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Chemistry — Class 12 — L13: Electrochemistry

NIOS Code 313 · Module 5 · Chemical Dynamics

Notes extracted from NIOS Chemistry Course (313), Lesson 13 — Electrochemistry (313_Chemistry_Eng_Lesson13.pdf). Content covers sections 13.1–13.15.
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Overview — Module 5: Electrochemistry

Electrochemistry deals with the conversion of electrical energy into chemical energy and vice versa. Passing current through solutions or molten salts drives chemical change (electrolysis). Dry cells, button cells, and lead-acid batteries reverse the process — chemical reactions produce electricity. This lesson covers redox as electron transfer, oxidation numbers, balancing redox equations, electrolytic conduction and molar conductivity, Kohlrausch's law, Faraday's laws, electrolytic and galvanic cells, electrode potentials, electrochemical series, Nernst equation, batteries and fuel cells, ΔG–emf relation, and corrosion.

Mastering these ideas links L12 ionic equilibria (ions that move) with L9 thermodynamics (ΔG = −nFE°) and explains everyday technology from car batteries to rust prevention.

Section 1: Redox and Oxidation Number (13.1–13.3)

Oxidation: loss of electrons. Reduction: gain of electrons. They occur together as redox reactions. The species that loses electrons is the reductant; the species that gains electrons is the oxidant. Na → Na⁺ + e⁻ (oxidation); Cl + e⁻ → Cl⁻ (reduction).

Oxidation number (ON) is the apparent charge when shared electrons are assigned to the more electronegative atom. Rules: elemental form = 0; monatomic ion = charge; O usually −2 (peroxides −1, superoxides −½); H is +1 with non-metals, −1 with metals; alkali metals +1; sum of ON in neutral compound = 0; in polyatomic ion = ion charge.

Examples: S in H₂SO₄ is +6; N in NO₃⁻ is +5; Cl in ClO₄⁻ is +7. Balancing redox uses oxidation-number method or ion-electron method (half-reactions in acidic or basic medium). Example 13.1 balances skeletal equations by writing oxidation and reduction half-reactions, equalizing electrons, then combining.

In the oxidation-number method you identify atoms that change ON, find the total increase and decrease in ON, and multiply half-reactions so the electron loss equals electron gain. The ion-electron method is preferred for aqueous ionic reactions: write half-reactions, balance atoms (H₂O/H⁺ in acid; H₂O/OH⁻ in base), balance charge with electrons, then add. Both methods must give the same balanced equation — a useful check in exams.

Potassium permanganate (KMnO₄) in acid is a classic oxidant: Mn goes from +7 to +2 (Mn²⁺). Dichromate Cr₂O₇²⁻ goes from +6 to +3. Recognizing these common changes speeds up balancing and helps when writing half-cell reactions for electrochemical series problems later in the chapter.

Redox as Electron Transfer Oxidation Loss of e⁻ Reductant e⁻ Reduction Gain of e⁻ Oxidant Redox = simultaneous oxidation + reduction
Electrons flow from reductant to oxidant — the basis of all electrochemical cells.

Section 2: Electrolytic Conduction (13.4–13.5)

Electrolytes conduct electricity in solution by movement of ions. Solutions obey Ohm's law: V = IR. Conductance L = 1/R (siemens, S). Conductivity κ (kappa) = 1/ρ; κ = L × (l/A) where l/A is the cell constant.

κ = L × cell constant  |  Λm = 1000 κ / M
Λm = molar conductivity (S cm² mol⁻¹) · M = molarity · κ decreases on dilution; Λm increases

Strong electrolytes (KCl) are fully dissociated — Λm rises gradually on dilution as inter-ionic forces weaken. Weak electrolytes (CH₃COOH) show sharp increase in Λm on dilution because degree of ionization α increases. Conductivity rises with temperature (~2–3% per degree) as viscosity falls and ion mobility increases.

13.5.1 Kohlrausch's Law

At infinite dilution, each ion contributes a fixed amount to molar conductivity independent of other ions: Λm° = ν₊λ₊° + ν₋λ₋°. This allows calculation of Λm° for weak electrolytes from strong ones.

Example 13.2: Λ°(NaCl)=126, HCl=426, CH₃COONa=91 → Λ°(CH₃COOH) = 91+426−126 = 391 S cm² mol⁻¹.

Table 13.1 compares conductivities: pure water is a very poor conductor (6×10⁻⁸ S cm⁻¹), 0.1 M HCl is much better (3.5×10⁻²), metals are orders of magnitude higher. Glass is an insulator (~10⁻¹⁴). This ranking explains why conductivity measurements distinguish electrolytes from non-electrolytes and strong from weak acids — exactly as in L12 ionic equilibrium.

Factors affecting conductivity: (i) nature of electrolyte — weak vs strong, ion charge, ion mobility; (ii) temperature — higher T lowers viscosity, speeds ions, and increases α for weak electrolytes; (iii) concentration — dilution decreases κ but increases Λ_m. Graph of Λ_m vs √c is nearly linear for strong electrolytes (Debye–Hückel–Onsager type behaviour at dilute limit) and curved for weak electrolytes.

Section 3: Electrolytic Cells and Faraday's Laws (13.6–13.7)

Electrolytic cells use electrical energy to drive non-spontaneous reactions. Anode is positive (oxidation); cathode is negative (reduction). Molten NaCl: Na⁺ + e⁻ → Na at cathode; 2Cl⁻ → Cl₂ + 2e⁻ at anode.

w = zIt  |  1 F = 96500 C = 1 mol e⁻
First law: mass ∝ charge · Second law: masses ∝ equivalent masses (series cells) · Ex 13.3: 500 C → 0.56 g Ag

Faraday I: mass liberated ∝ quantity of electricity (w = zQ = zIt). Faraday II: when same charge passes different electrolytes in series, masses liberated ∝ chemical equivalent masses. Example 13.4: 32.4 g Ag deposited → copper deposited = 9.53 g (Eq Cu/Eq Ag ratio).

Electrolysis of Aqueous NaCl (Pt) Cathode (−) H₂O + 2e⁻ → H₂ + 2OH⁻ H₂ evolved (not Na) Anode (+) 2Cl⁻ → Cl₂ + 2e⁻ Cl₂ (overpotential of O₂) Overpotential can change which product forms
Products of electrolysis depend on discharge potential and overpotential — especially for gases.

Products of electrolysis: Prefer process with easiest discharge. Aqueous NaCl: H₂ at cathode (water easier than Na⁺), Cl₂ at anode (O₂ overpotential). Aqueous CuSO₄ on Pt: Cu deposited, O₂ evolved. On Cu electrodes: refining — Cu dissolves at anode and deposits at cathode.

Discharge (decomposition) potential is the minimum voltage needed for an electrode process. Overpotential is the extra voltage often needed for gas evolution — H₂ overpotential is ~0 on Pt but ~1.5 V on mercury. That is why industrial NaOH manufacture uses a mercury cathode: Na⁺ is reduced instead of water, forming sodium amalgam, which then reacts with water to give NaOH and H₂. Exam questions love comparing molten vs aqueous NaCl and Pt vs Cu electrodes for CuSO₄.

Equivalent mass = molar mass / number of electrons lost or gained. For Ag⁺ + e⁻ → Ag, Eq = 108. For Cu²⁺ + 2e⁻ → Cu, Eq = 63.5/2 = 31.75. Faraday's second law then becomes a simple proportion when cells are in series — same Q, so m_A/m_B = Eq_A/Eq_B.

Section 4: Galvanic Cells and Electrode Potential (13.8–13.11)

Galvanic (voltaic) cells convert spontaneous redox into electricity. Classic Daniell cell: Zn rod in ZnSO₄ and Cu rod in CuSO₄, connected by a wire and salt bridge.

Anode (oxidation): Zn → Zn²⁺ + 2e⁻. Cathode (reduction): Cu²⁺ + 2e⁻ → Cu. Electrons flow external circuit from Zn to Cu. Salt bridge completes ionic circuit and prevents charge build-up without mixing solutions.

Cell notation: Zn | Zn²⁺(aq) || Cu²⁺(aq) | Cu · Double line = salt bridge · Left = anode · Right = cathode.

Daniell Cell (Galvanic) Anode (−) Zn → Zn²⁺ + 2e⁻ Oxidation e⁻ SB Cathode (+) Cu²⁺ + 2e⁻ → Cu Reduction E°_cell ≈ 1.1 V · Spontaneous
Chemical energy → electrical energy. Salt bridge (SB) maintains charge neutrality.

Standard electrode potential E° is measured against the Standard Hydrogen Electrode (SHE), assigned E° = 0. Electrodes with E° < 0 are better reductants than H₂; E° > 0 better oxidants.

cell = E°cathode − E°anode
Must be positive for spontaneous cell · Electrochemical series: increasing E°_red · Li top (strong reductant) · F₂ bottom (strong oxidant)

Electrochemical series applications: (1) Predict redox — metal reduces ions of metals above it in series (more positive E°). (2) Calculate E°_cell. (3) Feasibility: positive E°_cell required. (4) Metals above H₂ liberate H₂ from acids. Example: Cu²⁺ + 2Ag → Cu + 2Ag⁺ has E° = −0.46 V — not feasible; reverse is spontaneous.

Cell emf is the potential difference measured in an open circuit (no current); under load with current flowing it is called potential difference and is lower due to internal resistance. Standard cell emf E°_cell requires both half-cells in standard states (1 M ions, 1 bar gases, pure solids). Measurement of unknown E° uses a galvanic cell with SHE: if the study electrode is positive, E° is positive; if negative (as for Zn), E° is negative.

Key E° values to remember: Zn²⁺/Zn −0.76 V, Fe²⁺/Fe −0.44 V, SHE 0, Cu²⁺/Cu +0.34 V, Ag⁺/Ag +0.80 V, Cl₂/Cl⁻ +1.36 V, F₂/F⁻ +2.87 V. Lithium (−3.05 V) is the strongest metallic reductant; fluorine is the strongest oxidant. Zn displaces Fe²⁺ (Example 13.5); Mg|Mg²⁺||Ag⁺|Ag gives E°_cell = 3.165 V.

Section 5: Nernst Equation (13.12)

Electrode potential depends on ion concentration. For Mn+ + ne⁻ → M:

E = E° − (0.0591/n) log(1/[Mn+])   (298 K)
General: E = E° − (0.0591/n) log Q · Solids/liquids concentration = 1 · Ex 13.7: Ag⁺(0.1M) → E = 0.741 V

For cell reaction aA + bB → xX + yY: E_cell = E°_cell − (0.0591/n) log([X]x[Y]y/[A]a[B]b). Example 13.8: Ni|Ni²⁺(0.001)||Ag⁺(0.1)|Ag — first find E°_cell = 1.05 V, then apply Nernst with n=2.

Nernst equation shows how concentration cells and non-standard conditions shift emf — critical for batteries as they discharge and for corrosion under varying oxygen/pH conditions.

At 298 K the factor 2.303RT/F simplifies to 0.0591 V. For n = 2, each 10-fold change in Q shifts E by about 0.03 V. Only aqueous ions and gases (partial pressures) appear in Q — pure solids and pure liquids are taken as unit activity. Writing half-reactions first is the safest way to find n: the number of electrons cancelled when anode and cathode half-reactions are combined.

As a battery discharges, ion concentrations change, Q increases, and E_cell falls toward zero. When E_cell = 0 the cell is dead (equilibrium). Recharging a secondary cell drives the reverse electrolysis, restoring original concentrations and reactants — Faraday's laws then tell you how much charge is needed to restore a given mass of active material.

Section 6: Batteries, Fuel Cells, ΔG and Corrosion (13.13–13.15)

Primary cells (dry cell / Leclanché): Zn anode, graphite cathode, NH₄Cl–MnO₂ paste; ~1.5 V; not rechargeable. Secondary cells (lead storage): Pb anode, PbO₂ cathode, H₂SO₄ electrolyte; discharge forms PbSO₄; recharge reverses reaction. Fuel cells: continuous fuel feed; H₂–O₂ cell with KOH electrolyte produces water (Apollo program, E ≈ 0.9 V).

Gibbs energy: ΔG° = −nFE°. Negative ΔG° means spontaneous. Example 13.9: Daniell cell n=2, E°=1.1 V → ΔG° = −212.3 kJ.

Corrosion of Iron (Rusting) Anodic spot Fe → Fe²⁺ + 2e⁻ Cathodic spot O₂ + 4H⁺ + 4e⁻ → 2H₂O E°_cell ≈ 1.67 V → spontaneous · Needs O₂ + H₂O Protect: paint, oil, plating, sacrificial anode (Zn)
Rusting is an electrochemical process. Protection: coating or cathodic protection.

Corrosion: deterioration of metals by environment. Iron rusts when one spot acts as anode (Fe → Fe²⁺) and another as cathode (O₂ reduction). Fe²⁺ oxidizes further to Fe₂O₃·xH₂O (rust). Accelerated by CO₂, SO₂ (acidic); needs moisture and oxygen. Protection: paint, grease, phosphate coatings, tinning, galvanizing (Zn sacrificial anode), impressed current.

Cathodic protection makes the entire metal surface the cathode of an electrochemical cell so oxidation of the structural metal is suppressed. Zinc coating on iron (galvanizing) works because Zn has more negative E° — zinc corrodes preferentially and protects iron even if the coating is scratched. Tin plating protects only while the coating is intact; if scratched, iron becomes the anode and rusts faster. Completely homogeneous iron rusts less — local anodic/cathodic sites need surface differences or impurities.

Fuel cells combine continuous fuel supply (like a power plant) with high efficiency of electrochemical conversion. The H₂–O₂ cell overall reaction is simply water formation; porous carbon electrodes with Pt/Pd catalyst and aqueous KOH electrolyte give ~0.9 V. Dry cells cannot be recharged because Zn is consumed irreversibly; lead-acid batteries reverse cleanly because PbSO₄ can be converted back to Pb and PbO₂ on charging.

Exam Connections and Chapter Summary

Key numerical skills: oxidation numbers; Λm and Kohlrausch; Faraday mass calculations; E°_cell from series; Nernst at non-standard concentrations; ΔG° = −nFE°. Conceptual: electrolytic vs galvanic polarity; products of electrolysis with overpotential; salt bridge role; ECS applications; battery types; corrosion mechanism.

Intext questions test: conduction factors, Faraday laws, cell notation, SHE, series feasibility, Nernst setup. Terminal exercises combine multi-step problems — e.g. calculate E_cell then ΔG, or mass deposited then series cell partner.

Electrochemistry unifies redox chemistry with electrical work: ions move (conductivity), charge transfers mass (Faraday), spontaneous redox drives emf (galvanic), concentration shifts potential (Nernst), and thermodynamics links energy (ΔG). From industrial electrolysis of NaCl to preventing ship hull corrosion, these principles power modern technology.

Module 5 path: chemical equilibrium → ionic equilibrium (L12) → electrochemistry (L13). Conductivity and Faraday connect to lab quantitative analysis; Nernst and ΔG connect to chemical thermodynamics; batteries and corrosion connect to materials science and engineering. When you see a redox problem, ask: which half is oxidized? What is n? Is E° positive? Does concentration change E via Nernst? That decision tree covers almost every NIOS numerical and conceptual question on this chapter.

MCQ Quiz — L13 Electrochemistry

0 / 10 correct

Flashcards — L13

1 / 18

Golden Rules — L13 Electrochemistry

Most exam-important points from this chapter:

Redox & ON

Oxidation = loss of e⁻; reduction = gain. Assign ON by rules; balance half-reactions by ion-electron method.

Conductance

κ = L × cell constant; Λ_m = 1000κ/M. Dilution: κ↓, Λ_m↑. Kohlrausch for Λ_m° of weak electrolytes.

Faraday & electrolysis

w = zIt; 1 F = 96500 C. Series: m ∝ equivalent mass. Products depend on discharge potential + overpotential.

Cells & series

Galvanic: anode oxidation (−). E°_cell = E°_c − E°_a > 0 for spontaneous. ECS: metals above H₂ liberate H₂; positive E° = feasible redox.

Nernst & ΔG

E = E° − (0.0591/n) log Q. ΔG° = −nFE°. Batteries: primary/secondary/fuel. Corrosion = local galvanic cells on metal surface.

κ = L × (l/A)
Λm = 1000 κ / M
Λm° = ν₊λ₊° + ν₋λ₋°
w = zIt
1 F = 96500 C
cell = E°cath − E°an
E = E° − (0.0591/n) log Q
ΔG° = −nFE°
Anode = oxidation

Section 1: Formulas with Derivations

All key equations from NIOS Chemistry 313, Module 5 — Electrochemistry (sections 13.1–13.15).

Conductivity — κ = L × (l/A) = L × cell constant

Molar conductivity: Λm = 1000 κ / M  (S cm² mol⁻¹)

Where

• L = conductance (S) · κ = conductivity (S cm⁻¹)
• M = molarity · l/A = cell constant

Dilution

κ decreases on dilution (fewer ions per cm³). Λm increases — strong electrolytes gradually; weak electrolytes sharply (degree of ionization rises).

❌ Confusing κ with Λm.
✓ κ = bulk conductivity; Λm = conductivity per mole of electrolyte.

Kohlrausch's Law — Λm° = ν₊λ₊° + ν₋λ₋°

At infinite dilution, each ion contributes independently to molar conductivity.

Use

Find Λm° of weak electrolytes from strong ones: Λ°(CH₃COOH) = Λ°(CH₃COONa) + Λ°(HCl) − Λ°(NaCl)

Exam

Ex 13.2: Λ°(NaCl)=126, HCl=426, CH₃COONa=91. Find Λ°(CH₃COOH).

91 + 426 − 126

Answer: 391 S cm² mol⁻¹

Faraday's Laws — w = zIt  |  1 F = 96500 C mol⁻¹

First law: mass liberated ∝ charge passed (w = zQ = zIt)
Second law: masses of different substances ∝ equivalent masses (same Q)

Examples

Ex 13.3: 500 C → 0.56 g Ag · Ex 13.4: Ag and Cu in series — m_Cu/m_Ag = Eq_Cu/Eq_Ag

Cell emf — E°cell = E°cathode − E°anode

Always write reduction potentials. E°_cell must be positive for spontaneous cell.

Notation

Anode | anode ion || cathode ion | cathode · Salt bridge separates half-cells, maintains charge neutrality

Intermediate

Q: Mg|Mg²⁺||Ag⁺|Ag; E°_Mg=−2.365, E°_Ag=0.80

E°=0.80−(−2.365)

Answer: 3.165 V

Nernst Equation (298 K) — E = E° − (0.0591/n) log Q

Half-cell: E = E° − (0.0591/n) log(1/[Mn+])

Cell: E_cell = E°_cell − (0.0591/n) log([products]/[reactants])

Memory aid: "0.0591 over n" at 25°C — solids/liquids = 1.

Exam

Ex 13.7: Ag⁺(0.1 M)|Ag; E°=0.80 V

E = 0.80 − 0.0591 log(1/0.1)

Answer: E = 0.741 V

Gibbs Energy — ΔG° = −nFE°

W_max = −nFE° · Negative ΔG° → spontaneous cell reaction

Example 13.9

Daniell cell E°=1.1 V, n=2 → ΔG° = −2×96500×1.1 = −212.3 kJ

Section 2: Detailed Definitions

Oxidation / Reduction: Loss / gain of electrons. Oxidant accepts e⁻; reductant donates e⁻.

Oxidation number: Apparent charge when shared electrons assigned to more electronegative atom. Elemental form = 0; sum in neutral compound = 0.

Electrolytic cell: Electrical energy → chemical (non-spontaneous). Anode (+), cathode (−).

Galvanic (voltaic) cell: Chemical energy → electrical (spontaneous). Anode (−), cathode (+).

SHE: Standard hydrogen electrode; E° = 0 by definition. Reference for all electrode potentials.

Electrochemical series: Electrodes arranged by increasing E°_red. Top = strong reductants (Li); bottom = strong oxidants (F₂).

Section 3: Diagrams & Visuals

Electrochemistry Map — L13 κ, Λ_m, Kohlrausch Faraday w=zIt E°_cell Nernst Galvanic: Zn|Zn²⁺||Cu²⁺|Cu · Anode oxid · Cathode red ΔG° = −nFE° · Batteries · Fuel cells · Corrosion (rust) ECS: metal above H₂ displaces acid H₂ · positive E°_cell = feasible

Conductance → electrolysis → galvanic cells → Nernst & ΔG

ELECTROLYSIS PRODUCTS (aqueous, Pt) ═══════════════════════════════════════ aq NaCl: cathode H₂ · anode Cl₂ (overpotential) aq CuSO₄: cathode Cu · anode O₂ Cu electrodes + CuSO₄: refining (Cu dissolves anode) molten NaCl: Na + Cl₂ ═══════════════════════════════════════

Section 5: Comprehensive Q&A (12 Questions)

Q1: Oxidation number of S in H₂SO₄?

2(+1) + x + 4(−2) = 0 → x = +6.

Q2: Why does Λ_m increase on dilution for weak acids?

Degree of ionization α increases (Ostwald) → more ions → sharp rise in Λ_m. Strong electrolytes rise only due to weaker inter-ionic attraction.

Q3: Faraday's first law?

Mass of substance liberated ∝ quantity of electricity passed. w = zIt.

Q4: Difference between electrolytic and galvanic cell?

Electrolytic: electricity drives non-spontaneous reaction (anode +). Galvanic: spontaneous redox produces electricity (anode −).

Q5: Role of salt bridge?

Completes circuit ionically; prevents charge build-up; keeps solutions electrically neutral without mixing half-cell electrolytes.

Q6: Why is E° of SHE zero?

By convention — reference electrode. All other potentials measured relative to 2H⁺ + 2e⁻ ⇌ H₂ at 1 bar, [H⁺]=1 M.

Q7: Can Cu displace H₂ from acid?

No — Cu is below H₂ in electrochemical series (E°_Cu > 0). Zn, Mg above H₂ can liberate H₂.

Q8: Feasibility of Cu²⁺ + 2Ag → Cu + 2Ag⁺?

E°_cell = 0.34 − 0.80 = −0.46 V → not feasible. Reverse reaction is spontaneous.

Q9: Nernst for Daniell cell?

E = E° − (0.0591/2) log([Zn²⁺]/[Cu²⁺]). Only ions in Q; solids = 1.

Q10: Lead storage battery net discharge?

Pb + PbO₂ + 2H₂SO₄ → 2PbSO₄ + 2H₂O. Recharge reverses reaction.

Q11: ΔG° and cell emf?

ΔG° = −nFE°. Negative ΔG° (positive E°) means spontaneous galvanic cell.

Q12: Corrosion of iron — electrochemical?

Anodic: Fe → Fe²⁺ + 2e⁻. Cathodic: O₂ + 4H⁺ + 4e⁻ → 2H₂O. E°_cell ≈ 1.67 V spontaneous. Needs O₂ + H₂O; accelerated by acids.

Section 6: Tips, Tricks & Exam Hacks

Memory Aids

  • Anode: "AN OX" — anode oxidation
  • Cathode: "RED CAT" — reduction cathode
  • 1 F: 96500 C ≈ 1 mol e⁻
  • Nernst 25°C: 0.0591/n
  • ECS: top metals reduce bottom ions

Exam Tips

  • Always use reduction potentials for E°_cell
  • n = electrons transferred in balanced cell reaction
  • Overpotential: aq NaCl gives Cl₂ not O₂ at anode
  • Kohlrausch: only at infinite dilution
  • ΔG in J: F=96500, E in V

Section 7: Connections & Relationships

Builds on: L12 ionic equilibrium (ions, conductivity), L9 thermodynamics (ΔG), L7 solutions.
Leads to: industrial electrolysis, batteries, corrosion prevention, metallurgy (Cu refining).
Related: Faraday links mass to charge; Nernst links concentration to emf; ΔG links chemistry to electricity.

Section 8: Complete Quick Reference

• κ = L × cell constant · Λm = 1000κ/M · Λm° = Σ ionic contributions (Kohlrausch)

• w = zIt · 1 F = 96500 C · m₁/m₂ = E₁/E₂ (series)

• E°_cell = E°_cath − E°_an · E = E° − (0.0591/n) log Q

• ΔG° = −nFE° · Anode = oxidation · Cathode = reduction

Cells: Dry cell ~1.5 V · Lead-acid · H₂–O₂ fuel cell · Daniell Zn–Cu

Remember: ✓ Positive E°_cell spontaneous ✓ SHE = 0 ✓ Metals above H₂ liberate H₂ ✓ Corrosion needs O₂+H₂O

PYQ — Previous Year Questions

Extracted from NIOS Chemistry (313) board exam papers in your PDF. Chapter L13 — Electrochemistry only. Use Model Answer for marking points; Explanation for concept clarity.

L13 — Electrochemistry

8 question(s) · Sources: 313/MAY/205A, 313/MAY/205B, 313/MAY/205C, 313/TUS/105A

Section A — MCQ / Objective (from papers)

PYQ1. Write True (T) for correct statement and False (F) for incorrect statement (out of four attempt any two) : In a galvanic cell, electrons always flow from cathode to anode. Salt bridge is a contact between two half-cells without any mixing of electrolytes. Higher the valency of the ion, greater is its conducting power. Conductivity of a cell is the product of conductance and cell constant. ghr

2 marks · Q20 · 313/TUS/105A

Model Answer

Answer using key concepts from L13 (definitions, equations, and one example where useful). Stay within the suggested word range for a 2-mark NIOS question.

Explanation

Cross-check with L13 notes. Structure: definition/law → working → conclusion. Partial marks for correct equations even if explanation is short.

How to write for NIOS: Use 30–50 words (VSA) or short objective. Open with definition/equation, then reason, end with conclusion. Paper 313/TUS/105A · Q20 · 2 mark(s) · L13.

Section B — Short / Long answer (from papers)

PYQ2. The conductivity of 0·00241 M acetic acid is 7·896 × 105 S cm–1. Calculate the molar conductivity. Calculate the standard cell potential of the galvanic cell in which the following reaction takes place : 2Cr(s) 3Cd(s) Given, 0·00241 M Egr{Q>H$ Aåb

2 marks · Q33 · 313/MAY/205A

Model Answer

Answer using key concepts from L13 (definitions, equations, and one example where useful). Stay within the suggested word range for a 2-mark NIOS question.

Explanation

Cross-check with L13 notes. Structure: definition/law → working → conclusion. Partial marks for correct equations even if explanation is short.

How to write for NIOS: Use 30–50 words (VSA) or short objective. Open with definition/equation, then reason, end with conclusion. Paper 313/MAY/205A · Q33 · 2 mark(s) · L13.

PYQ3. Define corrosion. Which environmental conditions cause rusting? Write the reactions for anodic process and cathodic process taking place during rusting. Write the expression for standard Gibbs’ energy for the reaction occurring in a Daniell cell Explain the meaning of all the quantities represented in it. g§jmaU

3 marks · Q40 · 313/MAY/205A

Model Answer

State the precise definition from the L13 notes in 1–2 sentences, include formula/example if marks ≥ 2, and avoid extra theory beyond the ask.

Explanation

Definition questions score for accuracy of wording + one supporting point/example. Do not write full chapter summaries.

How to write for NIOS: Use 50–80 words with equation + reason. Open with definition/equation, then reason, end with conclusion. Paper 313/MAY/205A · Q40 · 3 mark(s) · L13.

PYQ4. The conductivity of 0·00241 M acetic acid is 7·896 × 105 S cm–1. Calculate the molar conductivity. Calculate the standard cell potential of the galvanic cell in which the following reaction takes place : 2Cr(s) 3Cd(s) Given, 0·00241 M Egr{Q>H$ Aåb

2 marks · Q35 · 313/MAY/205B

Model Answer

Answer using key concepts from L13 (definitions, equations, and one example where useful). Stay within the suggested word range for a 2-mark NIOS question.

Explanation

Cross-check with L13 notes. Structure: definition/law → working → conclusion. Partial marks for correct equations even if explanation is short.

How to write for NIOS: Use 30–50 words (VSA) or short objective. Open with definition/equation, then reason, end with conclusion. Paper 313/MAY/205B · Q35 · 2 mark(s) · L13.

PYQ5. Define corrosion. Which environmental conditions cause rusting? Write the reactions for anodic process and cathodic process taking place during rusting. Write the expression for standard Gibbs’ energy for the reaction occurring in a Daniell cell. Explain the meaning of all the quantities represented in it. g§jmaU

3 marks · Q38 · 313/MAY/205B

Model Answer

State the precise definition from the L13 notes in 1–2 sentences, include formula/example if marks ≥ 2, and avoid extra theory beyond the ask.

Explanation

Definition questions score for accuracy of wording + one supporting point/example. Do not write full chapter summaries.

How to write for NIOS: Use 50–80 words with equation + reason. Open with definition/equation, then reason, end with conclusion. Paper 313/MAY/205B · Q38 · 3 mark(s) · L13.

PYQ6. The conductivity of 0·00241 M acetic acid is 7·896 × 105 S cm–1. Calculate the molar conductivity. Calculate the standard cell potential of the galvanic cell in which the following reaction takes place : 2Cr(s) 3Cd(s) Given, 0·00241 M Egr{Q>H$ Aåb

2 marks · Q35 · 313/MAY/205C

Model Answer

Answer using key concepts from L13 (definitions, equations, and one example where useful). Stay within the suggested word range for a 2-mark NIOS question.

Explanation

Cross-check with L13 notes. Structure: definition/law → working → conclusion. Partial marks for correct equations even if explanation is short.

How to write for NIOS: Use 30–50 words (VSA) or short objective. Open with definition/equation, then reason, end with conclusion. Paper 313/MAY/205C · Q35 · 2 mark(s) · L13.

PYQ7. Define corrosion. Which environmental conditions cause rusting? Write the reactions for anodic process and cathodic process taking place during rusting. Write the expression for standard Gibbs’ energy for the reaction occurring in a Daniell cell. Explain the meaning of all the quantities represented in it. g§jmaU

3 marks · Q38 · 313/MAY/205C

Model Answer

State the precise definition from the L13 notes in 1–2 sentences, include formula/example if marks ≥ 2, and avoid extra theory beyond the ask.

Explanation

Definition questions score for accuracy of wording + one supporting point/example. Do not write full chapter summaries.

How to write for NIOS: Use 50–80 words with equation + reason. Open with definition/equation, then reason, end with conclusion. Paper 313/MAY/205C · Q38 · 3 mark(s) · L13.

PYQ8. Explain the electrolysis of aqueous copper sulphate using platinum electrodes. ßb¡{Q>Z‘ Bbo³Q´>moS>m|

2 marks · Q34 · 313/TUS/105A

Model Answer

Give the chemical reason linked to structure/bonding/equilibrium. Start with the principle, then apply to the species named in the question.

Explanation

Reasoning marks require principle + application. Cite electron effects, stability, or Le Chatelier as relevant.

How to write for NIOS: Use 30–50 words (VSA) or short objective. Open with definition/equation, then reason, end with conclusion. Paper 313/TUS/105A · Q34 · 2 mark(s) · L13.

Problem Solving — L13 Electrochemistry

Six problems spanning this chapter’s NIOS syllabus. Every question is built from the notes and formula sheet: solve with equations first, then read the formal textbook-style write-up, the easy explanation, and the topic in depth (formulas, meaning, exam tips). If the question says draw, a labelled pencil sketch is provided. Explanations open by default.

Question 1 of 6Cell

Draw a labelled sketch of a galvanic cell (anode, cathode, salt bridge). Where does oxidation occur?

Oxidation at anode
E°_cell = E°_red(cathode) − E°_red(anode)

Pencil sketch (labelled)

Galvanic cell (schematic) wire / V anode (−) cathode (+) salt bridge
Pencil sketch: galvanic cell electrodes labelled

Solution — step by step with formulas

  1. Oxidation at anode; reduction at cathode; salt bridge maintains charge balance.

Final answer: Oxidation at anode

Formulas used in this problem

Oxidation at anode
E°_cell = E°_red(cathode) − E°_red(anode)

Textbook formal language

A galvanic cell converts free energy of a spontaneous redox reaction into electrical work.

Working formulas: Oxidation at anode; E°_cell = E°_red(cathode) − E°_red(anode). State the definition or law first (NIOS style), use SI units, and box the final numerical answer with unit.

Easy language (same idea, plain words)

Anode is the ‘source’ of electrons (metal dissolves/oxidises); cathode is where electrons are used to reduce ions.

Read once for the idea, once for the numbers. Write the formula, substitute, then simplify. Check whether you used moles, grams, or litres correctly.

Topic in depth — Galvanic cell

Electron flow: anode → cathode in the external circuit.

Linked to chapter notes (L13). Remember: Oxidation at anode; E°_cell = E°_red(cathode) − E°_red(anode). Most exam errors are unit mix-ups (g vs mol, mL vs L) or wrong mole ratios from the equation.

Exam tip

Write Oxidation at anode; E°_cell = E°_red(cathode) − E°_red(anode) before substituting. Keep three significant figures until the end when data allow.

Common mistakes

  • Confusing mass (g) with amount of substance (mol).
  • Forgetting Avogadro’s number unit mol⁻¹ or STP volume 22.7 L mol⁻¹ (1 bar).
  • Using wrong mole ratio from the balanced equation.
  • Mixing up empirical and molecular formulas.
Question 2 of 6Nernst

Write the Nernst equation at 25°C for a cell reaction involving n electrons.

E = E° − (0.059/n) log Q (25°C)

Solution — step by step with formulas

  1. E = E° − (0.059/n) log Q.

Final answer: E = E° − (0.059/n) log Q

Formulas used in this problem

E = E° − (0.059/n) log Q (25°C)

Textbook formal language

Nernst equation gives emf when concentrations/pressures are non-standard.

Working formulas: E = E° − (0.059/n) log Q (25°C). State the definition or law first (NIOS style), use SI units, and box the final numerical answer with unit.

Easy language (same idea, plain words)

If Q is large (lots of products), voltage drops from E°.

Read once for the idea, once for the numbers. Write the formula, substitute, then simplify. Check whether you used moles, grams, or litres correctly.

Topic in depth — Nernst equation

At equilibrium E = 0 and Q = K.

Linked to chapter notes (L13). Remember: E = E° − (0.059/n) log Q (25°C). Most exam errors are unit mix-ups (g vs mol, mL vs L) or wrong mole ratios from the equation.

Exam tip

Write E = E° − (0.059/n) log Q (25°C) before substituting. Keep three significant figures until the end when data allow.

Common mistakes

  • Confusing mass (g) with amount of substance (mol).
  • Forgetting Avogadro’s number unit mol⁻¹ or STP volume 22.7 L mol⁻¹ (1 bar).
  • Using wrong mole ratio from the balanced equation.
  • Mixing up empirical and molecular formulas.
Question 3 of 6Conductance

Define molar conductivity and its SI unit idea.

Λ_m = κ / c

Solution — step by step with formulas

  1. Λ_m = κ/c; conductivity of solution per unit concentration.

Final answer: Λ_m = κ/c

Formulas used in this problem

Λ_m = κ / c

Textbook formal language

Molar conductivity characterises electrolyte conduction per mole of solute.

Working formulas: Λ_m = κ / c. State the definition or law first (NIOS style), use SI units, and box the final numerical answer with unit.

Easy language (same idea, plain words)

How well one mole of salt helps carry current when dissolved in a given volume.

Read once for the idea, once for the numbers. Write the formula, substitute, then simplify. Check whether you used moles, grams, or litres correctly.

Topic in depth — Molar conductivity

Λ_m rises on dilution for strong electrolytes (Kohlrausch).

Linked to chapter notes (L13). Remember: Λ_m = κ / c. Most exam errors are unit mix-ups (g vs mol, mL vs L) or wrong mole ratios from the equation.

Exam tip

Write Λ_m = κ / c before substituting. Keep three significant figures until the end when data allow.

Common mistakes

  • Confusing mass (g) with amount of substance (mol).
  • Forgetting Avogadro’s number unit mol⁻¹ or STP volume 22.7 L mol⁻¹ (1 bar).
  • Using wrong mole ratio from the balanced equation.
  • Mixing up empirical and molecular formulas.
Question 4 of 6Faraday

State Faraday’s first law of electrolysis.

m ∝ I t
F ≈ 96500 C mol⁻¹

Solution — step by step with formulas

  1. Mass of substance deposited/liberated is proportional to charge passed (m ∝ It).

Final answer: m ∝ charge (I×t)

Formulas used in this problem

m ∝ I t
F ≈ 96500 C mol⁻¹

Textbook formal language

Faraday related chemical change to quantity of electricity.

Working formulas: m ∝ I t; F ≈ 96500 C mol⁻¹. State the definition or law first (NIOS style), use SI units, and box the final numerical answer with unit.

Easy language (same idea, plain words)

More current and more time plate more metal.

Read once for the idea, once for the numbers. Write the formula, substitute, then simplify. Check whether you used moles, grams, or litres correctly.

Topic in depth — Faraday’s laws

1 F deposits one gram-equivalent.

Linked to chapter notes (L13). Remember: m ∝ I t; F ≈ 96500 C mol⁻¹. Most exam errors are unit mix-ups (g vs mol, mL vs L) or wrong mole ratios from the equation.

Exam tip

Write m ∝ I t; F ≈ 96500 C mol⁻¹ before substituting. Keep three significant figures until the end when data allow.

Common mistakes

  • Confusing mass (g) with amount of substance (mol).
  • Forgetting Avogadro’s number unit mol⁻¹ or STP volume 22.7 L mol⁻¹ (1 bar).
  • Using wrong mole ratio from the balanced equation.
  • Mixing up empirical and molecular formulas.
Question 5 of 6Battery

Distinguish primary and secondary cells with one example each.

Solution — step by step with formulas

  1. Primary non-rechargeable (dry cell); secondary rechargeable (lead-acid, Li-ion).

Final answer: Primary: dry cell; secondary: lead storage

Textbook formal language

Secondary cells reverse the cell reaction during charging.

Working formulas: (see solution steps). State the definition or law first (NIOS style), use SI units, and box the final numerical answer with unit.

Easy language (same idea, plain words)

Throwaway vs plug-in recharge.

Read once for the idea, once for the numbers. Write the formula, substitute, then simplify. Check whether you used moles, grams, or litres correctly.

Topic in depth — Cells

Fuel cells: continuous fuel/oxidant supply.

Linked to chapter notes (L13). Remember: (see solution steps). Most exam errors are unit mix-ups (g vs mol, mL vs L) or wrong mole ratios from the equation.

Exam tip

Write (see solution steps) before substituting. Keep three significant figures until the end when data allow.

Common mistakes

  • Confusing mass (g) with amount of substance (mol).
  • Forgetting Avogadro’s number unit mol⁻¹ or STP volume 22.7 L mol⁻¹ (1 bar).
  • Using wrong mole ratio from the balanced equation.
  • Mixing up empirical and molecular formulas.
Question 6 of 6Corrosion

Explain rusting of iron as an electrochemical process briefly.

Solution — step by step with formulas

  1. Anodic Fe → Fe²⁺ + 2e⁻; cathodic O₂ reduction in presence of water; rust forms.

Final answer: Anodic oxidation of Fe with cathodic O₂/H₂O reduction

Textbook formal language

Corrosion sets up local galvanic cells on the metal surface.

Working formulas: (see solution steps). State the definition or law first (NIOS style), use SI units, and box the final numerical answer with unit.

Easy language (same idea, plain words)

Wet iron loses electrons and turns to rust while oxygen is reduced.

Read once for the idea, once for the numbers. Write the formula, substitute, then simplify. Check whether you used moles, grams, or litres correctly.

Topic in depth — Rusting

Prevention: paint, galvanising, sacrificial anode.

Linked to chapter notes (L13). Remember: (see solution steps). Most exam errors are unit mix-ups (g vs mol, mL vs L) or wrong mole ratios from the equation.

Exam tip

Write (see solution steps) before substituting. Keep three significant figures until the end when data allow.

Common mistakes

  • Confusing mass (g) with amount of substance (mol).
  • Forgetting Avogadro’s number unit mol⁻¹ or STP volume 22.7 L mol⁻¹ (1 bar).
  • Using wrong mole ratio from the balanced equation.
  • Mixing up empirical and molecular formulas.