Which one of the following is the weakest oxidizing agent?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(Li^+\)
The standard electrode (reduction) potential \( E^\circ \) measures the tendency of a species to gain electrons. The higher the \( E^\circ \), the more readily the oxidised form is reduced, and the stronger it is as an oxidising agent; the lowest \( E^\circ \) marks the weakest oxidising agent.
\( \displaystyle \ce{Li^+ + e^- -> Li}\quad E^\circ = -3.05\ \mathrm{V} \)
\( \displaystyle \ce{K^+ + e^- -> K}\quad E^\circ = -2.93\ \mathrm{V} \)
Lithium has the most negative electrode potential, so the \( \ce{Li+} \) ion is the weakest oxidising agent.
Why the other options are incorrect
Option (1) \(F^-\) — Incorrect. Fluoride is the reduced form of the couple \( \ce{F2/F-} \) (\( E^\circ = +2.87 \) V). Fluorine gas is the strongest oxidising agent, and the fluoride ion is the weakest reducing agent; it is not compared as an oxidising agent at all.
Option (2) \(H_2\) — Incorrect. Hydrogen gas is also a reduced form (\( \ce{2H+ + 2e- -> H2} \), \( E^\circ = 0.00 \) V); it acts as a reducing agent.
Option (4) \(K^+\) — Incorrect. Its \( E^\circ \) (−2.93 V) is a little higher than that of \( \ce{Li+} \), so \( \ce{K+} \) is a slightly stronger oxidising agent than \( \ce{Li+} \).
NCERT (2026-27 edition):
Measurement of Electrode Potential – “… It can be seen that the standard electrode potential for fluorine is the highest in the Table indicating that fluorine gas (F2) has the maximum tendency to get reduced to fluoride ions (F–) and therefore fluorine gas is the strongest oxidising agent and fluoride ion is the weakest reducing agent. Lithium has the lowest electrode potential indicating that lithium ion is the weakest oxidising agent while lithium metal is the most powerful reducing agent in an aqueous solution.”