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The oxidation potentials of \(A\) and \(B\) are \[ +2.37\ \mathrm{V} \quad \text{and} \quad +1.66\ \mathrm{V} \] respectively. In chemical reactions

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Higher oxidation potential: \[ \Rightarrow \] greater tendency to lose electrons \[ \Rightarrow \] stronger reducing agent.
Updated On: May 30, 2026
  • \(A\) will be replaced by \(B\)
  • \(A\) will replace \(B\)
  • \(A\) will not replace \(B\)
  • \(A\) and \(B\) will not replace each other
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The Correct Option is B

Solution and Explanation

Step 1: Understanding the Concept:
Electrochemistry provides a quantitative way to determine the reactivity of elements using standard potentials.
The standard oxidation potential (\( E^{\circ}_{\text{ox}} \)) measures the tendency of a chemical species to lose electrons and become oxidized.
The higher the oxidation potential, the greater the tendency of the element to lose electrons.
Elements that lose electrons easily are termed "active metals" and act as strong reducing agents.
In a displacement reaction, a more active metal (higher oxidation potential) will displace a less active metal (lower oxidation potential) from its salt solution.
Step 2: Detailed Explanation:
We are provided with the following data:
Oxidation potential of A: \( E^{\circ}_{\text{ox}}(A) = +2.37 \text{ V} \).
Oxidation potential of B: \( E^{\circ}_{\text{ox}}(B) = +1.66 \text{ V} \).
Comparing these values:
\( 2.37 \text{ V}>1.66 \text{ V} \).
This comparison tells us that A is more easily oxidized than B.
In other words, A is more reactive and is a stronger reducing agent than B.
When metal A is added to a solution containing ions of metal B (\( B^{n+} \)), metal A will lose electrons to become \( A^{n+} \) and will give these electrons to \( B^{n+} \) ions.
The \( B^{n+} \) ions are reduced to solid metal B and precipitate out of the solution.
The reaction can be represented as:
\[ A(s) + B^{n+}(aq) \rightarrow A^{n+}(aq) + B(s) \]
This process is a spontaneous redox reaction because the overall cell potential (\( E_{\text{cell}} \)) is positive:
\[ E_{\text{cell}} = E^{\circ}_{\text{ox}}(A) - E^{\circ}_{\text{ox}}(B) = 2.37 - 1.66 = +0.71 \text{ V} \).
Since \( E_{\text{cell}}>0 \), the reaction proceeds in the forward direction.
This means A successfully "replaces" or "displaces" B from its compound.
Conversely, B cannot replace A because its oxidation potential is lower, and the reverse reaction would have a negative cell potential.
Step 3: Final Answer:
Since A has a higher oxidation potential than B, it has a stronger tendency to enter the ionic state by displacing B.
Therefore, A will replace B in chemical reactions.
The correct option is (B).
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