NCERT Exemplar: Electrochemistry
Multiple Choice Questions - I
Q.1. Which cell will measure standard electrode potential of copper electrode?
(i) Pt (s) | H2 (g,0.1 bar) | H+ (aq.,1 M) ║ Cu2+(aq.,1M) | Cu
(ii) Pt (s) | H2 (g, 1 bar) | H+ (aq.,1 M) ║ Cu2+ (aq.,2 M) | Cu
(iii) Pt (s) | H2 (g, 1 bar) | H+ (aq.,1 M) ║ Cu2+ (aq.,1 M) | Cu
(iv) Pt(s) | H2 (g, 1 bar) | H+ (aq.,0.1 M) ║ Cu2+ (aq.,1 M) | Cu
Ans. (iii)
Sol.
When measuring a standard electrode potential, each half-cell must be under standard conditions (1 bar for gases; 1 M for solutes). Constructing the cell with both hydrogen electrode and copper electrode at standard conditions gives the standard electrode potential directly. Therefore option (iii) is correct.
Q.2. Electrode potential for Mg electrode varies according to the equation
(i)
(ii)
(iii)
(iv)
Ans. (ii)
Sol.
Compare the Nernst form E = E° - (0.0591/n) log[Red/Ox] with y = mx + c. E vs log[Mg2+] is a straight line. The slope sign depends on the sign before the log term; for Mg2+/Mg the slope is positive in the plotted convention, giving option (ii).
Q.3. Which of the following statement is correct?
(i) ECell and ∆rG of cell reaction both are extensive properties.
(ii) ECell and ∆rG of cell reaction both are intensive properties.
(iii) ECell is an intensive property while ∆rG of cell reaction is an extensive property.
(iv) ECell is an extensive property while ∆rG of cell reaction is an intensive property.
Ans. (iii)
Sol.
ECell is intensive - it does not depend on the amount of substance. ∆rG is extensive - it scales with the amount (number of moles) of reactants/products. Thus (iii) is correct.
Q.4. The difference between the electrode potentials of two electrodes when no current is drawn through the cell is called ___________.
(i) Cell potential
(ii) Cell emf
(iii) Potential difference
(iv) Cell voltage
Ans. (ii)
Sol.
The term cell emf (electromotive force) specifically denotes the potential difference between electrodes when no current flows (i.e., open-circuit).
Q.5. Which of the following statement is not correct about an inert electrode in a cell?
(i) It does not participate in the cell reaction.
(ii) It provides surface either for oxidation or for reduction reaction.
(iii) It provides surface for conduction of electrons.
(iv) It provides surface for redox reaction
Ans. (iv)
Sol.
An inert electrode supplies a surface for electron transfer and conduction but does not itself take part in redox chemistry. It does not provide a surface that undergoes redox (i.e., the electrode material remains chemically unchanged). Thus statement (iv) as written (implying the electrode itself undergoes redox) is incorrect.
Q.6. An electrochemical cell can behave like an electrolytic cell when ____________.
(i) Ecell = 0
(ii) Ecell > Eext
(iii) Eext > Ecell
(iv) Ecell = Eext
Ans. (iii)
Sol.
If an external potential opposing the spontaneous cell emf is applied and the external potential exceeds the cell emf, the spontaneous direction is reversed and the cell acts as an electrolytic cell. Therefore when Eext > Ecell, electrolysis occurs.
Q.7. Which of the statements about solutions of electrolytes is not correct?
(i) Conductivity of solution depends upon size of ions.
(ii) Conductivity depends upon viscosiy of solution.
(iii) Conductivity does not depend upon solvation of ions present in solution.
(iv) Conductivity of solution increases with temperature.
Ans. (iii)
Sol.
Factors affecting ionic conductivity include ionic size (smaller ions move faster), viscosity of the solvent (higher viscosity → lower mobility), solvation (strong solvation increases effective hydrodynamic radius and lowers mobility), concentration, and temperature (higher T → higher mobility). Thus statement (iii) is incorrect.
Q.8. Using the data given below find out the strongest reducing agent.
(i) Cl-
(ii) Cr
(iii) Cr3+
(iv) Mn2+
Ans. (ii)
Sol.
Reducing power increases as the standard reduction potential becomes more negative (i.e., the metal more readily undergoes oxidation). Chromium metal (Cr) has the most negative standard reduction potential among the options, so Cr is the strongest reducing agent.
Q.9. Use the data given in Q.8 and find out which of the following is the strongest oxidising agent.
(i) Cl-
(ii) Mn2+
(iii) MnO4-
(iv) Cr3+
Ans. (iii)
Sol.
Oxidising agents have large positive standard reduction potentials. Permanganate ion (MnO4-) has a very high positive E°, so it is the strongest oxidising agent among the choices.
Q.10. Using the data given in Q.8 find out in which option the order of reducing power is correct.
(i) Cr3+ < Cl- < Mn2+ < Cr
(ii) Mn2+ < Cl- < Cr3+ < Cr
(iii) Cr3+ < Cl- < Cr2O72- < MnO4-
(iv) Mn2+ < Cr3+ < Cl- < Cr
Ans. (ii)
Sol.
On moving down the electrochemical series, E° decreases (becomes more negative) and reducing power increases. Therefore the order in option (ii) correctly lists increasing reducing power.
Q.11. Use the data given in Q.8 and find out the most stable ion in its reduced form.
(i) Cl-
(ii) Cr3+
(iii) Cr
(iv) Mn2+
Ans. (iv)
Sol.
A species with a very positive reduction potential is most stable in its reduced form. Mn2+ (from MnO4-/Mn2+) has a high positive E°, indicating Mn2+ is a stable reduced form among the options.
Q.12. Use the data of Q.8 and find out the most stable oxidised species.
(i) Cr3+
(ii) MnO4-
(iii) Cr2O72-
(iv) Mn2+
Ans. (i)
Sol.
Species with very negative reduction potentials correspond to stable oxidised forms. Cr3+/Cr has a strongly negative E° (-0.74 V in given data), indicating Cr3+ is a stable oxidised species among the choices.
(i) 1F
(ii) 6F
(iii) 3F
(iv) 2F
Ans. (iii)
Sol.
Write the reduction for 2Al3+ → 2Al; each Al3+ requires 3 e-. Total electrons per mole of Al produced = 3 mol e-, which corresponds to 3F of charge. Hence 3F is required to obtain one mole of Al.
Q.14. The cell constant of a conductivity cell _________.
(i) Changes with change of electrolyte.
(ii) Changes with change of concentration of electrolyte.
(iii) Changes with temperature of electrolyte.
(iv) Remains constant for a cell.
Ans. (iv)
Sol.
The cell constant G* = l/A depends only on the geometry of the cell (distance between electrodes l and electrode area A). For a particular cell these dimensions are fixed, so the cell constant remains constant.
Q.15. While charging the lead storage battery _________.
(i) PbSO4 anode is reduced to Pb.
(ii) PbSO4 cathode is reduced to Pb.
(iii) PbSO4 cathode is oxidised to Pb.
(iv) PbSO4 anode is oxidised to PbO2.
Ans. (i)
Sol.
During charging the discharge reactions are reversed. At cathode: PbSO4(s) + 2e- → Pb(s) + SO42-(aq). At anode: PbSO4(s) + 2H2O → PbO2(s) + SO42- + 4H+ + 2e-. Overall: 2PbSO4(s) + 2H2O → Pb(s) + PbO2(s) + 4H+ + 2SO42-. Thus PbSO4 at the cathode is reduced to Pb during charging.
Q.16.
(i)
(ii)
(iii)
(iv)
Ans. (ii)
Sol.
Kohlrausch's law: limiting molar conductivity of a salt equals the sum of limiting ionic molar conductivities of its cation and anion. Using that decomposition leads to the expression shown in option (ii).
(i) Na+ (aq) + e- → Na (s); E°Cell = -2.71V
(ii) 2H2O (l) → O2 (g) + 4H+ (aq) + 4e- ; E°Cell = 1.23V
(iii) H+ (aq) + e- → 1/2 H2 (g); E°Cell = 0.00 V
(iv) Cl- (aq) → 1/2 Cl2 (g) + e- ; E°Cell = 1.36 V
Ans. (iv)
Sol.
Thermodynamically water oxidation (ii) has lower E° (1.23 V) than chloride oxidation (1.36 V) and should be preferred. However, kinetics and overvoltage make water oxidation sluggish; chloride is oxidised preferentially under practical conditions, producing Cl2. Thus (iv) is the observed anode reaction in concentrated NaCl electrolysis.
Multiple Choice Questions - II
Note : In the following questions two or more than two options may be correct.
Q.18. The positive value of the standard electrode potential of Cu2+/Cu indicates that ____________.
(i) This redox couple is a stronger reducing agent than the H+/H2 couple.
(ii) This redox couple is a stronger oxidising agent than H+/H2.
(iii) Cu can displace H2 from acid.
(iv) Cu cannot displace H2 from acid.
Ans. (ii,iv)
Sol.
E°(Cu2+/Cu) = +0.34 V while E°(H+/H2) = 0.00 V. A higher (more positive) reduction potential means the species is a better oxidising agent. Therefore Cu2+/Cu is a stronger oxidising couple than H+/H2, and metallic Cu cannot displace H2 from acids. Thus (ii) and (iv) are correct.
Q.19. E°Cell for some half cell reactions are given below. On the basis of these mark the correct answer.
(i) H+ (aq) + e- → 1/2 H2 (g); E°Cell = 0.00 V
(ii) 2H2O (l) → O2 (g) + 4H+ (aq) + 4e- ; E°Cell = 1.23V
(c) 2SO2-4 (aq) → S2O82- (aq) + 2e- ; E°Cell = 1.96 V
(i) In dilute sulphuric acid solution, hydrogen will be reduced at cathode.
(ii) In concentrated sulphuric acid solution, water will be oxidised at anode.
(iii) In dilute sulphuric acid solution, water will be oxidised at anode.
(iv) In dilute sulphuric acid solution, SO42- ion will be oxidised to tetrathionate ion at anode.
Ans. (i, iii)
Sol.
In dilute H2SO4, the cathodic reduction is of H+ to H2, so (i) is correct. At the anode, among oxidation options, water oxidation (to O2, E° = 1.23 V) is kinetically favoured over sulfate oxidation (E° = 1.96 V), so in dilute acid water is oxidised - (iii) is correct.
Q.20. E°Cell = 1.1V for Daniel cell. Which of the following expressions are correct description of state of equilibrium in this cell?
(i) 1.1 = Kc
(ii)
(iii)
(iv) log Kc = 1.1
Ans. (ii,iii)
Sol.
At equilibrium ΔG° = -RT ln K and ΔG° = -nFE°. Therefore -nFE° = -RT ln K and hence expressions linking E°, n and log K are obtained. For Daniel cell n = 2 and substituting E° = 1.1 V gives the correct logarithmic relationships shown in options (ii) and (iii).
Q.21. Conductivity of an electrolytic solution depends on ________.
(i) Nature of electrolyte.
(ii) Concentration of electrolyte.
(iii) Power of AC source.
(iv) Distance between the electrodes.
Ans. (i,ii)
Sol.
Conductivity derives from mobile ions; it depends on the nature of the electrolyte (type and mobility of ions) and the concentration (number of ions per unit volume). Power of the source and electrode separation do not change intrinsic conductivity.
Q.22.
(i)
(ii)
(iii)
(iv)
Ans. (i,iii)
Sol.
This is based on Kohlrausch's law of independent migration: limiting molar conductivity of a salt is the sum of limiting ionic conductivities of its cation and anion. The decompositions shown in options (i) and (iii) are valid applications of that law.
Q.23. What will happen during the electrolysis of aqueous sodium sulphate solution by using platinum electrodes?
(i) Copper will deposit at cathode.
(ii) Copper will deposit at anode.
(iii) Oxygen will be released at anode.
(iv) Copper will dissolve at anode.
Ans. (i,iii)
Sol.
For electrolysis of CuSO4 with Pt electrodes: At cathode: Cu2+ + 2e- → Cu(s). At anode: water is oxidised to O2 (from OH-/H2O) when inert platinum is used, so O2 evolves. Thus copper is deposited at cathode and oxygen is released at the anode.
Q.24. What will happen during the electrolysis of aqueous solution of CuSO4 in the presence of Cu electrodes?
(i) Copper will deposit at cathode.
(ii) Copper will dissolve at anode.
(iii) Oxygen will be released at anode.
(iv) Copper will deposit at anode.
Ans. (i,ii)
Sol.
When copper electrodes are used, the anode reaction is Cu(s) → Cu2+ + 2e-, replenishing Cu2+ in solution. At cathode Cu2+ is reduced to Cu(s) and deposits. Oxygen evolution is not the primary oxidation in this case.
Q.25. Conductivity κ , is equal to ____________.
(ii) G*/R
(iii) ∧m
(iv) l/A
Ans. (i,ii)
Sol.
Conductivity κ is related to measured conductance G by κ = G × (l/A) = G × G*, where G* = l/A is the cell constant. Thus κ = G* / R when G = 1/R is used.
Q.26. Molar conductivity of ionic solution depends on ___________.
(i) Temperature.
(ii) Distance between electrodes.
(iii) Concentration of electrolytes in solution.
(iv) Surface area of electrodes.
Ans. (i,iii)
Sol.
Molar conductivity ∧m depends on temperature (mobility changes with T) and on concentration (∧m increases with dilution). It is independent of electrode geometry (distance or area).
Q.27. For the given cell, Mg|Mg2+|| Cu2+|Cu
(i) Mg is cathode
(ii) Cu is cathode
(iii) The cell reaction is Mg + Cu2+→ Mg2+ + Cu
(iv) Cu is the oxidising agent
Ans. (ii,iii)
Sol.
Left side denotes oxidation: Mg → Mg2+ + 2e-. Right side denotes reduction: Cu2+ + 2e- → Cu. Therefore Cu electrode is the cathode (reduction site) and the overall cell reaction is Mg + Cu2+ → Mg2+ + Cu.
Short Answer Type Questions
Q.28. Can absolute electrode potential of an electrode be measured?
Ans. No, only the difference in potential between two electrodes can be measured. Oxidation or reduction cannot occur in isolation; a reference electrode (e.g., SHE) is required to define electrode potential.
Q.29. Can E°Cell V or ∆rG° for cell reaction ever be equal to zero?
Ans. No, otherwise the reaction becomes non-feasible. Reaction is spontaneous when E°cell > 0 (∆rG° < 0). When E° = 0 and ∆rG° = 0 the system is at equilibrium (no net driving force).
Q.30. Under what condition is E°Cell = 0 or ∆rG° = 0?
Ans. At chemical equilibrium in the cell. E°cell = 0 and ∆rG° = -nFE°cell = 0.
Q.31. What does the negative sign in the expression E°Zn2+/Zn = - 0.76 V mean?
Ans. A negative standard reduction potential indicates the metal is a stronger reducing agent than hydrogen. Zinc is more reactive (more easily oxidised) than hydrogen, so when Zn is paired with standard hydrogen electrode, Zn is oxidised and H+ is reduced. Zinc electrode becomes the anode and hydrogen electrode the cathode.
Q.32. Aqueous copper sulphate solution and aqueous silver nitrate solution are electrolysed by 1 ampere current for 10 minutes in separate electrolytic cells. Will the mass of copper and silver deposited on the cathode be same or different? Explain your answer.
Ans. Different. Sol.
Faraday's second law: masses of substances liberated by the same quantity of electricity are proportional to their chemical equivalents (molar mass / n, where n = electrons involved). Quantity of charge Q = I × t is same for both. Different equivalent weights for Cu and Ag lead to different deposited masses.
Q.33. Depict the galvanic cell in which the cell reaction is Cu + 2Ag+ → 2Ag + Cu2+
Ans. Cell notation: Cu | Cu2+ || Ag+ | Ag. Oxidation at anode: Cu → Cu2+ + 2e-. Reduction at cathode: Ag+ + e- → Ag.
Q.34. Value of standard electrode potential for the oxidation of Cl- ions is more positive than that of water, even then in the electrolysis of aqueous sodium chloride, why is Cl- oxidised at anode instead of water?
Ans. Oxidation of water requires a higher overpotential (kinetic barrier) than chloride oxidation. Although the thermodynamic E° for water oxidation is lower, the kinetic overvoltage needed for O2 evolution is large, so Cl- is preferentially oxidised to Cl2. Possible anode reactions: Cl- → 1/2 Cl2 + e-, E° = 1.36 V. 2H2O → O2 + 4H+ + 4e-, E° = 1.23 V (but requires large overpotential).
Q.35. What is electrode potential?
Ans. The potential difference between an electrode and the surrounding electrolyte (solution) in an electrochemical cell is called the electrode potential.
Q.36. Consider the following diagram in which an electrochemical cell is coupled to an electrolytic cell. What will be the polarity of electrodes 'A' and 'B' in the electrolytic cell?
Ans. Electrons flow from zinc to copper in the galvanic cell (Zn | Zn2+ || Cu2+ | Cu). Electrons arrive at electrode A from the external circuit, so electrode A becomes positive (+). Electrode B receives electrons (is connected to the negative side), so electrode B is negative (-).
Q.37. Why is alternating current used for measuring resistance of an electrolytic solution?
Ans. AC is used to avoid net electrolysis at the electrodes; with DC the ion concentrations near electrodes change (electrolytic depletion/ build-up), altering resistance. AC prevents concentration polarization and provides a stable measurement of resistance.
Q.38. A galvanic cell has electrical potential of 1.1V. If an opposing potential of 1.1V is applied to this cell, what will happen to the cell reaction and current flowing through the cell?
Ans. When an opposing potential equal to the cell emf is applied, the net driving force becomes zero, the cell reaction stops, and no net current flows.
Q.39. How will the pH of brine (aq. NaCl solution) be affected when it is electrolysed?
Ans. pH will rise because NaOH is produced in the bulk during electrolysis of brine. Overall reactions: at cathode H2O + e- → 1/2 H2 + OH-; at anode Cl- → 1/2 Cl2 + e-. Net formation of OH- (and NaOH) increases pH.
Q.40. Unlike dry cell, the mercury cell has a constant cell potential throughout its useful life. Why?
Ans. In mercury cell the active species are metals and metal oxides that do not appreciably change the ionic composition of the electrolyte during discharge; ions in solution are not consumed in the same way as in dry cells. Thus the cell potential remains nearly constant over the life of the cell.
Q.41. 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.
Ans. Electrolyte B is the strong electrolyte. Sol.
Strong electrolytes are already fully dissociated; dilution mainly reduces interionic interactions, so ∧m increases only slightly. Weak electrolytes dissociate more on dilution, causing a large increase in ∧m. Since B's ∧m increases only 1.5× (small change), B is the strong electrolyte; A (25× increase) is weak.
Q.42. When acidulated water (dil.H2SO4 solution) is electrolysed, will the pH of the solution be affected? Justify your answer.
Ans. pH remains essentially unchanged. Sol.
At anode: 2H2O → O2 + 4H+ + 4e-. At cathode: 4H+ + 4e- → 2H2. Net H+ produced at anode is consumed at cathode, so [H+] remains approximately constant.
Q.43. In an aqueous solution how does specific conductivity of electrolytes change with addition of water?
Ans.Specific conductivity (κ) decreases on dilution because the number of ions per unit volume decreases as water is added.
Q.44. Which reference electrode is used to measure the electrode potential of other electrodes?
Ans.Standard Hydrogen Electrode (SHE) is used as the reference electrode; its potential is assigned as 0.00 V under standard conditions.
Q.45. Consider a cell given below
Cu|Cu2+|| Cl-|Cl2,Pt
Write the reactions that occur at anode and cathode
Ans. At anode: Cu → Cu2+ + 2e-. At cathode: Cl2 + 2e- → 2Cl-. Copper is oxidised at the anode and chlorine is reduced at the cathode.
Q.46. Write the Nernst equation for the cell reaction in the Daniel cell. How will the ECell be affected when concentration of Zn2+ ions is increased?
Ans. Cell reaction: Zn + Cu2+ → Zn2+ + Cu.
Sol.
Nernst equation for the cell: E = E° - (0.0591/n) log Q. For Zn/Cu cell Q = [Zn2+]/[Cu2+], and n = 2. Therefore E = E° - (0.0591/2) log ([Zn2+]/[Cu2+]). Increasing [Zn2+] increases Q, so E decreases.
Q.47. What advantage do the fuel cells have over primary and secondary batteries?
Ans. Fuel cells can run continuously as long as fuel and oxidant are supplied and products removed; they are not limited to the initial stored reactants like primary cells, and they avoid the long recharge cycles of secondary batteries.
Q.48. Write the cell reaction of a lead storage battery when it is discharged. How does the density of the electrolyte change when the battery is discharged?
Ans. Discharge reaction: Pb + PbO2 + 2H2SO4 → 2PbSO4 + 2H2O. Sol.
During discharge sulphuric acid is consumed and water is produced; thus the density of the electrolyte decreases.
Q.49. Why on dilution the Λm of CH3COOH increases drastically, while that of CH3COONa increases gradually?
Ans.CH3COOH is a weak electrolyte; on dilution its degree of dissociation increases markedly, increasing the number of ions and hence ∧m sharply. CH3COONa is a strong electrolyte and is nearly fully dissociated; dilution reduces interionic interactions so ∧m increases only gradually.
Matching Type
Note : Match the items of Column I and Column II in the following questions.
Q.50. Match the terms given in Column I with the units given in Column II.
| Column I | Column II |
| (i) ∧m | (a) S cm-1 |
| (ii) ECell | (b) m-1 |
| (iii) κ | (c) S cm2 mol-1 |
| (iv) G* = l/a | (d) V |
Ans. (i) → c, (ii) → d, (iii) → a, (iv)→ b
Sol.
Molar conductivity ∧m has units S cm2 mol-1. ECell is measured in volts (V). Conductivity κ has units S cm-1. Cell constant G* = l/A has units m-1.
Q.51. Match the terms given in Column I with the items given in Column II.
| Column I | Column II |
| (i) Λm | (a) Intensive property |
| (ii) EºCell | (b) Depends on number of ions/volume |
| (iii) κ | (c) Extensive property |
| (iv) ∆rGCell | (d) Increases with dilution |
Ans. (i) → d, (ii) → a, (iii)→ b, (iv) → c
Sol.
(i) Λm increases with dilution. (ii) E°Cell is intensive. (iii) κ (specific conductivity) depends on ions per unit volume. (iv) ΔrG is extensive (depends on amount of substance).
Q.52. Match the items of Column I and Column II.
| Column I | Column II |
| (i) Lead storage battery | (a) maximum efficiency |
| (ii) Mercury cell | (b) prevented by galvanisation |
| (iii) Fuel cell | (c) gives steady potential |
| (iv) Rusting | (d) Pb is anode, PbO2 is cathode |
Ans. (i)→ d, (ii) → c, (iii)→ a, (iv)→ b
Sol.
(i) Lead storage battery: Pb acts as anode and PbO2 as cathode in the cell reactions. (ii) Mercury cell gives a steady potential (ions not involved significantly). (iii) Fuel cell can achieve high efficiency by continuous reactant supply. (iv) Rusting (corrosion) can be prevented by galvanisation.
Q.53. Match the items of Column I and Column II.
| Column I | Column II |
| (i) κ | (a) I × t |
| (ii) Λm | (b) Λm/Λ0m |
| (iii) α | (c) κ/c |
| (iv) Q | (d) G*/R |
Ans. (i) → d, (ii) → c, (iii)→ b, (iv) → a
Sol.
(i) κ = G* / R (where G* is cell constant and R is resistance). (ii) Λm = κ/c. (iii) Degree of dissociation α = Λm/Λm0. (iv) Q = I × t.
Q.54. Match the items of Column I and Column II.
| Column I | Column II |
| (i) Lechlanche cell | (a) cell reaction 2H2 + O2 → 2H2O |
| (ii) Ni-Cd cell | (b) does not involve any ion in solution and is used in hearing aids. |
| (iii) Fuel cell | (c) rechargeable |
| (iv) Mercury cell | (d) reaction at anode, Zn → Zn2+ + 2e- |
Ans. (i)→ d, (ii) → c, (iii) → a, e (iv) → b
Sol.
(i) Leclanché cell anode reaction is Zn → Zn2+ + 2e-. (ii) Ni-Cd is a rechargeable battery. (iii) Fuel cell converts combustion energy to electrical energy (e.g., H2 + 1/2 O2 → H2O). (iv) Mercury cell produces a steady potential and is used in devices like hearing aids.
Q.55. Match the items of Column I and Column II on the basis of data given below:
| Column I | Column II |
| (i) F2 | (a) metal is the strongest reducing agent |
| (ii) Li | (b) metal ion which is the weakest oxidising agent |
| (iii) Au3+ | (c) non metal which is the best oxidising agent. |
| (iv) Br- | (d) unreactive metal |
| (v) Au | (e) anion that can be oxidised by Au3+ |
| (vi) Li+ | (f) anion which is the weakest reducing agent |
| (vii) F- | (g) metal ion which is an oxidising agent |
Ans. (i) → c, (ii) → a, (iii) → g, (iv) → e, (v) → d, (vi) → b, (vii) → f
Sol.
F2 is the strongest oxidising non-metal (E° ≈ +2.87 V). Li metal is the strongest reducing agent (most negative E°). Au3+ is a strong oxidising ion (positive E°). Br- can be oxidised by Au3+. Au metal is relatively unreactive. Li+ is a weak oxidising ion (most negative E°). F- is a poor reducing agent.
Assertion and Reason Type
Note : In the following questions a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices.
(i) Both assertion and reason are true and the reason is the correct explanation of assertion.
(ii) Both assertion and reason are true and the reason is not the correct explanation of assertion.
(iii) Assertion is true but the reason is false.
(iv) Both assertion and reason are false.
(v) Assertion is false but reason is true.
Q.56. Assertion : Cu is less reactive than hydrogen.
Reason : E°Cu2+/Cu is negative.
Ans. (iii)
Sol.
The assertion is true: copper is less reactive than hydrogen. The reason is false as stated: E°(Cu2+/Cu) is +0.34 V (positive), not negative. Thus (iii) is correct.
Q.57. Assertion : ECell should have a positive value for the cell to function.
Reason : Ecathode < Eanode
Ans. (iii)
Sol.
The assertion is true: for spontaneous (galvanic) operation Ecell > 0. The reason is false as written; the correct relation for spontaneous cell is Ecathode > Eanode.
Q.58. Assertion : Conductivity of all electrolytes decreases on dilution.
Reason : On dilution number of ions per unit volume decreases.
Ans. (i)
Sol.
Both assertion and reason are true and the reason correctly explains the assertion: specific conductivity κ depends on ion concentration per unit volume, which falls on dilution.
Q.59. Assertion : Λm for weak electrolytes shows a sharp increase when the electrolytic solution is diluted.
Reason : For weak electrolytes degree of dissociation increases with dilution of solution.
Ans. (i)
Sol.
Both assertion and reason are true; the increase in degree of dissociation on dilution causes the sharp rise in molar conductivity for weak electrolytes.
Q.60. Assertion : Mercury cell does not give steady potential.
Reason : In the cell reaction, ions are not involved in solution.
Ans. (v)
Sol.
Assertion is false (mercury cells do give a steady potential). Reason is true (ions are not involved appreciably). Hence (v) is correct.
Q.61. Assertion : Electrolysis of NaCl solution gives chlorine at anode instead of O2.
Reason : Formation of oxygen at anode requires overvoltage.
Ans. (i)
Sol.
Both assertion and reason are true and the reason correctly explains why Cl2 is formed at the anode: overvoltage for O2 evolution makes chloride oxidation kinetically favoured.
Q.62. Assertion : For measuring resistance of an ionic solution an AC source is used.
Reason : Concentration of ionic solution will change if DC source is used.
Ans. (i)
Sol.
Both assertion and reason are true and the reason correctly explains the assertion: DC causes electrolysis and concentration changes; AC avoids net electrolysis and gives stable resistance measurement.
Q.63. Assertion : Current stops flowing when ECell = 0.
Reason : Equilibrium of the cell reaction is attained.
Ans. (i)
Sol.
At Ecell = 0 the driving force for net reaction vanishes and the cell reaction is at equilibrium (no net current). Both statements are true and the reason explains the assertion.
Q.64. Assertion : EAg+/Ag increases with increase in concentration of Ag+ ions.
Reason : EAg+/Ag E has a positive value.
Ans. (ii)
Sol.
Both assertion and reason are true but the reason is not the correct explanation. Nernst equation explains the increase: E = E° + (0.0591/n) log[Ag+]; the positive E° is not the direct cause of the concentration dependence.
Q.65. Assertion : Copper sulphate can be stored in zinc vessel.
Reason : Zinc is less reactive than copper.
Ans. (iv)
Sol.
Both assertion and reason are false. Zinc is more reactive than copper and would be oxidised by Cu2+ ions; copper sulphate should not be stored in zinc vessels.
Long Answer Type Questions
Q.66. Consider the Fig. 3.2 and answer the following questions.
(i) Cell 'A' has ECell = 2V and Cell 'B' has ECell = 1.1V which of the two cells 'A' or 'B' will act as an electrolytic cell. Which electrode reactions will occur in this cell?
(ii) If cell 'A' has ECell = 0.5V and cell 'B' has ECell = 1.1V then what will be the reactions at anode and cathode?
Ans.
(i)
If cell A (2.0 V) is connected to cell B (1.1 V) such that A opposes B, the external higher emf (A) will drive the lower emf cell B in non-spontaneous direction; cell B will act as electrolytic cell. At the electrodes of B the spontaneous directions are reversed: At cathode of B (reduction in electrolytic mode): Zn2+ + 2e- → Zn. At anode of B (oxidation in electrolytic mode): Cu → Cu2+ + 2e-. (ii)
If A has smaller emf (0.5 V) and B has higher emf (1.1 V), then B will act as the galvanic (driving) cell and push electrons into A. Thus in this arrangement on the cell with Zn/Cu: Anode (oxidation): Zn → Zn2+ + 2e-. Cathode (reduction): Cu2+ + 2e- → Cu.
Q.67. Consider Fig. 3.2 and answer the questions (i) to (vi) given below.
(i) Redraw the diagram to show the direction of electron flow.
(ii) Is silver plate the anode or cathode?
(iii) What will happen if salt bridge is removed?
(iv) When will the cell stop functioning?
(v) How will concentration of Zn2+ ions and Ag+ ions be affected when the cell functions?
(vi) How will the concentration of Zn2+ ions and Ag+ ions be affected after the cell becomes 'dead'?
Ans.
(i)
Electrons flow from the more negative electrode (Zn) to the more positive electrode (Ag).
Silver (Ag) plate is the cathode because Ag+ ions are reduced to Ag(s) there. (iii)
If the salt bridge is removed, the circuit becomes incomplete in terms of ion migration balancing charge, charge separation builds up, and the cell quickly stops functioning. (iv)
The cell stops when Ecell becomes zero - when reaction reaches equilibrium or concentrations change such that no driving force remains. (v)
While functioning, Zn is oxidised to Zn2+, so [Zn2+] increases. Ag+ is reduced to Ag(s), so [Ag+] decreases. (vi)
After the cell is dead (equilibrium), concentrations remain constant (no net change).
Q.68. 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?
Ans. Relationship: ΔrG = -nFEcell. Sol.
If species are in standard states, ΔrG° = -nFE°cell. Maximum (non-PV, electrical) work from a galvanic cell is obtained when the process is carried out reversibly (no internal losses); in that reversible limit, the work equals the decrease in Gibbs free energy and is given by wmax = -ΔG = nFEcell. Example: For Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s) with n = 2, ΔrG = -2FEcell.
FAQs on NCERT Exemplar: Electrochemistry
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| 2. What are the applications of electrochemistry in everyday life? | |
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