Question:

The standard electrode potential \((\frac{M^{3+}}{M^{2+}})\) for V, Cr, Mn & Co are –0.26 V, –0.41 V, +1.57 V and +1.97 V, respectively. The metal ions which can liberate H2 from a dilute acid are

Show Hint

For questions involving \(H_2\) liberation:
• Compare the reduction potential of the metal ion with the standard hydrogen electrode (SHE).
• Negative reduction potentials indicate the ability to liberate \(H_2\) gas.

Updated On: Jan 9, 2025
  • V2+ and Cr2+

  • V2+ and Mn2+

  • Cr2+ and Co2+

  • Mn2+ and Co2+

Hide Solution
collegedunia
Verified By Collegedunia

The Correct Option is A

Solution and Explanation

 - Metal cations with negative values of reduction potential (\(\text{M}^{3+}/\text{M}^{2+}\)) or positive values of oxidation potential (\text{M}^{2+}/\text{M}^{3+}\)) can reduce H\(^+\) ions and liberate H\(_2\) gas from dilute acid.
- For the given metals:
    V\(^{2+}\) has a reduction potential of \(-0.26~\text{V}\).
    Cr\(^{2+}\) has a reduction potential of \(-0.41~\text{V}\).
- Both values are negative, meaning V\(^{2+}\) and Cr\(^{2+}\) can reduce H\(^+\) ions to liberate H\(_2\) gas.
Final Answer: \((3)\) V\(^{2+}\) and Cr\(^{2+}\).

Was this answer helpful?
0
0

Top Questions on Electrochemistry

View More Questions

Questions Asked in JEE Main exam

View More Questions