NCERT Solutions for Class 12 Chemistry Chapter 2 – Electrochemistry (English Medium)
⚡ Get NCERT Solutions for Class 12 Chemistry Chapter 2 – Electrochemistry with accurate answers and step-by-step explanations in an exam-friendly format. This chapter covers redox reactions, electrode potential, electrochemical cells (galvanic/voltaic), EMF and cell notation, Nernst equation, conductance (κ, Λm), electrolysis, and batteries & corrosion. Perfect for CBSE boards and NEET/JEE practice. ✅
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Q.1: Arrange the following metals in the order in which they displace each other from the solution of their salts:
Al, Cu, Fe, Mg and Zn.
Answer
✅ 👉 Displacement (reactivity) order:📌 Mg > Al > Zn > Fe > Cu
Explanation (Step by Step)
🤔 Rule:
A more reactive metal displaces a less reactive metal
from the solution of its salt.📌 So we simply arrange them according to the reactivity series :
Mg
is more reactive than
Al, Zn, Fe, Cu
→ displaces all of them from their salt solutions.
Al
can displace
Zn, Fe, Cu
(but not
Mg
).
Zn
can displace
Fe, Cu
(but not
Mg or Al
).
Fe
can displace
Cu
only.
Cu
is least reactive here →
cannot displace any of the others
.
👉 That’s why the correct order is:
Mg > Al > Zn > Fe > Cu
Mg > Al > Zn > Fe > Cu
Did You Know?
✅ Even though
Al is very reactive
,
it often appears “less reactive” in daily life because it forms a thin protective layer of
Al2O3
on its surface, which prevents further reaction
(passivation)
.
Q.2: Given the standard electrode potentials:
K+/K = −2.93 V, Ag+/Ag = +0.80 V, Hg2+/Hg = +0.79 V, Mg2+/Mg = −2.37 V, Cr3+/Cr = −0.74 V.
Arrange these metals in their increasing order of reducing power.
K+/K = −2.93 V, Ag+/Ag = +0.80 V, Hg2+/Hg = +0.79 V, Mg2+/Mg = −2.37 V, Cr3+/Cr = −0.74 V.
Arrange these metals in their increasing order of reducing power.
Answer
✅ 👉 Increasing order of reducing power:📌 Ag < Hg < Cr < Mg < K
Explanation (Step by Step)
🤔 Idea: The metal with more
negative standard reduction potential (E°)
gets oxidised more easily, so it is a
stronger reducing agent
.👉 So, as E° becomes more negative , reducing power increases .
Given E° values (in V)
• Ag+/Ag = +0.80 V
• Hg2+/Hg = +0.79 V
• Cr3+/Cr = −0.74 V
• Mg2+/Mg = −2.37 V
• K+/K = −2.93 V
• Ag+/Ag = +0.80 V
• Hg2+/Hg = +0.79 V
• Cr3+/Cr = −0.74 V
• Mg2+/Mg = −2.37 V
• K+/K = −2.93 V
📌 Arrange from
highest E° (weak reducer)
to
most negative E° (strongest reducer)
:
✅ Ag (+0.80) < Hg (+0.79) < Cr (−0.74) < Mg (−2.37) < K (−2.93)
✅ Ag (+0.80) < Hg (+0.79) < Cr (−0.74) < Mg (−2.37) < K (−2.93)
Did You Know?
✅
Potassium (K)
is such a strong reducing agent that it reacts violently with water to form
KOH
and
H2 gas
.👉 That’s why it is stored under kerosene/oil .
Q.3: Depict the galvanic cell in which the reaction
Zn(s) + 2Ag+(aq) → Zn2+(aq) + 2Ag(s)
takes place. Further show:
(i) Which electrode is negatively charged?
(ii) The carriers of current in the cell
(iii) Individual reaction at each electrode
Zn(s) + 2Ag+(aq) → Zn2+(aq) + 2Ag(s)
takes place. Further show:
(i) Which electrode is negatively charged?
(ii) The carriers of current in the cell
(iii) Individual reaction at each electrode
Answer
📌 Cell representation (cell notation):
👉 Zn(s) | Zn2+(aq) || Ag+(aq) | Ag(s)
👉 Zn(s) | Zn2+(aq) || Ag+(aq) | Ag(s)
(i) Negatively charged electrode
👉 Zinc electrode (Zn) = Anode (−)
👉 Zinc electrode (Zn) = Anode (−)
(ii) Carriers of current in the cell
📌 Electrons (e−) in the external wire
📌 Ions in the solutions/salt bridge (cations and anions)
📌 Electrons (e−) in the external wire
📌 Ions in the solutions/salt bridge (cations and anions)
(iii) Half-cell reactions
👉 At anode (oxidation):
Zn(s) → Zn2+(aq) + 2e−
👉 At cathode (reduction):
2Ag+(aq) + 2e− → 2Ag(s)
👉 At anode (oxidation):
Zn(s) → Zn2+(aq) + 2e−
👉 At cathode (reduction):
2Ag+(aq) + 2e− → 2Ag(s)
Explanation (Step by Step)
🤔 In a galvanic cell, the metal that gets oxidised becomes the
anode
and releases electrons, so it becomes
negative
.📌 Here, Zn changes to Zn2+ , so Zn is oxidised:
👉 Zn(s) → Zn2+(aq) + 2e− ✅ (Anode, −)
📌 Ag+ gains electrons and deposits as Ag(s), so Ag is reduced:
👉 Ag+(aq) + e− → Ag(s) ✅ (Cathode, +)
📌 Electron flow direction:
👉 Zn (anode) → Ag (cathode) through the external circuit.
👉 Zn (anode) → Ag (cathode) through the external circuit.
📌 Ion flow (inside cell):
👉 In salt bridge, anions move towards anode (to balance Zn2+) and cations move towards cathode (to balance loss of Ag+).
👉 In salt bridge, anions move towards anode (to balance Zn2+) and cations move towards cathode (to balance loss of Ag+).
Did You Know?
✅ The double line
“||”
in cell notation shows a
salt bridge
,
which is like a
“charge balancer”
.👉 Without a salt bridge, charge builds up quickly, and the cell stops working even though reactants are still present.
Q.4: Calculate the standard cell potential (E°cell) of the galvanic cell for the reactions:
(i) 2Cr(s) + 3Cd2+(aq) → 2Cr3+(aq) + 3Cd(s)
(ii) Fe2+(aq) + Ag+(aq) → Fe3+(aq) + Ag(s)
Also calculate ΔrG° and equilibrium constant (Kc).
(i) 2Cr(s) + 3Cd2+(aq) → 2Cr3+(aq) + 3Cd(s)
(ii) Fe2+(aq) + Ag+(aq) → Fe3+(aq) + Ag(s)
Also calculate ΔrG° and equilibrium constant (Kc).
Answer
(i) 2Cr(s) + 3Cd2+(aq) → 2Cr3+(aq) + 3Cd(s)
📌 Given standard reduction potentials:
• E°(Cr3+/Cr) = −0.74 V
• E°(Cd2+/Cd) = −0.40 V
✅ E°cell = +0.34 V
✅ ΔrG° = −196.86 kJ mol−1
✅ Kc ≈ 3.2 × 1034
📌 Given standard reduction potentials:
• E°(Cr3+/Cr) = −0.74 V
• E°(Cd2+/Cd) = −0.40 V
✅ E°cell = +0.34 V
✅ ΔrG° = −196.86 kJ mol−1
✅ Kc ≈ 3.2 × 1034
(ii) Fe2+(aq) + Ag+(aq) → Fe3+(aq) + Ag(s)
📌 Given standard reduction potentials:
• E°(Ag+/Ag) = +0.80 V
• E°(Fe3+/Fe2+) = +0.77 V
✅ E°cell = +0.03 V
✅ ΔrG° = −2.895 kJ mol−1
✅ Kc ≈ 3.22
📌 Given standard reduction potentials:
• E°(Ag+/Ag) = +0.80 V
• E°(Fe3+/Fe2+) = +0.77 V
✅ E°cell = +0.03 V
✅ ΔrG° = −2.895 kJ mol−1
✅ Kc ≈ 3.22
Explanation (Step by Step)
Common formulas (use in both) 📌
👉 E°cell = E°cathode − E°anode (in V)
👉 ΔrG° = −n F E°cell (in J mol−1)
👉 ΔrG° = −RT ln Kc ⇒ ln Kc = −ΔrG° / (RT)
Constants:
• F = 96500 C mol−1
• R = 8.314 J mol−1 K−1
• T = 298 K
👉 E°cell = E°cathode − E°anode (in V)
👉 ΔrG° = −n F E°cell (in J mol−1)
👉 ΔrG° = −RT ln Kc ⇒ ln Kc = −ΔrG° / (RT)
Constants:
• F = 96500 C mol−1
• R = 8.314 J mol−1 K−1
• T = 298 K
(i) Calculation 🤔
Step 1: Identify cathode and anode
• Cd2+ → Cd happens at cathode (reduction), E° = −0.40 V
• Cr → Cr3+ happens at anode (oxidation), E°(Cr3+/Cr) = −0.74 V
Step 2: E°cell
👉 E°cell = (−0.40) − (−0.74) = +0.34 V
Step 3: n (electrons transferred)
Cr(s) → Cr3+ + 3e−
For 2 Cr: electrons = 2 × 3 = 6, so n = 6
Step 4: ΔrG°
👉 ΔrG° = −nFE°cell
= −(6 mol) × (96500 C mol−1) × (0.34 V)
= −196860 J mol−1 = −196.86 kJ mol−1
Step 5: Kc
👉 ln Kc = −ΔrG°/(RT)
= 196860 J mol−1 / (8.314 × 298)
≈ 196860 / 2477.6 ≈ 79.45
So,
👉 Kc = e79.45 ≈ 3.2 × 1034 (dimensionless)
Step 1: Identify cathode and anode
• Cd2+ → Cd happens at cathode (reduction), E° = −0.40 V
• Cr → Cr3+ happens at anode (oxidation), E°(Cr3+/Cr) = −0.74 V
Step 2: E°cell
👉 E°cell = (−0.40) − (−0.74) = +0.34 V
Step 3: n (electrons transferred)
Cr(s) → Cr3+ + 3e−
For 2 Cr: electrons = 2 × 3 = 6, so n = 6
Step 4: ΔrG°
👉 ΔrG° = −nFE°cell
= −(6 mol) × (96500 C mol−1) × (0.34 V)
= −196860 J mol−1 = −196.86 kJ mol−1
Step 5: Kc
👉 ln Kc = −ΔrG°/(RT)
= 196860 J mol−1 / (8.314 × 298)
≈ 196860 / 2477.6 ≈ 79.45
So,
👉 Kc = e79.45 ≈ 3.2 × 1034 (dimensionless)
(ii) Calculation 🤔
Step 1: Cathode and anode
• Ag+ + e− → Ag(s) is cathode, E° = +0.80 V
• Fe2+ → Fe3+ + e− is anode (reverse of Fe3+/Fe2+), E°(Fe3+/Fe2+) = +0.77 V
Step 2: E°cell
👉 E°cell = 0.80 − 0.77 = 0.03 V
Step 3: n
Only 1 electron is transferred, so n = 1
Step 4: ΔrG°
👉 ΔrG° = −(1) × 96500 × 0.03
= −2895 J mol−1 = −2.895 kJ mol−1
Step 5: Kc
👉 ln Kc = 2895 / (8.314 × 298)
= 2895 / 2477.6 ≈ 1.168
So,
👉 Kc = e1.168 ≈ 3.22
Step 1: Cathode and anode
• Ag+ + e− → Ag(s) is cathode, E° = +0.80 V
• Fe2+ → Fe3+ + e− is anode (reverse of Fe3+/Fe2+), E°(Fe3+/Fe2+) = +0.77 V
Step 2: E°cell
👉 E°cell = 0.80 − 0.77 = 0.03 V
Step 3: n
Only 1 electron is transferred, so n = 1
Step 4: ΔrG°
👉 ΔrG° = −(1) × 96500 × 0.03
= −2895 J mol−1 = −2.895 kJ mol−1
Step 5: Kc
👉 ln Kc = 2895 / (8.314 × 298)
= 2895 / 2477.6 ≈ 1.168
So,
👉 Kc = e1.168 ≈ 3.22
Did You Know?
✅ If
E°cell is positive
,
then
ΔrG° is negative
,
and the reaction is spontaneous under standard conditions.✅ A very large Kc (like 1034 ) means the reaction goes almost completely to products, so it’s practically “one-way” in the lab.
Q.5: Write the Nernst equation and calculate emf (Ecell) of the following cells at 298 K:
(i) Mg(s) | Mg2+(0.001 M) || Cu2+(0.0001 M) | Cu(s)
(ii) Fe(s) | Fe2+(0.001 M) || H+(1 M) | H2(g, 1 bar) | Pt(s)
(iii) Sn(s) | Sn2+(0.050 M) || H+(0.020 M) | H2(g, 1 bar) | Pt(s)
(iv) Pt(s) | Br−(0.010 M) | Br2(l) || H+(0.030 M) | H2(g, 1 bar) | Pt(s)
Given (E° values):
E°(Mg2+/Mg) = −2.37 V, E°(Cu2+/Cu) = +0.34 V, E°(Fe2+/Fe) = −0.44 V,
E°(Sn2+/Sn) = −0.14 V, E°(Br2/Br−) = +1.08 V
(i) Mg(s) | Mg2+(0.001 M) || Cu2+(0.0001 M) | Cu(s)
(ii) Fe(s) | Fe2+(0.001 M) || H+(1 M) | H2(g, 1 bar) | Pt(s)
(iii) Sn(s) | Sn2+(0.050 M) || H+(0.020 M) | H2(g, 1 bar) | Pt(s)
(iv) Pt(s) | Br−(0.010 M) | Br2(l) || H+(0.030 M) | H2(g, 1 bar) | Pt(s)
Given (E° values):
E°(Mg2+/Mg) = −2.37 V, E°(Cu2+/Cu) = +0.34 V, E°(Fe2+/Fe) = −0.44 V,
E°(Sn2+/Sn) = −0.14 V, E°(Br2/Br−) = +1.08 V
Answer
Nernst equation at 298 K (general) 📌
👉 Ecell = E°cell − (0.0591/n) log10 Q
where n = electrons transferred , and Q = reaction quotient .
👉 Ecell = E°cell − (0.0591/n) log10 Q
where n = electrons transferred , and Q = reaction quotient .
✅ Final emf values:
(i) 2.68 V
(ii) 0.529 V (≈ 0.53 V)
(iii) 0.078 V (≈ 0.08 V)
(iv) −1.288 V (negative)
(i) 2.68 V
(ii) 0.529 V (≈ 0.53 V)
(iii) 0.078 V (≈ 0.08 V)
(iv) −1.288 V (negative)
Explanation
(i) Mg(s)|Mg2+(10−3 M) || Cu2+(10−4 M)|Cu(s)
Cell reaction: Mg(s) + Cu2+(aq) → Mg2+(aq) + Cu(s) , n = 2
• E°cell = E°cathode − E°anode
= (+0.34) − (−2.37) = 2.71 V
• Q = [Mg2+]/[Cu2+] = 10−3/10−4 = 10
👉 Ecell = 2.71 − (0.0591/2) log(10)
= 2.71 − 0.02955 × 1
= 2.68 V
✅ emf = 2.68 V
Cell reaction: Mg(s) + Cu2+(aq) → Mg2+(aq) + Cu(s) , n = 2
• E°cell = E°cathode − E°anode
= (+0.34) − (−2.37) = 2.71 V
• Q = [Mg2+]/[Cu2+] = 10−3/10−4 = 10
👉 Ecell = 2.71 − (0.0591/2) log(10)
= 2.71 − 0.02955 × 1
= 2.68 V
✅ emf = 2.68 V
(ii) Fe(s)|Fe2+(10−3 M) || H+(1 M)|H2(1 bar)|Pt(s)
Cell reaction: Fe(s) + 2H+(aq) → Fe2+(aq) + H2(g) , n = 2
(Here, a(H2) = 1 because pressure = 1 bar)
• E°cell = E°(H+/H2) − E°(Fe2+/Fe)
= 0.00 − (−0.44) = 0.44 V
• Q = [Fe2+] / [H+]2 = 10−3 / (1)2 = 10−3
👉 Ecell = 0.44 − (0.0591/2) log(10−3)
= 0.44 − 0.02955 × (−3)
= 0.529 V
✅ emf = 0.529 V (≈ 0.53 V)
Cell reaction: Fe(s) + 2H+(aq) → Fe2+(aq) + H2(g) , n = 2
(Here, a(H2) = 1 because pressure = 1 bar)
• E°cell = E°(H+/H2) − E°(Fe2+/Fe)
= 0.00 − (−0.44) = 0.44 V
• Q = [Fe2+] / [H+]2 = 10−3 / (1)2 = 10−3
👉 Ecell = 0.44 − (0.0591/2) log(10−3)
= 0.44 − 0.02955 × (−3)
= 0.529 V
✅ emf = 0.529 V (≈ 0.53 V)
(iii) Sn(s)|Sn2+(0.050 M) || H+(0.020 M)|H2(1 bar)|Pt(s)
Cell reaction: Sn(s) + 2H+(aq) → Sn2+(aq) + H2(g) , n = 2
• E°cell = 0.00 − (−0.14) = 0.14 V
• Q = [Sn2+] / [H+]2 = 0.050 / (0.020)2
= 0.050 / 0.0004 = 125
👉 Ecell = 0.14 − (0.0591/2) log(125)
= 0.14 − 0.02955 × 2.0969
= 0.0785 V
✅ emf = 0.078 V (≈ 0.08 V)
Cell reaction: Sn(s) + 2H+(aq) → Sn2+(aq) + H2(g) , n = 2
• E°cell = 0.00 − (−0.14) = 0.14 V
• Q = [Sn2+] / [H+]2 = 0.050 / (0.020)2
= 0.050 / 0.0004 = 125
👉 Ecell = 0.14 − (0.0591/2) log(125)
= 0.14 − 0.02955 × 2.0969
= 0.0785 V
✅ emf = 0.078 V (≈ 0.08 V)
(iv) Pt|Br−(0.010 M)|Br2(l) || H+(0.030 M)|H2(1 bar)|Pt
As written cell reaction: 2Br−(aq) + 2H+(aq) → Br2(l) + H2(g) , n = 2
• E°cell = E°(H+/H2) − E°(Br2/Br−)
= 0.00 − 1.08 = −1.08 V
• Q = 1 / ( [Br−]2 [H+]2 )
= 1 / (0.0102 × 0.0302)
= 1 / (10−4 × 9×10−4) = 1 / (9×10−8)
= 1.11 × 107
👉 Ecell = −1.08 − (0.0591/2) log(1.11×107)
= −1.08 − 0.02955 × 7.0457
= −1.288 V
✅ emf = −1.288 V (negative)
As written cell reaction: 2Br−(aq) + 2H+(aq) → Br2(l) + H2(g) , n = 2
• E°cell = E°(H+/H2) − E°(Br2/Br−)
= 0.00 − 1.08 = −1.08 V
• Q = 1 / ( [Br−]2 [H+]2 )
= 1 / (0.0102 × 0.0302)
= 1 / (10−4 × 9×10−4) = 1 / (9×10−8)
= 1.11 × 107
👉 Ecell = −1.08 − (0.0591/2) log(1.11×107)
= −1.08 − 0.02955 × 7.0457
= −1.288 V
✅ emf = −1.288 V (negative)
Did You Know?
✅ If
Ecell comes out negative
,
it means the cell reaction as written is
not spontaneous
under those conditions.👉 In such a case, simply reverse the cell notation . The reaction will proceed in the reverse direction, and the emf becomes positive.
📌 For part (iv), to get a positive emf, write the cell as:
👉 Pt | H2(1 bar) | H+(0.030 M) || Br−(0.010 M) | Br2(l) | Pt
Q.6: In button cells (watches etc.) the reaction is:
Zn(s) + Ag2O(s) + H2O(l) → Zn2+(aq) + 2Ag(s) + 2OH−(aq)
Determine E° and ΔrG° for the reaction.
[Given, E°(Zn/Zn2+) = 0.76 V and E°(Ag+/Ag) = +0.34 V]
Zn(s) + Ag2O(s) + H2O(l) → Zn2+(aq) + 2Ag(s) + 2OH−(aq)
Determine E° and ΔrG° for the reaction.
[Given, E°(Zn/Zn2+) = 0.76 V and E°(Ag+/Ag) = +0.34 V]
Answer
✅ 👉 Standard cell potential (E°cell) =
1.10 V
✅ 👉 Standard Gibbs free energy change (ΔrG°) = −2.12 × 105 J mol−1 = −212.3 kJ mol−1
Explanation (Step by Step)
Half reactions (standard conditions) 📌
Anode (oxidation)
Zn(s) → Zn2+(aq) + 2e−
E°(Zn2+/Zn) = −0.76 V
Anode (oxidation)
Zn(s) → Zn2+(aq) + 2e−
E°(Zn2+/Zn) = −0.76 V
Cathode (reduction)
Ag2O(s) + H2O(l) + 2e− → 2Ag(s) + 2OH−(aq)
E°(Ag2O/Ag) = +0.34 V
Ag2O(s) + H2O(l) + 2e− → 2Ag(s) + 2OH−(aq)
E°(Ag2O/Ag) = +0.34 V
✅
Step 1: Calculate E°cell
👉 E°cell = E°cathode − E°anode
= (+0.34) − (−0.76)
= +1.10 V
👉 E°cell = E°cathode − E°anode
= (+0.34) − (−0.76)
= +1.10 V
✅
Step 2: Calculate ΔrG°
🤔 Electrons transferred n = 2
👉 ΔrG° = −n F E°cell
= −(2 mol) × (96500 C mol−1) × (1.10 V)
= −212300 J mol−1
= −212.3 kJ mol−1
🤔 Electrons transferred n = 2
👉 ΔrG° = −n F E°cell
= −(2 mol) × (96500 C mol−1) × (1.10 V)
= −212300 J mol−1
= −212.3 kJ mol−1
Did You Know?
✅ Button cells using
Ag2O–Zn
are popular because they give a nearly
constant voltage
during most of their life.👉 That stable output is why they’re perfect for watches, calculators, hearing aids, and small medical devices.
Q.7: Define conductivity and molar conductivity for the solution of an electrolyte. Discuss their variation with concentration.
Answer
(i) Conductivity (κ)
📌 Conductivity (κ) is the ability of an electrolyte solution to conduct electricity.
It is the reciprocal of resistivity (ρ).
👉 κ = 1/ρ
Also,
👉 κ = (1/R) × (l/A) = G × (l/A)
Where:
• R = resistance (unit: Ω )
• G = 1/R = conductance (unit: S or Ω−1 )
• l = distance between electrodes (unit: cm )
• A = area of electrodes (unit: cm2 )
📌 If l = 1 cm and A = 1 cm2 , then κ = G .
✅ Unit of κ: S cm−1 (or Ω−1 cm−1 )
📌 Conductivity (κ) is the ability of an electrolyte solution to conduct electricity.
It is the reciprocal of resistivity (ρ).
👉 κ = 1/ρ
Also,
👉 κ = (1/R) × (l/A) = G × (l/A)
Where:
• R = resistance (unit: Ω )
• G = 1/R = conductance (unit: S or Ω−1 )
• l = distance between electrodes (unit: cm )
• A = area of electrodes (unit: cm2 )
📌 If l = 1 cm and A = 1 cm2 , then κ = G .
✅ Unit of κ: S cm−1 (or Ω−1 cm−1 )
(ii) Molar conductivity (Λm)
📌 Molar conductivity (Λm) is the conductance of the solution containing 1 mole of electrolyte .
👉 Λm = (κ × 1000) / C
Where:
• κ = conductivity (S cm−1)
• C = concentration in mol L−1
• 1000 converts L to cm3 (since 1 L = 1000 cm3 )
✅ Unit of Λm: S cm2 mol−1 (or Ω−1 cm2 mol−1 )
📌 Molar conductivity (Λm) is the conductance of the solution containing 1 mole of electrolyte .
👉 Λm = (κ × 1000) / C
Where:
• κ = conductivity (S cm−1)
• C = concentration in mol L−1
• 1000 converts L to cm3 (since 1 L = 1000 cm3 )
✅ Unit of Λm: S cm2 mol−1 (or Ω−1 cm2 mol−1 )
Explanation (Variation with Concentration)
1) Variation of Conductivity (κ) with concentration 🤔
📌 When a solution is diluted, the number of ions per unit volume decreases.
👉 So, conductivity generally decreases on dilution (for both strong and weak electrolytes).
✅ On dilution (concentration ↓): κ ↓
📌 When a solution is diluted, the number of ions per unit volume decreases.
👉 So, conductivity generally decreases on dilution (for both strong and weak electrolytes).
✅ On dilution (concentration ↓): κ ↓
2) Variation of Molar Conductivity (Λm) with concentration 📌
📌 On dilution, the volume of solution containing 1 mole of electrolyte becomes larger.
👉 Also, ions get more space to move, and in weak electrolytes, ionisation increases.
✅ On dilution (concentration ↓): Λm ↑
Reason (simple):
• κ goes down because fewer ions exist per cm3
• Λm goes up because it counts conduction for the whole solution volume that contains 1 mole .
📌 On dilution, the volume of solution containing 1 mole of electrolyte becomes larger.
👉 Also, ions get more space to move, and in weak electrolytes, ionisation increases.
✅ On dilution (concentration ↓): Λm ↑
Reason (simple):
• κ goes down because fewer ions exist per cm3
• Λm goes up because it counts conduction for the whole solution volume that contains 1 mole .
Did You Know?
✅ For strong electrolytes (like NaCl, HCl),
Λm increases slowly
on dilution because they are already almost fully ionised.✅ For weak electrolytes (like CH3COOH), Λm increases sharply on dilution because ionisation increases a lot as concentration decreases.