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Rajasthan Board Class 12 (RBSE) — free MCQ practice

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Try a Rajasthan Board Class 12 (RBSE) question

Which half-reaction from the notes represents reduction?Read the answer and worked solution →The potential developed between a metal rod and its solution due to oxidation is called:Read the answer and worked solution →The situation in which the charge buildup near the Zn rod prevents further oxidation is called:Read the answer and worked solution →Which half-reaction from the given notes represents oxidation?Read the answer and worked solution →

Worked practice questions

Which half-reaction from the notes represents reduction?

  1. $Zn \rightarrow Zn^{2+} + 2e^{-}$
  2. $Cu^{2+} + 2e^{-} \rightarrow Cu$
  3. $Cu \rightarrow Cu^{2+} + 2e^{-}$
  4. $Zn^{2+} \rightarrow Zn + 2e^{-}$
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Correct answer: B. $Cu^{2+} + 2e^{-} \rightarrow Cu$

Reduction involves gain of electrons, so electrons must appear on the reactant side. In $Cu^{2+} + 2e^{-} \rightarrow Cu$, copper ions gain electrons to form copper metal.

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The potential developed between a metal rod and its solution due to oxidation is called:

  1. Reduction potential
  2. Oxidation potential
  3. Solubility potential
  4. Neutralization potential
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Correct answer: B. Oxidation potential

The notes define the potential developed between the rod and the solution in this case as oxidation potential, represented by $E(Zn/Zn^{2+})$.

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The situation in which the charge buildup near the Zn rod prevents further oxidation is called:

  1. Equivalence point
  2. Triple point
  3. Saturation point
  4. End point
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Correct answer: C. Saturation point

The notes explicitly state that when the positive charge layer around the rod prevents further oxidation, that condition is known as the saturation point.

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