5
$\begingroup$

Strontium has a very low standard electrode potential and fluorine has a very high one.

\begin{align} \ce{F2 + 2e^- &<=> 2F^-} &\quad E^\circ &= \pu{+2.87 V} \tag{R1} \\ \ce{Sr &<=> Sr^+ + e^-} &\quad E^\circ &= \pu{-4.10 V} \tag{R2} \end{align}

In theory, a strontium–fluorine battery would have a voltage of $\pu{6.97 V},$ although there are many practical reasons such as danger and rarity of materials for such batteries not to be made.

Is a strontium–fluorine battery the theoretically highest voltage chemical battery using pure elements, or is it possible to obtain a higher one?

$\endgroup$
1
  • 1
    $\begingroup$ You could always change the concentrations to change the EMF. But, if fluorine and strontium are indeed at the extreme opposites, you are right. $\endgroup$
    – Yunfei Ma
    Mar 31, 2016 at 3:18

2 Answers 2

8
$\begingroup$

The half-potential you've given for strontium is only for the first ionisation. The half-potential you'd actually get is $\pu{-2.899 V}$ for the stable dication to give a cell potential of $\pu{5.769 V}$. From the CRC Handbook [1], lithium has the lowest element to stable ion potential of $\pu{-3.0401 V},$ which is why it is common in batteries.

Reference

  1. Lide, David R., ed. CRC Handbook of Chemistry and Physics, 87th ed. Boca Raton, FL: CRC Press. 2006. ISBN 0-8493-0487-3.
$\endgroup$
1
  • $\begingroup$ Can more exotic stuff such as electrides, halides and alkalides achieve higher potentials? The only standard electrode potential I can find for one of these strange anions is for the hydride anion which is 2.23V which is still not better than flourine. $\endgroup$ Apr 7, 2016 at 0:03
5
$\begingroup$

gsurfer04 is right that it wouldn't be strontium. If you're going really outlandish and willing to use explosive compounds, then the highest possible cell voltage would be obtained with the following redox reactions, using nitrogen to azide as the anode and krypton difluoride as the cathode:

\begin{align} &\text{Anode:} &\quad \ce{3 N2 + 2 H+ + 2 e- &<=> 2 HN3} &\quad &\pu{-3.09 V} \tag{R1}\\ &\text{Cathode:} &\quad \ce{KrF2 + 2e- &<=> Kr + 2 F-} &\quad &\pu{+3.50 V} \tag{R2} \end{align}

This would give $\pu{6.59 V}.$

More practical is to use lithium for the anode which is nearly as good at $\pu{-3.04 V}$ for $\pu{6.54 V}$ total. Hirashige et al. [1] tried using a $\ce{Li}-\ce{XeF2}$ cell (theoretically $\ce{6.49 V}),$ however they only managed $\pu{4.2 V}.$

Reference

  1. Hirashige, T.; Hagiwara, R.; Ito, Y. Chemical Stability and Electrochemical Activity of Xenon Difluoride in Propylene Carbonate. Journal of Fluorine Chemistry 2000, 106 (2), 205–209. DOI: 10.1016/S0022-1139(00)00332-8.
$\endgroup$
1
  • $\begingroup$ Please visit this page, this page and this one on how to format your future posts better with MathJax and Markdown. Note that DOI links and complete references (ACS style) are preferred over publisher URLs which rot over time. Also, you might want to correct anode reaction and provide reputable reference(s) for the given reduction potentials. $\endgroup$
    – andselisk
    Apr 12, 2022 at 3:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.