Class 12 Chemistry · Chapter 2
Short answer: NCERT Class 12 Chemistry Part I Chapter 2 Electrochemistry me 18 exercise questions hain, jo electrochemical cells, electrode potential, Nernst equation, conductance, Kohlrausch's law, aur electrolysis (Faraday's laws) tak poora coverage dete hain. Har numerical ka step-by-step solution — given, formula, substitution, answer — neeche diya gaya hai.
Electrochemistry NEET aur JEE dono me har saal aata hai, aur board me bhi 5+ marks ka guaranteed chapter hai. Ye chapter thermodynamics (Class 11) aur redox reactions (Class 11) ko jodta hai — cell ka EMF, Gibbs energy, aur equilibrium constant sab ek formula (ΔG° = −nFE°) se connected hain. Nernst equation aur conductivity ke numericals scoring hote hain agar formula aur units clear ho.
Chapter 2 Summary — 5 Minute Revision
Electrochemical cell (Galvanic / Voltaic cell)
- Ek device jo spontaneous redox reaction ki chemical energy ko electrical energy me convert karta hai.
- Anode — oxidation hoti hai, negative terminal (galvanic cell me). Cathode — reduction hoti hai, positive terminal.
- Cell notation convention: Anode (oxidation half) left me, salt bridge ke liye ||, Cathode (reduction half) right me. Udaharan (Daniell cell):
Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu(s)
Electrode potential aur EMF
- Standard electrode potential (E°) — 298 K, 1 atm pressure, 1 M concentration par measure kiya gaya reduction potential, hydrogen electrode (SHE, E° = 0) ke reference se.
- E°cell = E°cathode − E°anode (dono reduction potentials hi lein — subtract karna hai, add nahi).
- E°cell > 0 → reaction spontaneous (feasible).
EMF aur Gibbs free energy ka relation
Electrical work jo cell kar sakta hai, Gibbs energy ki kami ke barabar hota hai:
ΔG° = −nFE°cell
Isi se equilibrium constant nikalta hai:
ΔG° = −RT ln K = −nFE°cell ⇒ log K = nE°cell/0.0591 (298 K par)
Nernst equation
Standard conditions se hatte hi seedha E° use nahi kar sakte — concentration ke saath EMF badalta hai:
Ecell = E°cell − (RT/nF) ln Q
298 K par, log base 10 me (2.303RT/F = 0.0591 V):
Ecell = E°cell − (0.0591/n) log Q
Yahan Q reaction quotient hai — products/reactants ka concentration ratio (pure solids/liquids Q me nahi aate).
Conductance — specific aur molar conductivity
- Conductance (G) = 1/R, unit siemens (S).
- Specific conductivity (κ) = conductance × cell constant (l/A), unit S cm−1 ya S m−1.
- Molar conductivity (Λm) = ek mole electrolyte ki total conducting power, jab electrolyte ko 1 cm apart do electrodes ke beech rakha jaye:
Λm = κ × 1000 / M (M = molarity in mol/L, κ in S cm⁻¹)
- Dilution badhane par κ ghatta hai (ions kam concentrated), par Λm badhta hai (dissociation zyada hoti hai).
- Strong electrolyte (HCl, KCl): Λm dilution ke saath dheere-dheere, seedhi line me badhta hai — infinite dilution par extrapolate ho sakta hai.
- Weak electrolyte (CH3COOH): Λm infinite dilution ke paas tezi se badhta hai kyunki dissociation degree badh jaata hai — Λ°m extrapolation se nahi, Kohlrausch's law se milta hai.
Kohlrausch's law
Infinite dilution par, molar conductivity electrolyte ke ions ke independent contribution ka sum hota hai:
Λ°m(electrolyte) = ν+λ°+ + ν−λ°−
Iska use weak electrolytes ka Λ°m nikalne me hota hai (jo direct extrapolate nahi ho sakta), aur degree of dissociation α = Λm/Λ°m se milta hai.
Electrolysis aur Faraday's laws
- Faraday's 1st law: deposited/liberated mass, current aur time ke directly proportional hai — m = Zit (Z = electrochemical equivalent).
- Faraday's 2nd law: same charge se deposit hui masses unke equivalent weights ke proportional hoti hain.
- 1 Faraday (F) = 96,500 C/mol = 1 mole electrons ka charge.
m = (M / nF) × I × t
Batteries, fuel cells aur corrosion (conceptual)
- Primary battery (dry cell) — ek baar use ke baad recharge nahi ho sakti. Secondary battery (lead storage) — recharge ho sakti hai (reverse current se).
- Fuel cell (H2-O2) — reactants continuously supply hote hain, pollution-free electricity, spacecraft me use hota hai.
- Corrosion (rusting) — metal ka electrochemical oxidation hai; iron surface par galvanic cell jaisa mechanism banta hai (anode aur cathode regions), moisture aur O2 zaroori hain.
Poore Class 12 Chemistry ke handwritten colour notes
IITian & district toppers ke banaye short notes — revision-ready, diagram ke saath. Board se pehle poora syllabus 3 din me revise.
Exercise Questions — Solutions (Q1–Q18)
Q1. Daniell cell (Zn–Cu) ka cell notation likhiye aur cell reaction batayiye.
Convention: anode (oxidation) left, cathode (reduction) right.
Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu(s)
Overall cell reaction:
Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)
Anode par oxidation (Zn → Zn2+ + 2e−), cathode par reduction (Cu2+ + 2e− → Cu).
Q2. E°(Zn2+/Zn) = −0.76 V aur E°(Cu2+/Cu) = +0.34 V diye hain. Daniell cell ka E°cell calculate kijiye.
Given: Cu2+/Cu cathode hai (reduction, higher E°), Zn2+/Zn anode hai (oxidation, lower E°).
E°cell = E°cathode − E°anode = 0.34 − (−0.76)
E°cell = 1.10 V
Q3. Q2 ke Daniell cell ke liye ΔG° calculate kijiye (n = 2, F = 96500 C/mol).
Given: n = 2, F = 96500 C/mol, E°cell = 1.10 V
ΔG° = −nFE°cell = −2 × 96500 × 1.10
ΔG° = −212300 J/mol = −212.3 kJ/mol
Negative ΔG° confirm karta hai ki reaction spontaneous hai — jo positive E°cell se bhi consistent hai.
Q4. Daniell cell me [Zn2+] = 1 M aur [Cu2+] = 0.01 M ho, to Ecell nikaliye (E°cell = 1.10 V, n = 2).
Given: E°cell = 1.10 V, n = 2, [Zn2+] = 1 M, [Cu2+] = 0.01 M
Nernst equation (298 K):
Ecell = E°cell − (0.0591/n) log([Zn2+]/[Cu2+])
Ecell = 1.10 − (0.0591/2) log(1/0.01) = 1.10 − 0.02955 × 2
Ecell = 1.10 − 0.0591 = 1.041 V
Q5. Q2 ke Daniell cell ka equilibrium constant K nikaliye (298 K, n = 2, E°cell = 1.10 V).
Given: n = 2, E°cell = 1.10 V
log K = nE°cell/0.0591 = (2 × 1.10)/0.0591 = 37.23
K = 1037.23 ≈ 1.7 × 1037
Itna bada K matlab reaction practically complete hone tak aage badhta hai — equilibrium bahut products ki taraf hai.
Q6. Ek conductivity cell me resistance 100 Ω hai aur cell constant 1.5 cm−1 hai. Specific conductivity nikaliye.
Given: R = 100 Ω, cell constant (l/A) = 1.5 cm−1
κ = cell constant / R = 1.5 / 100
κ = 0.015 S cm⁻¹ = 1.5 × 10⁻² S cm⁻¹
Q7. Q6 ke solution ka concentration 0.5 mol/L hai. Molar conductivity Λm nikaliye.
Given: κ = 1.5 × 10⁻² S cm⁻¹, M = 0.5 mol/L
Λm = κ × 1000 / M = (1.5 × 10⁻² × 1000) / 0.5
Λm = 15 / 0.5
Λm = 30 S cm² mol⁻¹
Q8. Kohlrausch's law se CH3COOH ka Λ°m nikaliye. Diya hai: Λ°m(HCl) = 425.9, Λ°m(CH3COONa) = 91.0, Λ°m(NaCl) = 126.4 S cm² mol⁻¹.
Given: Λ°m(HCl) = 425.9, Λ°m(CH3COONa) = 91.0, Λ°m(NaCl) = 126.4 S cm² mol⁻¹
Kohlrausch's law se ion additivity use karte hain (Na+ aur Cl− cancel ho jaate hain):
Λ°m(CH3COOH) = Λ°m(HCl) + Λ°m(CH3COONa) − Λ°m(NaCl)
= 425.9 + 91.0 − 126.4
Λ°m(CH3COOH) = 390.5 S cm² mol⁻¹
Q9. Q8 ke CH3COOH solution ka ek concentration par Λm = 39.05 S cm² mol⁻¹ mila. Degree of dissociation α nikaliye.
Given: Λm = 39.05 S cm² mol⁻¹, Λ°m = 390.5 S cm² mol⁻¹ (Q8 se)
α = Λm / Λ°m = 39.05 / 390.5
α = 0.1 (yaani 10% dissociation)
Q10. CuSO4 solution ka electrolysis 2 A current se 1 ghante (3600 s) ke liye kiya jata hai. Cathode par kitna copper deposit hoga? (Cu ka M = 63.5 g/mol, n = 2)
Given: I = 2 A, t = 3600 s, M(Cu) = 63.5 g/mol, n = 2, F = 96500 C/mol
Charge:
Q = It = 2 × 3600 = 7200 C
Moles of electrons:
= 7200 / 96500 = 0.0746 mol
Cu2+ + 2e− → Cu, isliye moles of Cu = 0.0746/2 = 0.0373 mol
mass = 0.0373 × 63.5
mass ≈ 2.37 g
Q11. AgNO3 solution se 2 A current pass karke 1.0 g silver deposit karna hai. Kitna time lagega? (Ag ka M = 108 g/mol, n = 1)
Given: mass = 1.0 g, M(Ag) = 108 g/mol, n = 1, I = 2 A, F = 96500 C/mol
Moles of Ag:
= 1.0 / 108 = 0.00926 mol
Charge required:
Q = 0.00926 × 1 × 96500 = 893.7 C
Time:
t = Q/I = 893.7 / 2
t ≈ 446.9 s ≈ 7.45 min
Q12. Molten Al2O3 ke electrolysis se 5.4 g aluminium (M = 27 g/mol, n = 3) deposit karne ke liye kitna charge (coulombs) chahiye?
Given: mass = 5.4 g, M(Al) = 27 g/mol, n = 3, F = 96500 C/mol
Moles of Al:
= 5.4 / 27 = 0.2 mol
Moles of electrons:
= 0.2 × 3 = 0.6 mol
Charge:
Q = 0.6 × 96500
Q = 57900 C
Q13. Aqueous NaCl ke electrolysis me cathode aur anode par kaunsi gas milti hai, aur kyun (Na nahi)?
Cathode par H2 gas milti hai, Na dhaatu nahi — kyunki H2O ka reduction potential Na+ se zyada favourable hai (Na+ ka E° bahut negative hai, isliye water preferentially reduce hoti hai).
Anode par Cl2 gas milti hai (O2 ki jagah), kyunki O2 banne me overvoltage zyada hota hai, isliye Cl− ka oxidation kinetically favoured ho jaata hai.
2NaCl(aq) + 2H2O(l) → 2NaOH(aq) + H2(g) + Cl2(g)
Q14. E°(Cu2+/Cu) = 0.34 V aur E°(Fe2+/Fe) = −0.44 V hai. Kya Fe metal, Cu2+ ions ko displace kar sakta hai?
Given: E°(Cu2+/Cu) = 0.34 V, E°(Fe2+/Fe) = −0.44 V
Fe anode (oxidation) hoga kyunki uska E° kam hai, Cu cathode (reduction).
E°cell = E°cathode − E°anode = 0.34 − (−0.44) = 0.78 V
E°cell > 0, isliye reaction spontaneous hai — Fe, Cu2+ ko displace karke Cu deposit karega:
Fe(s) + Cu2+(aq) → Fe2+(aq) + Cu(s)
Q15. Acidified water se electrolysis me 2 A current 965 s ke liye pass ki gayi. STP par kitna H2 gas (volume me) cathode par nikalega?
Given: I = 2 A, t = 965 s, F = 96500 C/mol
Charge:
Q = It = 2 × 965 = 1930 C
Moles of electrons:
= 1930 / 96500 = 0.02 mol
2H+ + 2e− → H2, isliye moles of H2 = 0.02/2 = 0.01 mol
Volume at STP (1 mol gas = 22400 mL):
V = 0.01 × 22400 = 224 mL
V = 224 mL = 0.224 L
Q16. E°(Ag+/Ag) = 0.80 V aur E°(Fe3+/Fe2+) = 0.77 V diye hain. Reaction Fe2+ + Ag+ → Fe3+ + Ag ki feasibility check kijiye.
Given: E°(Ag+/Ag) = 0.80 V (cathode/reduction), E°(Fe3+/Fe2+) = 0.77 V (anode/oxidation)
E°cell = E°cathode − E°anode = 0.80 − 0.77 = 0.03 V
E°cell thoda positive hai (0.03 V), isliye reaction spontaneous hai lekin bahut zyada favourable nahi — equilibrium constant bhi bahut bada nahi hoga (chhoti E°cell ka matlab hai reaction poora complete nahi hoga).
Q17. Primary aur secondary battery me kya farak hai? Ek-ek udaharan dijiye.
Primary battery: reaction irreversible hoti hai, ek baar discharge hone ke baad recharge nahi ho sakti — feenk deni padti hai. Udaharan: dry cell (Leclanché cell).
Secondary battery: reaction reversible hoti hai, external current pass karke recharge ki jaa sakti hai. Udaharan: lead storage battery (car battery).
Q18. H2–O2 fuel cell ka basic working principle samjhaiye — iske do fayde bhi bataiye.
Fuel cell me H2 gas anode par supply hoti hai (oxidation hokar H+ banata hai), O2 gas cathode par supply hoti hai (reduction hokar OH−/H2O banata hai). Overall reaction:
2H2(g) + O2(g) → 2H2O(l)
Reactants continuously bahar se supply hote rehte hain (battery ki tarah storage nahi hoti), isliye current continuously milta rehta hai.
Fayde: (i) pollution-free — sirf paani banta hai, (ii) efficiency zyada hoti hai combustion se, isliye spacecraft me electricity ke saath drinking water ke liye bhi use hota hai.
Important Equations — Ek Nazar Me
| Concept | Formula | Yaad rakhein |
|---|---|---|
| Cell EMF | E°cell = E°cathode − E°anode | Subtract karna hai, add nahi |
| Gibbs energy relation | ΔG° = −nFE°cell | n = electrons transferred, F = 96500 C/mol |
| Equilibrium constant | log K = nE°cell/0.0591 (298 K) | ΔG° = −RT ln K se derive hota hai |
| Nernst equation (general) | Ecell = E°cell − (RT/nF) ln Q | Non-standard conditions ke liye |
| Nernst equation (298 K, log base 10) | Ecell = E°cell − (0.0591/n) log Q | 0.0591 sirf 298 K par valid |
| Specific conductivity | κ = G × (l/A) = cell constant / R | Unit: S cm⁻¹ |
| Molar conductivity | Λm = κ × 1000 / M | M in mol/L, κ in S cm⁻¹ |
| Kohlrausch's law | Λ°m = ν+λ°+ + ν−λ°− | Weak electrolyte ka Λ°m nikalne ke liye |
| Degree of dissociation | α = Λm / Λ°m | Weak electrolyte ke liye |
| Faraday's 1st law | m = Zit = (M/nF) × I × t | Z = electrochemical equivalent |
| Faraday's constant | 1 F = 96,500 C/mol | 1 mole electrons ka charge |
↔ Table ko side me swipe karein
Common Mistakes — Yahan Marks Kat te Hain
- E°cell me sign error. E°cell = E°cathode − E°anode hota hai, dono ko add nahi karte. Anode/cathode identify karne me galti se poora answer ulta ho jata hai.
- Nernst equation me 'n' galat lena. n balanced equation me transfer hue electrons ki sankhya hai, reactant ke moles nahi. Zn + Cu2+ → Zn2+ + Cu me n = 2 hai, 1 nahi.
- Specific conductivity (κ) aur molar conductivity (Λm) confuse karna. κ ek fixed volume ka conductance hai, Λm ek mole electrolyte ka — formula aur unit dono alag hain. Λm nikalte waqt ×1000/M lagana bhool jaana bahut common mistake hai.
- Faraday's constant ki value galat likhna. F = 96,500 C/mol hai, isse J ya kJ na samjhein — ye charge hai, energy nahi.
- Cell notation me anode/cathode swap karna. Convention fix hai: anode (oxidation) hamesha left, cathode (reduction) hamesha right, salt bridge '||' beech me. Ulta likhne se poora notation galat ho jaata hai.
- Λm ka dilution trend ulta samajhna. Strong electrolyte (KCl, HCl) me Λm dilution ke saath dheere badhta hai (ionic interactions kam hoti hain), par weak electrolyte (CH3COOH) me infinite dilution ke paas tezi se badhta hai kyunki dissociation degree badh jaati hai — dono ka graph ek jaisa nahi hota.
Board-Style Important Questions
- 1 mark: Molar conductivity ki definition dijiye, uska SI unit bhi likhiye.
- 1 mark: Ek general electrode reaction Mn+ + ne− → M ke liye Nernst equation likhiye.
- 2 marks: E°cell = 1.10 V aur n = 2 diya hai. Cell ke ΔG° ki calculation kijiye.
- 3 marks: Kohlrausch's law state kijiye aur iska ek application (weak electrolyte ka Λ°m nikalna) samjhaiye.
- 3 marks: ΔG°, E°cell, aur equilibrium constant K ke beech relation derive kijiye.
- 5 marks: H2–O2 fuel cell ka construction, working aur do fayde likhiye. Ek electrolysis-based numerical (Faraday's law) bhi solve kijiye.
Aksar Poochhe Jaane Wale Sawaal
Class 12 Chemistry Chapter 2 Electrochemistry me kitne exercise questions hain?
Is chapter me 18 exercise questions hain, jo cell notation, EMF, Nernst equation, conductivity, Kohlrausch's law, aur Faraday's laws — sabhi sub-topics cover karte hain, jinke step-by-step solutions upar diye gaye hain.
Nernst equation kya hai aur kab use hoti hai?
Nernst equation hume batata hai ki concentration badalne se cell ka EMF kaise change hota hai — standard conditions (1 M, 298 K) se hatte hi seedha E° use nahi kar sakte. 298 K par formula hai: E_cell = E°_cell − (0.0591/n) log Q.
E°_cell nikalne ka formula kya hai?
E°_cell = E°_cathode − E°_anode. Cathode par reduction hoti hai (higher/more positive E°), anode par oxidation (lower E°). Dono ko subtract karna hai, add nahi — ye sabse common mistake hai.
Specific conductivity aur molar conductivity me kya farak hai?
Specific conductivity (κ) ek unit volume electrolyte ka conductance hai (S cm⁻¹), jabki molar conductivity (Λ_m) ek mole electrolyte ka conductance hai jab 1 cm apart electrodes ke beech rakha jaye. Relation: Λ_m = κ × 1000 / M.
Kohlrausch's law kis kaam aata hai?
Kohlrausch's law weak electrolytes (jaise CH3COOH) ka Λ°_m (infinite dilution par molar conductivity) nikalne me use hota hai, kyunki weak electrolytes ka graph extrapolate karke Λ°_m nahi mil sakta — sirf strong electrolytes ke liye ye seedha extrapolation se milta hai.
Faraday's laws of electrolysis kya batate hain?
Pehla law: deposited mass current aur time ke directly proportional hai (m = Zit). Doosra law: same charge se deposit hui masses unke equivalent weights ke proportional hoti hain. 1 Faraday = 96,500 C/mol electrons ka charge hai.
Class 12 Chemistry — Saare Chapters
Board exam tak sirf revision karna hai?
Class 12 Chemistry ke saare chapters ke colour handwritten short notes — diagrams, formulas aur important points ek jagah.
NCERT Kaksha ek swatantra shaikshik platform hai aur NCERT ya CBSE se aadhikarik roop se sambaddh (officially affiliated) nahi hai. Kisi galti ki jaankari dene ke liye contact kijiye.