Question

The number of ions from the following that have the ability to liberate hydrogen from a dilute acid is _______. Ti²⁺, Cr²⁺ and V²⁺.

• A. 0
• B. 2
• C. 3
• D. 1

Answer 1

The Correct Option is C

To determine the number of ions that can liberate hydrogen from a dilute acid, we need to analyze their reducing abilities. Strong reducing agents can donate electrons to \(\text{H}^+\), reducing it to \(\text{H}_2\).
Step 1: Evaluate \(\text{Ti}^{2+}\)
Titanium(II) (\(\text{Ti}^{2+}\)) has a strong tendency to get oxidized to \(\text{Ti}^{3+}\), making it a strong reducing agent.  
\(\text{Ti}^{2+}\) can react with \(\text{H}^+\) from dilute acids to liberate \(\text{H}_2\).
Step 2: Evaluate \(\text{Cr}^{2+}\)  
Chromium(II) (\(\text{Cr}^{2+}\)) is a strong reducing agent and can be oxidized to \(\text{Cr}^{3+}\).  
\(\text{Cr}^{2+}\) reacts with \(\text{H}^+\) from dilute acids to liberate \(\text{H}_2\):
\[2\text{Cr}^{2+} (\text{aq}) + 2\text{H}^+ (\text{aq}) \rightarrow 2\text{Cr}^{3+} (\text{aq}) + \text{H}_2 (\text{g}).\]
Step 3: Evaluate \(\text{V}^{2+}\)  
Vanadium(II) (\(\text{V}^{2+}\)) is also a strong reducing agent and can be oxidized to \(\text{V}^{3+}\).  
\(\text{V}^{2+}\) reacts with \(\text{H}^+\) from dilute acids to liberate \(\text{H}_2\).
Conclusion: 
All three ions (\(\text{Ti}^{2+}\), \(\text{Cr}^{2+}\), and \(\text{V}^{2+}\)) can liberate \(\text{H}_2\) from dilute acids.
Final Answer: (3).