Question

what is the e_cell for the electrolysis of molten lif?
half-reaction | e°(v)
li⁺ + e⁻ → li | -3.05
f₂ + 2e⁻ → 2f⁻ | +2.87
options: 5.92 v, -0.18 v, +0.18 v, -5.92 v

Explanation:

Step1: Identify reactions in electrolysis

In electrolysis of molten LiF, oxidation (anode) and reduction (cathode) occur. For LiF, \( \text{Li}^+ \) is reduced and \( \text{F}^- \) is oxidized. But the given half - reactions: \( \text{Li}^+ + e^-
ightarrow \text{Li} \) (reduction, \( E^{\circ}_{red}=- 3.05\space V \)) and \( \text{F}_2 + 2e^-
ightarrow2\text{F}^- \) (reduction, \( E^{\circ}_{red}= + 2.87\space V \)). For oxidation of \( \text{F}^- \), the reverse reaction is \( 2\text{F}^-
ightarrow\text{F}_2 + 2e^- \), so \( E^{\circ}_{ox}=-E^{\circ}_{red}=- 2.87\space V \)

Step2: Calculate \( E_{cell} \) for electrolysis

For electrolysis, \( E_{cell}=E^{\circ}_{cathode}(reduction)-E^{\circ}_{anode}(reduction) \) (or \( E_{cell}=E^{\circ}_{red,cathode}+E^{\circ}_{ox,anode} \)). The reduction of \( \text{Li}^+ \) is at cathode (\( E^{\circ}_{red,cathode}=-3.05\space V \)) and oxidation of \( \text{F}^- \) (reverse of \( \text{F}_2 \) reduction) is at anode. So \( E_{cell}=E^{\circ}_{red,cathode}+E^{\circ}_{ox,anode}=-3.05+( - 2.87)=- 5.92\space V \) (or using \( E_{cell}=E^{\circ}_{cathode}-E^{\circ}_{anode} \), where \( E^{\circ}_{cathode}=-3.05\space V \) and \( E^{\circ}_{anode}= + 2.87\space V \), so \( E_{cell}=-3.05 - 2.87=-5.92\space V \))

Answer:

-5.92 V