Explanation: The standard cell potential is calculated using:
E°_cell = E°_cathode – E°_anode
From the cell representation, copper acts as the cathode (right side) and zinc acts as the anode (left side).
E°_cell = +0.34 V – (-0.76 V) = +1.10 V.

Explanation: The maximum electrical work done by a galvanic cell is equal to the decrease in Gibbs free energy of the system.
ΔG° = -nFE°_cell
Where n is the number of moles of electrons transferred, and F is the Faraday constant (~96500 C mol⁻¹). For a spontaneous reaction, E°_cell must be positive, making ΔG° negative.

Explanation: At chemical equilibrium, the cell potential becomes zero (E_cell = 0) and the reaction quotient equals the equilibrium constant (Q_c = K_c).
Substituting these into the Nernst equation:
0 = E°_cell – (0.0591 / n) log K_c ⇒ E°_cell = (0.0591 / n) log K_c.
Rearranging the equation yields: log K_c = nE°_cell / 0.0591. Thus, both statements A and B are mathematically equivalent and correct.

Explanation: Under standard state conditions, the Standard Hydrogen Electrode (SHE) utilizes an acidic solution containing hydronium/hydrogen ions at a unit activity/concentration of **1.0 M**.
Since [H⁺] = 1.0 M, the pH = -log(1) = 0. The electrode potential of SHE is assigned a value of 0.00 V at all temperatures.

Explanation: The quantity l / A is constant for a given conductivity cell and is called the **Cell Constant**, denoted by G*.
Relation: G* = R × κ (where R is resistance and κ is conductivity). Its unit is cm⁻¹ or m⁻¹.

Explanation: Conductivity (κ) is the conductance of a unit volume of cell solution. On dilution, the number of current-carrying ions present per unit volume of the solution decreases. Consequently, the conductivity/specific conductance **decreases** upon dilution.

Explanation: Molar conductivity is defined as Λ_m = κ × V (where V is the volume containing 1 mole of electrolyte). On dilution, although conductivity (κ) decreases, the volume (V) increases much more significantly. Thus, the product Λ_m **increases** with dilution (decrease in concentration).

Explanation: The formula for molar conductivity is Λ_m = (κ × 1000) / M.
Since conductivity (κ) is in S cm⁻¹ and molarity (M) is in mol L⁻¹ (or mol cm⁻³ equivalence), the units resolve to **S cm² mol⁻¹** (or **Ω⁻¹ cm² mol⁻¹**).

Explanation: Kohlrausch’s law states that the limiting molar conductivity of an electrolyte is the sum of the individual ionic conductivities of its cations and anions.
For acetic acid: Λ°_CH₃COOH = λ°_CH₃COO⁻ + λ°_H⁺.
We can obtain this by combining strong electrolytes:
Λ°_CH₃COONa + Λ°_HCl – Λ°_NaCl = (λ°_CH₃COO⁻ + λ°_Na⁺) + (λ°_H⁺ + λ°_Cl⁻) – (λ°_Na⁺ + λ°_Cl⁻) = λ°_CH₃COO⁻ + λ°_H⁺ = Λ°_CH₃COOH.

Explanation: At any concentration c, only a fraction of a weak electrolyte dissociates. Molar conductivity (Λ_m) is proportional to this active fraction. At infinite dilution, dissociation is complete (100%), and molar conductivity reaches its maximum value (Λ°_m).
Therefore, the degree of dissociation α is given by the ratio: α = Λ_m / Λ°_m.

Explanation: Passing 1 Faraday of charge (96500 C) deposits exactly 1 equivalent weight (E) of a substance.
The electrochemical equivalent (z) is the mass deposited by passing exactly 1 Coulomb of charge.
Therefore, z = E / 96500 (where E is the chemical equivalent weight of the substance).

Explanation:
– The reduction half-reaction in acidic medium is: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O.
– According to the stoichiometry, 1 mole of MnO₄⁻ requires 5 moles of electrons for complete reduction.
– Since the charge of 1 mole of electrons is 1 Faraday (1 F), reducing 1 mole of MnO₄⁻ requires exactly **5 F** of electricity.

Explanation:
– Moles of Aluminium required = given mass / atomic mass = 27 g / 27 g/mol = 1 mole.
– The electrode reaction is: Al³⁺ + 3e⁻ → Al(s).
– Depositing 1 mole of Al requires 3 moles of electrons.
– Thus, the charge required is **3 Faradays** (3 F).

Explanation: Only metals situated above hydrogen in the electrochemical series (with negative standard reduction potentials) can displace hydrogen from dilute acids.
Copper has a positive standard reduction potential (E°_Cu²⁺/Cu = +0.34 V) and lies below hydrogen, meaning it is less reactive than hydrogen and **cannot displace** it from dilute acids.

Explanation: Rusting of iron behaves as a miniature electrochemical cell on the metal surface:
– At the **anode (oxidation spot)**, metallic iron dissolves to form ferrous ions: Fe(s) → Fe²⁺(aq) + 2e⁻.
– At the **cathode (reduction spot)**, oxygen gas is reduced in the presence of H⁺ ions: O₂(g) + 4H⁺(aq) + 4e⁻ → 2H₂O(l).

Explanation: During the discharging of a lead accumulator, the overall cell reaction is:
Pb(s) + PbO₂(s) + 2H₂SO₄(aq) → 2PbSO₄(s) + 2H₂O(l).
This reaction consumes **sulfuric acid (H₂SO₄)**, decreasing the density of the electrolyte. Lead sulfate (PbSO₄) is deposited on both electrodes as a product.

Explanation: Fuel cells convert the chemical energy of combustion of fuels directly into electrical energy. In the H₂ – O₂ fuel cell, hydrogen and oxygen gases are bubbled through porous carbon electrodes. The net chemical reaction is the formation of water:
2H₂(g) + O₂(g) → 2H₂O(l).

Explanation: In aqueous NaCl solution, competition occurs between ions:
– At the **cathode**, H⁺ (from water, E° = -0.83 V under basic conditions or 0.00 V) has a much higher reduction potential than Na⁺ (E° = -2.71 V). Thus, **H₂ gas** is evolved.
– At the **anode**, due to the overvoltage of oxygen evolution, chloride ions are oxidized preferentially to produce **Cl₂ gas**.
(Note: Electrolysis of molten NaCl yields Na metal at the cathode and Cl₂ gas at the anode).

Explanation: The Nernst equation for the Daniell cell is:
E_cell = E°_cell – (0.0591 / 2) log ([Zn²⁺] / [Cu²⁺]).
Increasing the concentration of reactant [Cu²⁺] decreases the value of the logarithmic ratio term. Since this term is subtracted, the overall cell potential E_cell **increases**.

Explanation: For strong electrolytes, molar conductivity decreases linearly with the square root of concentration (√c). This relationship is represented by the Debye-Huckel-Onsager equation:
Λ_m = Λ°_m – A √c.
A plot of Λ_m vs √c yields a straight line with y-intercept equal to Λ°_m and slope equal to -A.

Explanation: One Faraday (1 F) is defined as the total electric charge carried by one mole of electrons.
Calculation:
– Charge of 1 electron = 1.602 × 10⁻¹⁹ C
– Avogadro’s number = 6.022 × 10²³ mol⁻¹
– Total charge per mole = 1.602 × 10⁻¹⁹ × 6.022 × 10²³ ≈ 96487 C mol⁻¹ (approximated as **96500 C** for calculations).

Explanation: Cathodic protection involves connecting iron to a more active metal like **Zinc (Zn)** or **Magnesium (Mg)**, which has a more negative reduction potential. This active metal oxidizes preferentially, acting as a “sacrificial anode” and protecting the iron structure from rusting. Coating iron with zinc is specifically termed galvanization.

Explanation: A galvanic cell operates as long as there is a thermodynamic driving force. Over time, reactants are consumed and products accumulate until the system reaches chemical equilibrium (ΔG = 0). At this point, the current ceases to flow, and the cell potential E_cell drops to **0.00 V** (dead/completely discharged cell).

Explanation: Faraday’s Second Law of Electrolysis states that when the same quantity of charge is passed through different electrolytes connected in series, the masses of substances liberated at the electrodes are directly proportional to their **chemical equivalent weights**.
Mathematically: w₁ / w₂ = E₁ / E₂.

Explanation: Formulated by Friedrich Kohlrausch in 1876, the law is formally termed the **”Law of independent migration of ions”**. It states that at infinite dilution (where inter-ionic attractions become negligible), each ion migrates independently of its co-ion and makes a definite contribution to the total molar conductivity of the electrolyte.