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Rajasthan Board Class 12 (RBSE) Electrochemical Series — practice questions

Explore 6 free Electrochemical Series questions in Rajasthan Board Class 12 (RBSE) 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 1–6 of 6.

In a GALVANIC cell, the ANODE is:Read the answer and worked solution →The standard electrode potential of the Standard Hydrogen Electrode (SHE) is defined as:Read the answer and worked solution →The Daniell cell (Zn-Cu) has E°cell = +1.10 V. Calculate ΔG° (F = 96500 C/mol):Read the answer and worked solution →The Nernst equation at 25°C for a cell reaction is:Read the answer and worked solution →Faraday's first law of electrolysis states that the mass deposited is:Read the answer and worked solution →The byproduct of a hydrogen-oxygen fuel cell (used in space + electric cars) is:Read the answer and worked solution →

Worked practice questions

In a GALVANIC cell, the ANODE is:

  1. The positive electrode where reduction occurs
  2. The positive electrode where oxidation occurs
  3. The negative electrode where oxidation occurs
  4. The negative electrode where reduction occurs
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Correct answer: C. The negative electrode where oxidation occurs

In a galvanic cell, the anode is the NEGATIVE electrode where OXIDATION occurs (electrons released go to external circuit). The cathode is the POSITIVE electrode where REDUCTION occurs. **Mnemonic:** AN-OX (anode oxidation).

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The standard electrode potential of the Standard Hydrogen Electrode (SHE) is defined as:

  1. $+1.10$ V (the Daniell cell value)
  2. $0$ V exactly (the universal reference)
  3. $+0.34$ V (the Cu electrode value)
  4. $-0.76$ V (the Zn electrode value)
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Correct answer: B. $0$ V exactly (the universal reference)

SHE is the UNIVERSAL REFERENCE with E° defined as 0 V exactly. All other electrode potentials are measured relative to SHE. Cu/Cu²⁺ = +0.34 V; Zn/Zn²⁺ = −0.76 V.

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The Daniell cell (Zn-Cu) has E°cell = +1.10 V. Calculate ΔG° (F = 96500 C/mol):

  1. $-106 kJ$ (using wrong n value)
  2. $-53 kJ$ (using n = 1)
  3. $+212 kJ$ (positive sign incorrect)
  4. $-212 kJ/mol$ (using n = 2)
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Correct answer: D. $-212 kJ/mol$ (using n = 2)

ΔG° = −nFE° = −2 × 96500 × 1.10 = −212,300 J/mol = −212 kJ/mol. Negative ΔG° confirms spontaneous reaction (galvanic cell works). n = 2 for Zn → Zn²⁺ + 2e⁻.

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