Which of the following ions does not liberate hydrogen gas on reaction with dilute acids?

We have to check whether the given divalent transition-metal ions can act as reducing agents strong enough to convert hydrogen ions from a dilute acid into molecular hydrogen. In other words, we must test whether the reaction

$$M^{2+}\;+\;2\,H^+\;\rightarrow\;M^{3+}\;+\;H_2$$

is spontaneous for each ion. Spontaneity is decided with standard electrode (reduction) potentials. First, recall the definition:

For any redox couple $$Ox + ne^- \rightarrow Red$$ the standard reduction potential is denoted $$E^\circ$$. If two half-reactions are combined so that

$$\text{cell reaction}\;=\;\text{reduction half-reaction}\;+\;\text{oxidation half-reaction},$$

then the overall standard cell potential is

$$E^\circ_{cell}=E^\circ_{\text{red}}(\text{reduction})-E^\circ_{\text{red}}(\text{oxidation}).$$

If $$E^\circ_{cell}\gt 0,$$ the reaction is spontaneous in the forward direction.

In the present case the reduction half-reaction is the hydrogen electrode:

$$2\,H^+ + 2e^- \;\rightarrow\; H_2,\qquad E^\circ_{\text{red}}=0.00\;{\rm V}.$$

The oxidation half-reaction is simply the reverse of the $$M^{3+}/M^{2+}$$ reduction couple:

$$M^{2+}\;\rightarrow\;M^{3+}+e^-.$$

Hence, for each ion,

$$E^\circ_{cell}=0.00\;{\rm V}-E^\circ_{\text{red}}(M^{3+}/M^{2+}).$$

The requirement $$E^\circ_{cell}\gt 0$$ simplifies to

$$0.00\;{\rm V}-E^\circ_{\text{red}}(M^{3+}/M^{2+})\gt 0\;\;\Longrightarrow\;\;E^\circ_{\text{red}}(M^{3+}/M^{2+})\lt 0.00\;{\rm V}.$$

So, only those $$M^{2+}$$ ions whose $$M^{3+}/M^{2+}$$ reduction potential is negative will be able to displace hydrogen from dilute acids.

Now we list the standard reduction potentials (all values at 298 K, 1 M):

$$\begin{aligned} Ti^{3+}+e^- &\rightarrow Ti^{2+}, & E^\circ &= -0.37\;{\rm V},\\[2pt] V^{3+}+e^- &\rightarrow V^{2+},\; & E^\circ &= -0.26\;{\rm V},\\[2pt] Cr^{3+}+e^- &\rightarrow Cr^{2+}, & E^\circ &= -0.41\;{\rm V},\\[2pt] Mn^{3+}+e^- &\rightarrow Mn^{2+},\; & E^\circ &= +1.51\;{\rm V}. \end{aligned}$$

We see that

• $$E^\circ(Ti^{3+}/Ti^{2+})=-0.37\;{\rm V}\lt 0,$$ so $$Ti^{2+}$$ will reduce $$H^+$$ to $$H_2.$$
• $$E^\circ(V^{3+}/V^{2+})=-0.26\;{\rm V}\lt 0,$$ so $$V^{2+}$$ will also liberate $$H_2.$$$
• $$E^\circ(Cr^{3+}/Cr^{2+})=-0.41\;{\rm V}\lt 0,$$ so $$Cr^{2+}$$ likewise produces $$H_2.$$$
• $$E^\circ(Mn^{3+}/Mn^{2+})=+1.51\;{\rm V}$$ which is positive. Substituting into the earlier expression gives

  $$E^\circ_{cell}=0.00\;{\rm V}-(+1.51\;{\rm V})=-1.51\;{\rm V}\lt 0.$$

  The negative value means the forward reaction is non-spontaneous; therefore $$Mn^{2+}$$ cannot reduce $$H^+$$, and no hydrogen gas is evolved.

Hence, among the given options, only $$Mn^{2+}$$ fails to liberate hydrogen gas with dilute acids.

Hence, the correct answer is Option C.