What voltage can theoretically be achieved in a battery in which lithium metal is oxidized and fluorine gas is reduced, and why might such a battery be difficult to produce?

Verified step by step guidance

1. Identify the half-reactions for the oxidation and reduction processes. For lithium, the oxidation half-reaction is: \( \text{Li} \rightarrow \text{Li}^+ + \text{e}^- \). For fluorine, the reduction half-reaction is: \( \text{F}_2 + 2\text{e}^- \rightarrow 2\text{F}^- \).
2. Look up the standard reduction potentials for each half-reaction. The standard reduction potential for \( \text{Li}^+ + \text{e}^- \rightarrow \text{Li} \) is approximately -3.04 V, and for \( \text{F}_2 + 2\text{e}^- \rightarrow 2\text{F}^- \) is approximately +2.87 V.
3. Calculate the standard cell potential (\( E^\circ_{\text{cell}} \)) by using the formula: \( E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \). Substitute the values: \( E^\circ_{\text{cell}} = 2.87 \text{ V} - (-3.04 \text{ V}) \).
4. Discuss why such a battery might be difficult to produce. Consider factors like the reactivity of lithium metal, which is highly reactive and can pose safety risks, and the handling of fluorine gas, which is corrosive and toxic.
5. Conclude by considering the practical challenges in manufacturing and safely operating a battery with such reactive materials, despite the high theoretical voltage.