1. Identify what “strong reducing agent” means

A strong reducing agent is a species that readily loses electrons and gets oxidized.

So we must find the species that is most easily oxidized to a more stable oxidation state.

2. Use the special stability of lanthanide electronic configurations

Among lanthanides, certain oxidation states are especially stable due to electronic configuration:

• $\mathrm{Ce}^{4+}$ is stable because Ce has configuration: $\mathrm{Ce}: [Xe]4f^15d^16s^2$ and $\mathrm{Ce}^{4+} : [Xe]$ which is a noble-gas configuration.

• $\mathrm{Eu}^{2+}$ is stable because: $\mathrm{Eu}: [Xe]4f^76s^2$ so $\mathrm{Eu}^{2+} : [Xe]4f^7$ which is a half-filled $4f$ shell, especially stable.

• $\mathrm{Gd}^{3+}$ is stable because: $\mathrm{Gd}: [Xe]4f^75d^16s^2$ so $\mathrm{Gd}^{3+} : [Xe]4f^7$ again half-filled and very stable.

• $\mathrm{Lu}^{3+}$ is stable because: $\mathrm{Lu}: [Xe]4f^{14}5d^16s^2$ so $\mathrm{Lu}^{3+} : [Xe]4f^{14}$ which is a completely filled $4f$ shell.

3. Check which species can act as a reducing agent

A reducing agent gets oxidized.

Option A: $\mathrm{Eu}^{2+}$

It can be oxidized to: $\mathrm{Eu}^{2+} \rightarrow \mathrm{Eu}^{3+} + e^-$ Since lanthanides generally prefer the $+3$ oxidation state, $\mathrm{Eu}^{2+}$ readily gives up one electron. Thus $\mathrm{Eu}^{2+}$ acts as a strong reducing agent.

Option B: $\mathrm{Gd}^{3+}$

$\mathrm{Gd}^{3+}$ is already a highly stable common oxidation state. It is not easily oxidized further, so it is not a strong reducing agent.

Option C: $\mathrm{Lu}^{3+}$

$\mathrm{Lu}^{3+}$ is also a very stable state with $4f^{14}$. It does not readily lose electrons further, so it is not a strong reducing agent.

Option D: $\mathrm{Ce}^{4+}$

$\mathrm{Ce}^{4+}$ tends to gain an electron to become $\mathrm{Ce}^{3+}$, so it behaves as an oxidizing agent, not a reducing agent.

4. Conclusion

The species that acts as a strong reducing agent is: $\boxed{\mathrm{Eu}^{2+}}$

So the correct option is A.