Singlet Oxygen during Cycling of the Aprotic Sodium–O₂ Battery

Abstract

Aprotic sodium–O₂ batteries require the reversible formation/dissolution of sodium superoxide (NaO₂) on cycling. Poor cycle life has been associated with parasitic chemistry caused by the reactivity of electrolyte and electrode with NaO₂, a strong nucleophile and base. Its reactivity can, however, not consistently explain the side reactions and irreversibility. Herein we show that singlet oxygen (¹O₂) forms at all stages of cycling and that it is a main driver for parasitic chemistry. It was detected in‐ and ex‐situ via a ¹O₂ trap that selectively and rapidly forms a stable adduct with ¹O₂. The ¹O₂ formation mechanism involves proton‐mediated superoxide disproportionation on discharge, rest, and charge below ca. 3.3 V, and direct electrochemical ¹O₂ evolution above ca. 3.3 V. Trace water, which is needed for high capacities also drives parasitic chemistry. Controlling the highly reactive singlet oxygen is thus crucial for achieving highly reversible cell operation.

The need to advance batteries beyond the limits of current technology in terms of energy, sustainability, and cost has generated immense interest in rechargeable aprotic metal–O₂ batteries.1 They store charge at the cathode by reversibly forming/decomposing Li₂O₂ or NaO₂ in the Li–O₂ or Na–O₂ cell, respectively. Despite the lower theoretical specific capacity of 488 mAh g⁻¹ of NaO₂ and lower voltage of 2.27 V (vs. 1168 mAh g⁻¹ at 2.96 V for Li–O₂), the Na–O₂ cell has been reported to have significant advantages over Li–O₂ with respect to rechargeability and energy efficiency.2 Realizing the Na–O₂ cell, however, still faces many challenges in practice, including the Na‐metal anode, lower than theoretical cathode capacity, and perhaps most importantly, insufficient cycle life associated with parasitic chemistry (that is, side reactions) at the cathode.2, 3

Since the very first papers published on Na–O₂ batteries, superoxide has been perceived responsible for parasitic chemistry with electrolyte and electrode.1c, 3, 4 A key measure for parasitic chemistry is the ratio of e⁻ passed to O₂ consumed/evolved. During discharge, this ratio is typically at the ideal value of one despite approximately 5 % of the expected NaO₂ being missing and replaced by side products, such as Na₂CO₃, Na acetate, and Na formate. During subsequent resting and charge, more of these side products form and typically the e⁻/O₂ ratio deviates by several percent from one.2a, 3c, 4b, 5 Although less side products form than in the Li–O₂ cell, cyclability is similarly poor: restricted capacity can often be maintained for up to hundreds of cycles albeit at the expense of true energy, but at full discharge cells fail within some 10 cycles and capacity fading becomes significantly worse with rising charge cut‐off voltage.1c, 2a,2c,2d, 3a,3d, 6

Superoxide's potential reactivity towards organic substrates stems from its nucleophilicity, basicity, and radical nature, which may cause nucleophilic substitutions, H⁺ and H‐atom abstraction.7 Theoretical work, however, has revealed that all these reactions are unlikely with commonly used ether electrolytes owing to the high activation energies and strong endothermicity.8 Also, the extent of parasitic chemistry at the various stages of cycling does not match the abundance of superoxide. Overall, the reactivity of superoxide cannot consistently explain the observed parasitic chemistry, which thus may only be inhibited with better knowledge about reactive species and their formation mechanism.

Herein we show that singlet oxygen (¹Δ_g or ¹O₂), the first excited state of ground state triplet oxygen (³Σ_g ⁻ or ³O₂), is the reactive species, which drives parasitic reactions. It is generated at all stages of cycling in relative quantities resembling the occurrence of side reactions: relatively little during discharge, rest, and low charging voltages, and strongly increasing amounts at higher charging voltages.

Methods to sensitively detect ¹O₂ rely on chemical probes, which selectively react with ¹O₂. Probes include spin traps and fluorophores, which become EPR active or fluorescing upon reaction with ¹O₂.9 However, these probes are not electrochemically inert in the relevant potential range between approximately 2 and 3.6 V versus Na/Na⁺ and may react with superoxide. Previously, we have shown that 9,10‐dimethylanthracene (DMA) fulfills all the requirements;10 it is stable in contact with superoxide, reacts rapidly with ¹O₂ to its endoperoxide (DMA‐O₂), and has a sufficiently wide potential window (Figure S1 and Scheme S1 in the Supporting Information). ¹O₂ can either be monitored by DMA consumption via its absorbance or fluorescence between 300 and 500 nm (Figure S2), or by detecting DMA and DMA‐O₂ via HPLC.

To probe whether ¹O₂ would form during discharge and charge we constructed Na–O₂ cells as detailed in the Methods Section of the Supporting Information with 0.5 m NaSO₃CF₃ (NaOTf) in diglyme containing 40 ppm H₂O and 30 mm DMA as the electrolyte. Water was used because it is required as a phase‐transfer catalyst to allow large NaO₂ particles to grow.2c To specifically probe reactions at the cathode and to exclude unwanted reactions of electrolyte components with a Na‐metal anode, we used the Na intercalation material Na_3−xV₂(PO₄)₃ as the counter electrode. It operates at approximately 3.4 V versus Na/Na⁺ and is thus well within the stability window of the electrolyte.11 Cells were run at constant current to various states of discharge or charge, then stopped, and the electrolyte extracted and analyzed using HPLC (Figure 1).

Figure 1.

a) Load curve for galvanostatic cycling of a carbon‐paper cathode at 90 μA cm⁻² in 0.5 m NaOTf in diglyme containing 40 ppm H₂O and 30 mm DMA. b) HPLC runs of electrolyte samples taken at the points ②, ③, and ④ in (a), showing ¹O₂ to have formed as indicated by the conversion of DMA into DMA‐O₂. The blank sample was extracted from a cell that rested for 5 h.

The cycle curve shows flat discharge and charge plateaus at approximately 2.2 and 2.3 V, respectively, with a sharp rise in voltage when recharge approaches about 55 % (Figure 1 a). X‐ray diffraction confirms that the cube‐shaped discharge product is NaO₂ (Figure S3, S4) in accord with many previous reports on similar cells.1b,1c, 2a,2b, 3a After discharge or recharge to 2.8 and 3.65 V, 4.1, 4.3, and 7.2 %, respectively, of the DMA at the sample points ②, ③, and ④ was converted into DMA‐O₂ (Figure 1 b). The amounts of DMA‐O₂ equate to 1.4, 1.3, and 2.1 % of the O₂ involved based on the charge at these stages of cycling to have turned into ¹O₂. ¹O₂ is thus generated both on discharge and charge, and charging to higher voltage yields significantly more ¹O₂.

Operando spectroscopy is well suited to probe in detail onset potentials and reaction rates. An operando fluorescence set‐up as detailed in the Supporting Information was constructed. Briefly, it consisted of a gas‐tight quartz cuvette with a carbon paper working electrode and Na_3−xV₂(PO₄)₃ as counter and reference electrode and an O₂ filled head space. The O₂‐saturated electrolyte was the same as before except for a lower DMA concentration (1.6×10⁻⁵  m), which best suits fluorescence detection (see the Supporting Information). Excitation and emission wavelengths were chosen according to the maxima in the respective spectra (Figure S2).

While cells as in Figure 1 comprising a porous separator sandwiched between working and counter electrode typically yield NaO₂, cycling the cathode in the operando setup yielded Na₂O₂⋅H₂O as can be inferred from the high charging voltage (Figure S5). Rather similar thermodynamics of NaO₂ and Na₂O₂ ($E_{{\mathrm{O}}_2{/\mathrm{NaO}}_2}^0$ =2.27 V, $E_{{\mathrm{O}}_2{/\mathrm{Na}}_2{\mathrm{O}}_2}^0$ =2.33 V) were used to explain that the Na–O₂ cell could yield both as a discharge product.1b, 3b Proton sources decisively influence the discharge product, yet the precise governing factors are unknown.1b The reason for peroxide rather than superoxide as a discharge product in the operando setup may be the large electrolyte‐to‐electrode volume ratio and the stirred electrolyte, both of which are essential for the method. The electrolyte after discharge in the operando setup also contains DMA‐O₂, revealing that discharge to Na₂O₂ also forms ¹O₂ (Figure S6).

To investigate charging of NaO₂, we first discharged a cathode in the standard cell and then placed it in the operando cell. Figure 2 shows the DMA concentration and consumption rate for stepwise potentiostatic charging up to 3.7 V. As soon as the cell was polarized to charging potentials, DMA was consumed at an initially moderate rate, which sharply increased beyond a voltage of about 3.3 V. Thermodynamically, ¹O₂ can be expected to form upon electrochemically oxidizing NaO₂ above $E_{{\mathrm{O}}_2{/\mathrm{NaO}}_2}^0+E({}^1{\Delta }_g\leftarrow {}^3{\Sigma }_g^{-})$ . With 0.97 eV energy difference between ¹O₂ and ³O₂ the thermodynamic threshold to evolve ¹O₂ can be estimated to be 3.24 V. ¹O₂ forming at high rate above approximately 3.3 V can thus be explained by the reaction NaO₂→Na⁺+e⁻+¹O₂ beyond the thermodynamic threshold of 3.24 V. Up to 3.3 V the ¹O₂ fraction is around 0.75 to 1.1 % of the expected O₂ (Figure S7). Beyond 3.3 V this fraction increases steadily to around 4 %.

Figure 2.

Operando fluorescence detection of ¹O₂ during potentiostatic charging of a Na–O₂ cathode. The carbon‐paper cathode was first discharged in a Swagelok type cell to 75 mAh cm⁻² in 0.5 m NaOTf in diglyme containing 40 ppm H₂O and then introduced into the operando setup containing the same electrolyte and additionally 1.6×10⁻⁵  m DMA. a) Voltage and current profile. b) DMA concentration. c) DMA consumption rate.

Previously it was reported that most parasitic chemistry occurs during discharge and less on charge.2a, 3a Charging above around 3.5 V caused the number of O₂ per e⁻ evolved to deviate more significantly from one than at lower voltages. Correlating these findings with the results in Figure 1, Figure 2, and Figure S7 suggests that parasitic chemistry in the Na–O₂ cathode is closely related to ¹O₂ formation since the extent of side reactions follows the occurrence of ¹O₂: ¹O₂ is formed at a higher rate during discharge than at charging below 3.3 V and at much higher rate above 3.3 V. In quantitative terms the 1.4 % ¹O₂ during discharge and an approximately 95 % yield of NaO₂ suggests a significant fraction of the side products stem from ¹O₂. Measured ¹O₂ amounts represents a lower bound since at the low DMA concentration not necessarily all ¹O₂ will react with DMA and may undergo other decay routes, such as reactions with cell components.

While thermodynamics directly explains how ¹O₂ forms when NaO₂ is oxidized above 3.3 V, its formation during discharge and recharge below 3.3 V is more surprising. With the Li–O₂ cell, disproportionation of the LiO₂ intermediate to Li₂O₂ was suggested as a major ¹O₂ source during discharge and low charging voltages.10 This pathway is inactive in the Na–O₂ cell since discharge stops at NaO₂. When, however, water or other proton sources are present, oxygen reduction will lead to HOO^..12 This soluble species enables proton‐assisted phase‐transfer catalysis, which allows the typical micron‐sized NaO₂ cubes to grow on discharge and to be oxidized on charge; HOO^. is soluble and mobile in the electrolyte and precipitates NaO₂ via the metathesis reaction HOO^.+Na⁺ $\overleftarrow{\underset{}{\to }}$ NaO₂+H⁺.2c, 13 HOO^., however, can be reduced by superoxide or disproportionate according to Equation (1), (2),12 which both have been shown to release ¹O₂.14

$${\displaystyle \mathrm{HOO}{}^{•}+\mathrm{O}{}_2{}^{-}\to \mathrm{HO}{}_4{}^{-}\to \mathrm{HOO}{}^{-}+{}^1\mathrm{O}{}_2}$$ (1)

$${\displaystyle 2\mathrm{HOO}{}^{•}+\mathrm{H}{}_2\mathrm{O}\to \mathrm{H}{}_2\mathrm{O}{}_4+\mathrm{H}{}_2\mathrm{O}\to \mathrm{H}{}_2\mathrm{O}{}_2+\mathrm{H}{}_2\mathrm{O}+{}^1\mathrm{O}{}_2}$$ (2)

That water is driving these reactions is supported by significantly less ¹O₂ being formed with dry electrolyte as compared to wet electrolyte (Figure S8).

These reactions may not only take place during discharge, where new superoxide is generated, but also at rest since the metathesis reaction is an equilibrium and allows NaO₂ to redissolve into HOO^.. A number of studies have shown instability of NaO₂ upon prolonged rest.3c, 4, 5 Although details vary, they generally report gradual conversion of NaO₂ into Na₂O₂⋅H₂O accompanied by electrolyte degradation to form NaOH, Na₂CO₃, and organic compounds including Na formate and acetate. Explanations included nucleophilic attack, H⁺ and H‐atom abstraction by superoxide. To probe how ¹O₂ is involved in forming the parasitic products we first discharged electrodes in DMA‐free electrolyte and then placed them for various times in electrolyte containing 30 mm DMA. X‐ray diffraction after storage shows decreased NaO₂ and side products to have formed in accord with previous studies (Figure S9). The electrolyte was then analyzed for DMA‐O₂ and the electrodes were analyzed for Na₂CO₃ content by immersing them in 0.1 m H₂SO₄ and monitoring the CO₂ evolved by mass spectrometry (Figure 3). ¹O₂ as indicated by the presence of DMA‐O₂ continuously increases with resting time as does the Na₂CO₃ content, which serves as an indication for the extent of parasitic products.

Figure 3.

¹O₂ generation and parasitic chemistry during prolonged contact of the discharge product NaO₂ with the electrolyte. Cells were first discharged in DMA‐free electrolyte (0.5 m NaOTf in diglyme containing 40 ppm H₂O), then the cathodes washed and immersed in the same electrolyte containing additionally 30 mm DMA. After the times indicated the electrolyte was analyzed for DMA‐O₂ and the electrodes for Na₂CO₃. The dotted lines are linear fits versus the square root of rest time.

DMA‐O₂ and Na₂CO₃ roughly increase proportional to the square root of the resting time, which means that the formation rate decreases with time. The formed parasitic products thus partly passivate the NaO₂ surface and slow down further decomposition. A larger amount of DMA‐O₂ than additional Na₂CO₃ formed, which can be explained by two phenomena. First, ¹O₂ trapped by DMA does not react with cell components and would thus have led to even more parasitic products without DMA. Second, other compounds than Na₂CO₃ will form which are not expelled by acid. Overall, Figure 3 shows that the previously reported electrolyte degradation upon resting is significantly driven by ¹O₂ formation.

Pathways to ¹O₂ during discharge, rest, and charge are summarized in Scheme 1. We suggest proton mediated superoxide disproportionation or reduction to interfere with the metathesis equilibrium HOO^.+Na⁺ $\overleftarrow{\underset{}{\to }}$ NaO₂+H⁺ as a universal path to ¹O₂ during all stages of cycling. Upon charge at voltages exceeding the thermodynamic threshold of 3.24 V ¹O₂ evolves by direct electrochemical oxidation according to NaO₂→Na⁺+e⁻+¹O₂.

Scheme 1.

Pathways leading to ¹O₂ during discharge, rest, and charge. “Disp”=disproportionation and “red”=reduction.

In conclusion we demonstrate that singlet oxygen forms at the cathode of the Na–O₂ cell during all stages of cycling and during rest, and accounts for a significant fraction of the side‐products formed. ¹O₂ is thus a major hurdle in cycling the Na–O₂ cell via the reversible formation/decomposition of NaO₂. Water takes up an ambivalent role by being required for high capacity via solution‐mediated growth and dissolution of large NaO₂ deposits, and at the same time appearing to be the main driver for ¹O₂ formation during discharge and at low charging voltages. Future work should thus focus on finding ways to either prevent ¹O₂ formation or to efficiently eliminate it.

Conflict of interest

The authors declare no conflict of interest.

Supporting information

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Supplementary

L. Schafzahl, N. Mahne, B. Schafzahl, M. Wilkening, C. Slugovc, S. M. Borisov, S. A. Freunberger, Angew. Chem. Int. Ed. 2017, 56, 15728.