Method for repairing valence imbalance of all-vanadium electrolyte

Abstract

The invention discloses a method for repairing valence imbalance of an all-vanadium electrolyte, and relates to the field of batteries. The all-vanadium electrolyte valence state unbalance repairing method comprises the steps that valence state unbalance electrolytes to be repaired are mixed, the valence state of vanadium in the mixed solution is detected, whether valence state unbalance is valence state positive deviation or valence state negative deviation is judged according to the valence state of vanadium, the valence state positive deviation is repaired in an electrolysis, reduction and remixing repairing mode, and the valence state negative deviation is repaired in an electrolysis, reduction and remixing repairing mode. For valence state negative deviation, an oxidation method is adopted, the repairing efficiency is high, operation is easy, and the obtained target electrolyte has good electrochemical performance and can be used as positive and negative electrolyte again.

Description

Technical Field

[0001] This application relates to the field of batteries, and in particular to a method for repairing valence imbalance in an all-vanadium electrolyte. Background Technology

[0002] The electrolyte used in vanadium redox flow batteries (VRFB) at the factory is typically referred to as a 3.5±0.02 valence electrolyte, meaning that the electrolyte contains trivalent vanadium ions (V³⁺) and tetravalent vanadium ions (V⁴⁺). 4 The molar concentration ratio of ⁺ is 1:1. During long-term operation, due to side reactions (such as hydrogen evolution or oxygen evolution), the electrolyte valence state is prone to imbalance, deviating from 3.5 ± 0.02. This imbalance reduces battery capacity and efficiency. Therefore, it is necessary to restore the valence state of the electrolyte to 3.5 ± 0.02.

[0003] Currently, commonly used methods for repairing valence state imbalances in vanadium-containing electrolytes include electrolytic regeneration and electrolyte mixing compensation. When using electrolytic regeneration to treat electrolytes with valence state imbalances, the positive and negative electrodes are in opposition during electrolysis; an increase in the valence state on one side inevitably leads to a decrease on the other. Therefore, one side will always be incompletely treated, resulting in a portion of the electrolyte remaining untreated and low repair efficiency. The electrolyte mixing compensation method is even more demanding, requiring the identification of a suitable substance with a proper valence state difference and in appropriate quantities. It is complex to operate and difficult to implement in practice. Summary of the Invention

[0004] The main objective of this application is to propose a method for repairing valence imbalance in vanadium electrolytes, aiming to solve the problems of low repair efficiency and complex operation in existing methods for repairing valence imbalance in vanadium electrolytes.

[0005] This application provides a method for repairing valence imbalance in a vanadium-containing electrolyte, comprising the following steps: S1. Provide the positive electrolyte and the negative electrolyte to be repaired, and mix the positive electrolyte and the negative electrolyte to obtain a mixture; S2. Detect the valence state K value of vanadium in the mixture obtained in step S1, and determine the repair method of the mixture based on the K value. The determination criteria and repair methods of the mixture include: (1) If K>3.52, the mixture is divided into two parts. One part is used as the positive electrode electrolyte and the other part is used as the negative electrode electrolyte. Electrolysis is performed using an electrolysis device to reduce the valence state of vanadium in the negative electrode electrolyte to 3.0±0.02. A reducing agent is added to the positive electrode electrolyte to reduce the valence state of vanadium in the positive electrode electrolyte to 4.0±0.02. The negative electrode electrolyte with a valence state of 3.0±0.02 is mixed with the positive electrode electrolyte with a valence state of 4.0±0.02 to obtain the target electrolyte with a valence state of vanadium of 3.5±0.02. (2) If K < 3.48, add an oxidant to the mixture to raise the valence state of vanadium in the mixture to 3.5 ± 0.02, and obtain the target electrolyte; In step (1), the reducing agent is selected from at least one of ethylene glycol, glycerol, oxalic acid, citric acid and ascorbic acid; In step (2), the oxidant is selected from at least one of hydrogen peroxide, potassium permanganate, potassium dichromate and ammonium persulfate.

[0006] By employing the above technical solution, for the vanadium electrolyte with valence imbalance to be repaired, the positive and negative electrode electrolytes are first mixed and stirred to achieve thorough mixing and obtain a relatively homogeneous mixture. Samples are taken from the mixture and the molar concentrations of trivalent and tetravalent vanadium ions are measured to obtain the valence state K value of vanadium in the mixture. If the K value is greater than 3.52, it indicates that the valence state of vanadium in the mixture is positively deviated; conversely, if the K value is less than 3.48, it indicates that the valence state of vanadium in the mixture is negatively deviated.

[0007] For cases where the vanadium valence state deviates positively, the mixture is divided into two parts. One part is used as the positive electrolyte, and the other part as the negative electrolyte. Electrolysis is performed using an electrolysis device. By controlling the electrolysis time, the vanadium valence state in the negative electrolyte is reduced to 3.0 ± 0.02. At this point, the vanadium valence state in the positive electrolyte increases. A reducing agent is added to the positive electrolyte to reduce the vanadium valence state to 4.0 ± 0.02. The negative electrolyte with a valence state of 3.0 ± 0.02 and the positive electrolyte with a valence state of 4.0 ± 0.02 are mixed and stirred thoroughly to obtain the target electrolyte with a vanadium valence state of 3.5 ± 0.02. In the technical solution of this application, for cases where the valence state deviates positively, a repair method of mixing-electrolysis-reduction-remixing is adopted to obtain a target electrolyte with a valence state of 3.5±0.02. This method has high repair efficiency, is simple to operate, and the obtained target electrolyte has better electrochemical performance.

[0008] For cases where the valence state deviates negatively, a specific oxidant is directly added to the mixture to oxidize the excess trivalent vanadium ions to tetravalent vanadium ions, thereby raising the valence state of vanadium in the mixture to 3.5 ± 0.02, thus obtaining the target electrolyte. The resulting target electrolyte exhibits superior electrochemical performance.

[0009] In the technical solution of this application, the positive and negative electrode electrolytes with valence imbalance are first mixed. The valence state of vanadium in the mixture is used to determine whether the mixture is in a positive or negative deviation from the valence state. Then, it is classified and processed. The repair efficiency is high and the operation is simple. The target electrolyte obtained after repair can be used again as the positive and negative electrode electrolytes of the vanadium redox flow battery, so as to give the vanadium redox flow battery high energy efficiency and capacity retention.

[0010] Optionally, in the step (1), the electrolysis time is T, and the calculation formula of T is as follows: ,

[0011] In the formula, n1 is the amount of substance of total vanadium in the negative electrolyte, with the unit of mol; K1 is the valence state value of vanadium in the negative electrolyte to be electrolyzed; K2 is the target valence state value of vanadium after electrolysis of the negative electrolyte; I is the current used during electrolysis, with the unit of A; m is the total number of battery sheets included in the electrolysis stack; the unit of T is h.

[0012] By adopting the above technical solution, for the case of positive valence deviation, the repair method of electrolysis-reduction-re-mixing is adopted. The electrolysis step in it is relatively crucial. It is necessary to avoid over-electrolysis and prevent insufficient electrolysis. Therefore, it is necessary to strictly control the electrolysis time to obtain the electrolyte with the target valence state. In the technical solution of this application, by substituting the values of the parameters n1, K1, K2, I, and m corresponding to the above dimensions into the above formula of time T, the electrolysis time T required can be obtained, achieving the effect of relatively accurate electrolysis.

[0013] It can be understood that during electrolysis, the electrolysis stack includes multiple stack units connected in series. Each stack unit includes multiple battery sheets. m is the sum of the number of battery sheets in all stack units. For example, if the electrolysis stack includes 10 stack units connected in series, and each stack unit includes 45 battery sheets, then m = 10×45 = 450.

[0014] Optionally, in the step (1), if 3.52 < K ≤ 3.7, one-step electrolysis is adopted during electrolysis, and the value of K2 is 3.0 ± 0.02; If 3.7 < K ≤ 4.0, two-step electrolysis is adopted during electrolysis. When performing the first-step electrolysis, the value of K2 is 3.5 ± 0.02; when performing the second-step electrolysis, the value of K2 is 3.0 ± 0.02; Among them, when two-step electrolysis is adopted, after each step of electrolysis, a reducing agent needs to be added to the positive electrolyte to reduce the valence state of vanadium in the positive electrolyte to 4.0 ± 0.02.

[0015] By adopting the above technical solution, the K value of the valence state of vanadium in the mixed solution is further subdivided. If 3.52 < K ≤ 3.7, that is, the case where the positive valence deviation is not very high, one-step electrolysis can be adopted; if 3.7 < K ≤ 4.0, that is, the case where the positive valence deviation is relatively high, two-step electrolysis needs to be adopted. The step-by-step electrolysis method is adopted, and after each step of electrolysis, a reducing agent needs to be added to the positive electrolyte to reduce the valence state of vanadium in the positive electrolyte to 4.0 ± 0.02 to prevent precipitation due to excessive pentavalent vanadium content and block the stack, ensuring the electrolysis efficiency and electrolysis effect, and further ensuring the electrochemical performance of the obtained target electrolyte.

[0016] Optionally, in step (1), after electrolysis, the mass of reducing agent added to the positive electrode electrolyte is m1, and the formula for calculating m1 is as follows: ,

[0017] In the formula, n2 is the amount of vanadium pentavalent ions after electrolysis of the positive electrode electrolyte, in mol; M1 is the molar mass of the reducing agent, in g / mol; Z1 is the ratio of the molar coefficient of vanadium pentavalent ions to that of the reducing agent when vanadium pentavalent ions are completely reduced to tetravalent vanadium ions; ω1 is the purity of the reducing agent, in wt%; and m1 is in kg.

[0018] By adopting the above technical solution, for cases where the valence state deviates positively, electrolysis is used. After each electrolysis step, a specific amount of reducing agent needs to be added to the positive electrode electrolyte to accurately reduce the valence state of vanadium in the positive electrode electrolyte to 4.0±0.02.

[0019] Optionally, in step (2), the oxidant is ammonium persulfate, and the mass of the oxidant added to the mixture is m2. The formula for calculating m2 is as follows: ,

[0020] In the formula, C is the molar concentration of total vanadium in the mixture, in mol / L; V is the total volume of the mixture, in L; M2 is the molar mass of the oxidant, in g / mol; ω2 is the purity of the oxidant, in wt%; Z2 is the ratio of the molar coefficient of trivalent vanadium ions to that of the oxidant when trivalent vanadium ions are completely oxidized to tetravalent vanadium ions; and m2 is in g.

[0021] By adopting the above technical solution, for cases where the valence state deviates negatively, an oxidation method is used. A specific amount of oxidant is directly added to the mixture to raise the valence state of vanadium in the mixture to 3.5 ± 0.02, thus obtaining the target electrolyte. Regarding the selection of the oxidant, hydrogen peroxide is added to dilute the concentration of the electrolyte to be repaired, and potassium permanganate and potassium dichromate are added to introduce new metallic impurities. Therefore, hydrogen peroxide, potassium permanganate, and potassium dichromate all affect the charge-discharge efficiency of the electrolyte. Ammonium persulfate is preferred as the oxidant because: First, ammonium persulfate has a high standard oxidation potential (around 2.01V), which is much higher than the oxidation potential of tetravalent and trivalent vanadium ions (around 0.34V). Therefore, from the perspective of reaction feasibility, it is perfectly suitable, and its ability to oxidize trivalent vanadium ions is also strong.

[0022] 2. When ammonium persulfate is used as an oxidant, the sulfuric acid produced is one of the components of the vanadium electrolyte. The increase of sulfate ions has a positive effect on the conductivity and stability of the electrolyte, inhibits the precipitation of pentavalent vanadium, and enhances the electrode reactivity.

[0023] Third, when ammonium persulfate is used as an oxidant, the ammonium ions produced by its hydrolysis can significantly improve the stability of vanadium electrolyte (especially low-temperature stability) and also increase the solubility of vanadium ions. While dealing with the negative valence imbalance, it is equivalent to adding an additive in disguise, thereby improving the performance of the electrolyte.

[0024] Optionally, in step S2, the formula for calculating the valence state K value of vanadium in the mixture is as follows: ,

[0025] In the formula, C(V) 4+ ) represents the molar concentration of tetravalent vanadium ions in the mixture, in mol / L; C(V) represents the sum of the molar concentrations of trivalent and tetravalent vanadium ions in the mixture, in mol / L.

[0026] By adopting the above technical solution, the valence state K value of vanadium in the mixture can be obtained by calculating the ratio of the molar concentration of tetravalent vanadium ions to the molar concentration of total vanadium in the mixture, and then adding it to the constant term 3.

[0027] Optionally, in step S2, the molar concentrations of trivalent vanadium ions and tetravalent vanadium ions in the mixture are detected by ultraviolet-visible spectrophotometry or potentiometric titration.

[0028] By adopting the above technical solution, the molar concentrations of trivalent vanadium ions and tetravalent vanadium ions in the mixed solution are sampled and detected. Substituting these molar concentrations into the above formula for calculating the valence state K value, the valence state K value of vanadium in the mixed solution can be obtained.

[0029] Optionally, in step S2, the molar concentration of total vanadium in the mixture is 1.5~3.0 mol / L.

[0030] By adopting the above technical solution, the remediation method provided in this application is applicable to a total vanadium molar concentration with an upper limit of 3.0 mol / L, which is more widely applicable than the upper limit of 2 mol / L applicable to traditional methods.

[0031] Optionally, steps S1 and S2 are performed at a temperature of 5~40°C.

[0032] By adopting the above technical solution, the repair method provided in this application can not only be operated at room temperature, but also is applicable to low temperature environment. The repair method provided in this application has better environmental adaptability.

[0033] In summary, this application includes at least one of the following beneficial technical effects: 1. In the technical solution of this application, for cases where the valence state deviates positively, a repair method of mixing-electrolysis-reduction-remixing is adopted to obtain a target electrolyte with a valence state of 3.5±0.02. The repair efficiency is high, the operation is simple, and the obtained target electrolyte has better electrochemical performance and can be used again as a positive and negative electrode electrolyte.

[0034] 2. For cases where the valence state deviates negatively, a specific oxidant is directly added to the mixture to oxidize the excess trivalent vanadium ions to tetravalent vanadium ions, thereby raising the valence state of vanadium in the mixture to 3.5 ± 0.02, thus obtaining the target electrolyte. The resulting target electrolyte exhibits better electrochemical performance and can be reused as both positive and negative electrode electrolytes. Attached Figure Description

[0035] Figure 1 This is a capacity effect diagram corresponding to the electrolyte obtained by repair in Example 1 of this application; Figure 2 This is a capacity effect diagram corresponding to the electrolyte obtained by repair in Example 2 of this application; Figure 3 This is a capacity effect diagram corresponding to the electrolyte obtained by repair in Example 3 of this application; Figure 4 This is a diagram showing the capacity effect of the electrolyte obtained after repair in Example 4 of this application. Detailed Implementation

[0036] The present application will be further described in detail below with reference to the accompanying drawings and embodiments. Example 1

[0037] A method for repairing valence imbalance in a vanadium-containing electrolyte, the method being operated at 30°C, and comprising the following steps: S1. Provide 5.01L of positive electrolyte and 5.01L of negative electrolyte to be repaired, and mix the 5.01L positive electrolyte and 5.01L negative electrolyte. Stir at 200rpm for 20min to obtain a mixture.

[0038] S2. Take a 0.02 L sample from the mixture obtained in step S1 and use potentiometric titration to determine the molar concentrations of trivalent vanadium ions and tetravalent vanadium ions in the sample. The molar concentrations of trivalent vanadium ions are 0.6 mol / L and tetravalent vanadium ions are 1.4 mol / L. Therefore, the total molar concentration of vanadium in the mixture is 2.0 mol / L. Thus, the valence state K of vanadium in the mixture is 3 + (1.4 / 2.0) = 3.7.

[0039] S3. Divide the remaining mixture after sampling in step S2 into two equal volumes. One portion will be used as the positive electrolyte, and the other as the negative electrolyte. Electrolyze the mixture using an electrolysis device under the following conditions: charging cut-off voltage of 1.55V, discharging cut-off voltage of 1.00V, and current density of 160mA / cm³. 2 The area of ​​both the positive and negative electrodes is 50 cm². 2 The electrolysis time T (h) is controlled, and the formula for calculating T is as follows: ,

[0040] In the formula, n1 is the total amount of vanadium in the negative electrode electrolyte, in mol; K1 is the valence state of vanadium in the negative electrode electrolyte to be electrolyzed; K2 is the target valence state of vanadium after electrolysis; I is the current used during electrolysis, in A; m is the total number of cells in the electrolytic stack; and T is in hours (h). Specifically, in this embodiment, n1 = 5L × 2.0 mol / L = 10.0 mol, K1 = 3.7, K2 = 3.0, I = 8A, and m = 100. Substituting the values ​​of n1, K1, K2, I, and m into the above formula, the electrolysis time T = 0.2345 h is obtained.

[0041] After electrolysis for 0.2345 hours, the electrolysis was completed. A 0.02 L sample was taken from the negative electrode electrolyte after electrolysis, and the molar concentration of vanadium ions in the negative electrode electrolyte sample was determined by potentiometric titration. The valence state of vanadium is 3+[C(V 4+ ) / C(V)], where C(V) 4+ ) represents the molar concentration of tetravalent vanadium ions in the negative electrode electrolyte sample, in mol / L; C(V) represents the sum of the molar concentrations of trivalent and tetravalent vanadium ions in the negative electrode electrolyte sample, in mol / L. After detection and calculation, the valence state of vanadium in the electrolyzed negative electrode electrolyte is 3.0.

[0042] S4. Add glycerol (purity 98.0 wt%) as a reducing agent to the positive electrode electrolyte after electrolysis. The mass of glycerol added is m1 (kg), and the formula for calculating m1 is as follows: ,

[0043] In the formula, n2 is the amount of pentavalent vanadium ions in the positive electrode electrolyte after electrolysis, in mol; M1 is the molar mass of the reducing agent, in g / mol; Z1 is the ratio of the molar coefficient of pentavalent vanadium ions to that of the reducing agent when pentavalent vanadium ions are completely reduced to tetravalent vanadium ions; ω1 is the purity of the reducing agent, in wt%; and m1 is in kg. Specifically, in this embodiment, after electrolysis, 0.02 L of the positive electrode electrolyte after electrolysis is taken, and the molar concentration (n2) of pentavalent vanadium ions in the positive electrode electrolyte sample is detected by potentiometric titration. The result shows that n2 = 0.8 mol / L, where M1 = 92 g / mol, Z1 = 14, ω1 = 0.98, and m1 = 5.364 × 10⁻⁶. -3 kg.

[0044] Add 5.364 × 10⁻⁶ to the positive electrode electrolyte after electrolysis. -3 kg of glycerol was used to reduce vanadium ions in the 5-valent state to vanadium ions in the 4-valent state. A 0.02 L sample was taken from the reduced positive electrode electrolyte, and the molar concentration of vanadium ions in the sample was determined by potentiometric titration. The valence state of vanadium in the reduced positive electrode electrolyte was 4+[C(V... 5+ ) / C(V)], where C(V) 5+ ) represents the molar concentration of pentavalent vanadium ions in the positive electrode electrolyte sample, in mol / L; C(V) represents the sum of the molar concentrations of pentavalent and tetravalent vanadium ions in the positive electrode electrolyte sample, in mol / L. After detection and calculation, the valence state of vanadium in the reduced positive electrode electrolyte is 4.0.

[0045] S5. The positive electrode electrolyte with a valence of 4.0 obtained in step S4 and the negative electrode electrolyte with a valence of 3.0 obtained in step S3 are mixed at a volume ratio of 1:1 and stirred at a stirring rate of 200 rpm for 20 min to obtain the target electrolyte with a valence of 3.5. Example 2

[0046] A method for repairing valence imbalance in a vanadium-containing electrolyte, the method being operated at 30°C, and comprising the following steps: S1. Provide 5.01L of positive electrolyte and 5.01L of negative electrolyte to be repaired, and mix the 5.01L positive electrolyte and 5.01L negative electrolyte. Stir at 200rpm for 20min to obtain a mixture.

[0047] S2. Take a 0.02 L sample from the mixture obtained in step S1 and use potentiometric titration to determine the molar concentrations of trivalent vanadium ions and tetravalent vanadium ions in the sample. The molar concentrations of trivalent vanadium ions are 0.6 mol / L and tetravalent vanadium ions are 1.4 mol / L. Therefore, the total molar concentration of vanadium in the mixture is 2.0 mol / L. Thus, the valence state K of vanadium in the mixture is 3 + (1.4 / 2.0) = 3.7.

[0048] S3. Divide the remaining mixture after sampling in step S2 into two equal volumes. One portion will be used as the positive electrolyte, and the other as the negative electrolyte. Electrolyze the mixture using an electrolysis device under the same conditions as in Example 1. Control the electrolysis time T (h), and the formula for calculating T is the same as in Example 1.

[0049] After electrolysis for 0.2345 hours, the electrolysis was completed. A 0.02 L sample was taken from the negative electrode electrolyte after electrolysis, and the molar concentration of vanadium ions in the negative electrode electrolyte sample was detected by potentiometric titration to calculate the valence state of vanadium in the negative electrode electrolyte. The calculation method was the same as in Example 1. After detection and calculation, the valence state of vanadium in the negative electrode electrolyte after electrolysis was 3.0.

[0050] S4. Add anhydrous oxalic acid (purity 98.0 wt%) as a reducing agent to the positive electrode electrolyte after electrolysis. The mass of anhydrous oxalic acid added is m1 (kg), and the calculation formula for m1 is the same as in Example 1. Specifically, in this example, after electrolysis, 0.02 L of the positive electrode electrolyte after electrolysis is taken, and the molar concentration (n2) of pentavalent vanadium ions in the positive electrode electrolyte sample is detected by potentiometric titration. The detection result is n2 = 0.8 mol / L, where M1 = 90 g / mol, Z1 = 2, ω1 = 0.98. The calculated value of m1 is 36.735 × 10⁻⁶. -3 kg. Add 36.735 × 10⁻⁶ kg to the positive electrode electrolyte after electrolysis. -3 kg of anhydrous oxalic acid was used to reduce vanadium ions in the 5-valent state to vanadium ions in the 4-valent state. A 0.02 L sample was taken from the reduced positive electrode electrolyte, and the molar concentration of vanadium ions in the sample was determined by potentiometric titration. The valence state of vanadium in the positive electrode electrolyte was calculated using the same method as in Example 1. The results showed that the valence state of vanadium in the reduced positive electrode electrolyte was 4.0.

[0051] S5. The positive electrode electrolyte with a valence of 4.0 obtained in step S4 and the negative electrode electrolyte with a valence of 3.0 obtained in step S3 are mixed at a volume ratio of 1:1 and stirred at a stirring rate of 200 rpm for 20 min to obtain the target electrolyte with a valence of 3.5. Example 3

[0052] A method for repairing valence imbalance in a vanadium-based electrolyte, the method being operated at 5°C, and comprising the following steps: S1. Provide 5.01L of positive electrolyte and 5.01L of negative electrolyte to be repaired, and mix the 5.01L positive electrolyte and 5.01L negative electrolyte. Stir at 200rpm for 20min to obtain a mixture.

[0053] S2. Take a 0.02 L sample from the mixture obtained in step S1 and use potentiometric titration to determine the molar concentrations of trivalent vanadium ions and tetravalent vanadium ions in the sample. The molar concentrations of trivalent vanadium ions are 0.6 mol / L and tetravalent vanadium ions are 1.4 mol / L. Therefore, the total molar concentration of vanadium in the mixture is 2.0 mol / L. Thus, the valence state K of vanadium in the mixture is 3 + (1.4 / 2.0) = 3.7.

[0054] S3. Divide the remaining mixture after sampling in step S2 into two equal volumes. One portion will be used as the positive electrolyte, and the other as the negative electrolyte. Electrolyze the mixture using an electrolysis device under the same conditions as in Example 1. Control the electrolysis time T (h), and the formula for calculating T is the same as in Example 1.

[0055] After electrolysis for 0.2345 hours, the electrolysis was completed. A 0.02 L sample was taken from the negative electrode electrolyte after electrolysis, and the molar concentration of vanadium ions in the negative electrode electrolyte sample was detected by potentiometric titration to calculate the valence state of vanadium in the negative electrode electrolyte. The calculation method was the same as in Example 1. After detection and calculation, the valence state of vanadium in the negative electrode electrolyte after electrolysis was 3.0.

[0056] S4. Add glycerol (purity 98.0 wt%) as a reducing agent to the electrolyzed positive electrode electrolyte. The mass of glycerol added is m1 (kg), and the calculation formula for m1 is the same as in Example 1. Specifically, in this example, after electrolysis, 0.02 L of the electrolyzed positive electrode electrolyte is taken, and the molar concentration (n2) of pentavalent vanadium ions in the positive electrode electrolyte sample is detected by potentiometric titration. The detection result is n2 = 0.8 mol / L, where M1 = 92 g / mol, Z1 = 14, ω1 = 0.98. The calculated value of m1 is 5.364 × 10⁻⁶. - 3 kg. Add 5.364 × 10⁻⁶ kg to the positive electrode electrolyte after electrolysis. -3 kg of glycerol was used to reduce pentavalent vanadium ions to tetravalent vanadium ions. A 0.02 L sample was taken from the reduced positive electrode electrolyte, and the molar concentration of vanadium ions in the sample was determined by potentiometric titration. The valence state of vanadium in the positive electrode electrolyte was calculated using the same method as in Example 1. The results showed that the valence state of vanadium in the reduced positive electrode electrolyte was 4.0.

[0057] S5. The positive electrode electrolyte with a valence of 4.0 obtained in step S4 and the negative electrode electrolyte with a valence of 3.0 obtained in step S3 are mixed at a volume ratio of 1:1 and stirred at a stirring rate of 200 rpm for 20 min to obtain the target electrolyte with a valence of 3.5. Example 4

[0058] A method for repairing valence imbalance in a vanadium-containing electrolyte, the method being operated at 30°C, and comprising the following steps: S1. Provide 5.01L of positive electrolyte and 5.01L of negative electrolyte to be repaired, and mix the 5.01L positive electrolyte and 5.01L negative electrolyte. Stir at 200rpm for 20min to obtain a mixture.

[0059] S2. Take a 0.02 L sample from the mixture obtained in step S1 and use potentiometric titration to determine the molar concentrations of trivalent vanadium ions and tetravalent vanadium ions in the mixture sample. The results show that the molar concentration of trivalent vanadium ions is 0.5 mol / L and the molar concentration of tetravalent vanadium ions is 2.0 mol / L. That is, the total molar concentration of vanadium in the mixture is 2.5 mol / L, and the valence state K value of vanadium in the mixture is 3 + (2.0 / 2.5) = 3.8.

[0060] S3. Divide the remaining mixture after sampling in step S2 into two equal volumes. One portion is used as the positive electrolyte, and the other portion is used as the negative electrolyte. Electrolyze the mixture in two steps using an electrolysis device. The electrolysis conditions for each step are the same as in Example 1. Control the electrolysis time T (h) of the first step. The calculation formula for T is the same as in Example 1. In this example, n1 = 5L × 2.5mol / L = 12.5mol, K1 = 3.8, K2 = 3.5, I = 8A, and m = 100. Substituting the values ​​corresponding to n1, K1, K2, I, and m into the formula, we obtain the electrolysis time of the first step, T = 0.125625h. After the first electrolysis step is completed, 0.02 L of sample is taken from the negative electrode electrolyte after electrolysis, and the molar concentration of vanadium ions in the negative electrode electrolyte sample is detected by potentiometric titration. The valence state of vanadium in the negative electrode electrolyte is calculated according to the method in Example 1. After detection and calculation, the valence state of vanadium in the negative electrode electrolyte after electrolysis is 3.51.

[0061] S4. Add glycerol (98.0 wt% purity) as a reducing agent to the positive electrode electrolyte after the first electrolysis step. The mass of glycerol added is m1 (kg). The calculation formula for m1 is the same as in Example 1. In this example, after the first electrolysis step is completed, 0.02 L of the positive electrode electrolyte is taken and the molar concentration (n2) of pentavalent vanadium ions in the positive electrode electrolyte sample is detected by potentiometric titration. The detection result is n2 = 0.225 mol / L, where M1 = 92 g / mol, Z1 = 14. The calculated value of m1 is 1.509 × 10⁻⁶. -3 kg. Add 1.509 × 10⁻⁶ kg to the positive electrode electrolyte after the first electrolysis step. -3 kg of glycerol was used to reduce vanadium ions in the 5-valent state to vanadium ions in the 4-valent state. A 0.02 L sample was taken from the reduced positive electrode electrolyte, and the molar concentration of vanadium ions in the positive electrode electrolyte sample was determined by potentiometric titration. The valence state of vanadium in the reduced positive electrode electrolyte was calculated to be 4.0 valence according to the method in Example 1.

[0062] S5. The positive electrode electrolyte with a valence state of 4.0 obtained in step S4 and the negative electrode electrolyte with a valence state of 3.51 obtained in step S3 are subjected to a second electrolysis using the same electrolysis equipment and conditions as in step S3. The second electrolysis time T (h) is controlled and the calculation formula for T is the same as in Example 1. In this example, n1 = 4.98L × 2.5mol / L = 12.45 mol, K1 = 3.51, K2 = 3.0, I = 8A, m = 100. Substituting the values ​​corresponding to n1, K1, K2, I, and m into the formula, the second electrolysis time T = 0.2127h is obtained. After the second electrolysis is completed, 0.02L of the negative electrode electrolyte is sampled, and the molar concentration of vanadium ions in the negative electrode electrolyte sample is detected by potentiometric titration. The valence state of vanadium in the negative electrode electrolyte after electrolysis is detected and calculated to be 3.0 according to the method in Example 1. The mass of the re-added reducing agent glycerol (purity 98.0 wt%) was detected and calculated using the method in step S4: m1 = 8.55 × 10⁻⁶. -3 kg, add 8.55 × 10 kg to the positive electrode electrolyte after the second electrolysis step. -3 kg of glycerol (98.0 wt% purity) was used as a reducing agent to reduce vanadium ions in the 5-valent state to vanadium ions in the 4-valent state. A 0.02 L sample was taken from the reduced positive electrode electrolyte, and the molar concentration of vanadium ions in the positive electrode electrolyte sample was determined by potentiometric titration. The valence state of vanadium in the reduced positive electrode electrolyte was calculated to be 4.0 valence according to the method in Example 1.

[0063] S6. The positive electrode electrolyte with a valence of 4.0 and the negative electrode electrolyte with a valence of 3.0 obtained in step S5 are mixed at a volume ratio of 1:1 and stirred at a stirring rate of 200 rpm for 20 min to obtain the target electrolyte with a valence of 3.5. Example 5

[0064] A method for repairing valence imbalance in a vanadium-containing electrolyte, the method being operated at 30°C, and comprising the following steps: S1. Provide 5.01L of positive electrolyte and 5.01L of negative electrolyte to be repaired, and mix the 5.01L positive electrolyte and 5.01L negative electrolyte. Stir at 200rpm for 20min to obtain a mixture.

[0065] S2. Take a 0.02 L sample from the mixture obtained in step S1 and use potentiometric titration to determine the molar concentrations of trivalent vanadium ions and tetravalent vanadium ions in the sample. The results show that the molar concentration of trivalent vanadium ions is 1.6 mol / L and the molar concentration of tetravalent vanadium ions is 0.4 mol / L. Therefore, the total molar concentration of vanadium in the mixture is 2.0 mol / L, and the valence state K of vanadium in the mixture is 3 + (0.4 / 2.0) = 3.2.

[0066] S3. Add ammonium persulfate (99wt% purity) as an oxidant to the remaining mixture after sampling in step S2. The mass of ammonium persulfate added is m2 (g), and the formula for calculating m2 is as follows: ,

[0067] In the formula, C is the molar concentration of total vanadium in the mixture, in mol / L; V is the total volume of the mixture, in L; M2 is the molar mass of the oxidant, in g / mol; ω2 is the purity of the oxidant, in wt%; Z2 is the ratio of the molar coefficient of trivalent vanadium ions to that of the oxidant when trivalent vanadium ions are completely oxidized to tetravalent vanadium ions; and m2 is in g.

[0068] In this example, K=3.2, C=2.0mol / L, V=10L, M2=228g / mol, Z2=2, ω=0.99, and the calculated m2=1842.42g. 1842.42g of ammonium persulfate (99wt% purity) was added to the mixture to oxidize excess trivalent vanadium ions to tetravalent vanadium ions. After oxidation, a 0.02L sample was taken, and the molar concentrations of trivalent and tetravalent vanadium ions were determined by potentiometric titration. The measured molar concentrations of trivalent and tetravalent vanadium ions were 1.01mol / L and 0.99mol / L, indicating that the vanadium valence state in the resulting mixture after oxidation was 3.495, thus obtaining the target electrolyte. Example 6

[0069] A method for repairing valence imbalance in a vanadium-containing electrolyte, the method being operated at 30°C, and comprising the following steps: S1. Provide 5.01L of positive electrolyte and 5.01L of negative electrolyte to be repaired, and mix the 5.01L positive electrolyte and 5.01L negative electrolyte. Stir at 200rpm for 20min to obtain a mixture.

[0070] S2. Take a 0.02 L sample from the mixture obtained in step S1 and use potentiometric titration to determine the molar concentrations of trivalent vanadium ions and tetravalent vanadium ions in the sample. The results show that the molar concentration of trivalent vanadium ions is 1.6 mol / L and the molar concentration of tetravalent vanadium ions is 0.4 mol / L. Therefore, the total molar concentration of vanadium in the mixture is 2.0 mol / L, and the valence state K of vanadium in the mixture is 3 + (0.4 / 2.0) = 3.2.

[0071] S3. Add potassium permanganate (purity 99wt%) as an oxidant to the remaining mixture after sampling in step S2. The mass of potassium permanganate added is m2 (g), and the formula for calculating m2 is as follows: ,

[0072] In the formula, C is the molar concentration of total vanadium in the mixture, in mol / L; V is the total volume of the mixture, in L; M2 is the molar mass of the oxidant, in g / mol; ω2 is the purity of the oxidant, in wt%; Z2 is the ratio of the molar coefficient of trivalent vanadium ions to that of the oxidant when trivalent vanadium ions are completely oxidized to tetravalent vanadium ions; and m2 is in g.

[0073] In this embodiment, K=3.2, C=2.0mol / L, V=10L, M2=158g / mol, Z2=5, ω=0.99, and the calculated m2=510.71g. 510.71g of potassium permanganate (99wt% purity) was added to the mixture to oxidize excess trivalent vanadium ions to tetravalent vanadium ions. After oxidation, a 0.02L sample was taken, and the molar concentrations of trivalent and tetravalent vanadium ions in the sample were determined by potentiometric titration. The measured molar concentrations of trivalent and tetravalent vanadium ions were 1.0mol / L and 0.99mol / L, respectively. This indicates that the vanadium valence state in the resulting mixture after oxidation is 3.5, thus yielding the target electrolyte. Comparative Example 1

[0074] The vanadium electrolyte in this comparative example is the unrepaired electrolyte with valence imbalance from Example 1. That is, the molar concentration of trivalent vanadium ions in the electrolyte is 0.6 mol / L, the molar concentration of tetravalent vanadium ions is 1.4 mol / L, the molar concentration of total vanadium is 2.0 mol / L, and the valence K value of vanadium is 3 + (1.4 / 2.0) = 3.7. Performance Test 1

[0075] Take 130 mL of each of the electrolytes obtained from Examples 1-4, distribute them evenly into the positive and negative electrode containers, assemble a single cell, and test its charge and discharge performance.

[0076] Among them, 160mA / cm 2 The current density was tested at a constant current of 8A for 50 effective charge-discharge cycles, with charge-discharge cutoff voltages of 1.65V and 0.8V respectively, using nitrogen as the protective gas; the test results are as follows. Figures 1-4 As shown. By Figures 1-4 It is known that after repairing the vanadium electrolyte with valence imbalance according to the repair method provided in this application, the capacity of the obtained electrolyte can be restored to more than 95% of the initial discharge capacity at the time of manufacture.

[0077] in, Figures 1-4 The initial charging capacity and initial discharging capacity refer to the initial capacity obtained by testing the electrolyte at the time of manufacture. Performance Test 2

[0078] The electrolytes obtained from the repair in Examples 1-6, and the electrolyte with valence imbalance provided in Comparative Example 1, were used as positive and negative electrode electrolytes to assemble an all-vanadium redox flow battery, and the constant current charge-discharge performance of the repaired electrolyte was tested.

[0079] Among them, 160mA / cm 2 The current density was tested at a constant current of 8A for 100 effective charge-discharge cycles, with charge and discharge cutoff voltages of 1.65V and 0.8V, respectively, using nitrogen as the protective gas. The battery's energy efficiency (%) and capacity retention (%) after 100 effective charge-discharge cycles were tested, and the results are shown in Table 1 below.

[0080] Wherein, the battery capacity retention rate = (Q 100 / Q)*100%, where Q 100 Q refers to the discharge energy at the 100th effective cycle, while Q refers to the maximum discharge energy obtained in 100 effective cycles.

[0081] Table 1 Electrolyte Constant Current Charge-Discharge Performance

[0082] As shown in Table 1, the target electrolyte obtained by this application can be used again as the positive and negative electrode electrolyte of the all-vanadium redox flow battery, and the electrolyte obtained by the application has better electrochemical performance.

Claims

1. A method for repairing valence imbalance in a vanadium-containing electrolyte, characterized in that, It includes the following steps: S1. Provide the positive electrode electrolyte and the negative electrode electrolyte to be repaired, mix the positive electrode electrolyte and the negative electrode electrolyte to obtain a mixed solution; S2. Detect the valence state K value of vanadium in the mixed solution obtained in step S1, and determine the repair method of the mixed solution according to the K value. The determination criteria and the repair methods of the mixed solution include: (1) If K > 3.52, divide the mixed solution into two parts. One part of the mixed solution is used as the positive electrode electrolyte, and the other part of the mixed solution is used as the negative electrode electrolyte. Use an electrolysis device for electrolysis to reduce the valence state of vanadium in the negative electrode electrolyte to 3.0 ± 0.02, and add a reducing agent to the positive electrode electrolyte to reduce the valence state of vanadium in the positive electrode electrolyte to 4.0 ± 0.

02. Mix the negative electrode electrolyte with a valence state of 3.0 ± 0.02 and the positive electrode electrolyte with a valence state of 4.0 ± 0.02 to obtain a target electrolyte with a valence state of vanadium of 3.5 ± 0.02; (2) If K < 3.48, add an oxidizing agent to the mixed solution to increase the valence state of vanadium in the mixed solution to 3.5 ± 0.02 to obtain a target electrolyte; Among them, in step (1), the reducing agent is selected from at least one of ethylene glycol, glycerol, oxalic acid, citric acid, and ascorbic acid; In step (2), the oxidizing agent is selected from at least one of hydrogen peroxide, potassium permanganate, potassium dichromate, and ammonium persulfate.

2. The method for repairing the valence imbalance of the all-vanadium electrolyte according to claim 1, characterized in that, In step (1), the electrolysis time is T, and the calculation formula of T is as follows: ， In the formula, n1 is the amount of substance of total vanadium in the negative electrode electrolyte, with the unit of mol; K1 is the valence state value of vanadium in the negative electrode electrolyte to be electrolyzed; K2 is the target valence state value of vanadium in the negative electrode electrolyte after electrolysis; I is the current used during electrolysis, with the unit of A; m is the total number of battery sheets included in the electrolysis stack; the unit of T is h.

3. The method for repairing the valence imbalance of the all-vanadium electrolyte according to claim 2, characterized in that, In step (1), if 3.52 < K ≤ 3.7, one-step electrolysis is adopted during electrolysis, and the value of K2 is 3.0 ± 0.02; If 3.7 < K ≤ 4.0, two-step electrolysis is adopted during electrolysis. When performing the first-step electrolysis, the value of K2 is 3.5 ± 0.02; when performing the second-step electrolysis, the value of K2 is 3.0 ± 0.02; Among them, when two-step electrolysis is adopted, after each step of electrolysis, a reducing agent needs to be added to the positive electrode electrolyte to reduce the valence state of vanadium in the positive electrode electrolyte to 4.0 ± 0.

02.

4. The method for repairing the valence imbalance of the all-vanadium electrolyte according to claim 1, characterized in that, In step (1), after electrolysis, the mass of the reducing agent added to the positive electrode electrolyte is m1, and the calculation formula of m1 is as follows: ， In the formula, n2 is the amount of substance of vanadium ions with a valence state of 5 in the positive electrode electrolyte after electrolysis, with the unit of mol; M1 is the molar mass of the reducing agent, with the unit of g / mol; Z1 is the ratio of the molar coefficient of vanadium ions with a valence state of 5 to the reducing agent when vanadium ions with a valence state of 5 are completely reduced to vanadium ions with a valence state of 4; ω1 is the purity of the reducing agent, with the unit of wt%; the unit of m1 is kg.

5. The method for repairing the valence imbalance of the all-vanadium electrolyte according to claim 1, characterized in that, In step (2), when the oxidizing agent is ammonium persulfate, the mass of the oxidizing agent added to the mixed solution is m2, and the calculation formula of m2 is as follows: ， In the formula, C is the molar concentration of total vanadium in the mixture, in mol / L; V is the total volume of the mixture, in L; M2 is the molar mass of the oxidant, in g / mol; ω2 is the purity of the oxidant, in wt%; Z2 is the ratio of the molar coefficient of trivalent vanadium ions to that of the oxidant when trivalent vanadium ions are completely oxidized to tetravalent vanadium ions; and m2 is in g.

6. The method for repairing the valence imbalance of the all-vanadium electrolyte according to claim 1, characterized in that, In step S2, the formula for calculating the valence state K value of vanadium in the mixture is as follows: ， In the formula, C(V) 4+ ) represents the molar concentration of tetravalent vanadium ions in the mixture, in mol / L; C(V) represents the sum of the molar concentrations of trivalent and tetravalent vanadium ions in the mixture, in mol / L.

7. The method for repairing the valence imbalance of a vanadium-containing electrolyte according to claim 6, characterized in that, In step S2, the molar concentrations of trivalent vanadium ions and tetravalent vanadium ions in the mixed solution are detected by ultraviolet-visible spectrophotometry or potentiometric titration.

8. The method for repairing the valence imbalance of a full vanadium electrolyte according to claim 1, characterized in that, In step S2, the total vanadium molar concentration in the mixture is 1.5~3.0 mol / L.

9. The method for repairing the valence imbalance of a full vanadium electrolyte according to claim 1, characterized in that, Steps S1 and S2 are performed at a temperature of 5~40℃.

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