The standard reduction potentials at 25°C of Li^+\ — Electrochemistry Chemistry Question

💡 Solution & Explanation

Step 1 - Understanding the Concept of Reducing Power A reducing agent donates electrons to another reactant and undergoes oxidation: $$\ce{M(s) -> M^{n+}(aq) + n e^{-}}$$ The reducing power of a metal is inversely proportional to its standard reduction potential: $$\text{Reducing Power} \propto \frac{1}{E^\circ_{\text{red}}}$$ Step 2 - Analyzing the Given Standard Reduction Potentials We are given the standard reduction potentials ($E^\circ$) at $25^\circ\text{C}$: * $E^\circ(\ce{Li^+/Li}) = -3.05\text{ V}$ * $E^\circ(\ce{Ba^{2+}/Ba}) = -2.73\text{ V}$ * $E^\circ(\ce{Na^+/Na}) = -2.71\text{ V}$ * $E^\circ(\ce{Mg^{2+}/Mg}) = -2.37\text{ V}$ Step 3 - Determining the Correct Order of Reducing Power Ordering from lowest (most negative) to highest (least negative): $$-3.05\text{ V} < -2.73\text{ V} < -2.71\text{ V} < -2.37\text{ V}$$ Since reducing power increases as the reduction potential becomes more negative: $$\text{Reducing Power: } \ce{Li} > \ce{Ba} > \ce{Na} > \ce{Mg}$$ Step 4 - Explanation of Options * **Option (A) is correct:** $\ce{Li}$ has the most negative reduction potential ($-3.05\text{ V}$), giving it the highest reducing power — it is the strongest reducing agent. * **Option (B) is incorrect:** Ba ($-2.73\text{ V}$) is less negative than Li. * **Option (C) is incorrect:** Na ($-2.71\text{ V}$) is less negative than both Li and Ba. * **Option (D) is incorrect:** Mg ($-2.37\text{ V}$) has the least negative potential — weakest reducing agent. $$\text{Correct Answer: } \boxed{\text{A}}$$