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      Class 12 CHEMISTRY – JEE

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      • Class 12 CHEMISTRY – JEE
      CoursesClass 12ChemistryClass 12 CHEMISTRY – JEE
      • 1. Solid State
        11
        • Lecture1.1
          Crystalline & Amorphous Solid 50 min
        • Lecture1.2
          Law of Crystallography 01 hour
        • Lecture1.3
          Bravius lattice & Important Terms of solid state 48 min
        • Lecture1.4
          Type of Cubic crystal & Closest packed St. 01 hour
        • Lecture1.5
          Tetrahedral & Octahedral Void 38 min
        • Lecture1.6
          Type of Voids & Radius Ratio 44 min
        • Lecture1.7
          Type of ionic solid 59 min
        • Lecture1.8
          Defect in Solid 48 min
        • Lecture1.9
          Metallic Bonding 52 min
        • Lecture1.10
          Chapter Notes – Solid State
        • Lecture1.11
          NCERT Solutions – Solid State
      • 2. Solution and its C.P
        9
        • Lecture2.1
          Condition of solution formation, TD of Solution, Factors affecting solubility-Henary’s Law 55 min
        • Lecture2.2
          Colligative Properties, Raoult’s Law 49 min
        • Lecture2.3
          Relative lowering of V.P. & Problems 45 min
        • Lecture2.4
          Non ideal solution, Azeotropic Solution 46 min
        • Lecture2.5
          Elevation in B.P., Depression in F.P. 47 min
        • Lecture2.6
          Osmotic Pressure, Abnormal C.P. & Van’t Hoff Factor 59 min
        • Lecture2.7
          Solution – Ostwald Walker Exp. 13 min
        • Lecture2.8
          Chapter Notes – Solution and its C.P
        • Lecture2.9
          NCERT Solutions – Solution and its C.P
      • 3. Chemical Kinetics
        10
        • Lecture3.1
          Rate of reaction 37 min
        • Lecture3.2
          Differential Rate Law 38 min
        • Lecture3.3
          Integrated Rate Law 56 min
        • Lecture3.4
          Integrated Rate problems 53 min
        • Lecture3.5
          Pseudo order Reaction 40 min
        • Lecture3.6
          Reaction Mechanism 47 min
        • Lecture3.7
          Collision Model 34 min
        • Lecture3.8
          Arhenius Equation 34 min
        • Lecture3.9
          Chapter Notes – Chemical Kinetics
        • Lecture3.10
          NCERT Solutions – Chemical Kinetics
      • 4. Electrochemistry
        13
        • Lecture4.1
          Introduction & Galvanic cell 32 min
        • Lecture4.2
          Cell Notation & Cell Reaction 35 min
        • Lecture4.3
          Electrode & Cell Potential 38 min
        • Lecture4.4
          Electrochemical series 39 min
        • Lecture4.5
          The Nernst Equation 39 min
        • Lecture4.6
          Concentration cell, Battery, Corrosion 52 min
        • Lecture4.7
          Electrolysis 20 min
        • Lecture4.8
          Faraday Law 45 min
        • Lecture4.9
          Resistance & Conductance 40 min
        • Lecture4.10
          Molar & Eq. Conductance, Kohlraush’s Law 29 min
        • Lecture4.11
          Problems on Resistance & Conductance 23 min
        • Lecture4.12
          Chapter Notes – Electrochemistry
        • Lecture4.13
          NCERT Solutions – Electrochemistry
      • 5. Surface Chemistry
        11
        • Lecture5.1
          Introduction & Surface tension & surface energy 33 min
        • Lecture5.2
          Adsorption 47 min
        • Lecture5.3
          Factors affecting Adsorption 39 min
        • Lecture5.4
          Catalysis 34 min
        • Lecture5.5
          Type of Catalysis & Enzyme Catalysis 41 min
        • Lecture5.6
          Colloidal Solution 57 min
        • Lecture5.7
          Type of Colloidal Solution 43 min
        • Lecture5.8
          Properties of Colloidal Solution 50 min
        • Lecture5.9
          Protective Colloids 58 min
        • Lecture5.10
          Chapter Notes – Surface Chemistry
        • Lecture5.11
          NCERT Solutions – Surface Chemistry
      • 6. Alcohol & Ether
        8
        • Lecture6.1
          Preparation 35 min
        • Lecture6.2
          Physical Properties & Oxidation Of Alcohol 29 min
        • Lecture6.3
          Hydrates, Acetal, Ketal 38 min
        • Lecture6.4
          Tests Of Alcohol 47 min
        • Lecture6.5
          Ether Preparation & Its Properties 33 min
        • Lecture6.6
          Thiol & Thioether 16 min
        • Lecture6.7
          Chapter Notes – Alcohol & Ether
        • Lecture6.8
          NCERT Solutions – Alcohol & Ether
      • 7. Aldehyde & Ketone
        10
        • Lecture7.1
          Preparation 33 min
        • Lecture7.2
          Physical Properties, Beckmann Rearrangement, Witting Reaction 46 min
        • Lecture7.3
          Schmidt Reaction, Bayer Villegar Oxidation 22 min
        • Lecture7.4
          Aldol Condensation Reaction 40 min
        • Lecture7.5
          Cannizzaro Reaction 32 min
        • Lecture7.6
          Acyloin, Benzoin, Clasien, Perkin Condensation 28 min
        • Lecture7.7
          Reformasky Reaction, Tischenko Reaction 20 min
        • Lecture7.8
          Tests-8 40 min
        • Lecture7.9
          Chapter Notes – Aldehyde & Ketone
        • Lecture7.10
          NCERT Solutions – Aldehyde & Ketone
      • 8. Acid & derivatives
        4
        • Lecture8.1
          Preparation 31 min
        • Lecture8.2
          Chemical Reactions Of Acids 31 min
        • Lecture8.3
          Arndt Eistert, Curtius, Hvz, Hoffmann Reaction 19 min
        • Lecture8.4
          Acid Derivatives 38 min
      • 9. Nitrogen containing compounds
        4
        • Lecture9.1
          Alkyl Nitrites, Nitro Alkane 27 min
        • Lecture9.2
          Alkane Nitrile & Isonitrile 20 min
        • Lecture9.3
          Amine Preparation 24 min
        • Lecture9.4
          Properties Of Amines 13 min
      • 10. Aromatic Compounds
        7
        • Lecture10.1
          Benzene 41 min
        • Lecture10.2
          Aromatic Hydrocarbon 29 min
        • Lecture10.3
          Aryl Halides 18 min
        • Lecture10.4
          Phenol 40 min
        • Lecture10.5
          Aromatic Aldehyde 39 min
        • Lecture10.6
          Aniline 32 min
        • Lecture10.7
          Phenyl Diazonium Salts 37 min
      • 11. Biomolecules
        14
        • Lecture11.1
          Introduction & Types Of Carbohydrates 47 min
        • Lecture11.2
          D-glucose & D-fructose 50 min
        • Lecture11.3
          Reactions Of D-glucose & D-fructose 32 min
        • Lecture11.4
          Reactions Of D-glucose & D-fructose 23 min
        • Lecture11.5
          Sucrose, Maltose, Lactose 31 min
        • Lecture11.6
          Starch, Cellulose, Glycogen 27 min
        • Lecture11.7
          Reducing Sugar, Mutarotation, Osazone Formation 40 min
        • Lecture11.8
          Problems On Carbohydrates 41 min
        • Lecture11.9
          Amino Acids 48 min
        • Lecture11.10
          Peptides 47 min
        • Lecture11.11
          Proteins 18 min
        • Lecture11.12
          Enzyme & Vitamins 30 min
        • Lecture11.13
          Nucleic Acid 36 min
        • Lecture11.14
          Chapter Notes – Biomolecules
      • 12. Polymer Chemistry
        6
        • Lecture12.1
          Polymerisation Addition Reaction 32 min
        • Lecture12.2
          Coordination Addition, Condensation Reaction 24 min
        • Lecture12.3
          Division Of Polymer 41 min
        • Lecture12.4
          Examples Of Polymer 31 min
        • Lecture12.5
          Examples Of Polymer 31 min
        • Lecture12.6
          Chapter Notes – Polymer Chemistry
      • 13. Practical Organic Chemistry
        4
        • Lecture13.1
          Poc Qualitative Analysis 23 min
        • Lecture13.2
          Poc Qualitative Analysis 20 min
        • Lecture13.3
          Poc Quantitative Analysis 29 min
        • Lecture13.4
          Poc Quantitative Analysis 20 min
      • 14. P block elements II
        13
        • Lecture14.1
          VA – Elemental Properties of N family 51 min
        • Lecture14.2
          VA – Compounds of N family 43 min
        • Lecture14.3
          VA – N & Its compounds 45 min
        • Lecture14.4
          VA – Oxides & Oxyacids of Nitrogen 55 min
        • Lecture14.5
          VA – P & its compounds 31 min
        • Lecture14.6
          VA – Oxides & Oxyacids of P 31 min
        • Lecture14.7
          VIA 1 – Elemental Properties of O-Family 36 min
        • Lecture14.8
          VIA 2 – compounds of VIA elements 41 min
        • Lecture14.9
          VIA 3 – Oxygen & Ozone 47 min
        • Lecture14.10
          VIA 4 – Sulphur & oxides of Sulphur 37 min
        • Lecture14.11
          VIA 5 – Sulphuric Acid 25 min
        • Lecture14.12
          Chapter Notes – P block elements
        • Lecture14.13
          NCERT Solutions – P block elements
      • 15. P block elements III
        5
        • Lecture15.1
          VIIA 1 – elemental properties of Halogen 40 min
        • Lecture15.2
          VIIA 2 – Compounds of Halogen 49 min
        • Lecture15.3
          VIIA 3 – Chlorine & its Compounds 41 min
        • Lecture15.4
          VIIIA 1 – Properties of Noble Gas 34 min
        • Lecture15.5
          VIIIA 2 – Compounds of Noble Gas 34 min
      • 16. D block metals
        8
        • Lecture16.1
          D block – Elemental Properties 55 min
        • Lecture16.2
          Elemental Properties 01 hour
        • Lecture16.3
          Elemental Properties 53 min
        • Lecture16.4
          KMnO4 & K2Cr2O7 47 min
        • Lecture16.5
          Problems 40 min
        • Lecture16.6
          Problems 20 min
        • Lecture16.7
          Chapter Notes – The d-and f-Block Elements
        • Lecture16.8
          NCERT Solutions – The d-and f-Block Elements
      • 17. F block metals
        3
        • Lecture17.1
          Lanthanoids 52 min
        • Lecture17.2
          Actinoids 48 min
        • Lecture17.3
          Problems 42 min
      • 18. Co-ordination compounds
        17
        • Lecture18.1
          Introduction of Complex Compound, Ligands 42 min
        • Lecture18.2
          Classification of Ligands, Denticity 35 min
        • Lecture18.3
          Nomenclature of Complex Compounds 46 min
        • Lecture18.4
          Nomenclature of Complex Compounds 2 40 min
        • Lecture18.5
          Bonding in Complex Compound, Primary & Secondary Valency 44 min
        • Lecture18.6
          Concept of EAN 29 min
        • Lecture18.7
          VBT in Complex Compounds 58 min
        • Lecture18.8
          Examples on VBT in complex compounds 31 min
        • Lecture18.9
          CFT in Complex Compounds 43 min
        • Lecture18.10
          CFT for Octahedral & Tetrahedral Complex 35 min
        • Lecture18.11
          Colour & Stability of Complex Compounds 28 min
        • Lecture18.12
          Structural Isomerism in Complex Compounds 49 min
        • Lecture18.13
          Geometrical Isomerism in Complex Compounds 43 min
        • Lecture18.14
          Optical Isomerism in Complex Compounds, use of Complex 01 hour
        • Lecture18.15
          Organometallic Compounds 29 min
        • Lecture18.16
          Chapter Notes – Co-ordination compounds
        • Lecture18.17
          NCERT Solutions – Co-ordination compounds
      • 19. Environmental Chemistry
        4
        • Lecture19.1
          Introduction & Air Pollution 35 min
        • Lecture19.2
          Air Pollution 20 min
        • Lecture19.3
          Water Pollution 23 min
        • Lecture19.4
          Soil Pollution, Prevention of Pollution 16 min

        NCERT Solutions – Electrochemistry

        3.1. 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

        The following is the order in which the given metals displace each other from the solution of their salts.

        Mg, Al, Zn, Fe, Cu

        3.2. Given the standard electrode potentials,

        K+/K = −2.93V, Ag+/Ag = 0.80V,

        Hg2+/Hg = 0.79V

        Mg2+/Mg = −2.37 V, Cr3+/Cr = − 0.74V

        Arrange these metals in their increasing order of reducing power.

        Answer

        The lower the reduction potential, the higher is the reducing power. The given standard electrode potentials increase in the order of K+/K < Mg2+/Mg < Cr3+/Cr < Hg2+/Hg < Ag+/Ag.

        Hence, the reducing power of the given metals increases in the following order:

        Ag < Hg < Cr < Mg < K

        3.3. Depict the galvanic cell in which the reaction Zn(s) + 2Ag+(aq) → Zn2+(aq) + 2Ag(s) takes place. Further show:

        (i) Which of the electrode is negatively charged?

        (ii) The carriers of the current in the cell.

        (iii) Individual reaction at each electrode.

        Answer

        The galvanic cell in which the given reaction takes place is depicted as:

        (i) Zn electrode (anode) is negatively charged.

        (ii) Ions are carriers of current in the cell and in the external circuit, current will flow from silver to zinc.

        (iii) The reaction taking place at the anode is given by,

        The reaction taking place at the cathode is given by,

        3.4. Calculate the standard cell potentials of galvanic cells in which the following reactions take place:

        (i) 2Cr(s) + 3Cd2+(aq) → 2Cr3+(aq) + 3Cd

        (ii) Fe2+(aq) + Ag+(aq) → Fe3+(aq) + Ag(s)

        Calculate the ΔrGθ and equilibrium constant of the reactions.

        Answer

        (i) 

        The galvanic cell of the given reaction is depicted as:

        Now, the standard cell potential is

        In the given equation,

        n = 6

        F = 96487 C mol−1

         = +0.34 V

        Then,  = −6 × 96487 C mol−1 × 0.34 V

        = −196833.48 CV mol−1

        = −196833.48 J mol−1

        = −196.83 kJ mol−1

        Again,

         = −RT ln K

        = 34.496

         K = antilog (34.496)

        = 3.13 × 1034

        (ii) 

        The galvanic cell of the given reaction is depicted as:

        Now, the standard cell potential is

        Here, n = 1.

        Then, 

        = −1 × 96487 C mol−1 × 0.03 V

        = −2894.61 J mol−1

        = −2.89 kJ mol−1

        Again, 

        = 0.5073

        K = antilog (0.5073)

        = 3.2 (approximately)

         

        3.5. Write the Nernst equation and emf of the following cells at 298 K:

        (i) Mg(s) | Mg2+(0.001M) || Cu2+(0.0001 M) | Cu(s)

        (ii) Fe(s) | Fe2+(0.001M) || H+(1M)|H2(g)(1bar) | Pt(s)

        (iii) Sn(s) | Sn2+(0.050 M) || H+(0.020 M) | H2(g) (1 bar) | Pt(s)

        (iv) Pt(s) | Br2(l) | Br−(0.010 M) || H+(0.030 M) | H2(g) (1 bar) | Pt(s).

        Answer

        (i) For the given reaction, the Nernst equation can be given as:

        = 2.7 − 0.02955

        = 2.67 V (approximately)

        (ii) For the given reaction, the Nernst equation can be given as:

        = 0.52865 V

        = 0.53 V (approximately)

        (iii) For the given reaction, the Nernst equation can be given as:

        = 0.14 − 0.0295 × log125

        = 0.14 − 0.062

        = 0.078 V

        = 0.08 V (approximately)

        (iv) For the given reaction, the Nernst equation can be given as:

         

        3.6. In the button cells widely used in watches and other devices the following reaction takes place:

        Zn(s) + Ag2O(s) + H2O(l) → Zn2+(aq) + 2Ag(s) + 2OH−(aq)

        Determine and  for the reaction.

        Answer

        = 1.104 V

        We know that,

        = −2 × 96487 × 1.04

        = −213043.296 J

        = −213.04 kJ

        3.7. Define conductivity and molar conductivity for the solution of an electrolyte. Discuss their variation with concentration.

        Answer

        Conductivity of a solution is defined as the conductance of a solution of 1 cm in length and area of cross-section 1 sq. cm. The inverse of resistivity is called conductivity or specific conductance. It is represented by the symbolκ. If ρ is resistivity, then we can write:

        The conductivity of a solution at any given concentration is the conductance (G) of one unit volume of solution kept between two platinum electrodes with the unit area of cross-section and at a distance of unit length.

        i.e., 

        (Since a = 1, l = 1)

        Conductivity always decreases with a decrease in concentration, both for weak and strong electrolytes. This is because the number of ions per unit volume that carry the current in a solution decreases with a decrease in concentration.

        Molar conductivity:

        Molar conductivity of a solution at a given concentration is the conductance of volume V of a solution containing 1 mole of the electrolyte kept between two electrodes with the area of cross-section A and distance of unit length.

        Now, l = 1 and A = V (volume containing 1 mole of the electrolyte).

        Molar conductivity increases with a decrease in concentration. This is because the total volume V of the solution containing one mole of the electrolyte increases on dilution.

        The variation of withfor strong and weak electrolytes is shown in the following plot:

        3.8.The conductivity of 0.20 M solution of KCl at 298 K is 0.0248 Scm−1. Calculate its molar conductivity.

        Answer

        Given,

        κ = 0.0248 S cm−1

        c = 0.20 M

         Molar conductivity, 

        = 124 Scm2mol−1

        3.9. The resistance of a conductivity cell containing 0.001M KCl solution at 298 K is 1500 Ω. What is the cell constant if conductivity of 0.001M KCl solution at 298 K is 0.146 × 10−3 S cm−1.

        Answer

        Given,

        Conductivity, κ = 0.146 × 10−3 S cm−1

        Resistance, R = 1500 Ω

         Cell constant = κ × R

        = 0.146 × 10−3 × 1500

        = 0.219 cm−1

        3.10. The conductivity of sodium chloride at 298 K has been determined at different concentrations and the results are given below:

        Concentration/M 0.001 0.010 0.020 0.050 0.100

        102 × κ/S m−1 1.237 11.85 23.15 55.53 106.74

        Calculate for all concentrations and draw a plot between and c½. Find the value of.

        Answer

        Given,

        κ = 1.237 × 10−2 S m−1, c = 0.001 M

        Then, κ = 1.237 × 10−4 S cm−1, c½ = 0.0316 M1/2

        = 123.7 S cm2 mol−1

        Given,

        κ = 11.85 × 10−2 S m−1, c = 0.010M

        Then, κ = 11.85 × 10−4 S cm−1, c½ = 0.1 M1/2

        = 118.5 S cm2 mol−1

        Given,

        κ = 23.15 × 10−2 S m−1, c = 0.020 M

        Then, κ = 23.15 × 10−4 S cm−1, c1/2 = 0.1414 M1/2

        = 115.8 S cm2 mol−1

        Given,

        κ = 55.53 × 10−2 S m−1, c = 0.050 M

        Then, κ = 55.53 × 10−4 S cm−1, c1/2 = 0.2236 M1/2

        = 111.1 1 S cm2 mol−1

        Given,

        κ = 106.74 × 10−2 S m−1, c = 0.100 M

        Then, κ = 106.74 × 10−4 S cm−1, c1/2 = 0.3162 M1/2

        = 106.74 S cm2 mol−1

        Now, we have the following data:

        0.0316

        0.1

        0.1414

        0.2236

        0.3162

        123.7

        118.5

        115.8

        111.1

        106.74

        Since the line interruptsat 124.0 S cm2 mol−1, = 124.0 S cm2 mol−1.

        3.11. Conductivity of 0.00241 M acetic acid is 7.896 × 10−5 S cm−1. Calculate its molar conductivity and if for acetic acid is 390.5 S cm2 mol−1, what is its dissociation constant?

        Answer

        Given, κ = 7.896 × 10−5 S m−1

        c = 0.00241 mol L−1

        Then, molar conductivity, 

        = 32.76S cm2 mol−1

        Again,  = 390.5 S cm2 mol−1

        Now, 

        = 0.084

        Dissociation constant, 

        = 1.86 × 10−5 mol L−1

        3.12. How much charge is required for the following reductions:

        (i) 1 mol of Al3+ to Al.

        (ii) 1 mol of Cu2+ to Cu.

        (iii) 1 mol of  to Mn2+.

        Answer

        (i) 

         Required charge = 3 F

        = 3 × 96487 C

        = 289461 C

        (ii) 

        Required charge = 2 F

        = 2 × 96487 C

        = 192974 C

        (iii) 

        i.e.,

         Required charge = 5 F

        = 5 × 96487 C

        = 482435 C

        3.13. How much electricity in terms of Faraday is required to produce

        (i) 20.0 g of Ca from molten CaCl2.

        (ii) 40.0 g of Al from molten Al2O3.

        Answer

        (i) According to the question,

        Electricity required to produce 40 g of calcium = 2 F

        Therefore, electricity required to produce 20 g of calcium

        = 1 F

        (ii) According to the question,

        Electricity required to produce 27 g of Al = 3 F

        Therefore, electricity required to produce 40 g of Al 

        = 4.44 F

        3.14. How much electricity is required in coulomb for the oxidation of

        (i) 1 mol of H2O to O2.

        (ii) 1 mol of FeO to Fe2O3.

        Answer

        (i) According to the question,

        Now, we can write:

        Electricity required for the oxidation of 1 mol of H2O to O2 = 2 F

        = 2 × 96487 C

        = 192974 C

        (ii) According to the question,

        Electricity required for the oxidation of 1 mol of FeO to Fe2O3 = 1 F

        = 96487 C

        3.15. A solution of Ni(NO3)2 is electrolysed between platinum electrodes using a current of 5 amperes for 20 minutes. What mass of Ni is deposited at the cathode?

        Answer

        Given,

        Current = 5A

        Time = 20 × 60 = 1200 s

        Charge = current × time

        = 5 × 1200

        = 6000 C

        According to the reaction,

        Nickel deposited by 2 × 96487 C = 58.71 g

        Therefore, nickel deposited by 6000 C 

        = 1.825 g

        Hence, 1.825 g of nickel will be deposited at the cathode.

        3.16. Three electrolytic cells A,B,C containing solutions of ZnSO4, AgNO3 and CuSO4, respectively are connected in series. A steady current of 1.5 amperes was passed through them until 1.45 g of silver deposited at the cathode of cell B. How long did the current flow? What mass of copper and zinc were deposited?

        Answer

        According to the reaction:

        i.e., 108 g of Ag is deposited by 96487 C.

        Therefore, 1.45 g of Ag is deposited by =

        = 1295.43 C

        Given,

        Current = 1.5 A

        Time 

        = 863.6 s

        = 864 s

        = 14.40 min

        Again,

        i.e., 2 × 96487 C of charge deposit = 63.5 g of Cu

        Therefore, 1295.43 C of charge will deposit 

        = 0.426 g of Cu

        i.e., 2 × 96487 C of charge deposit = 65.4 g of Zn

        Therefore, 1295.43 C of charge will deposit 

        = 0.439 g of Zn

        3.17. Using the standard electrode potentials given in Table 3.1, predict if the reaction between the following is feasible:

        (i) Fe3+(aq) and I−(aq)

        (ii) Ag+ (aq) and Cu(s)

        (iii) Fe3+ (aq) and Br− (aq)

        (iv) Ag(s) and Fe3+ (aq)

        (v) Br2 (aq) and Fe2+ (aq).

        Answer

        Sincefor the overall reaction is positive, the reaction between Fe3+(aq) and I−(aq) is feasible.

        Since  for the overall reaction is positive, the reaction between Ag+ (aq) and Cu(s) is feasible.

        Since for the overall reaction is negative, the reaction between Fe3+(aq) and Br−(aq) is not feasible.

        Since E for the overall reaction is negative, the reaction between Ag (s) and Fe3+(aq) is not feasible.

        Since for the overall reaction is positive, the reaction between Br2(aq) and Fe2+(aq) is feasible.

         

        3.18. Predict the products of electrolysis in each of the following:

        (i) An aqueous solution of AgNO3 with silver electrodes.

        (ii) An aqueous solution of AgNO3with platinum electrodes.

        (iii) A dilute solution of H2SO4with platinum electrodes.

        (iv) An aqueous solution of CuCl2 with platinum electrodes.

        Answer

        (i) At cathode:

        The following reduction reactions compete to take place at the cathode.

        The reaction with a higher value of  takes place at the cathode. Therefore, deposition of silver will take place at the cathode.

        At anode:

        The Ag anode is attacked by ions. Therefore, the silver electrode at the anode dissolves in the solution to form Ag+.

        (ii) At cathode:

        The following reduction reactions compete to take place at the cathode.

        The reaction with a higher value of  takes place at the cathode. Therefore, deposition of silver will take place at the cathode.

        At anode:

        Since Pt electrodes are inert, the anode is not attacked by  ions. Therefore, OH− or  ions can be oxidized at the anode. But OH− ions having a lower discharge potential and get preference and decompose to liberate O2.

        (iii) At the cathode, the following reduction reaction occurs to produce H2 gas.

        At the anode, the following processes are possible.

        For dilute sulphuric acid, reaction (i) is preferred to produce O2 gas. But for concentrated sulphuric acid, reaction (ii) occurs.

        (iv) At cathode:

        The following reduction reactions compete to take place at the cathode.

        The reaction with a higher value of  takes place at the cathode. Therefore, deposition of copper will take place at the cathode.

        At anode:
        The following oxidation reactions are possible at the anode.

        At the anode, the reaction with a lower value of  is preferred. But due to the over-potential of oxygen, Cl− gets oxidized at the anode to produce Cl2 gas

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