Chemistry

Problem

Four half-reactions are listed below. Fe

^{3+}

+ e

^{-}

→ Fe

^{2+}

E^{\circ}=+0.77\,\text{V}

n=1

^{+}

+ e

^{-}

→ Ag(s)

E^{\circ}=+0.80\,\text{V}

n=1

^{3+}

+ 3 e

^{-}

→ Cr(s)

E^{\circ}=-0.74\,\text{V}

n=3

^{3+}

+ 3 e

^{-}

→ Al(s)

E^{\circ}=-1.66\,\text{V}

n=3

Design a galvanic cell that • transfers exactly 3 mol e

^{-}

per mole of overall reaction, and • has the smallest positive

E^{\circ}_{\text{cell}}

that is at least +1.00 V. Which pairing of half-reactions satisfies both criteria when the cell operates spontaneously under standard conditions? a)

\mathrm{Ag^{+}/Ag}

as the cathode and

\mathrm{Al/Al^{3+}}

as the anode b)

\mathrm{Cr^{3+}/Cr}

as the cathode and

\mathrm{Fe^{2+}/Fe^{3+}}

as the anode c)

\mathrm{Fe^{3+}/Fe^{2+}}

as the cathode and

\mathrm{Cr/Cr^{3+}}

as the anode d)

\mathrm{Ag^{+}/Ag}

as the cathode and

\mathrm{Cr/Cr^{3+}}

as the anode

Intuitive explanation

The key is recognizing that $E°_{\text{cell}} = E°_{\text{cathode}} - E°_{\text{anode}}$ , and you need to balance electron transfer to exactly 3 moles while keeping the cell voltage positive but minimal. Start by identifying which half-reactions can be paired so their electrons match without extra scaling—notice that Cr and Al both consume 3 electrons, while Fe and Ag each consume 1. The cathode (reduction, higher $E°$ ) minus anode (oxidation, lower $E°$ ) must give a voltage ≥ +1.00 V. Your job is to test combinations systematically: which pairing naturally transfers 3 electrons and yields the smallest positive voltage that still meets the minimum?

Intuitive explanation

Browse the problem library