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Electrochemistry

NCERT Class 12 · Chemistry Based on NCERT Class 12 Chemistry textbook · Free CBSE study kit

Chapter Notes

Electrochemical Cells

**Definition**: An electrochemical cell is a device that either converts chemical energy into electrical energy (galvanic cell) or uses electrical energy to drive non-spontaneous chemical reactions (electrolytic cell).

**Key distinction**: In a galvanic (voltaic) cell, a spontaneous redox reaction produces electrical energy. In an electrolytic cell, external electrical energy drives a non-spontaneous reaction.

**Example**: The Daniell cell converts the spontaneous reaction Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s) into electrical energy, generating an emf of 1.1 V at standard conditions.

**Important exam points**:

  • Galvanic cells have positive emf values and operate spontaneously
  • Electrolytic cells require external voltage greater than the cell's emf to operate
  • When applied external voltage equals the cell emf, no reaction occurs (equilibrium state)
  • Electron flow is always from anode to cathode through external circuit
  • Current flow (conventional) is opposite to electron flow
  • Galvanic Cells: Construction and Function

    **Cell components**:

  • Two half-cells, each containing a metallic electrode immersed in an electrolyte
  • External circuit: metallic wire connecting electrodes through voltmeter and switch
  • Internal connection: salt bridge connecting the two electrolyte solutions (or single electrolyte if both electrodes are in same solution)
  • **Half-cell reactions** (using Daniell cell example):

  • **Reduction half-reaction (Cathode)**: Cu²⁺(aq) + 2e⁻ → Cu(s)
  • **Oxidation half-reaction (Anode)**: Zn(s) → Zn²⁺(aq) + 2e⁻
  • **Electrode definition**:

  • **Anode**: Electrode where oxidation occurs; negative with respect to solution in galvanic cell
  • **Cathode**: Electrode where reduction occurs; positive with respect to solution in galvanic cell
  • **Cell notation convention** (IUPAC):

  • Format: Anode(left) | Anode electrolyte || Cathode electrolyte | Cathode(right)
  • Example: Zn(s)|Zn²⁺(aq)||Cu²⁺(aq)|Cu(s)
  • Double vertical line (||) represents salt bridge
  • **Cell potential (emf)**:

    The potential difference between cathode and anode, measured in volts when no current flows:

    **E°cell = E°cathode − E°anode = E°right − E°left**

    **Exam-critical points**:

  • Salt bridge prevents charge accumulation and maintains electrical neutrality
  • Electron flow direction: anode → cathode (through external wire)
  • Current direction: cathode → anode (opposite to electron flow)
  • Standard conditions: all ion concentrations = 1 M, gas pressure = 1 bar, T = 298 K
  • Standard Electrode Potential and Standard Hydrogen Electrode

    **Standard Electrode Potential (E°)**: The potential of a half-cell measured against the standard hydrogen electrode under standard conditions (all concentrations 1 M, gases 1 bar pressure, 298 K).

    **Standard Hydrogen Electrode (SHE)**:

  • Representation: Pt(s) | H₂(g, 1 bar) | H⁺(aq, 1 M)
  • Half-reaction: H⁺(aq) + e⁻ → ½H₂(g)
  • **Assigned potential: 0.00 V at all temperatures** (reference electrode)
  • Construction: Platinum electrode coated with platinum black, dipped in 1 M H⁺ solution with pure H₂ gas bubbling through
  • **Measuring standard electrode potential**:

    When a half-cell is connected with SHE as anode and the half-cell as cathode, the measured emf equals the standard electrode potential of that half-cell:

    **E°cell = E°half-cell − E°SHE = E°half-cell − 0 = E°half-cell**

    **Examples from standard table**:

  • Cu²⁺(aq) + 2e⁻ → Cu(s): E° = +0.34 V (Cu²⁺ more easily reduced than H⁺)
  • Zn²⁺(aq) + 2e⁻ → Zn(s): E° = −0.76 V (H⁺ more easily reduced than Zn²⁺)
  • **Interpretation of electrode potential values**:

  • **Positive E°**: Reduced form more stable than H₂ gas; species is better oxidizing agent than H⁺
  • **Negative E°**: Oxidized form more stable; species is stronger reducing agent than H⁺/H₂ couple
  • **Trends in standard electrode potentials table**:

  • F₂/F⁻ has highest E° (+2.87 V): strongest oxidizing agent in aqueous solution
  • Li⁺/Li has lowest E° (−3.05 V): strongest reducing agent
  • As E° increases: oxidizing power increases, reducing power of right-side species decreases
  • As E° decreases (going down): reducing power of left-side species increases
  • **Real-life application**: Cu does not dissolve in dilute HCl because E°(H⁺/H₂) = 0.00 V > E°(Cu²⁺/Cu) = +0.34 V, meaning H⁺ cannot oxidize Cu. However, HNO₃ oxidizes Cu because NO₃⁻ is a stronger oxidizing agent than H⁺.

    **Exam-important fact**: Standard electrode potentials are reduction potentials—they always represent the reduction half-reaction. To find cell emf: E°cell = E°(cathode, reduction) − E°(anode, oxidation)

    Nernst Equation

    **Definition**: The Nernst equation relates electrode potential under non-standard conditions to the standard electrode potential and the concentration of reactants and products.

    **For a single electrode** (reduction half-reaction: Mⁿ⁺ + ne⁻ → M):

    **E = E° − (RT/nF) ln(1/[Mⁿ⁺])**

    Or equivalently:

    **E = E° − (RT/nF) ln[M]/[Mⁿ⁺]** (concentration of solid M = 1)

    **For a complete cell reaction**:

    **Ecell = E°cell − (RT/nF) ln Q**

    Where:

  • **R** = Gas constant = 8.314 J K⁻¹ mol⁻¹
  • **F** = Faraday constant = 96,487 C mol⁻¹ (charge of 1 mole of electrons)
  • **T** = Temperature in Kelvin
  • **n** = Number of electrons transferred
  • **Q** = Reaction quotient = [Products]/[Reactants] with stoichiometric coefficients
  • **At 298 K, converting to base-10 logarithm**:

    **E = E° − (0.0592/n) log Q**

    **Nernst equation for Daniell cell** (n = 2):

    **Ecell = 1.10 − (0.0592/2) log([Zn²⁺]/[Cu²⁺])**

    **Nernst equation for Ni|Ni²⁺||Ag⁺|Ag cell** (n = 2):

    **Ecell = E°cell − (0.0592/2) log([Ni²⁺]/[Ag⁺]²)**

    **Important observations**:

  • Ecell increases with increasing [Cu²⁺] and decreasing [Zn²⁺] in Daniell cell
  • When Q = 1 (standard conditions), Ecell = E°cell
  • When Q = K (equilibrium), Ecell = 0
  • The equation shows how concentration changes affect cell potential
  • **Example calculation**: For Mg(s) + 2Ag⁺(0.0001 M) → Mg²⁺(0.130 M) + 2Ag(s)

  • Q = [Mg²⁺]/[Ag⁺]² = 0.130/(0.0001)² = 1.3 × 10⁷
  • E°cell = E°(Ag⁺/Ag) − E°(Mg²⁺/Mg) = 0.80 − (−2.36) = 3.16 V
  • Ecell = 3.16 − (0.0592/2) log(1.3 × 10⁷) = 3.16 − 0.224 = 2.94 V
  • Relationship Between E°cell, ΔG°, and Equilibrium Constant

    **Thermodynamic relationship**:

    The Gibbs free energy change is directly related to the electrical work done by the cell:

    **ΔG° = −nFE°cell**

    Where:

  • **ΔG°** = Standard Gibbs free energy change
  • **n** = Number of moles of electrons transferred
  • **F** = Faraday constant
  • **E°cell** = Standard cell potential
  • **Derivation of relationship with Kc**:

    At equilibrium, Ecell = 0 and Q = Kc, so:

    0 = E°cell − (0.0592/n) log Kc

    Rearranging:

    **E°cell = (0.0592/n) log Kc**

    Or in terms of natural logarithm:

    **E°cell = (2.303RT/nF) ln Kc**

    **Interpretation**:

  • **E°cell > 0**: ΔG° < 0; reaction is spontaneous; Kc > 1
  • **E°cell = 0**: ΔG° = 0; system at equilibrium; Kc = 1
  • **E°cell < 0**: ΔG° > 0; reaction is non-spontaneous; Kc < 1
  • **Example**: For Daniell cell reaction:

  • E°cell = 1.10 V, n = 2
  • log Kc = (1.10 × 2)/0.0592 = 37.16
  • Kc = 1.45 × 10³⁷ (extremely large, reaction goes nearly to completion)
  • **Exam importance**: This relationship allows calculation of equilibrium constants from electrochemical data—useful when direct measurement is difficult.

    Electrolytic Cells

    **Definition**: An electrolytic cell uses external electrical energy to drive non-spontaneous chemical reactions.

    **Comparison with galvanic cells**:

    | Feature | Galvanic | Electrolytic |

    |---------|----------|-------------|

    | **Spontaneity** | Spontaneous (ΔG° < 0) | Non-spontaneous (ΔG° > 0) |

    | **E°cell** | Positive | Negative |

    | **Anode** | Negative (oxidation occurs) | Positive (oxidation occurs) |

    | **Cathode** | Positive (reduction occurs) | Negative (reduction occurs) |

    | **External voltage** | None needed | Eext > E°cell required |

    | **Power source** | Produces electrical energy | Consumes electrical energy |

    **Operating principle**: When external voltage applied equals cell emf, reaction stops (I = 0). When external voltage exceeds cell emf (Eext > E°cell), reaction reverses, and the cell now functions as an electrolytic cell.

    **Electrolysis of ionic solutions**: The actual products depend on the relative positions of ions in the electrochemical series and their concentrations. Ions that are easier to oxidize/reduce (higher/lower E° values) preferentially react.

    **Exam-critical concept**: In an electrolytic cell, the anode is connected to the positive terminal of the external source, and the cathode to the negative terminal—opposite to the galvanic cell configuration.

    Faraday's Laws of Electrolysis

    **First Law of Electrolysis**: The amount of substance deposited (or dissolved) at an electrode during electrolysis is directly proportional to the quantity of electricity passed through the cell.

    Mathematically: **m ∝ Q**, or **m = (M × Q)/(n × F)**

    Where:

  • **m** = mass of substance deposited (in grams)
  • **M** = Molar mass of the substance (g/mol)
  • **Q** = Total charge passed (in coulombs)
  • **n** = Number of electrons involved in the reduction/oxidation
  • **F** = Faraday constant = 96,487 C/mol
  • Also expressed in terms of current and time:

    **m = (M × I × t)/(n × F)**

    Where **I** = current (in amperes) and **t** = time (in seconds)

    **Second Law of Electrolysis**: When the same amount of charge is passed through different electrolytes, the masses of substances deposited at the electrodes are proportional to their chemical equivalent weights (molar mass/number of electrons).

    **m₁/m₂ = (M₁/n₁)/(M₂/n₂)**

    Or equivalently: **m₁/m₂ = (Eq₁)/(Eq₂)** where Eq = equivalent weight

    **Practical applications**:

  • Electrorefining of metals (purification)
  • Electroplating (coating objects with metal)
  • Electrolytic production of chemicals (NaOH, Cl₂, F₂)
  • Metal extraction from ores
  • **Example problem**: Calculate mass of Cu deposited when 5 A current passes through CuSO₄ solution for 2 hours.

  • n = 2 (Cu²⁺ + 2e⁻ → Cu)
  • Q = I × t = 5 A × (2 × 3600 s) = 36,000 C
  • m = (64 × 36,000)/(2 × 96,487) = 11.9 g
  • Conductance and Conductivity

    **Electrical conductivity (κ)**: The ability of a solution to conduct electric current; measured in siemens per meter (S m⁻¹) or siemens per centimeter (S cm⁻¹).

    **Resistance (R)** and **Resistivity (ρ)** relationship:

    **R = ρ × (l/A)**

    Where:

  • **l** = length of conductor (cm)
  • **A** = cross-sectional area (cm²)
  • **ρ** = resistivity (Ω cm)
  • **Conductance (G)**: Reciprocal of resistance:

    **G = 1/R** (measured in siemens, S)

    **Conductivity (κ)**: Reciprocal of resistivity:

    **κ = 1/ρ** (measured in S cm⁻¹)

    Therefore: **κ = G × (l/A)** or **G = κ × (A/l)**

    **Molar conductivity (Λₘ)**: The conductivity of solution containing 1 mole of electrolyte; measured in S cm² mol⁻¹.

    **Λₘ = κ × (1000/c)**

    Where:

  • **c** = concentration of electrolyte in mol L⁻¹ (or M)
  • The factor 1000 converts cm³ to mL in a liter
  • Alternative expression:

    **Λₘ = (κ/c) × 1000** (when c is in mol dm⁻³)

    **Measurement of conductivity**:

  • Use **conductivity cell** containing two platinum electrodes separated by fixed distance
  • Cell constant (l/A) is determined using standard KCl solution: **Cell constant = κ(KCl) × G(measured)**
  • Then for unknown solution: κ = G(measured) × Cell constant
  • **Variation with concentration**:

  • **Conductivity (κ)** decreases with dilution (fewer ions per unit volume)
  • **Molar conductivity (Λₘ)** increases with dilution (ions have more space, less inter-ionic interference)
  • At infinite dilution (c → 0), molar conductivity approaches maximum value **Λ°ₘ** (molar conductivity at infinite dilution)
  • **Exam-important distinctions**:

  • **Electrolytic (ionic) conductivity**: Due to movement of ions in solution
  • **Electronic conductivity**: Due to movement of free electrons in metals
  • Kohlrausch's Law

    **Statement**: The molar conductivity of an electrolyte at infinite dilution is equal to the sum of molar conductivities of its constituent ions at infinite dilution.

    **Mathematical form**: **Λ°ₘ = λ°₊ + λ°₋**

    Where:

  • **Λ°ₘ** = Molar conductivity at infinite dilution
  • **λ°₊** = Molar conductivity of cation at infinite dilution
  • **λ°₋** = Molar conductivity of anion at infinite dilution
  • **For compounds with polyatomic ions**:

    **Λ°ₘ(NaCl) = λ°(Na⁺) + λ°(Cl⁻)**

    **Λ°ₘ(CaCl₂) = λ°(Ca²⁺) + 2λ°(Cl⁻)**

    **Important applications**:

    1. **Determination of degree of dissociation (α)**:

    **α = Λₘ/Λ°ₘ** (at any concentration)

    Where Λₘ is molar conductivity at concentration c, and Λ°ₘ is at infinite dilution

    2. **Calculation of unknown molar conductivity**:

    If Λ°ₘ of an unknown electrolyte cannot be directly measured, it can be calculated from known ionic conductivities.

    3. **Determination of Ka for weak electrolytes**:

    For weak electrolyte HA:

    **Ka = (α² × c)/(1 − α)** where **α = Λₘ/Λ°ₘ**

    Or: **Ka = (Λₘ)² × c / (Λ°ₘ × (Λ°ₘ − Λₘ))**

    **Example**: For acetic acid (weak acid):

  • If Λₘ = 50 S cm² mol⁻¹ at 0.01 M and Λ°ₘ = 390.5 S cm² mol⁻¹
  • α = 50/390.5 = 0.128
  • Ka = (0.128)² × 0.01 / (1 − 0.128) = 1.87 × 10⁻⁵
  • **Exam-critical points**:

  • Kohlrausch's law applies only at infinite dilution
  • At finite concentrations, interionic forces cause deviations
  • Strong electrolytes follow the law more accurately than weak electrolytes
  • The law is additive—different electrolytes with common ions can be related
  • Batteries

    **Definition**: A battery is a galvanic cell (or combination of cells) that converts chemical energy directly into electrical energy through spontaneous redox reactions.

    Lead Storage Battery (Lead-Acid Battery)

    **Construction**:

  • **Anode**: Lead (Pb) plates containing spongy lead
  • **Cathode**: Lead dioxide (PbO₂) plates
  • **Electrolyte**: Dilute sulfuric acid (H₂SO₄), approximately 4 M
  • **Number of cells**: 6 cells in series (each producing ~2.1 V), total ~12 V
  • **Electrode reactions during discharge**:

    **At anode (oxidation)**:

    Pb(s) + SO₄²⁻ → PbSO₄(s) + 2e⁻

    **At cathode (reduction)**:

    PbO₂(s) + 4H⁺ + SO₄²⁻ + 2e⁻ → PbSO₄(s) + 2H₂O(l)

    **Overall cell reaction**:

    Pb(s) + PbO₂(s) + 2H₂SO₄(aq) → 2PbSO₄(s) + 2H₂O(l)

    **Standard cell potential**: E°cell = 2.1 V per cell

    **Charging process**: When connected to external source, reactions reverse (electrolytic mode):

  • PbSO₄ at anode → Pb + SO₄²⁻
  • PbSO₄ at cathode → PbO₂ + 4H⁺ + SO₄²⁻
  • **Advantages**:

  • High cell potential per unit
  • Can be recharged many times
  • Can deliver large currents due to low internal resistance
  • Relatively inexpensive
  • Reliable and long-lasting
  • **Disadvantages**:

  • Heavy (lead is dense)
  • Cannot be stored indefinitely without maintenance
  • Sulfation occurs if left uncharged
  • Produces hydrogen and oxygen gases during overcharging (explosion hazard)
  • **Exam importance**: Lead storage battery is the most common secondary battery (rechargeable cell) in vehicles.

    Hydrogen-Oxygen Fuel Cell

    **Definition**: A fuel cell is a galvanic cell in which chemical energy of fuel combustion is directly converted to electrical energy without heat generation.

    **Construction of H₂-O₂ fuel cell**:

  • **Anode (negative electrode)**: Porous carbon electrode where H₂ is oxidized
  • **Cathode (positive electrode)**: Porous carbon electrode where O₂ is reduced
  • **Electrolyte**: Acidic or alkaline aqueous solution
  • **Catalyst**: Platinum or other catalyst to increase reaction rate
  • **Electrode reactions in acidic medium**:

    **At anode (oxidation)**:

    H₂(g) → 2H⁺(aq) + 2e⁻ (E° = 0.00 V)

    **At cathode (reduction)**:

    O₂(g) + 4H⁺(aq) + 4e⁻ → 2H₂O(l) (E° = 1.23 V)

    **Overall cell reaction**:

    2H₂(g) + O₂(g) → 2H₂O(l)

    **Standard cell potential**:

    E°cell = 1.23 − 0.00 = 1.23 V

    **Electrode reactions in alkaline medium**:

    **At anode**:

    2H₂(g) + 4OH⁻(aq) → 4H₂O(l) + 4e⁻

    **At cathode**:

    O₂(g) + 2H₂O(l) + 4e⁻ → 4OH⁻(aq)

    **Overall reaction**: Same as acidic medium: 2H₂ + O₂ → 2H₂O

    **Advantages of fuel cells**:

  • High energy density
  • Environmentally clean—only product is water
  • Silent operation (no moving parts)
  • Can operate continuously as long as fuel supplied
  • High efficiency (up to 60-70%)
  • Produce no air pollution or greenhouse gases
  • **Disadvantages**:

  • High initial cost
  • Hydrogen storage and safety concerns
  • Need for catalyst
  • Slow start-up compared to batteries
  • Requires pure hydrogen (expensive purification)
  • **Real-world applications**:

  • Space programs (Apollo missions used H₂-O₂ fuel cells)
  • Electric vehicles
  • Portable power generation
  • Backup power systems
  • Future sustainable energy source
  • **Exam-critical concept**: Fuel cells are non-rechargeable galvanic cells that consume reactants and cannot be regenerated by reversing the reaction with external voltage.

    Corrosion and Its Prevention

    **Definition**: Corrosion is the spontaneous electrochemical degradation of metals and their alloys when exposed to environmental conditions (moisture, oxygen, etc.).

    **Types of corrosion**:

    1. **Electrochemical corrosion** (most common)

    2. **Direct chemical attack**

    **Mechanism of electrochemical corrosion** (rusting of iron example):

    Iron corrodes in presence of water and oxygen:

    **4Fe(s) + 3O₂(g) + 6H₂O(l) → 4Fe(OH)₃(s)** (overall)

    Or: 4Fe(s) + 3O₂(g) + 2H₂O(l) → 4FeO·OH(s) (hydrated iron oxide)

    This involves two half-reactions:

    **Anode reaction (oxidation)**:

    2Fe(s) → 2Fe²⁺(aq) + 4e⁻

    **Cathode reaction (reduction)**:

    O₂(g) + 2H₂O(l) + 4e⁻ → 4OH⁻(aq)

    The Fe²⁺ ions react with OH⁻ ions and oxygen to form rust (Fe₂O₃·nH₂O or FeO·OH).

    **Electrochemical cell formation**:

  • **Anode**: Iron (site of oxidation/corrosion)
  • **Cathode**: Oxide layer, impurities, or nobler metal (site of O₂ reduction)
  • **Electrolyte**: Water with dissolved ions (salt, acids, etc.)
  • **Galvanic corrosion** (when two different metals in contact):

  • Nobler metal (higher E°) acts as cathode (protected)
  • More active metal (lower E°) acts as anode (corroded)
  • This occurs even without oxygen due to potential difference
  • **Factors affecting corrosion rate**:

  • **Oxygen availability**: Increased O₂ increases corrosion (faster cathode reaction)
  • **pH**: Acidic conditions accelerate corrosion
  • **Presence of salts**: Increases conductivity; NaCl especially increases corrosion
  • **Temperature**: Higher temperature increases reaction rates
  • **Nature of metal**: Metals with more negative E° corrode faster
  • **Presence of impurities**: Creates local galvanic cells
  • Prevention of Corrosion

    **Method 1: Coating/Barrier protection**

  • **Paint**: Isolates metal from moisture and oxygen
  • **Oil/Grease**: Prevents water contact
  • **Plastic coating**: Protects from environment
  • **Enameling/Varnish**: Forms protective layer
  • Disadvantage: Must cover entire surface; even small breaks lead to corrosion
  • **Method 2: Sacrificial anode protection (Cathodic protection)**

  • **Principle**: Attach a more active metal (higher reducing power, lower E°) to the structure
  • The active metal oxidizes preferentially while the main metal (cathode) is protected
  • **Examples**:
  • Zinc or magnesium plates attached to iron pipes
  • Zinc coating on steel (galvanizing)
  • Sacrificial anodes on ship hulls
  • **Method 3: Impressed current cathodic protection**

  • **Method**: Apply external voltage to make the structure negative (cathode)
  • Used for pipelines, buried structures, and ships
  • Requires continuous power source
  • **Method 4: Alloying**

  • **Stainless steel**: Fe + Cr + Ni forms protective chromium oxide layer
  • **Aluminum alloys**: Form stable Al₂O₃ layer (self-passivating)
  • These metals develop passive films that prevent further corrosion
  • **Method 5: Corrosion inhibitors**

  • Chemical additives (e.g., silicates, phosphates) that:
  • Form protective films
  • Reduce electrochemical activity
  • Increase pH to reduce corrosion
  • Used in cooling systems and boilers
  • **Method 6: Environmental control**

  • Remove dissolved oxygen (deaeration)
  • Reduce salt concentration
  • Control pH (keep slightly alkaline)
  • Reduce temperature where possible
  • **Real-life example**: Protection of underground pipelines uses combination of:

  • External plastic coating (barrier)
  • Sacrificial magnesium anodes buried along pipeline
  • Impressed current cathodic protection at critical points
  • Corrosion inhibitors in the liquid being transported
  • **Exam-important points**:

  • Corrosion is spontaneous (E°cell > 0, ΔG° < 0) due to formation of lower energy products
  • Rusting of iron is not simple oxidation but an electrochemical process requiring both oxygen and moisture
  • Nobler metals (Au, Ag, Cu) corrode much more slowly than active metals (Fe, Zn)
  • Galvanic series shows relative activity: K > Na > Ca > Mg > Al > Zn > Fe > Ni > Sn > Pb > H > Cu > Hg > Ag > Au
  • ---

    **Summary of Key Formulae for Board Exam**:

    1. **Cell potential**: E°cell = E°cathode − E°anode

    2. **Nernst equation**: Ecell = E°cell − (0.0592/n) log Q (at 298 K)

    3. **Gibbs energy relation**: ΔG° = −nFE°cell

    4. **Equilibrium constant**: E°cell = (0.0592/n) log Kc

    5. **Conductivity**: κ = G × (l/A)

    6. **Molar conductivity**: Λₘ = κ × (1000/c)

    7. **Kohlrausch's law**: Λ°ₘ = λ°₊ + λ°₋

    8. **Degree of dissociation**: α = Λₘ/Λ°ₘ

    9. **Faraday's law**: m = (M × I × t)/(n × F)

    10. **Ka from conductivity**: Ka = (Λₘ)² × c / (Λ°ₘ × (Λ°ₘ − Λₘ))

    MCQs — 10 Questions with Answers

    Q1. In a Daniell cell, the standard cell potential E° = 1.1 V at 25°C. What is ΔG° for the reaction Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s)? (F = 96500 C/mol)

    • A. −212.3 kJ/mol ✓
    • B. −106.15 kJ/mol
    • C. +212.3 kJ/mol
    • D. +106.15 kJ/mol

    Answer: A — ΔG° = −nFE° = −2 × 96500 × 1.1 = −212,300 J/mol = −212.3 kJ/mol; negative value indicates spontaneity.

    Q2. A 0.1 M NaCl solution has conductivity κ = 1.2 × 10⁻² S/cm. What is its molar conductivity?

    • A. 120 S·cm²/mol ✓
    • B. 12 S·cm²/mol
    • C. 1.2 S·cm²/mol
    • D. 0.12 S·cm²/mol

    Answer: A — Molar conductivity Λm = (κ/c) × 1000 = (1.2 × 10⁻²/0.1) × 1000 = 120 S·cm²/mol (factor 1000 converts concentration from mol/cm³ to mol/L).

    Q3. According to Kohlrausch's law, the molar conductivity at infinite dilution Λ°m for an electrolyte depends on:

    • A. only the cation present
    • B. only the anion present
    • C. the sum of limiting molar conductivities of cations and anions ✓
    • D. the concentration of the solution

    Answer: C — Kohlrausch's law states Λ°m = Σ(λ° cation) + Σ(λ° anion); it is independent of concentration and depends on ionic mobilities.

    Q4. In the electrolysis of molten NaCl, 0.5 mol of electrons is passed through the cell. How many moles of Cl₂ gas are produced at the anode?

    • A. 0.25 mol ✓
    • B. 0.5 mol
    • C. 1.0 mol
    • D. 2.0 mol

    Answer: A — At the anode: 2Cl⁻ → Cl₂ + 2e⁻; 0.5 mol electrons produces 0.5/2 = 0.25 mol Cl₂.

    Q5. Which of the following statements about the Nernst equation is NOT correct? (I) Q represents the reaction quotient with all species at their instantaneous concentrations. (II) As Q increases, Ecell decreases. (III) At equilibrium, Ecell = E° because Q = K.

    • A. Only (I) is incorrect
    • B. Only (II) is incorrect
    • C. Only (III) is incorrect ✓
    • D. All three statements are correct

    Answer: C — At equilibrium, Ecell = 0 (not E°) because E° = (0.0592/n)log(K), so Ecell = E° − (0.0592/n)log(K) = 0; statements (I) and (II) are correct.

    Q6. A galvanic cell has E° = +0.34 V. Which statement is correct? (A) The cell reaction is non-spontaneous and ΔG° > 0. (B) The cell reaction is spontaneous and ΔG° < 0. (C) The cell reaction is at equilibrium with ΔG° = 0. (D) The cell potential will increase with temperature.

    • A. The cell reaction is non-spontaneous and ΔG° > 0
    • B. The cell reaction is spontaneous and ΔG° < 0 ✓
    • C. The cell reaction is at equilibrium with ΔG° = 0
    • D. The cell potential will increase with temperature

    Answer: B — Since E° is positive, ΔG° = −nFE° is negative, indicating the reaction is spontaneous; options (A) and (C) contradict this.

    Q7. Calculate the mass of copper deposited at the cathode when 2 A current passes through a copper sulfate solution for 30 minutes. (Atomic mass Cu = 64 g/mol, F = 96500 C/mol)

    • A. 1.19 g ✓
    • B. 2.38 g
    • C. 3.57 g
    • D. 4.76 g

    Answer: A — Charge Q = I × t = 2 × (30 × 60) = 3600 C; moles of Cu = Q/(nF) = 3600/(2 × 96500) = 0.01865 mol; mass = 0.01865 × 64 = 1.19 g.

    Q8. In a galvanic cell, why does molar conductivity increase as the solution becomes more dilute?

    • A. Because ions collide more frequently with solvent molecules and move slower
    • B. Because ions are farther apart and experience less interionic attraction, moving faster ✓
    • C. Because concentration increases and more ions carry charge
    • D. Because the solution becomes more resistive at lower ionic strength

    Answer: B — Upon dilution, ions are more dispersed, interionic attractions decrease, and ions move faster with higher mobility, increasing molar conductivity despite lower total ion concentration.

    Q9. HOTS: A galvanic cell is constructed using Zn and Cu electrodes. Initially, [Zn²⁺] = 0.1 M and [Cu²⁺] = 1.0 M. Given E°Zn²⁺|Zn = −0.76 V and E°Cu²⁺|Cu = +0.34 V, calculate Ecell using the Nernst equation at 25°C. (R = 8.314 J/(mol·K), F = 96500 C/mol)

    • A. 1.03 V ✓
    • B. 1.10 V
    • C. 1.17 V
    • D. 0.95 V

    Answer: A — E°cell = 0.34 − (−0.76) = 1.10 V; Q = [Zn²⁺]/[Cu²⁺] = 0.1/1.0 = 0.1; Ecell = 1.10 − (0.0592/2)log(0.1) = 1.10 − (0.0296 × −1) = 1.10 + 0.0296 ≈ 1.03 V.

    Q10. During the electrolysis of acidified water using inert electrodes, oxygen is produced at the anode and hydrogen at the cathode. If 500 cm³ of O₂ (at STP) is collected, what volume of H₂ (at STP) is produced? (Assume 2 mol H₂O → 2 mol H₂ + 1 mol O₂)

    • A. 250 cm³
    • B. 500 cm³
    • C. 1000 cm³ ✓
    • D. 1500 cm³

    Answer: C — Stoichiometry: 1 mol O₂ : 2 mol H₂; moles O₂ = 500/22400 mol; moles H₂ = 2 × (500/22400) = 1000/22400; volume H₂ = 1000 cm³ at STP.

    Flashcards

    What is the definition of a galvanic cell?

    A galvanic cell is an electrochemical device that converts the chemical energy of a spontaneous redox reaction into electrical energy.

    In a galvanic cell, which electrode is negative and where does oxidation occur?

    The anode is negative and oxidation occurs there; electrons flow from anode to cathode through the external circuit.

    State the Nernst equation and define each symbol.

    Ecell = E° − (RT/nF)ln(Q), where R = gas constant, T = temperature, n = moles of electrons, F = Faraday constant (96500 C/mol), Q = reaction quotient.

    What is molar conductivity and how does it vary with dilution?

    Molar conductivity Λm = κ/c (conductivity ÷ concentration) and it increases with dilution because ions move faster when dispersed.

    State Kohlrausch's law of independent ion migration.

    At infinite dilution, the molar conductivity of an electrolyte is the sum of limiting molar conductivities of its constituent cations and anions: Λ°m = Σ(λ° cation) + Σ(λ° anion).

    What is Faraday's first law of electrolysis?

    The amount of substance deposited or liberated at an electrode is directly proportional to the quantity of charge passed: moles = Q/(nF) where Q is charge and n is electrons transferred.

    How are ΔG° and E° cell related?

    ΔG° = −nFE°cell; if E° is positive, ΔG° is negative and the reaction is spontaneous.

    In an electrolytic cell, which electrode is the anode and what happens there?

    The anode is positive and oxidation occurs there; species lose electrons and are oxidized.

    What is the relationship between conductivity κ and resistivity ρ?

    Conductivity κ is the reciprocal of resistivity: κ = 1/ρ; conductivity measures how well a solution conducts electricity.

    How does applying an external voltage greater than the cell potential change the direction of a galvanic cell reaction?

    When external voltage exceeds cell potential, the reaction reverses and the cell functions as an electrolytic cell, converting electrical energy into chemical energy.

    Important Board Questions

    Define electrode potential and standard electrode potential. Give one example. [2 marks]

    Electrode potential is the potential difference at electrode-electrolyte interface; standard electrode potential is when all species are at unit concentration (1 M) or unit activity. Example: E°Cu²⁺|Cu = +0.34 V means Cu²⁺ + 2e⁻ → Cu at standard conditions.

    Derive the relationship between ΔG°, E° and equilibrium constant K. Show all steps and explain the significance when E° > 0, E° = 0, and E° < 0. [5 marks]

    Start with ΔG° = −nFE° and relate to ΔG° = −RT ln(K); set up equilibrium condition Ecell = 0 = E° − (RT/nF)ln(K); analyse sign of E° against ΔG° and K to interpret spontaneity and direction.

    A silver coulometer (containing AgNO₃ solution with silver electrodes) is connected in series with a copper sulfate cell during electrolysis. When 1.08 g of silver is deposited at the cathode of the coulometer, calculate: (i) the total charge passed through the circuit in coulombs, (ii) the mass of copper deposited at the cathode of the copper sulfate cell. (Atomic masses: Ag = 108 g/mol, Cu = 64 g/mol; F = 96500 C/mol) [6 marks]

    For (i): Ag⁺ + e⁻ → Ag; moles Ag = 1.08/108 = 0.01 mol; charge = 0.01 × F = 965 C. For (ii): In CuSO₄ cell, Cu²⁺ + 2e⁻ → Cu; same charge flows so moles Cu = 965/(2 × 96500) = 0.005 mol; mass Cu = 0.005 × 64 = 0.32 g (show stoichiometric ratio 1 mol e⁻ : Ag but 2 mol e⁻ : Cu).

    Next chapterChemical Kinetics →

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