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

Answer: (C)

3

Answer 2

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).