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PUCET (Punjab) Electrochemical Series — practice questions — page 2

Explore 110 free Electrochemical Series questions in PUCET (Punjab) Chemistry. Open each question or exercise to read its content, reference answers and available explanation.

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Electrochemical Series questions and exercises with answers

Showing questions 31–60 of 110.

The Faraday constant F = 96500 C mol⁻¹ represents:Read the answer and worked solution →A CuSO₄ solution is electrolysed with 1.5 A for 10 minutes. The mass of copper (M = 63.5) deposited at the cathode is approximately:Read the answer and worked solution →In the electrolysis of **molten** NaCl, the products at the cathode and anode respectively are:Read the answer and worked solution →During electrolysis of **aqueous** NaCl (brine), the products are:Read the answer and worked solution →The electrolyte in a lead storage battery (lead-acid) is approximately:Read the answer and worked solution →In the H₂–O₂ fuel cell used in the Apollo space programme, the electrolyte is:Read the answer and worked solution →Rusting of iron is an electrochemical phenomenon. At the cathodic spot, the reaction is:Read the answer and worked solution →Rust is chemically:Read the answer and worked solution →A device converting CHEMICAL energy to electrical energy via spontaneous redox is a:Read the answer and worked solution →In the Daniell cell, the cathode reaction is:Read the answer and worked solution →EMF of standard Daniell cell at 25°C is approximately:Read the answer and worked solution →Faraday's first law of electrolysis states mass deposited is proportional to:Read the answer and worked solution →The value of Faraday's constant F is approximately:Read the answer and worked solution →In a galvanic cell, electrons flow externally from:Read the answer and worked solution →The Nernst equation gives cell EMF at non-standard conditions. For a half-cell Mⁿ⁺ + ne⁻ → M, it is:Read the answer and worked solution →Relation between standard cell potential E°, ΔG°, and equilibrium constant K:Read the answer and worked solution →For ΔG° < 0 (spontaneous), E° must be:Read the answer and worked solution →The molar conductivity Λ_m of an electrolyte:Read the answer and worked solution →Kohlrausch's law states the limiting molar conductivity of an electrolyte is:Read the answer and worked solution →For the cell Zn | Zn²⁺ (1 M) || Cu²⁺ (1 M) | Cu, with E°(Zn) = -0.76 V, E°(Cu) = +0.34 V, E_cell:Read the answer and worked solution →Standard electrode potential of SHE (Standard Hydrogen Electrode) is taken as:Read the answer and worked solution →For Cu²⁺ + 2e⁻ → Cu, how much charge to deposit 0.5 mol Cu? (F = 96500 C/mol)Read the answer and worked solution →Conductivity κ of pure water is approximately:Read the answer and worked solution →At 25°C, ratio of Nernst factor 0.0591/n to RT/(nF) ln 10 in SI units:Read the answer and worked solution →A current of 2 A flows for 30 min through a CuSO4 solution. Mass of Cu deposited (M_Cu = 63.5):Read the answer and worked solution →For the cell: Zn|Zn²⁺(0.1 M)||Cu²⁺(0.01 M)|Cu, at 25°C, E_cell is:Read the answer and worked solution →Equilibrium constant K of the Daniell cell at 25°C (E° = 1.10 V, n = 2):Read the answer and worked solution →A 0.1 M weak acid HA has Λ_m = 10 S·cm²/mol; at infinite dilution Λ_m^0 = 400. Degree of dissociation α:Read the answer and worked solution →In electrolysis of CuSO4 with Pt electrodes, the cathode product is:Read the answer and worked solution →Standard electrode potentials: E°(Fe³⁺/Fe²⁺) = +0.77 V, E°(I2/I⁻) = +0.54 V. For the reaction 2Fe³⁺ + 2I⁻ → 2Fe²⁺ + I2, E° equals:Read the answer and worked solution →

Worked practice questions

The Faraday constant F = 96500 C mol⁻¹ represents:

  1. The cell potential of a standard cell
  2. The current required for electrolysis
  3. The charge on one mole of electrons (N_A × e = 6.022 × 10²³ × 1.602 × 10⁻¹⁹ C)
  4. The charge on a single electron
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Correct answer: C. The charge on one mole of electrons (N_A × e = 6.022 × 10²³ × 1.602 × 10⁻¹⁹ C)

**1 Faraday (F) = charge on 1 mole of electrons = N_A × e = 6.022 × 10²³ × 1.602 × 10⁻¹⁹ ≈ 96485 C ≈ 96500 C mol⁻¹**.

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A CuSO₄ solution is electrolysed with 1.5 A for 10 minutes. The mass of copper (M = 63.5) deposited at the cathode is approximately:

  1. 0.15 g
  2. 1.00 g
  3. 0.60 g
  4. 0.30 g
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Correct answer: D. 0.30 g

Q = It = 1.5 × 600 = 900 C. Cu²⁺ + 2e⁻ → Cu requires 2F = 193 000 C per mole (63.5 g). Mass = 63.5 × 900 / (2 × 96 500) = **≈ 0.296 g ≈ 0.30 g**.

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In the electrolysis of **molten** NaCl, the products at the cathode and anode respectively are:

  1. H₂ at cathode and O₂ at anode
  2. Na at cathode and Cl₂ at anode
  3. H₂ at cathode and Cl₂ at anode
  4. Na at cathode and O₂ at anode
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Correct answer: B. Na at cathode and Cl₂ at anode

Molten NaCl has only Na⁺ (reduced at cathode → Na) and Cl⁻ (oxidised at anode → Cl₂). No water, no H₂ or O₂. **Na metal and Cl₂ gas** — this is how Na is industrially produced (Down's process).

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