CBSE Class 12 Chemistry Electrochemistry Worksheet Set 04

CBSE Class 12 Chemistry Electrochemistry (1). Students can download these worksheets and practice them. This will help them to get better marks in examinations. Also refer to other worksheets for the same chapter and other subjects too. Use them for better understanding of the subjects.

1. The difference between the electrode potentials of two electrodes when no current is drawn through the cell is called ___________.

2. Under what condition an electrochemical cell can behave like an electrolytic cell ?

3. What is the quantity of charge in faraday is required to obtain one mole of aluminum from Al2O3 ?

4. How the cell constant of a conductivity cell changes with change of electrolyte, concentration and temperature?

5. What will happen at anode during the electrolysis of aqueous solution of CuSO4 in the presence of Cu electrodes?

6. Under what condition is ECell = 0 or ΔrG = 0 ? OR Give the condition for Daniell Cell in which there is no flow of electrons or current.

7. Why is alternating current used for measuring resistance of an electrolytic solution?

8. How will the pH of brine (aq. NaCl solution) be affected when it is electrolyzed ?

9. Unlike dry cell, the mercury cell has a constant cell potential throughout its useful life. Why?

10. Mention the purpose of salt-bridge placed between two half-cells of a galvanic cell?

11. Two metals A and B have electrode potential values of – 0.25V and 0.80V respectively. Which of these will liberate hydrogen gas from dilute H2SO4 ?

12. What is the effect of temperature on molar conductivity?

13. What is the role of ZnCl2 in the dry cell ?

14. Why is the equilibrium constant K, related to only E° cell and not Ecell ?

15.Rusting of iron is quicker in saline water than in ordinary water. Why is it so?

16. Why rusting of iron prevented in alkaline medium?

1. The difference between the electrode potentials of two electrodes when no current is drawn through the cell is called ___________.

2. Under what condition an electrochemical cell can behave like an electrolytic cell ?

3. What is the quantity of charge in faraday is required to obtain one mole of aluminum from Al2O3 ?

4. How the cell constant of a conductivity cell changes with change of electrolyte, concentration and temperature?

5. What will happen at anode during the electrolysis of aqueous solution of CuSO4 in the presence of Cu electrodes?

6. Under what condition is ECell = 0 or ΔrG = 0 ?

OR Give the condition for Daniell Cell in which there is no flow of electrons or current.

7. Why is alternating current used for measuring resistance of an electrolytic solution?

8. How will the pH of brine (aq. NaCl solution) be affected when it is electrolyzed ?

9. Unlike dry cell, the mercury cell has a constant cell potential throughout its useful life. Why?

10. Mention the purpose of salt-bridge placed between two half-cells of a galvanic cell?

11. Two metals A and B have electrode potential values of – 0.25V and 0.80V respectively. Which of these will liberate hydrogen gas from dilute H2SO4 ?

12. What is the effect of temperature on molar conductivity?

13. What is the role of ZnCl2 in the dry cell ?

14. Why is the equilibrium constant K, related to only E° cell and not Ecell ?

15.Rusting of iron is quicker in saline water than in ordinary water. Why is it so?

16. Why rusting of iron prevented in alkaline medium?

17. 1 mole KCl dissolved in 500 cc of the solution, Due to more mobility of ions and more degree of dissociation.

18. Recharging is possible in this case because PbSO4 formed during discharging is a sticky solid which sticks to the electrode. Therefore it can either take up or give up electrons during recharge.

19. Bisphenol

20. Coulomb/ mol

ELECTROCHEMISTRY

TWO Marks Each

1.Solutions of two electrolytes ‘A’ and ‘B’ are diluted. The Λm of ‘B’ increases 1.5 times while that of A increases 25 times. Which of the two is a strong electrolyte? Justify your answer.

2. When acidulated water (dil.H2SO4 solution) is electrolysed, will the pH of the solution be affected? Justify your answer.

3. What advantage do the fuel cells have over primary and secondary batteries?

4. How does the density of the electrolyte change when the lead storage battery is discharged ?

5. Why on dilution the Λm of CH3COOH increases drastically, while that of CH3COONa increases gradually?

6. What is the relationship between Gibbs free energy of the cell reaction in a galvanic cell and the emf of the cell? When will the maximum work be obtained from a galvanic cell?

7.Define corrosion. Write chemical formula of rust.

8.Can you store copper sulphate solutions in a zinc pot?

9.Write the cell reaction which occur in the lead storage battery

(a) when the battery is in use (b) when the battery is on charging.

10.Write the product of electrolysis of aqueous copper sulphate by using platinum electrode. Answer

1.Electrolyte ‘B’ is strong as on dilution the number of ions remains the same, only interionic attraction decreases therefore increase in ∧ is small.

2.pH of the solution will not be affected as[ H+] remains constant.

Important Questions for NCERT Class 12 Chemistry Electrochemistry 

Question. Standard free energies of formation (in kJ/mol) at 298 K are –237.2, –394.4 and –8.2 for H₂O(l), CO₂(g) and pentane(g) respectively. The value of E°cell for the pentane-oxygen fuel cell is
(a) 1.0968 V
(b) 0.0968 V
(c) 1.968 V
(d) 2.0968 V 

Question. The equilibrium constant of the reaction :
Cu(s) + 2Ag⁺_(aq) → Cu²⁺(aq) + 2Ag(s);
E° = 0.46 V at 298 K is
(a) 2.0 × 10¹⁰
(b) 4.0 × 10¹⁰
(c) 4.0 × 10¹⁵
(d) 2.4 × 10¹⁰ 

Question. The standard e.m.f. of a galvanic cell involving cell reaction with n = 2 is found to be 0.295 V at 25°C.
The equilibrium constant of the reaction would be
(a) 2.0 × 10¹¹
(b) 4.0 × 10¹²
(c) 1.0 × 10²
(d) 1.0 × 10¹⁰
(Given F = 96500 C mol–1, R = 8.314 J K^–1 mol^–1)

Question. On the basis of the information available from the reaction, 4/3Al + O₂ → 2/3Al₂O₃, ΔG = –827 kJ mol^–1 of O₂, the minimum e.m.f. required to carry out an electrolysis of Al2O3 is (F = 96500 C mol^–1)
(a) 2.14 V
(b) 4.28 V
(c) 6.42 V
(d) 8.56 V 

Question. For the disproportionation of copper
2Cu⁺ → Cu²⁺ + Cu, E° is (Given : E° for Cu²⁺/Cu is 0.34 V and E° for Cu²⁺/Cu+ is 0.15 V)
(a) 0.49 V
(b) –0.19 V
(c) 0.38 V
(d) –0.38 V

Question. E° for the cell, Zn | Zn²⁺_(aq) ||Cu²⁺_(aq) | Cu is 1.10 V at 25°C, the equilibrium constant for the reaction
Zn + Cu²⁺ _(aq) → Cu + Zn2+
(aq) is of the order
(a) 10⁺¹⁸ 
(b) 10⁺¹⁷
(c) 10^–28 
(d) 10⁺³⁷

Question. The molar conductivity of a 0.5 mol/dm³ solution of AgNO₃ with electrolytic conductivity of 5.76 × 10^–3 S cm^–1 at 298 K is
(a) 2.88 S cm²/mol
(b) 11.52 S cm²/mol
(c) 0.086 S cm²/mol
(d) 28.8 S cm²/mol

Question. At 25 °C molar conductance of 0.1 molar aqueous solution of ammonium hydroxide is 9.54 ohm^–1 cm² mol^–1 and at infinite dilution its molar conductance is 238 ohm^–1 cm² mol^–1. The degree of ionisation of ammonium hydroxide at the same concentration and temperature is
(a) 4.008%
(b) 40.800%
(c) 2.080%
(d) 20.800% 

Question. Limiting molar conductivity of NH₄OH [i.e., L°m(NH₄OH)] is equal to
(a) L°m(NH₄Cl) + L°m(NaCl) – L°m(NaOH)
(b) L°m(NaOH) + L°m(NaCl) – L°m(NH₄Cl)
(c) L°m(NH₄OH) + L°m(NH₄Cl) – L°m(HCl)
(d) L°m(NH4Cl) + L°m(NaOH) – L°m(NaCl)

Question. Molar conductivities (L°m) at infinite dilution of NaCl, HCl and CH₃COONa are 126.4, 425.9 and 91.0 S cm² mol^–1 respectively. (L°m) for CH₃COOH will be
(a) 425.5 S cm² mol^–1
(b) 180.5 S cm² mol^–1
(c) 290.8 S cm² mol^–1
(d) 390.5 S cm² mol^–1

Question. An increase in equivalent conductance of a strong electrolyte with dilution is mainly due to
(a) increase in ionic mobility of ions
(b) 100% ionisation of electrolyte at normal dilution
(c) increase in both i.e., number of ions and ionic mobility of ions
(d) increase in number of ions.

Question. The equivalent conductance of M/32 solution of a weak monobasic acid is 8.0 mho cm2 and at infinite dilution is 400 mho cm². The dissociation constant of this acid is
(a) 1.25 × 10^–6
(b) 6.25 × 10^–4
(c) 1.25 × 10^–4
(d) 1.25 × 10^–5

Question. Kohlrausch’s law states that at
(a) infinite dilution, each ion makes definite contribution to conductance of an electrolyte whatever be the nature of the other ion of the electrolyte
(b) infinite dilution, each ion makes definite contribution to equivalent conductance of an electrolyte, whatever be the nature of the other ion of the electrolyte
(c) finite dilution, each ion makes definite contribution to equivalent conductance of an electrolyte, whatever be the nature of the other ion of the electrolyte
(d) infinite dilution each ion makes definite contribution to equivalent conductance of an electrolyte depending on the nature of the other ion of the electrolyte. 

Question. Equivalent conductances of Ba²⁺ and Cl^– ions are 127 and 76 ohm^–1 cm^–1 eq^–1 respectively. Equivalent conductance of BaCl₂ at infinite dilution is
(a) 139.5
(b) 101.5
(c) 203
(d) 279

VERY SHORT ANSWER TYPE QUESTIONS 

Question. Why is it not possible to measure single electrode potential ?
Answer. Because the half cell containing single electrode cannot exist independently, as charge cannot flow on its own in a single electrode.

Question. Name the factor on which emf of a cell depends.
Answer. Emf of a cell depends on following factors :
(a) Nature of reactants
(b) Concentration of solution in two half cells
(c) Temperature

Question. What is the effect of temperature on the electrical conductance of metal ?
Answer. Temperature increases, electrical conductance decreases.

Question. What is the effect of temperature on the electrical conductance of electrolyte ?
Answer. Temperature increases, electrical conductance increases.

Question. What is the electrolyte used in a dry cell ?
Answer. A paste of NH₄Cl.

SHORT ANSWER-I TYPE QUESTIONS

(a) Increase in concentration of Mn+ ions in solution.
(b) By increasing the temperature.

Question. How much electricity in term of Faraday is required to produce 40 gram of Al from Al₂O₃ ? (Atomic mass of Al = 27 g/mol)
Answer. Al³⁺ + 3e– → Al
27 gram of Al require electricity = 3F
40 gram of Al require electricity = 3F/27× 40 = 4.44 F

Question. Predict the product of electrolysis of an aqueous solution of CuCl₂ with an inert electrode.
Answer. CuCl₂ (s) + Aq → Cu²⁺ + 2Cl^–
H₂O → H⁺ + OH–
At cathode (Reduction) : Cu²⁺ will be reduced in preference to H+ ions.
Cu²⁺ + 2e^– → Cu(s)
At anode (Oxidation) : Cl^– ions will be oxidized in preference to OH^– ions.
Cl– → ½Cl₂ + 1e^–
Thus, Cu will be deposited at cathode and Cl₂ will be liberated at anode.

Question. How much electricity is required in Coulomb for the oxidation of 1 mole of FeO to Fe₂O₃ ?
Answer. Fe²⁺ → Fe³⁺ + 1e^–
So, 1F = 1 × 96500 C = 96500 C

Question. The standard reduction potential for the Zn²⁺ (aq)/Zn (s) half cell is – 0.76V.
Write the reactions occurring at the electrodes when coupled with standard hydrogen electrode (SHE).
Answer. At anode : Zn (s) → Zn²⁺ (aq) + 2e^–
At cathode : 2H⁺ + 2e⁻ → H₂ (g)
Zn (s) + 2H⁺ (al) → Zn²⁺ (aq) + H₂ (g)

Question. Two metals A and B have reduction potential values – 0.76 V and + 0.34 V respectively. Which of these will liberate H₂ from dil. H₂SO₄ ?
Answer. Metal having higher oxidation potential will liberate H₂ from H₂SO₄. Thus, A will liberate H₂ from H₂SO₄.

Question. How does conc. of sulphuric acid change in lead storage battery when current is drawn from it ?
Answer. Concentration of sulphuric acid decreases.

Question. Why is alternating current used for measuring resistance of an electrolytic solution ?
Answer. The alternating current is used to prevent electrolysis so that the concentration of ions in the solution remains constant.

Question. E^θ values of MnO₄⁻, Ce₄⁺ and Cl₂ are 1.507, 1.61 and 1.358 V respectively.Arrange these in order of increasing strength as oxidizing agent.
Answer. Cl₂ < MnO₄⁻ < Ce⁴⁺

Question. Give products of electrolysis of an aqueous solution of AgNO3 with silver electrode.
Answer. At anode : Ag (s) → Ag⁺ + e^–
At cathode : Ag⁺ + e⁻ → Ag (s)

SHORT ANSWER-II TYPE QUESTIONS

Question. Depict the galvanic cell in which the reaction
Zn (s) + 2Ag⁺ → Zn²⁺ + 2Ag (s)
takes place. Further show :
(a) Which of the electrode is negatively charged ?
(b) The carriers of the current in the cell.
(c) Individual reaction at each electrode.
Answer. Zn (s)|Zn²⁺ (aq) || Ag⁺ (aq)|Ag (s)
(a) Zn electrode (anode)
(b) Ions are carriers of the current in the cell.
(c) At anode :
Zn (s) → Zn²⁺ + 2e^–
At cathode :
Ag⁺ + e− → Ag (s)

Question. 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. Cell constant = k × R
= 0.146 × 10^-3 × 1500
= 0.219 cm^-1

Question. Calculate the standard cell potentials of galvanic cells in which the following reaction take place :
2Cr (s) + 3Cd²⁺ (aq) → 2Cr³⁺ (aq) + 3Cd (s)
Also calculate ΔGº and equilibrium constant of the reaction.
Answer. E⁰_cell = E⁰_cathode − E⁰_anode
= − 0.40 − (− 0.74) = 0.34 V
ΔGº = − nFE⁰_cell = − 6 × 96500 × 0.34 = − 196860
= − 196860 J mol^-1 = − 196.86 kJ/mol
− ΔGº = 2.303 RT log K_c
196860 = 2.303 × 8.314 × 298 log K_c
Or log K_c = 34.5014
K_c = antilog 34.5014 = 3.192 × 10³⁴

Case Based MCQs

Case IV : Read the passage given below and answer the following questions.

The concentration of potassium ions inside a biological cell is at least twenty times higher than the outside. The resulting potential difference across the cell is important in several processes such as transmission of nerve impulses and maintaining the ion balance. A simple model for such a concentration cell involving a metal M is M_(s) | M⁺(aq.; 0.05 molar) || M⁺(aq; 1 molar) | M(s)

Question. If the 0.05 molar solution of M⁺ is replaced by a 0.0025 molar M⁺ solution, then the magnitude of the cell potential would be
(a) 130 mV
(b) 185 mV
(c) 154 mV
(d) 600 mV
Answer: c

Question. The potential of an electrode change with change in
(a) concentration of ions in solution
(b) position of electrodes
(c) voltage of the cell
(d) all of these.
Answer: a

Question. The value of equilibrium constant for a feasible cell reaction is
(a) < 1
(b) = 1
(c) > 1
(d) zero
Answer: c

Question. For the above cell,
(a) E_cell = 0 ; DG > 0
(b) E_cell > 0 ; DG < 0
(c) E_cell < 0 ; DG° > 0
(d) E_cell > 0 ; DG° = 0
Answer: b

Question. What is the emf of the cell when the cell reaction attains equilibrium?
(a) 1
(b) 0
(c) > 1
(d) < 1
Answer: b

Case  : Read the passage given below and answer the following questions.

All chemical reactions involve interaction of atoms and molecules. A large number of atoms/ molecules are present in a few gram of any chemical compound varying with their atomic/molecular masses. To handle such large number conveniently, the mole concept was introduced.All electrochemical cell reactions are also based on mole concept. For example, a 4.0 molar aqueous solution of NaCl is prepared and 500 mL of this solution is electrolysed. This leads to the evolution of chlorine gas at one of the electrode. The amount of products formed can be calculated by using mole concept.

Question. If cathode is a Hg electrode, then the maximum weight of amalgam formed from this solution is
(Given : Atomic mass of Na = 23u and Hg = 200.59 u)
(a) 300 g
(b) 446 g
(c) 396 g
(d) 296 g
Answer: b

Question. The total number of moles of chlorine gas evolved is
(a) 0.5
(b) 1.0
(c) 1.5
(d) 1.9
Answer: b

Question. In electrolysis of aqueous NaCl solution when Pt electrode is taken, then which gas is liberated at cathode?
(a) H₂ gas
(b) Cl₂ gas
(c) O₂ gas
(d) None of these
Answer: a

One Mark Questions

Question. The difference between the electrode potentials of two electrodes when no current is drawn through the cell is called ___________.
Answer: Cell emf

Question. Under what condition an electrochemical cell can behave like an electrolytic cell ?
Answer: When \( \text{E}_{\text{ext}} > \text{E}_{\text{cell}} \)

Question. What is the quantity of charge in faraday is required to obtain one mole of aluminum from \( \text{Al}_2\text{O}_3 \) ?
Answer: \( 3\text{F} \)

Question. How the cell constant of a conductivity cell changes with change of electrolyte, concentration and temperature?
Answer: Remain unchanged for a cell

Question. What will happen at anode during the electrolysis of aqueous solution of \( \text{CuSO}_4 \) in the presence of Cu electrodes?
Answer: Copper will dissolve at anode

Question. Under what condition is \( \text{E}_{\text{cell}} = 0 \) or \( \Delta_r\text{G} = 0 \) ? OR Give the condition for Daniell Cell in which there is no flow of electrons or current.
Answer: When the cell reaction reaches equilibrium

Question. Why is alternating current used for measuring resistance of an electrolytic solution?
Answer: Alternating current is used to prevent electrolysis so that concentration of ions in the solution remains constant. Otherwise if DC is used the ions will get discharged and electrolysis will occur

Question. How will the pH of brine (aq. NaCl solution) be affected when it is electrolyzed ?
Answer: The pH of the solution will rise as \( \text{NaOH} \) is formed in the electrolytic cell.

Question. Unlike dry cell, the mercury cell has a constant cell potential throughout its useful life. Why?
Answer: Ions are not involved in the overall cell reaction of mercury cells.

Question. Mention the purpose of salt-bridge placed between two half-cells of a galvanic cell?
Answer: Neutralize the two half cell.

Question. Two metals A and B have electrode potential values of -0.25V and 0.80V respectively. Which of these will liberate hydrogen gas from dilute \( \text{H}_2\text{SO}_4 \) ?
Answer: Metal - A

Question. What is the effect of temperature on molar conductivity?
Answer: Molar conductivity of an electrolyte increases with increase in temperature.

Question. What is the role of \( \text{ZnCl}_2 \) in the dry cell ?
Answer: \( \text{ZnCl}_2 \) absorbs ammonia produced in the reaction by forming a complex \( [\text{Zn}(\text{NH}_3)_4]^{2+} \)

Question. Why is the equilibrium constant K, related to only \( \text{E}^\circ_{\text{cell}} \) and not \( \text{E}_{\text{cell}} \) ?
Answer: This is because \( \text{E}_{\text{cell}} \) is zero at equilibrium.

Question. Rusting of iron is quicker in saline water than in ordinary water. Why is it so?
Answer: Due to presence of ions in saline water conductivity is more than the ordinary water. Hence in miniature electrochemical cell flow of electrons will increase, consequently rusting of iron is increased.

Question. Why rusting of iron prevented in alkaline medium?
Answer: In alkaline medium, atmospheric oxygen is unable to take electron which is given by the oxidation of Fe.

Question. Which will have greater molar conductivity and why? 1 mole KCl dissolved in 200 cc of the solution OR 1 mole KCl dissolved in 500 cc of the solution.
Answer: 1 mole KCl dissolved in 500 cc of the solution, Due to more mobility of ions and more degree of dissociation.

Question. Why Lead storage battery as a secondary cell can be recharged?
Answer: Recharging is possible in this case because \( \text{PbSO}_4 \) formed during discharging is a sticky solid which sticks to the electrode. Therefore it can either take up or give up electrons during recharge.

Question. Write the name of a chemical substance which is used to prevent corrosion.
Answer: Bisphenol

Question. Write the unit of Faraday constant.
Answer: Coulomb/ mol

Two Marks Questions

Question. Solutions of two electrolytes ‘A’ and ‘B’ are diluted. The \( \Lambda_m \) of ‘B’ increases 1.5 times while that of A increases 25 times. Which of the two is a strong electrolyte? Justify your answer.
Answer: Electrolyte ‘B’ is strong as on dilution the number of ions remains the same, only interionic attraction decreases therefore increase in \( \Lambda \) is small.

Question. When acidulated water (dil.\( \text{H}_2\text{SO}_4 \) solution) is electrolysed, will the \( \text{p}^{\text{H}} \) of the solution be affected? Justify your answer.
Answer: pH of the solution will not be affected as \( [ \text{H}^+ ] \) remains constant.
At anode: \( 2 \text{H}_2\text{O} \rightarrow \text{O}_2 + 4\text{H}^+ + 4\text{e}^- \)
At cathode: \( 4\text{H}^+ + 4\text{e}^- \rightarrow 2\text{H}_2 \)

Question. What advantage do the fuel cells have over primary and secondary batteries?
Answer: Primary batteries contain a limited amount of reactants and are discharged when the reactants have been consumed. Secondary batteries can be recharged but take a long time to recharge. Fuel cell runs continuously as long as the reactants are supplied to it and products are removed continuously.

Question. How does the density of the electrolyte change when the lead storage battery is discharged ?
Answer: Density of electrolyte decreases as water is formed and sulphuric acid is consumed as the product during discharge of the battery.
\[ \text{Pb} + \text{PbO}_2 + 2\text{H}_2\text{SO}_4 \rightarrow 2\text{PbSO}_4 + 2\text{H}_2\text{O} \]

Question. Why on dilution the \( \Lambda_m \) of \( \text{CH}_3\text{COOH} \) increases drastically, while that of \( \text{CH}_3\text{COONa} \) increases gradually?
Answer: In the case of \( \text{CH}_3\text{COOH} \), which is a weak electrolyte, the number of ions increase on dilution due to an increase in degree of dissociation. In the case of strong electrolyte the number of ions remains the same but the inter ionic attraction decreases.

Question. What is the relationship between Gibbs free energy of the cell reaction in a galvanic cell and the emf of the cell? When will the maximum work be obtained from a galvanic cell?
Answer: \( \Delta_r\text{G} = -n\text{FE}_{\text{(cell)}} \). Maximum work is obtained if the concentration of all the reacting species is unit.

Question. Define corrosion. Write chemical formula of rust.
Answer: Corrosion is a process of formation sulphides, oxides, carbonates, hydroxides, etc. of metal on its surface as a result of its reaction with air and water, surrounding it. Formula of rust- \( \text{Fe}_2\text{O}_3 \cdot x\text{H}_2\text{O} \)

Question. Can you store copper sulphate solutions in a zinc pot?
Answer: No, We cannot store \( \text{CuSO}_4 \) solution in zinc pot, because electrode potential of zinc is less than copper, so \( \text{Cu}^{2+} \) ions get replaced by \( \text{Zn}^{2+} \) ions in solution . Zn is more reactive metals than Cu . (Displacement reaction)

Question. Write the cell reaction which occur in the lead storage battery (a) when the battery is in use (b) when the battery is on charging.
Answer:
(a) When battery is in use
Anode reaction (Oxidation): \( \text{Pb} + \text{SO}_4^{2-} \rightarrow \text{PbSO}_4 + 2\text{e}^- \)
Cathode reaction (Reduction): \( \text{PbO}_2 + \text{SO}_4^{2-} + 4\text{H}^+ + 2\text{e}^- \rightarrow \text{PbSO}_4 + 2\text{H}_2\text{O} \)
Cell Reaction: \( \text{Pb} + \text{PbO}_2 + 2\text{SO}_4^{2-} + 4\text{H}^+ \rightarrow 2\text{PbSO}_4 + 2\text{H}_2\text{O} \)

(b) When the battery is on charging
Anode reaction (Oxidation): \( \text{PbSO}_4 + 2\text{H}_2\text{O} \rightarrow \text{PbO}_2 + \text{SO}_4^{2-} + 4\text{H}^+ + 2\text{e}^- \)
Cathode reaction (Reduction): \( \text{PbSO}_4 + 2\text{e}^- \rightarrow \text{Pb} + \text{SO}_4^{2-} \)
Cell Reaction: \( 2\text{PbSO}_4 + 2\text{H}_2\text{O} \rightarrow \text{Pb} + \text{PbO}_2 + 2\text{SO}_4^{2-} + 4\text{H}^+ \)

Question. Write the product of electrolysis of aqueous copper sulphate by using platinum electrode.
Answer: \( \text{CuSO}_4 \rightarrow \text{Cu}^{2+} + \text{SO}_4^{2-} \) and \( \text{H}_2\text{O} \rightleftharpoons \text{H}^+ + \text{OH}^- \)
At Anode [ \( \text{SO}_4^{2-} \), \( \text{OH}^- \) ]: \( 4\text{OH}^- \rightarrow 2\text{H}_2\text{O} + \text{O}_2 + 4\text{e}^- \)
At Cathode [ \( \text{Cu}^{2+} \), \( \text{H}^+ \) ]: \( \text{Cu}^{2+} + 2\text{e}^- \rightarrow \text{Cu} \)

Three Marks Questions

Question. Calculate the EMF of the cell in which the following reaction take place: \( \text{Ni}(\text{s}) + 2\text{Ag}^+(0.002\text{M}) \rightarrow \text{Ni}^{2+}(0.160\text{M}) + 2\text{Ag}(\text{s}) \) given that \( E^\circ_{\text{cell}} = 1.05\text{V} \).
Answer: Given that \( E^\circ_{\text{cell}} = 1.05\text{ V} \)
According to Nernst equation: \[ E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.059}{n}\log\frac{[\text{Ni}^{2+}]}{[\text{Ag}^+]^2} \] \[ E_{\text{cell}} = 1.05 - \frac{0.059}{2}\log\frac{0.160}{(0.002)^2} \] \[ E_{\text{cell}} = 1.05 - \frac{0.059}{2}\log\frac{0.160}{0.000004} \] \[ E_{\text{cell}} = 0.9143\text{ V} \]

Question. If a current of 0.5 ampere flows through a metallic wire for 2 hours, then how many electrons flow through the wire ?
Answer: Quantity of electricity (Q) = Current (ampere) × time (second) = \( 0.5 \times 2 \times 60 \times 60 = 3600\text{ C} \) (Coulombs).
A flow of 96487 C of electricity \( \approx 6.022 \times 10^{23} \) electrons.
\[ \therefore 3600\text{ C of electricity} = \frac{6.022 \times 10^{23}}{96487} \times 3600 = 2.246 \times 10^{22}\text{ electrons} \]

Question. Calculate the potential of hydrogen electrode in contact with a solution whose \( \text{p}^{\text{H}} \) is 10.
Answer: \( \text{p}^{\text{H}} = 10 \) means \( [\text{H}^+] = 10^{-10}\text{ M} \).
Now for the electrode; \( \text{H}^+ + \text{e}^- \rightarrow \frac{1}{2}\text{H}_2 \) (Here \( n = 1 \)) \[ E_{\text{H}^+/\text{H}_2} = E^\circ_{\text{H}^+/\text{H}_2} - \frac{0.059}{n}\log\frac{1}{[\text{H}^+]} \] \[ E_{\text{H}^+/\text{H}_2} = 0 - 0.059\log\frac{1}{10^{-10}} \] \[ E_{\text{H}^+/\text{H}_2} = -0.059 \log 10^{10} \] \[ E_{\text{H}^+/\text{H}_2} = -0.059 \times 10 = -0.59\text{ V} \]

Question. The molar conductivity of \( 0.025\text{ mol L}^{-1} \) methanoic acid is \( 46.1\text{ S cm}^2\text{ mol}^{-1} \). Calculate its degree of dissociation and dissociation constant. Given \( \lambda^\circ_{\text{H}^+} = 346.6\text{ S cm}^2\text{ mol}^{-1} \) and \( \lambda^\circ_{\text{HCOO}^-} = 54.6\text{ S cm}^2\text{ mol}^{-1} \).
Answer: \( \Lambda_m^c = 46.1\text{ S cm}^2\text{ mol}^{-1} \), \( C = 0.025\text{ mol L}^{-1} \).
\[ \Lambda_m^\circ = \lambda^\circ_{\text{H}^+} + \lambda^\circ_{\text{HCOO}^-} = 349.6 + 54.6 = 404.2\text{ S cm}^2\text{ mol}^{-1} \]
Degree of dissociation: \[ \alpha = \frac{\Lambda_m^c}{\Lambda_m^\circ} = \frac{46.1}{404.2} = 0.114 \]
Dissociation constant: \[ K_a = \frac{C\alpha^2}{1-\alpha} = \frac{0.025 \times (0.114)^2}{1 - 0.114} = 0.0003667 = 3.67 \times 10^{-4} \]

Question. If a current of 0.5 ampere flows through a metallic wire for 2 hours, then how many electrons flow through the wire ?
Answer: Quantity of electricity (Q) = Current (ampere) × time (second) = \( 0.5 \times 2 \times 60 \times 60 = 3600\text{ C} \) (Coulombs).
A flow of 96487 C of electricity \( \approx 6.022 \times 10^{23} \) electrons.
\[ \therefore 3600\text{ C of electricity} = \frac{6.022 \times 10^{23}}{96487} \times 3600 = 2.246 \times 10^{22}\text{ electrons} \]

Question. Calculate \( \Lambda^\circ_m \) for \( \text{CaCl}_2 \) and \( \text{MgSO}_4 \) from the data given in the table of Book.
Answer:
\[ \Lambda^\circ_m(\text{CaCl}_2) = \lambda^\circ_{\text{Ca}^{2+}} + 2\lambda^\circ_{\text{Cl}^-} = 119.0 + 2 \times 76.3 = 119.0 + 152.6 = 271.6\text{ S cm}^2\text{ mol}^{-1} \]
\[ \Lambda^\circ_m(\text{MgSO}_4) = \lambda^\circ_{\text{Mg}^{2+}} + \lambda^\circ_{\text{SO}_4^{2-}} = 106.0 + 160.0 = 266.0\text{ S cm}^2\text{ mol}^{-1} \]

Question. The Conductivity of \( 0.001028\text{ mol L}^{-1} \) acetic acid is \( 4.95 \times 10^{-5}\text{ S cm}^{-1} \). Calculate its dissociation constant if \( \Lambda^\circ_m \) for acetic acid is \( 390.5\text{ S cm}^2\text{ mol}^{-1} \).
Answer:
\[ \Lambda_m = \frac{\kappa \times 1000}{C} = \frac{4.95 \times 10^{-5} \times 1000}{0.001028} = 48.15\text{ S cm}^2\text{ mol}^{-1} \]
\[ \alpha = \frac{\Lambda_m}{\Lambda^\circ_m} = \frac{48.15}{390.0} = 0.1233 \]
Dissociation constant: \[ K_a = \frac{C\alpha^2}{1-\alpha} = \frac{0.001028 \times (0.1233)^2}{1 - 0.1233} = 1.78 \times 10^{-5} \]

Question. A solution of \( \text{CuSO}_4 \) is electrolysed for 10 minutes with a current of 1.5 amperes. What is the mass of copper deposited at the cathode ?
Answer: Quantity of electricity (Q) = Current × time = \( 1.5 \times 10 \times 60 = 900\text{ C} \).
According to the reaction : \( \text{Cu}^{2+} + 2\text{e}^- \rightarrow \text{Cu} \).
We required 2 F or \( 2 \times 96487\text{ C} \) of electricity to deposit 1 mol or 63 g of Cu.
\[ \therefore 900\text{ C electricity will deposit} = \frac{63}{2 \times 96487} \times 900 = 0.2938\text{ g of Cu at the cathode} \]

Question. The conductivity of 0.20 M solution of KCl at 298 K is \( 0.0248\text{ S cm}^{-1} \). Calculate its molar conductivity.
Answer: \( 0.2\text{ M} \Rightarrow 0.2\text{ moles KCl} \) present in 1 litre i.e. \( 1000\text{ cm}^3 \) of solution.
\[ \therefore 1\text{ mole KCl present in } \frac{1000}{0.2}\text{ cm}^3\text{ of solution} \]
\( \kappa = 0.0248\text{ S cm}^{-1} \Rightarrow \) Conductance of \( 1\text{ cm}^3 \) solution = \( 0.0248\text{ S} \).
\[ \therefore \text{Conductance of } \frac{1000}{0.2}\text{ cm}^3\text{ solution} = 0.0248 \times \frac{1000}{0.2} = 124\text{ S cm}^2\text{ mol}^{-1} \]
So Molar conductivity (\( \Lambda_m \)) = \( 124\text{ S cm}^2\text{ mol}^{-1} \).

Question. Write the Nernst equation and find emf of the following cells at 298 K: \( \text{Mg(s)} \mid \text{Mg}^{2+}(0.001\text{ M}) \parallel \text{Cu}^{2+}(0.0001\text{ M}) \mid \text{Cu(s)} \)
Answer:
Oxidation Half: \( \text{Mg} \rightarrow \text{Mg}^{2+} + 2\text{e}^- \)
Reduction Half: \( \text{Cu}^{2+} + 2\text{e}^- \rightarrow \text{Cu} \)
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Cell Reaction: \( \text{Mg} + \text{Cu}^{2+} \rightarrow \text{Mg}^{2+} + \text{Cu} \)
Here number of moles of electrons (n) = 2
\[ E^\circ_{\text{cell}} = E^\circ_{\text{Cu}^{2+}/\text{Cu}} - E^\circ_{\text{Mg}^{2+}/\text{Mg}} = 0.34 - (-2.37) = 2.71\text{ V} \]
The Nernst equation for the cell: \[ E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.059}{2}\log\frac{[\text{Mg}^{2+}]}{[\text{Cu}^{2+}]} \] \[ E_{\text{cell}} = 2.71 - \frac{0.059}{2}\log\frac{0.001}{0.0001} \] \[ E_{\text{cell}} = 2.71 - 0.0295 \log 10 = 2.71 - 0.0295 = 2.6805\text{ V} \]

Five Marks Questions

Question. (a) A Leclanche cell is also called dry cell. Why? (b)Why is the voltage of a mercury cell constant during its working? (c)Name two metals that can be used for cathodic protection of iron? (d)What do you mean by primary and secondary battery?
Answer:
(a) Leclanche cell consists of zinc anode (container) and carbon cathode. The electrolyte is a moist paste of \( \text{MnO}_2 \), \( \text{ZnCl}_2 \), \( \text{NH}_4\text{Cl} \) and carbon black. Because there is no free liquid in the cell, it is called dry cell.
(b) As all the products and reactants are either in solid or liquid state, their concentration does not change with the use of the cell.
(c) Names of the metals are - Zinc and Magnesium.
(d) In the primary batteries, the reaction occurs only once and after the use over a period of time battery becomes dead and cannot be reused again. A secondary battery, after used, can be recharged by passing current through it in the opposite direction so that it can be used again.

Question. (a)What do you understand by strong and weak electrolytes? (b)State Faraday’s Laws of electrolysis? (c)Silver is deposited on a metallic vessel by passing a current of 0.2 amps. for 3 hrs. Calculate the weight of silver deposited. (At mass of silver = 108 amu, F = 96500 C)?
Answer:
(a) An electrolyte that ionizes completely in solution is a strong electrolyte eg. \( \text{NaCl} \), \( \text{CaCl}_2 \) etc and an electrolyte that ionizes partially in solution is weak electrolyte eg \( \text{CH}_3\text{COOH} \box \), \( \text{NH}_4\text{OH} \) etc.
(b) Faraday’s Laws of electrolysis:
First Law: The amount of chemical reaction which occurs at any electrode during electrolysis by a current is proportional to the quantity of electricity passed through the electrolyte.
Second Law: The amount of different substances liberated by the same quantity of electricity passing through the electrolytic solution is proportional to their chemical equivalent weights.
(c) \( 2.417\text{ g} \) of silver.

Question. (a)Define the term resistivity and give its SI unit . (b) What are the factors on which conductivity of an electrolyte depend? (c) The molar conductivity of 0.1M \( \text{CH}_3\text{COOH} \) solution is \( 4.6\text{ cm}^2\text{ mol}^{-1} \). What is the conductivity and resistivity of the solution?
Answer:
(a) The resistivity of a substance is its resistance when it is one meter long and its area of cross Section is one \( \text{m}^2 \). Unit: ohm .meter
(b) The conductivity of an electrolyte depends upon:
i) The nature of electrolyte ii) Size of the ions produced iii) Nature of solvent and its viscosity. iv) Concentration of electrolyte. v) Temperature
(c)
\[ \Lambda_m = \frac{\kappa \times 1000}{C} \Rightarrow \kappa = \frac{\Lambda_m \times C}{1000} = \frac{4.6 \times 0.1}{1000} = 0.00046\text{ S cm}^{-1} \]
\[ \text{Resistivity } (\rho) = \frac{1}{\kappa} = \frac{1}{0.00046} = 2174\text{ }\Omega\text{cm} \]

Question. (a) State the factors that affect the value of electrode potential? (b) Write Nernst equation for a Al-\( \text{ZnSO}_4 \) cell? (c) write the chemistry of rusting of iron
Answer:
(a) Factors affecting electrode potential values are - a) Concentration of electrolyte b) Temperature.
(b) The Nernst equation for a Al-\( \text{ZnSO}_4 \) cell:
The Cell is : \( \text{Al} \mid \text{Al}^{3+} \parallel \text{Zn}^{2+} \mid \text{Zn} \)
Anode reaction : \( \text{Al} \rightarrow \text{Al}^{3+} + 3\text{e}^- \) [ × 2 ]
Cathode reaction : \( \text{Zn}^{2+} + 2\text{e}^- \rightarrow \text{Zn} \) [ × 3 ]
Cell reaction : \( 2\text{Al} + 3\text{Zn}^{2+} \rightarrow 2\text{Al}^{3+} + 3\text{Zn} \)
\[ E = E^\circ - \frac{0.059}{6}\log\frac{[\text{Al}^{3+}]^2}{[\text{Zn}^{2+}]^3} \]
(c) The chemistry of rusting of iron:
[1] Creation of Acidic medium: Atmospheric carbon dioxide and water vapour combine to form carbonic acid. \[ \text{CO}_2 + \text{H}_2\text{O} \rightarrow \text{H}_2\text{CO}_3 \]
[2] Iron will oxidise [ Anode- Oxidation half ]: \[ \text{Fe} \rightarrow \text{Fe}^{2+} + 2\text{e}^- \text{ (x 2)} \]
[3] In another spot, oxygen of air will take the electrons with help of \( \text{H}^+ \) ion and will be reduced to \( \text{H}_2\text{O} \) [ Cathode - Reduction half ]: \[ \text{O}_2 + 4\text{H}^+ + 4\text{e}^- \rightarrow 2\text{H}_2\text{O} \]
Cell Reaction: \[ 2\text{Fe} + \text{O}_2 + 4\text{H}^+ \rightarrow 2\text{Fe}^{2+} + 2\text{H}_2\text{O} \]
[4] Atmospheric oxygen further oxidises ferrous ion into ferric oxide: \[ 2\text{Fe}^{2+} + 2\text{H}_2\text{O} + \frac{1}{2}\text{O}_2 \rightarrow \text{Fe}_2\text{O}_3 + 4\text{H}^+ \]
[5] Ferric oxide will hydrolyse with water to form rust: \[ \text{Fe}_2\text{O}_3 + x\text{H}_2\text{O} \rightarrow \text{Fe}_2\text{O}_3 \cdot x\text{H}_2\text{O} \text{ [ Rust ]} \]

Question. (a) Can an electrochemical cell act as electrolytic cell? How? (b) Explain construction and working of standard Hydrogen electrode? (c) What is an electrochemical series? How does it predict the feasibility of a certain redox reaction?
Answer:
(a) Yes, An electrochemical cell can be converted into electrolytic cell by applying an external opposite potential greater than its own electrical potential.
(b) Standard Hydrogen electrode:
The Standard Hydrogen Electrode consists of a platinum electrode coated with platinum black. The electrode is dipped in an acidic solution and pure hydrogen gas is bubbled through it. The concentration of both the reduced and oxidised forms of hydrogen is maintained at unity. This implies that the pressure of hydrogen gas is one bar and the concentration of hydrogen ion in the solution is one molar.
Anode: \( \text{H}_2 \rightarrow 2\text{H}^+ + 2\text{e}^- \)
Cathode: \( 2\text{H}^+ + 2\text{e}^- \rightarrow \text{H}_2 \)
\[ E^\circ(2\text{H}^+/\text{H}_2) = 0 \]
If it acts as cathode: The maximum bubbling of hydrogen gas from the solution will evolve.
If it acts as anode: The minimum bubbling of hydrogen gas from the solution will evolve.

(c) The arrangement of metals and ions in increasing order of their electrode potential values is known as electrochemical series.
The reduction half reaction for which the reduction potential is lower than the other will act as anode and one with greater value will act as cathode. Reverse reaction will not occur.