Influence of Hydrofluoric Acid Formation on Lithium Ion Insertion in Nanostructured V₂O₅

Abstract

Vanadium oxide (V 2 O 5 ) is a multifaceted material possessing desirable redox properties, including accessibility to multiple valence states, which make it attractive as a cathode for lithium ion batteries and microbatteries. Studies show that performance of this electrode material is dependent on the electrolyte employed and that solid electrolyte interphase (SEI) layer formation is responsible for the fade in capacity with multiple cycling. Nanostructured V 2 O 5 thin films synthesized through reactive ballistic deposition (RBD) were studied with electrochemical methods, ex situ Raman and ex situ XPS in two widely used electrolytes: LiClO 4 /propylene carbonate (PC) and LiPF 6 / diethyl carbonate (DEC) + ethylene carbonate (EC). Films cycled in LiPF 6 /DEC+EC experienced a 32% greater capacity fade between the first and second lithiathion/delithiation cycles than those cycled in LiClO 4 /PC, due to a redox-induced change in the surface morphology and composition and an irreversible transformation into an amorphous state as monitored by ex situ Raman. From X-ray photoelectron spectroscopy (XPS), it was shown that V 2 O 5 cycled in LiPF 6 /DEC+EC contained a high atomic concentration percentage of fluoride (16.18%) in comparison with V 2 O 5 electrodes cycled in LiClO 4 /PC (3.94%). No significant amounts of carbonates, oxalates, or oxyfluorophosphates typically associated with SEI formation were found when V 2 O 5 was cycled in either electrolyte. The results obtained suggest instead that HF, formed upon water contamination of the electrolyte, reacts with V 2 O 5 through a self-catalyzed process both at open circuit and under applied potential. The formation of vanadium oxyfluorides causes active mass loss and severe capacity fade upon discharging/charging.

Flexible and Highly Scalable V2O5‐rGO Electrodes in an Organic Electrolyte for Supercapacitor Devices

“…However, the relative low intensity of the deconvoluted V 4+ (VO 2 ) peaks suggests that a majority of the Vanadium are of +5 oxidation state, indicating that most of the oxides present were V 2 O 5 . The O1s spectrum (Figure c) centered at 532 eV can be deconvoluted into three peaks: 531.7 eV (O1s O‐C) and 533.2 eV (O1s O = C‐O) which corresponds to the remaining oxygen functional groups on rGO; V‐O peak at 530.6 eV is attributed to the oxygen ions in V 2 O 5 , where the intensity is in the similar range to the V2p /3 peaks; the peak at 535 eV is attributed to the presence of H 2 O molecules, indicating the highly hydrated state of our free‐standing electrodes. The C1s peak (Figure d) can be deconvoluted into 5 peaks, each indicating the different carbon bonds: C=C (284.6 eV); C‐N and C‐O (285.2 eV); C=N (286 eV); C=O (287 eV); O‐C=O (288.2 eV).…”

Direct Evidence of a Chemical Conversion Mechanism of Atomic-Layer-Deposited TiO2 Anodes During Lithiation Using LiPF6 Salt

“…Electrodes were also cycled in a second electrolyte containing 2000 ppm TBA added to the standard as a HF scavenger. Previously, the presence of TBA in solution has been shown to decrease the F – concentration by almost half . The third electrolyte, containing LiClO 4 salt in place of LiPF 6 , was chosen as a non-fluorine-containing control.…”

Preparation of Hydrophobic PPy Coated V2O5 Yolk–Shell Nanospheres-Based Cathode Materials with Excellent Cycling Performance

“…LiPF 6 first reacts with the trace amount of water in electrodes and produces HF and PF 5 (reaction ). , PF 5 further reacts with the trace amount of water to produce HF (reaction ). , HF then reacts with V 2 O 5 to form vanadium oxyfluorides (VOF n ) and H 2 O, thus resulting in the first dissolution of vanadium (reaction ).…”

“… , HF then reacts with V 2 O 5 to form vanadium oxyfluorides (VOF n ) and H 2 O, thus resulting in the first dissolution of vanadium (reaction ). , Moreover, VOF n further reacts with the newly formed or residual H 2 O to produce more HF and VOF n – x (OH) x , thereby leading to the chain reactions of HF formation and V 2 O 5 dissolution (reaction ). Therefore, the loss and amorphization of V 2 O 5 can be considered as a self-catalyzed process .…”

“… , Moreover, VOF n further reacts with the newly formed or residual H 2 O to produce more HF and VOF n – x (OH) x , thereby leading to the chain reactions of HF formation and V 2 O 5 dissolution (reaction ). Therefore, the loss and amorphization of V 2 O 5 can be considered as a self-catalyzed process . Hence, shielding the effects of trace water (H 2 O) on LiPF 6 in the electrolyte and avoiding the formation of HF during the electrochemical process could be an effective approach to solve the problem of vanadium dissolution.…”