Question

what occurs during oxidation in the reaction for the electrolysis of 1.0 m lif (aq)?
half-reaction | e° (v)
li⁺ + e⁻ → li | - 3.05
2h₂o + 2e⁻ → h₂ + 2oh⁻ | - 0.83
f₂ + 2e⁻ → 2f⁻ | + 2.87
o₂ + 4h⁺ + 4e⁻ → 2h₂o | + 1.23
fluorine gas is produced.
hydrogen gas is produced.
lithium ions change to solid lithium.
oxygen gas is produced.

Explanation:

1. Identify Oxidation and Reduction: In electrolysis, oxidation (loss of electrons) occurs at the anode, reduction (gain of electrons) at the cathode. For aqueous LiF, species to consider: \( \text{Li}^+ \), \( \text{F}^- \), \( \text{H}_2\text{O} \).
2. Anode (Oxidation) Candidates:

• \( \text{F}^- \) oxidation: \( 2\text{F}^-

ightarrow \text{F}_2 + 2\text{e}^- \), \( E^\circ_{\text{ox}} = -2.87 \, \text{V} \) (reverse of \( \text{F}_2 + 2\text{e}^-
ightarrow 2\text{F}^- \), \( E^\circ_{\text{red}} = +2.87 \, \text{V} \)).

• \( \text{H}_2\text{O} \) oxidation: \( 2\text{H}_2\text{O}

ightarrow \text{O}_2 + 4\text{H}^+ + 4\text{e}^- \), \( E^\circ_{\text{ox}} = -1.23 \, \text{V} \) (reverse of \( \text{O}_2 + 4\text{H}^+ + 4\text{e}^-
ightarrow 2\text{H}_2\text{O} \), \( E^\circ_{\text{red}} = +1.23 \, \text{V} \)).

1. Compare Oxidation Potentials: The less negative (more positive) \( E^\circ_{\text{ox}} \) is more favorable. \( -1.23 \, \text{V} \) (water oxidation) is less negative than \( -2.87 \, \text{V} \) (fluoride oxidation), so water oxidizes preferentially.
2. Product of Water Oxidation: \( 2\text{H}_2\text{O}

ightarrow \text{O}_2 + 4\text{H}^+ + 4\text{e}^- \) produces \( \text{O}_2 \) gas.

Answer:

Oxygen gas is produced.