Classification of Elements and Periodicity in Properties
Which one among the following metals is the weakest reducing agent?
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Hint 1 of 2
What thermodynamic quantity determines the relative reducing strength of metals in aqueous solution?
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Step-by-step solutionView
Correct answer
Sodium is the weakest reducing agent among the alkali metals in aqueous solution because it has the least negative standard electrode potential (E∘=−2.71 V).
Option analysis
Why each option works or fails
A · Li
Believing that lithium's high ionization energy makes it the weakest reducing agent, ignoring its exceptionally high hydration enthalpy. Remember that reducing power in aqueous solution depends on standard reduction potentials; Li has the most negative potential (−3.04 V) due to its extremely large hydration energy, making it the strongest reducing agent in water, not the weakest.
B · K
Assuming reducing power follows the gas-phase ionization energy trend strictly down the group without checking electrode potentials. Compare the standard reduction potentials directly: potassium has an E∘ of −2.92 V, which is more negative than sodium's −2.71 V, so K is a stronger reducing agent than Na.
C · Rb
Confusing rubidium's position in Group 1 with low reactivity or assuming reducing power decreases toward heavier alkali metals. Note that rubidium has an E∘ of −2.93 V, making it a much stronger reducing agent in solution than sodium.
D · Na
None; this is the correct response. Sodium has the least negative standard reduction potential (E∘=−2.71 V) among all the alkali metals because its relatively lower hydration enthalpy does not offset its ionization energy as effectively as in lithium.
Reviewed route
Solution
StepWorking
01concept
Reducing power of alkali metals in aqueous medium depends on their standard reduction potentials (EM+/M∘). The more negative the reduction potential (or more positive the standard oxidation potential), the stronger the reducing agent. The trend in standard reduction potentials for alkali metals is: Li(−3.04 V)<Rb(−2.93 V)≈K(−2.93 V)<Na(−2.71 V). Thus, Na has the least negative E∘ value among them and is the weakest reducing agent.
02option_verdict
Li has the most negative standard reduction potential (−3.04 V) due to its exceptionally high hydration enthalpy, making it the strongest reducing agent in aqueous medium.
03option_verdict
K has a reduction potential of −2.93 V, which is more negative than that of Na (−2.71 V), making it a stronger reducing agent than Na.
04option_verdict
Rb has a reduction potential of −2.93 V, which is more negative than that of Na, making it a stronger reducing agent than Na.
05option_verdict
Na has the least negative reduction potential (−2.71 V) among all alkali metals because its high sublimation enthalpy and relatively moderate hydration enthalpy make oxidation less thermodynamically favorable than for Li, K, and Rb. Thus, Na is the weakest reducing agent.
✓discriminator
In aqueous solution, the reducing power order is Li>Rb≈K>Na. Sodium (Na) is universally recognized in Group 1 chemistry as the weakest reducing agent.
Hints that build this answer step by step
What thermodynamic quantity determines the relative reducing strength of metals in aqueous solution?
Standard reduction potential (E∘), where more negative values indicate stronger reducing agents
Among the alkali metals Li, Na, K, and Rb, which has the least negative (highest) standard reduction potential E∘?
Why isn't lithium the weakest reducing agent if it has the highest ionization energy?
In aqueous solution, reducing power is governed by the standard reduction potential, which accounts for sublimation energy, ionization energy, and hydration energy. Li+ has an extremely high hydration enthalpy due to its small size, more than compensating for its high ionization energy and making Li the strongest reducing agent, while Na ends up as the weakest.