Chapter 2: Electrochemistry Long Questions | chemistry Class 12 CBSE

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Chapter 2: Electrochemistry Long Questions | chemistry Class 12 CBSE | ByteNotes.in

Long Answer

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Question 1

Long Form

Compare the functioning of a Daniell cell under three conditions: external voltage less than, equal to, and greater than 1.1 V. What does this demonstrate about the nature of electrochemical cells?

When Eext < 1.1 V: electrons flow from Zn to Cu, current flows from Cu to Zn, Zn dissolves at anode and Cu deposits at cathode — the cell acts as a galvanic cell.
When Eext = 1.1 V: no current flows and no chemical reaction occurs — the opposing voltage exactly balances the cell potential, representing a state of dynamic equilibrium.
When Eext > 1.1 V: electrons flow from Cu to Zn, current reverses, Zn deposits at zinc electrode and copper dissolves — the cell now acts as an electrolytic cell.
This demonstrates the reversibility of electrochemical cells: a galvanic cell can be converted into an electrolytic cell by applying a sufficiently large opposing potential.
The condition Eext = E(cell) = 1.1 V represents the reversible limit, showing that EMF is the minimum external potential needed to reverse the spontaneous reaction.

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Question 2

Long Form

Derive the relationship between standard cell potential E°(cell), standard Gibbs energy ΔrG°, and equilibrium constant Kc. Illustrate with a numerical example.

The reversible work done by a galvanic cell equals the decrease in Gibbs energy: ΔrG = −nFE(cell), where n = moles of electrons and F = 96487 C mol−1.
Under standard conditions (all concentrations = 1): ΔrG° = −nFE°(cell).
Since ΔrG° = −RT ln Kc, equating gives: −nFE°(cell) = −RT ln Kc, hence E°(cell) = (RT/nF) ln Kc = (0.059/n) log Kc at 298 K.
Example: For Daniell cell, E°(cell) = 1.1 V, n = 2. Then log Kc = (1.1 × 2)/0.059 = 37.29, giving Kc = 2 × 10^37.
ΔrG° = −2 × 96487 × 1.1 = −212.27 kJ mol−1. The large negative ΔrG° and huge Kc confirm the reaction is highly spontaneous and essentially irreversible.

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Question 3

Long Form

Describe the construction, functioning, and electrode reactions of a hydrogen-oxygen fuel cell. Why are fuel cells considered superior to conventional batteries?

Construction: H2 and O2 are bubbled through porous carbon electrodes into concentrated NaOH solution. Finely divided Pt or Pd catalysts are incorporated into electrodes to speed up reactions.
Cathode reaction: O2(g) + 2H2O(l) + 4e− → 4OH−(aq); oxygen is reduced.
Anode reaction: 2H2(g) + 4OH−(aq) → 4H2O(l) + 4e−; hydrogen is oxidised. Overall: 2H2(g) + O2(g) → 2H2O(l).
The cell runs continuously as long as reactants are supplied, unlike batteries which become exhausted. The only product is water — fuel cells are pollution-free.
Efficiency is ~70% vs ~40% for thermal plants and batteries have fixed energy. Fuel cells were used in the Apollo programme; the water produced was used as drinking water for astronauts.

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Question 4

Long Form

Explain the variation of conductivity and molar conductivity with concentration for both strong and weak electrolytes. How is Λ°m determined for each type?

Conductivity always decreases with dilution for both strong and weak electrolytes because the number of ions per unit volume carrying current decreases.
Molar conductivity Λm = κ/c increases with dilution as the total volume containing one mole of electrolyte increases, more than compensating for the drop in κ.
For strong electrolytes: Λm increases slowly and linearly with √c; Λ°m is obtained by extrapolating the Λm vs √c graph to zero concentration (y-intercept).
For weak electrolytes (e.g., acetic acid): Λm increases steeply near very low concentrations due to increasing degree of dissociation; extrapolation is unreliable.
For weak electrolytes, Λ°m is calculated from Kohlrausch's law using tabulated ionic conductivities: e.g., Λ°m(HAc) = Λ°m(HCl) + Λ°m(NaAc) − Λ°m(NaCl) = 390.5 S cm2 mol−1.

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Question 5

Long Form

Analyse how standard electrode potentials from Table 2.1 can be used to predict the relative strengths of oxidising and reducing agents, and to determine whether a given reaction is feasible.

In Table 2.1, reduction half-reactions are listed in decreasing order of E°; species at the top (high E°) are strongest oxidising agents and their reduced forms are weakest reducing agents.
F2 (E° = +2.87 V) is the strongest oxidising agent and F− is the weakest reducing agent. Li (E° = −3.05 V) makes Li+ the weakest oxidising agent and Li metal the strongest reducing agent.
A reaction is feasible if E°cell = E°cathode − E°anode > 0, meaning ΔrG° = −nFE° < 0.
Example: Cu (E° = +0.34 V) does not dissolve in HCl because H+ cannot oxidise Cu; but Zn (E° = −0.76 V) dissolves in HCl since H+ can oxidise Zn.
Electrochemical cells are therefore used to determine pH, solubility products, equilibrium constants, and for potentiometric titrations.

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Question 6

Long Form

Describe the quantitative aspects of electrolysis through Faraday's laws. Calculate the charge required for the reduction of 1 mole of Al3+ to Al and deposit 63 g of Cu from CuSO4 solution.

Faraday's First Law: amount of chemical reaction is proportional to charge passed Q = It (coulombs = amperes × seconds).
One Faraday = charge on one mole of electrons = NA × 1.6021×10−19 C = 96487 C mol−1 ≈ 96500 C mol−1.
For Al3+ + 3e− → Al: 3 moles of electrons (3F = 3 × 96487 = 289461 C) are required to deposit 1 mole (27 g) of Al.
For Cu2+ + 2e− → Cu: 2 moles of electrons (2F = 192974 C) are required to deposit 1 mole (63 g) of Cu.
Faraday's Second Law: the same charge (2F) that deposits 63 g Cu would also deposit 108 g Ag (from Ag+ + e−, 1F per mole, so 2F for 2 moles = 216 g). Equivalent weights govern the ratio.

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Question 7

Long Form

Analyse the products of electrolysis of molten NaCl versus aqueous NaCl solution with inert electrodes. What factors determine the products?

Molten NaCl: only Na+ and Cl− ions present. Na+ is reduced at cathode (Na+ + e− → Na) and Cl− is oxidised at anode (Cl− → ½Cl2 + e−). Products: Na metal and Cl2 gas.
Aqueous NaCl: Na+, Cl−, H+, OH−, and H2O all present. At cathode, H+ (E° = 0 V) is reduced in preference to Na+ (E° = −2.71 V), producing H2 gas.
At the anode, Cl− (E° = 1.36 V) is oxidised in preference to water (E° = 1.23 V), despite lower thermodynamic preference, due to high overpotential for O2 evolution.
Net reaction in aqueous NaCl: NaCl + H2O → NaOH + ½H2 + ½Cl2; products are NaOH, H2 and Cl2.
Determining factors: number and types of ions present, standard electrode potentials of competing reactions, nature of electrodes (reactive vs inert), and overpotential effects.

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Question 8

Long Form

Compare primary and secondary batteries with examples, discussing their electrode reactions and practical applications.

Primary batteries: non-rechargeable; reaction occurs once and the battery becomes dead. Example: dry cell — Zn anode oxidised, MnO2 cathode reduced, E ≈ 1.5 V; used in clocks and transistors.
Mercury cell: Zn-Hg amalgam anode, HgO cathode, KOH/ZnO electrolyte; constant E ≈ 1.35 V because cell reaction produces only solids, not ions in solution. Used in hearing aids and watches.
Secondary batteries: rechargeable by passing current in opposite direction. Lead storage battery: Pb anode, PbO2 cathode, 38% H2SO4 electrolyte. Used in automobiles and invertors.
Lead battery discharge: Pb + PbO2 + 2H2SO4 → 2PbSO4 + 2H2O; charging reverses this, regenerating Pb, PbO2 and H2SO4.
Nickel-cadmium cell: longer life than lead storage but more expensive; Cd anode, Ni(OH)3 cathode; discharge reaction is Cd + 2Ni(OH)3 → CdO + 2Ni(OH)2 + H2O.

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Question 9

Long Form

Describe how Kohlrausch's law is applied to determine the dissociation constant of acetic acid from conductance measurements.

First, Λ°m for acetic acid is calculated using Kohlrausch's law: Λ°m(HAc) = Λ°m(HCl) + Λ°m(NaAc) − Λ°m(NaCl) = 425.9 + 91.0 − 126.4 = 390.5 S cm2 mol−1.
At a known concentration c, conductivity κ is measured; Λm = (κ × 1000)/c (in S cm2 mol−1).
Degree of dissociation α = Λm/Λ°m (e.g., if Λm = 48.15 S cm2 mol−1 at 0.001028 M, then α = 48.15/390.5 = 0.1233).
Dissociation constant Ka = cα2/(1 − α). With c = 0.001028 mol L−1, Ka = (0.001028 × 0.1233²)/(1 − 0.1233) = 1.78 × 10−5 mol L−1.
This method is significant because Ka for weak acids/bases can be determined accurately from conductance data without needing titration or pH measurement.

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Question 10

Long Form

Explain corrosion as an electrochemical process and describe the various methods used to prevent corrosion of iron.

Corrosion is an electrochemical phenomenon: on an iron surface, an anodic spot acts as anode where Fe is oxidised (2Fe → 2Fe2+ + 4e−, E° = −0.44 V), and a cathodic spot reduces O2 (O2 + 4H+ + 4e− → 2H2O, E° = +1.23 V), giving E°cell = 1.67 V (spontaneous).
Fe2+ ions are oxidised further by O2 to Fe3+, which forms hydrated ferric oxide Fe2O3·xH2O (rust). H+ required comes from H2CO3 (from dissolved CO2) and other dissolved acidic oxides.
Prevention method 1: Coating the surface with paint, lacquer, or chemicals like bisphenol prevents contact with atmosphere — physically blocking the electrochemical process.
Prevention method 2: Coating with less reactive metals like Sn or Zn (galvanisation); Zn acts as a sacrificial metal even if scratched because it is more reactive and corrodes preferentially.
Prevention method 3: Sacrificial electrode (cathodic protection) — attaching a more reactive metal (Mg or Zn block) to iron; this metal is oxidised preferentially, protecting the iron structure.

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Chapter 2: Electrochemistry | Long Questions

Compare the functioning of a Daniell cell under three conditions: external voltage less than, equal to, and greater than 1.1 V. What does this demonstrate about the nature of electrochemical cells?

Derive the relationship between standard cell potential E°(cell), standard Gibbs energy ΔrG°, and equilibrium constant Kc. Illustrate with a numerical example.

Describe the construction, functioning, and electrode reactions of a hydrogen-oxygen fuel cell. Why are fuel cells considered superior to conventional batteries?

Explain the variation of conductivity and molar conductivity with concentration for both strong and weak electrolytes. How is Λ°m determined for each type?

Analyse how standard electrode potentials from Table 2.1 can be used to predict the relative strengths of oxidising and reducing agents, and to determine whether a given reaction is feasible.

Describe the quantitative aspects of electrolysis through Faraday's laws. Calculate the charge required for the reduction of 1 mole of Al3+ to Al and deposit 63 g of Cu from CuSO4 solution.

Analyse the products of electrolysis of molten NaCl versus aqueous NaCl solution with inert electrodes. What factors determine the products?

Compare primary and secondary batteries with examples, discussing their electrode reactions and practical applications.

Describe how Kohlrausch's law is applied to determine the dissociation constant of acetic acid from conductance measurements.

Explain corrosion as an electrochemical process and describe the various methods used to prevent corrosion of iron.

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