.
Molar conductivity is related to the specific conductivity (electrolytic conductivity, κ) and molar concentration C (in molL−1) by:
Λm=Cκ×1000
…
- CBSE 2025Set ANNUAL1 markMCQQ.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 …
- CBSE 2025Set ANNUAL1 markMCQQ.The formula used to calculate molar conductivity of an electrolyte is _____.(a) Λ=k1000c(b) c=k1000Λ(c) Λ=c1000k(d) k=Λc1000
›Reveal solutionSolution
Λm=c1000κ.
Molar conductivity (Λm) relates to specific conductivity (κ) and molar concentration c (in moldm−3) by:
Λm=cκ×1000
…
- CBSE 2025Set ANNUAL1 markQ.What is the SI unit of molar conductivity? OR Write the relation between specific conductivity and molar conductivity.
›Reveal solutionSolution
Molar conductivity's SI unit follows from Λm=κ/C: siemens metre-squared per mole.
Molar conductivity is defined as Λm=Cκ, where κ (specific/electrical conductivity) has SI unit Sm−1 and C (molar concentration) has SI unit molm−3. Dividing, the SI unit of Λm works out to
molm−3Sm−1=Sm2mol−1.
(In practical lab work, where κ is often expressed in Scm−1 and C in molL−1, the commonly used relation is Λm=C1000κ, giving the c.g.s.-style unit Scm2mol−1.)
…
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