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).
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Approach Solution -2
Step 1: Understanding the concept.
An ion can liberate hydrogen gas from a dilute acid if it has a reduction potential lower than that of hydrogen (E° = 0.00 V). This means the ion is a stronger reducing agent than hydrogen ions (H⁺).
Step 2: Write the standard reduction potentials.
Ti²⁺ + 2e⁻ → Ti; E° = –1.63 V
V²⁺ + 2e⁻ → V; E° = –1.18 V
Cr²⁺ + 2e⁻ → Cr; E° = –0.91 V
Step 3: Compare with the hydrogen electrode.
All the above ions have negative reduction potentials compared to hydrogen (0.00 V), meaning they can easily reduce H⁺ ions to hydrogen gas (H₂).
Step 4: Conclusion.
All three ions (Ti²⁺, Cr²⁺, and V²⁺) can liberate hydrogen from a dilute acid because they are stronger reducing agents than H⁺.
