PRODUCTS OF ELECTROLYSIS — COMPLETE JEE ANSWERS

A. Molten salts

1. Molten NaCl

NaCl(l)→Na++Cl−\mathrm{NaCl(l)\rightarrow Na^+ + Cl^-}

Cathode:Na++e−→Na\mathrm{Na^+ +e^-\rightarrow Na}

Anode:2Cl−→Cl2+2e−\mathrm{2Cl^-\rightarrow Cl_2+2e^-}

Answer: A) Na, Cl₂

2. Molten CaCl₂

Cathode:Ca2++2e−→Ca\mathrm{Ca^{2+}+2e^-\rightarrow Ca}

Anode:2Cl−→Cl2+2e−\mathrm{2Cl^-\rightarrow Cl_2+2e^-}

Answer: A) Ca + Cl₂

3. Molten Al₂O₃

At cathode:Al3++3e−→Al\mathrm{Al^{3+}+3e^-\rightarrow Al}

At anode:2O2−→O2+4e−\mathrm{2O^{2-}\rightarrow O_2+4e^-}

So theoretically:Al+O2\boxed{\mathrm{Al+O_2}}

Answer: A

Important industrial note: In the Hall-Héroult process, carbon anodes are used, so oxygen reacts with carbon and mainly forms CO₂ (with some CO).

4. Why molten NaCl instead of aqueous NaCl for sodium?

In aqueous NaCl, both Na⁺ and water can be reduced at the cathode.

Water is preferentially reduced:2H2O+2e−→H2+2OH−\mathrm{2H_2O+2e^-\rightarrow H_2+2OH^-}

rather than:Na++e−→Na\mathrm{Na^++e^-\rightarrow Na}

Therefore sodium metal cannot normally be obtained from aqueous NaCl.

In molten NaCl, water is absent, so:Na+→Na\boxed{\mathrm{Na^+\rightarrow Na}}

B. Aqueous NaCl / Brine

5. Electrolysis of concentrated aqueous NaCl

Cathode:2H2O+2e−→H2+2OH−\mathrm{2H_2O+2e^-\rightarrow H_2+2OH^-}

Anode:2Cl−→Cl2+2e−\mathrm{2Cl^-\rightarrow Cl_2+2e^-}

Overall:2NaCl+2H2O→2NaOH+H2+Cl2\boxed{\mathrm{2NaCl+2H_2O\rightarrow 2NaOH+H_2+Cl_2}}

Answer: B) H₂ at cathode, Cl₂ at anode

6. Solution left after electrolysis of brine

Na⁺ remains in solution and combines with OH⁻:Na++OH−→NaOH\mathrm{Na^++OH^-\rightarrow NaOH}

Therefore:NaOH\boxed{\mathrm{NaOH}}

Answer: A

This is the basis of the chlor-alkali process.

7. Why isn’t Na⁺ discharged in aqueous NaCl?

Because reduction of water is thermodynamically/kinetically more favorable than reduction of Na⁺ under ordinary aqueous electrolysis conditions.2H2O+2e−→H2+2OH−\mathrm{2H_2O+2e^-\rightarrow H_2+2OH^-}

Hence:H2 rather than Na\boxed{\mathrm{H_2\ rather\ than\ Na}}

8. Dilute NaCl solution

At cathode:H2\boxed{H_2}

At anode, especially in dilute solution:O2\boxed{O_2}

because oxidation of water/OH⁻ competes strongly with chloride oxidation.

JEE caution: The anodic product from chloride solutions depends on concentration and electrode conditions. Concentrated brine favors Cl₂.

9. pH near cathode in NaCl electrolysis

Cathode reaction:2H2O+2e−→H2+2OH−\mathrm{2H_2O+2e^-\rightarrow H_2+2OH^-}

OH⁻ concentration increases.

Therefore:pH increases; solution becomes basic\boxed{\text{pH increases; solution becomes basic}}

C. Aqueous CuSO₄

10. CuSO₄ with Cu electrodes

Cathode:Cu2++2e−→Cu\mathrm{Cu^{2+}+2e^-\rightarrow Cu}

Anode:Cu→Cu2++2e−\mathrm{Cu\rightarrow Cu^{2+}+2e^-}

Thus:Cu deposits at cathode and Cu dissolves at anode\boxed{\text{Cu deposits at cathode and Cu dissolves at anode}}

Answer: A

11. CuSO₄ with Pt electrodes

Cathode:Cu2++2e−→Cu\mathrm{Cu^{2+}+2e^-\rightarrow Cu}

Anode:2H2O→O2+4H++4e−\mathrm{2H_2O\rightarrow O_2+4H^++4e^-}

Therefore:Cu at cathode, O2 at anode\boxed{\mathrm{Cu\ at\ cathode,\ O_2\ at\ anode}}

Answer: A

12. What happens to CuSO₄ concentration with Pt electrodes?

Cu²⁺ is removed:Cu2+→Cu\mathrm{Cu^{2+}\rightarrow Cu}

while sulfate remains in solution.

Therefore CuSO₄ concentration decreases.

Also H₂SO₄ effectively accumulates as electrolysis proceeds.[CuSO4]↓\boxed{[\mathrm{CuSO_4}]\downarrow}

13. CuSO₄ with Cu electrodes

Cathode removes Cu²⁺:Cu2++2e−→Cu\mathrm{Cu^{2+}+2e^-\rightarrow Cu}

Anode produces Cu²⁺:Cu→Cu2++2e−\mathrm{Cu\rightarrow Cu^{2+}+2e^-}

The rates are equal ideally.

Therefore:CuSO₄ concentration remains approximately unchanged\boxed{\text{CuSO₄ concentration remains approximately unchanged}}

14. Which electrode loses mass?

At anode:Cu→Cu2++2e−\mathrm{Cu\rightarrow Cu^{2+}+2e^-}

So copper anode dissolves.Anode\boxed{\text{Anode}}

D. Aqueous AgNO₃

15. AgNO₃ with Pt electrodes

Cathode:Ag++e−→Ag\mathrm{Ag^++e^-\rightarrow Ag}

Anode:2H2O→O2+4H++4e−\mathrm{2H_2O\rightarrow O_2+4H^++4e^-}

Therefore:Ag at cathode, O2 at anode\boxed{\mathrm{Ag\ at\ cathode,\ O_2\ at\ anode}}

Answer: A

16. AgNO₃ with Ag electrodes

Cathode:Ag++e−→Ag\mathrm{Ag^++e^-\rightarrow Ag}

Anode:Ag→Ag++e−\mathrm{Ag\rightarrow Ag^++e^-}

Thus:Ag deposits at cathode and Ag dissolves at anode\boxed{\text{Ag deposits at cathode and Ag dissolves at anode}}

Answer: A

17. Why is Ag⁺ discharged instead of water?

Reduction of Ag⁺:Ag++e−→Ag\mathrm{Ag^++e^-\rightarrow Ag}

has a much more favorable reduction potential than reduction of water under ordinary conditions.

Therefore:Ag+ is preferentially reduced\boxed{\mathrm{Ag^+\ is\ preferentially\ reduced}}

E. Aqueous KI / KBr / NaCl

18. Aqueous KI

Cathode:2H2O+2e−→H2+2OH−\mathrm{2H_2O+2e^-\rightarrow H_2+2OH^-}

Anode:2I−→I2+2e−\mathrm{2I^-\rightarrow I_2+2e^-}

Therefore:H2+I2\boxed{\mathrm{H_2+I_2}}

Answer: B

19. Species discharged at anode in KI

I⁻ is easily oxidized:2I−→I2+2e−\mathrm{2I^-\rightarrow I_2+2e^-}

Answer: B) I⁻

20. Increasing ease of oxidation

Halide oxidation tendency:F−<Cl−<Br−<I−\boxed{\mathrm{F^-<Cl^-<Br^-<I^-}}

Thus I⁻ is oxidized most easily.

21. Aqueous NaBr

Cathode:H2O→H2\mathrm{H_2O\rightarrow H_2}

Anode:2Br−→Br2+2e−\mathrm{2Br^-\rightarrow Br_2+2e^-}

Therefore:H2+Br2\boxed{\mathrm{H_2+Br_2}}

Answer: B

22. Why concentrated NaCl gives Cl₂ but dilute solution may give O₂?

At the anode, two processes compete:2Cl−→Cl2+2e−\mathrm{2Cl^-\rightarrow Cl_2+2e^-}

and2H2O→O2+4H++4e−\mathrm{2H_2O\rightarrow O_2+4H^++4e^-}

High Cl⁻ concentration favors chloride oxidation, while in dilute chloride solution oxygen evolution becomes more competitive.Concentrated Cl⁻ → Cl₂ favored\boxed{\text{Concentrated Cl⁻ → Cl₂ favored}}Dilute Cl⁻ → O₂ can become favored\boxed{\text{Dilute Cl⁻ → O₂ can become favored}}

F. Selective discharge

23. Which ion is preferentially discharged?

Among:Na+,Cu2+,Ag+,H+\mathrm{Na^+,Cu^{2+},Ag^+,H^+}

Ag⁺ is easiest to reduce.Ag++e−→Ag\boxed{\mathrm{Ag^++e^-\rightarrow Ag}}

Answer: Ag⁺

24. Increasing tendency to get reduced

Using standard reduction potentials:Na+<Mg2+<Zn2+<Cu2+<Ag+\boxed{\mathrm{Na^+<Mg^{2+}<Zn^{2+}<Cu^{2+}<Ag^+}}

So Ag⁺ has the greatest tendency to be reduced.

25. Most easily oxidized

F−,Cl−,Br−,I−\mathrm{F^-,Cl^-,Br^-,I^-}

Oxidation tendency:I−>Br−>Cl−>F−\boxed{\mathrm{I^->Br^->Cl^->F^-}}

Answer: I⁻

26. Why isn’t F⁻ discharged in aqueous solution?

Fluoride is extremely difficult to oxidize.

Water oxidation to O₂ occurs preferentially under ordinary aqueous electrolysis conditions.F− does not normally give F2\boxed{\mathrm{F^-\ does\ not\ normally\ give\ F_2}}

27. Aqueous Na₂SO₄ with inert electrodes

Na⁺ is not reduced.

Water is reduced:2H2O+2e−→H2+2OH−\mathrm{2H_2O+2e^-\rightarrow H_2+2OH^-}

At anode:2H2O→O2+4H++4e−\mathrm{2H_2O\rightarrow O_2+4H^++4e^-}

Therefore:H2+O2\boxed{H_2+O_2}

Answer: B

28. Aqueous KOH

At cathode:2H2O+2e−→H2+2OH−\mathrm{2H_2O+2e^-\rightarrow H_2+2OH^-}

At anode:4OH−→O2+2H2O+4e−\mathrm{4OH^-\rightarrow O_2+2H_2O+4e^-}

Therefore:H2+O2\boxed{H_2+O_2}

Answer: B

29. Dilute H₂SO₄

Cathode:H2\boxed{H_2}

Anode:O2\boxed{O_2}

Answer: B

Sulphate ion generally remains a spectator.

G. Mixed electrolytes

30. Cu²⁺ and Ag⁺ together

Ag⁺ has greater reduction tendency:Ag++e−→Ag\mathrm{Ag^++e^-\rightarrow Ag}

occurs before Cu²⁺ reduction.Ag+ first\boxed{\mathrm{Ag^+\ first}}

31. Ag⁺, Cu²⁺, H⁺ and Na⁺

General reduction preference:Ag+>Cu2+>H+>Na+\boxed{\mathrm{Ag^+>Cu^{2+}>H^+>Na^+}}

Thus:

  1. Ag⁺ → Ag
  2. Cu²⁺ → Cu
  3. H⁺/water → H₂
  4. Na⁺ generally remains in solution

32. Cl⁻, Br⁻ and I⁻

Oxidation preference:I−>Br−>Cl−\boxed{\mathrm{I^->Br^->Cl^-}}

Thus I₂ forms first, then Br₂, then Cl₂ as conditions change.

33. Cu²⁺, H⁺ and Na⁺

Cu²⁺ is preferentially reduced:Cu2++2e−→Cu\mathrm{Cu^{2+}+2e^-\rightarrow Cu}

After Cu²⁺ becomes sufficiently depleted, H₂ evolution may occur.Cu2+ first\boxed{\mathrm{Cu^{2+}\ first}}

34. Cu²⁺ and Zn²⁺

Cu²⁺ has much higher reduction potential:Cu2++2e−→Cu\mathrm{Cu^{2+}+2e^-\rightarrow Cu}

Therefore:Cu\boxed{\mathrm{Cu}}

is deposited preferentially.

35. When Cu²⁺ concentration becomes very low

The effective reduction potential for Cu²⁺ decreases according to the Nernst equation.

Eventually hydrogen evolution may become competitive:2H2O+2e−→H2+2OH−\mathrm{2H_2O+2e^-\rightarrow H_2+2OH^-}

Thus:H2 may start evolving\boxed{\mathrm{H_2\ may\ start\ evolving}}

H. Electrolysis of water

36. Volume ratio of H₂ : O₂

Overall:2H2O→2H2+O2\mathrm{2H_2O\rightarrow2H_2+O_2}

Therefore:H2:O2=2:1\boxed{H_2:O_2=2:1}

Answer: B

37. Gas at cathode

Reduction occurs at cathode:2H2O+2e−→H2+2OH−\mathrm{2H_2O+2e^-\rightarrow H_2+2OH^-}H2\boxed{H_2}

38. Gas at anode

Oxidation occurs:2H2O→O2+4H++4e−\mathrm{2H_2O\rightarrow O_2+4H^++4e^-}O2\boxed{O_2}

39. Why add electrolyte to water?

Pure water has very low electrical conductivity.

Adding a suitable electrolyte increases the number of ions and therefore increases conductivity.Electrolyte increases conductivity\boxed{\text{Electrolyte increases conductivity}}

I. Quantitative JEE questions

Use:m=MItnF\boxed{m=\frac{MIt}{nF}}

where

  • MM = molar mass
  • II = current
  • tt = time
  • nn = electrons
  • F=96500 C mol−1F=96500\ C\,mol^{-1}

40. 2 A through CuSO₄ for 965 s

For Cu:Cu2++2e−→Cu\mathrm{Cu^{2+}+2e^-\rightarrow Cu}m=63.5×2×9652×96500m=\frac{63.5\times2\times965}{2\times96500}m=0.635 g\boxed{m=0.635\ g}

41. 1.93 A through AgNO₃ for 1000 s

Ag++e−→Ag\mathrm{Ag^++e^-\rightarrow Ag}m=108×1.93×100096500m=\frac{108\times1.93\times1000}{96500}m=2.16 g\boxed{m=2.16\ g}

42. Faradays required for 1 mol Al

Al3++3e−→Al\mathrm{Al^{3+}+3e^-\rightarrow Al}

1 mol Al requires 3 mol electrons.3F\boxed{3F}

43. Faradays required for 1 mol H₂

2H++2e−→H2\mathrm{2H^++2e^-\rightarrow H_2}

2 mol electrons are needed.2F\boxed{2F}

44. Faradays required for 1 mol O₂

Oxidation:2H2O→O2+4H++4e−\mathrm{2H_2O\rightarrow O_2+4H^++4e^-}

Therefore:4F\boxed{4F}

45. 9650 C through acidified water

1F=96500C1F=96500C

Therefore:9650C=0.1F9650C=0.1F

For H₂:2F→1 mol H22F\rightarrow1\ mol\ H_2

So:0.1F→0.05 mol H20.1F\rightarrow0.05\ mol\ H_2

At STP:V=0.05×22.4V=0.05\times22.4V=1.12L\boxed{V=1.12L}

46. Ag and Cu cells in series

Given:mAg=1.08gm_{\mathrm{Ag}}=1.08g

Moles Ag:n=1.08108=0.01n=\frac{1.08}{108}=0.01

Ag requires 1 electron per ion:0.01 mol e−0.01\ mol\ e^-

For Cu²⁺:Cu2++2e−→Cu\mathrm{Cu^{2+}+2e^-\rightarrow Cu}

Therefore Cu deposited:0.005 mol0.005\ mol

Mass:0.005×63.50.005\times63.50.3175g\boxed{0.3175g}

47. Equal electricity through molten NaCl and AlCl₃

Na⁺ needs 1 electron:Na++e−→Na\mathrm{Na^++e^-\rightarrow Na}

Al³⁺ needs 3:Al3++3e−→Al\mathrm{Al^{3+}+3e^-\rightarrow Al}

Equivalent weights:ENa=23E_{\mathrm{Na}}=23EAl=273=9E_{\mathrm{Al}}=\frac{27}{3}=9

Therefore:mNa:mAl=23:9\boxed{m_{\mathrm{Na}}:m_{\mathrm{Al}}=23:9}

48. 5 A for 20 minutes through CuSO₄

t=20×60=1200st=20\times60=1200sm=63.5×5×12002×96500m=\frac{63.5\times5\times1200}{2\times96500}m≈1.975g\boxed{m\approx1.975g}

J. Electrolysis in series

49. 0.108 g Ag deposited

Moles Ag:0.108108=0.001\frac{0.108}{108}=0.001

Thus:0.001 mol e−0.001\ mol\ e^-

For Cu:nCu=0.0012=0.0005n_{\mathrm{Cu}}=\frac{0.001}{2}=0.0005

Mass:0.0005×63.50.0005\times63.50.03175g\boxed{0.03175g}

50. AgNO₃, CuSO₄ and AlCl₃ in series

For equal electricity:m∝Mnm\propto\frac{M}{n}

Ag:1081=108\frac{108}{1}=108

Cu:63.52=31.75\frac{63.5}{2}=31.75

Al:273=9\frac{27}{3}=9

Therefore:Ag:Cu:Al=108:31.75:9\boxed{\mathrm{Ag:Cu:Al=108:31.75:9}}

51. Which gives maximum mass?

For equal electricity:m∝Mnm\propto\frac{M}{n}

Ag has the largest equivalent weight.AgNO3\boxed{\mathrm{AgNO_3}}

52. Equal electricity through AgNO₃, CuSO₄ and Al₂(SO₄)₃

Equivalent weights:Ag=108Ag=108Cu=63.52=31.75Cu=\frac{63.5}{2}=31.75Al=273=9Al=\frac{27}{3}=9

Therefore:Ag>Cu>Al\boxed{Ag>Cu>Al}

K. Faraday + products

53. 0.1 mol Cu deposited from CuSO₄

Reaction:Cu2++2e−→Cu\mathrm{Cu^{2+}+2e^-\rightarrow Cu}

0.1 mol Cu requires:0.2 mol e−0.2\ mol\ e^-

At anode:2H2O→O2+4H++4e−\mathrm{2H_2O\rightarrow O_2+4H^++4e^-}

4 electrons → 1 mol O₂.

Therefore:0.2e−→0.05 mol O20.2e^-\rightarrow0.05\ mol\ O_20.05 mol O2\boxed{0.05\ mol\ O_2}

54. 2 mol H₂ produced

2H2O→2H2+O2\mathrm{2H_2O\rightarrow2H_2+O_2}

Ratio:H2:O2=2:1H_2:O_2=2:1

Therefore:1 mol O2\boxed{1\ mol\ O_2}

55. 1 mol Al produced

Al3++3e−→Al\mathrm{Al^{3+}+3e^-\rightarrow Al}

1 mol Al requires 3 mol electrons.

For O₂:4e−→1molO24e^-\rightarrow1mol O_2

Therefore:n(O2)=34n(O_2)=\frac340.75 mol O2\boxed{0.75\ mol\ O_2}

56. 2 mol Na produced

Na++e−→Na\mathrm{Na^++e^-\rightarrow Na}

2 mol Na requires 2 mol electrons.

At anode:2Cl−→Cl2+2e−\mathrm{2Cl^-\rightarrow Cl_2+2e^-}

Thus:1 mol Cl2\boxed{1\ mol\ Cl_2}

57. 2 mol H₂ produced in NaCl electrolysis

Cathode:2H2O+2e−→H2+2OH−\mathrm{2H_2O+2e^-\rightarrow H_2+2OH^-}

2 mol H₂ requires 4 mol electrons.

Anode:2Cl−→Cl2+2e−\mathrm{2Cl^-\rightarrow Cl_2+2e^-}

4 electrons produce:2 mol Cl2\boxed{2\ mol\ Cl_2}

L. Electrolysis + pH

58. Cathode region in aqueous NaCl

OH⁻ is produced:2H2O+2e−→H2+2OH−\mathrm{2H_2O+2e^-\rightarrow H_2+2OH^-}

Therefore:Basic\boxed{\text{Basic}}

Answer: B

59. Electrolysis of dilute H₂SO₄

Overall:2H2O→2H2+O2\mathrm{2H_2O\rightarrow2H_2+O_2}

H₂SO₄ mainly acts as an electrolyte and is not consumed overall.

Therefore its concentration can remain approximately constant if water loss/volume effects are neglected.pH does not change significantly ideally\boxed{\text{pH does not change significantly ideally}}

60. CuSO₄ + Pt electrodes

Cathode:Cu2++2e−→Cu\mathrm{Cu^{2+}+2e^-\rightarrow Cu}

Anode:2H2O→O2+4H++4e−\mathrm{2H_2O\rightarrow O_2+4H^++4e^-}

H⁺ accumulates.

Therefore:acidity increases\boxed{\text{acidity increases}}

61. AgNO₃ + Pt electrodes

Cathode:Ag++e−→Ag\mathrm{Ag^++e^-\rightarrow Ag}

Anode:2H2O→O2+4H++4e−\mathrm{2H_2O\rightarrow O_2+4H^++4e^-}

NO₃⁻ remains in solution while H⁺ is generated.

Thus HNO₃ effectively accumulates:[HNO3]↑\boxed{[\mathrm{HNO_3}]\uparrow}

M. Inert vs active electrodes

62. CuSO₄ with Cu electrodes

Cathode:Cu2++2e−→Cu\mathrm{Cu^{2+}+2e^-\rightarrow Cu}

Anode:Cu→Cu2++2e−\mathrm{Cu\rightarrow Cu^{2+}+2e^-}

Therefore:Cu deposition at cathode; Cu dissolution at anode\boxed{\text{Cu deposition at cathode; Cu dissolution at anode}}

63. CuSO₄ with Pt electrodes

Cathode:Cu2++2e−→Cu\mathrm{Cu^{2+}+2e^-\rightarrow Cu}

Anode:2H2O→O2+4H++4e−\mathrm{2H_2O\rightarrow O_2+4H^++4e^-}

Therefore:Cu+O2\boxed{\mathrm{Cu+O_2}}

64. Why does anode product change?

Cu is an active electrode and can itself undergo oxidation:Cu→Cu2++2e−\mathrm{Cu\rightarrow Cu^{2+}+2e^-}

Pt is essentially inert under these conditions, so water is oxidized:H2O→O2\mathrm{H_2O\rightarrow O_2}

Hence:Nature of electrode determines the anodic reaction\boxed{\text{Nature of electrode determines the anodic reaction}}

65. Which electrode can participate?

Pt and graphite are generally inert.

Cu can participate:Cu→Cu2++2e−\mathrm{Cu\rightarrow Cu^{2+}+2e^-}

Therefore the intended answer is:Cu / active electrode\boxed{\text{Cu / active electrode}}

Answer: D if “both B and sometimes other active electrodes” is the intended wording.

66. CuCl₂ with Cu electrodes

Cathode:Cu2++2e−→Cu\mathrm{Cu^{2+}+2e^-\rightarrow Cu}

Anode:Cu→Cu2++2e−\mathrm{Cu\rightarrow Cu^{2+}+2e^-}

Therefore:Cu deposited at cathode, Cu dissolves at anode\boxed{\mathrm{Cu\ deposited\ at\ cathode,\ Cu\ dissolves\ at\ anode}}

67. CuCl₂ with Pt electrodes

Cathode:Cu2++2e−→Cu\mathrm{Cu^{2+}+2e^-\rightarrow Cu}

Anode:2Cl−→Cl2+2e−\mathrm{2Cl^-\rightarrow Cl_2+2e^-}

Therefore:Cu+Cl2\boxed{\mathrm{Cu+Cl_2}}

N. JEE Advanced-type questions

68. NaCl + CuSO₄ solution, inert electrodes

Ions present:Na+,Cu2+,Cl−,SO42−Na^+,Cu^{2+},Cl^-,SO_4^{2-}

At cathode:

Cu²⁺ is reduced preferentially:Cu2++2e−→Cu\mathrm{Cu^{2+}+2e^-\rightarrow Cu}

After Cu²⁺ becomes depleted, water can produce H₂.

At anode:

In sufficiently concentrated chloride solution:2Cl−→Cl2+2e−\mathrm{2Cl^-\rightarrow Cl_2+2e^-}

With decreasing chloride concentration, oxygen evolution becomes increasingly important.

So the important sequence is:Cathode: Cu → H₂\boxed{\text{Cathode: Cu → H₂}}Anode: Cl₂ → potentially O₂\boxed{\text{Anode: Cl₂ → potentially O₂}}

69. Cu²⁺ and Cl⁻ solution

At cathode:Cu2+→Cu\mathrm{Cu^{2+}\rightarrow Cu}

At anode, Cl⁻ can be oxidized:2Cl−→Cl2+2e−\mathrm{2Cl^-\rightarrow Cl_2+2e^-}

Water can also be oxidized:2H2O→O2+4H++4e−\mathrm{2H_2O\rightarrow O_2+4H^++4e^-}

Which dominates depends on concentration and electrode conditions.Cl2 or O2\boxed{\mathrm{Cl_2\ or\ O_2}}

can therefore be possible anodic products.

70. Concentrated CuCl₂ + Pt

Cathode:Cu2++2e−→Cu\mathrm{Cu^{2+}+2e^-\rightarrow Cu}

Anode:2Cl−→Cl2+2e−\mathrm{2Cl^-\rightarrow Cl_2+2e^-}

Therefore:Cu at cathode, Cl2 at anode\boxed{\mathrm{Cu\ at\ cathode,\ Cl_2\ at\ anode}}

71. Dilute CuCl₂ + Pt

Cathode:Cu\boxed{\mathrm{Cu}}

At anode, Cl₂ formation competes with O₂ evolution.

For sufficiently dilute chloride:O2 can become the major anodic product\boxed{\mathrm{O_2\ can\ become\ the\ major\ anodic\ product}}

This is a classic concentration-based JEE concept.

72. Aqueous Na₂SO₄ + Pt

Na⁺ and SO₄²⁻ are essentially spectator ions.

Cathode:H2O→H2\mathrm{H_2O\rightarrow H_2}

Anode:H2O→O2\mathrm{H_2O\rightarrow O_2}

Thus Na⁺ and SO₄²⁻ are not consumed in the ideal overall reaction.Their amounts remain essentially unchanged\boxed{\text{Their amounts remain essentially unchanged}}

However, because water is consumed, the concentration of the dissolved Na₂SO₄ can increase as electrolysis proceeds.

73. Why aren’t Na⁺ and OH⁻ both discharged in aqueous NaOH?

At cathode, Na⁺ is not preferentially reduced.

Instead:2H2O+2e−→H2+2OH−\mathrm{2H_2O+2e^-\rightarrow H_2+2OH^-}

At anode:4OH−→O2+2H2O+4e−\mathrm{4OH^-\rightarrow O_2+2H_2O+4e^-}

Therefore:H2 at cathode and O2 at anode\boxed{H_2\text{ at cathode and }O_2\text{ at anode}}

74. Fe³⁺, Cu²⁺ and H⁺

Fe³⁺ can be reduced to Fe²⁺:Fe3++e−→Fe2+\mathrm{Fe^{3+}+e^-\rightarrow Fe^{2+}}

This occurs much more readily than reduction of Cu²⁺ to Cu under ordinary conditions.

Thus:Fe3+→Fe2+\boxed{\mathrm{Fe^{3+}\rightarrow Fe^{2+}}}

Important: Fe³⁺ does not mean iron metal is deposited first. The first reduction step is Fe³⁺ → Fe²⁺.

75. Cl⁻, OH⁻ and SO₄²⁻

At an inert anode, chloride can be oxidized preferentially under suitable/concentrated conditions:2Cl−→Cl2+2e−\mathrm{2Cl^-\rightarrow Cl_2+2e^-}

OH⁻/water can produce O₂.

Therefore, under typical chloride-containing suitable conditions:Cl−→Cl2\boxed{\mathrm{Cl^-\rightarrow Cl_2}}

But concentration and electrode conditions matter.

76. Aqueous CaCl₂ + inert electrodes

At cathode, Ca²⁺ is not normally deposited from aqueous solution.

Water is reduced:2H2O+2e−→H2+2OH−\mathrm{2H_2O+2e^-\rightarrow H_2+2OH^-}

At anode, in sufficiently concentrated chloride solution:2Cl−→Cl2+2e−\mathrm{2Cl^-\rightarrow Cl_2+2e^-}

Therefore:H2+Cl2\boxed{\mathrm{H_2+Cl_2}}

and Ca²⁺ combines with OH⁻ in the vicinity of the cathode:Ca2++2OH−→Ca(OH)2\mathrm{Ca^{2+}+2OH^-\rightarrow Ca(OH)_2}

77. Aqueous MgCl₂ + inert electrodes

Cathode:2H2O+2e−→H2+2OH−\mathrm{2H_2O+2e^-\rightarrow H_2+2OH^-}

Anode, especially concentrated chloride:2Cl−→Cl2+2e−\mathrm{2Cl^-\rightarrow Cl_2+2e^-}

Thus:H2+Cl2\boxed{\mathrm{H_2+Cl_2}}

Mg(OH)₂ can precipitate near the cathode:Mg2++2OH−→Mg(OH)2\mathrm{Mg^{2+}+2OH^-\rightarrow Mg(OH)_2}

78. Why isn’t Mg²⁺ deposited from aqueous MgCl₂?

Mg²⁺ requires very negative reduction conditions:Mg2++2e−→Mg\mathrm{Mg^{2+}+2e^-\rightarrow Mg}

In aqueous solution, water is reduced preferentially:2H2O+2e−→H2+2OH−\mathrm{2H_2O+2e^-\rightarrow H_2+2OH^-}

Hence:H2 forms instead of Mg\boxed{\mathrm{H_2\ forms\ instead\ of\ Mg}}

To obtain Mg metal, molten MgCl₂ is used.

79. Why can Cu²⁺ be deposited but Na⁺ cannot?

Cu²⁺ has a relatively high reduction potential:Cu2++2e−→Cu\mathrm{Cu^{2+}+2e^-\rightarrow Cu}

so it can compete successfully with water.

Na⁺ has a very negative reduction potential:Na++e−→Na\mathrm{Na^++e^-\rightarrow Na}

so water is reduced instead.

Thus:Cu2+→Cu\boxed{\mathrm{Cu^{2+}\rightarrow Cu}}

but:Na+↛Na in ordinary aqueous electrolysis\boxed{\mathrm{Na^+\not\rightarrow Na\ in\ ordinary\ aqueous\ electrolysis}}

80. Na⁺, H⁺, Cl⁻ and OH⁻

At cathode, reduction occurs.

Between Na⁺ and H₂O/H⁺, hydrogen formation is preferred:H+/H2O→H2\boxed{\mathrm{H^+ / H_2O\rightarrow H_2}}

At anode, oxidation occurs.

Cl⁻ is preferentially oxidized under suitable chloride concentration:Cl−→Cl2\boxed{\mathrm{Cl^-\rightarrow Cl_2}}

So the common JEE answer is:Cathode: H₂\boxed{\text{Cathode: H₂}}Anode: Cl₂\boxed{\text{Anode: Cl₂}}

🔥 JEE MASTER TABLE — MEMORISE THIS

ElectrolyteElectrodeMain product
Molten NaClCathodeNa
AnodeCl₂
Aqueous NaCl (concentrated)CathodeH₂
AnodeCl₂
Aqueous NaCl (dilute)CathodeH₂
AnodeO₂ can predominate
Molten CaCl₂CathodeCa
AnodeCl₂
Aqueous CaCl₂CathodeH₂
AnodeCl₂ under suitable conditions
Aqueous MgCl₂CathodeH₂
AnodeCl₂ under suitable conditions
Molten Al₂O₃CathodeAl
AnodeO₂ with inert anode
CuSO₄ + Cu electrodesCathodeCu
AnodeCu dissolves
CuSO₄ + Pt electrodesCathodeCu
AnodeO₂
AgNO₃ + Ag electrodesCathodeAg
AnodeAg dissolves
AgNO₃ + Pt electrodesCathodeAg
AnodeO₂
KI(aq) + inertCathodeH₂
AnodeI₂
KBr(aq) + inertCathodeH₂
AnodeBr₂
Na₂SO₄(aq) + inertCathodeH₂
AnodeO₂
H₂SO₄(aq) + inertCathodeH₂
AnodeO₂
KOH(aq) + inertCathodeH₂
AnodeO₂
CuCl₂ + PtCathodeCu
AnodeCl₂
CuCl₂ + CuCathodeCu
AnodeCu dissolves

⭐ 5 rules that solve most JEE questions

Rule 1 — Molten vs aqueous

Molten salt:metal ion can be discharged\boxed{\text{metal ion can be discharged}}

Aqueous solution:water competes with metal ion\boxed{\text{water competes with metal ion}}

Rule 2 — Active vs inert anode

Active metal anode:M→Mn++ne−\boxed{\mathrm{M\rightarrow M^{n+}+ne^-}}

Inert anode: water/anion is oxidized.

Rule 3 — Cathode

Think:easily reducible cation first\boxed{\text{easily reducible cation first}}

Typical:Ag+>Cu2+>H+>Zn2+>Na+\boxed{\mathrm{Ag^+>Cu^{2+}>H^+>Zn^{2+}>Na^+}}

Rule 4 — Anode

For halides:I−>Br−>Cl−>F−\boxed{\mathrm{I^->Br^->Cl^->F^-}}

So:I2>Br2>Cl2\boxed{I_2>Br_2>Cl_2}

in ease of formation by oxidation.

Rule 5 — Faraday

m=MItnF\boxed{m=\frac{MIt}{nF}}

and for the same quantity of electricity:m∝Mn\boxed{m\propto\frac{M}{n}}

This single relation handles a huge number of JEE electrolysis numericals.

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