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Which of the following half-cell reaction has the most negative standard electrode potential?

  • A.\(\mathrm{Li^{+}{}_{(aq)} + e^{-} \rightarrow Li_{(s)}}\)
  • B.\(\mathrm{F_{2(g)} + 2e^{-} \rightarrow 2F^{-}{}_{(aq)}}\)
  • C.\(\mathrm{Na^{+}{}_{(aq)} + e^{-} \rightarrow Na_{(s)}}\)
  • D.\(\mathrm{I_{2(aq)} + 2e^{-} \rightarrow 2I^{-}{}_{(aq)}}\)
  • E.\(\mathrm{Cu^{+}{}_{(aq)} + e^{-} \rightarrow Cu_{(s)}}\)

Answer: A

The most negative standard electrode potential belongs to Li+(aq) + e- → Li(s).

Lithium has the most negative standard reduction potential of all, about -3.05 V, because it loses its electron very readily in water (its small ion is strongly hydrated). It is therefore the strongest reducing agent in aqueous solution.

Na+/Na is also strongly negative (about -2.71 V) but less so than lithium. F2/F- is the most positive of all, about +2.87 V, and I2/I- (about +0.54 V) and Cu+/Cu (about +0.52 V) are positive.

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