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Section 2.4 Vanadium crossover b VO2+ ions diffuse into the negative half-cell and react to (4b + 2a) V3+ ions. Thereby, b V3+ and (a+2b) V2+ partake in the reactions. As b V3+ ions are required in the reactions, the VO2+ and VO2+ ions diffusing into the negative half-cell effectively react to (3b + 2a) V3+ ions in total. This process is depicted in Figure 2-2 a). The second order represents the simpler one, as shown in Figure 2-2 b). The VO2+ and VO2+ ions diffusing into the negative half-cell react to (3b + 2a) V3+ ions in total, which is equivalent to the first order. Because of its simpler sequence, the second order is implemented into the model. a) Negative half-cell, reaction order (2-17) → (2-18) (skipped) → (2-19) D5MemAMem/δMem(c5C+− c5C−) D4MemAMem/δMem(c4C+− c4C−) bVO2+ 2bVO2+ 4bV3+ bV3+ 2bV2+ aVO2+ 2aV3+ aV2+ Membrane Bulk electrolyte (2-17) (2-19) b) Negative half-cell, reaction order (2-18) → (2-17) (skipped) → (2-19) D5MemAMem/δMem(c5C+− c5C−) bVO2+ 3bV3+ 2bV2+ (2-18) D4MemAMem/δMem(c4C+− c4C−) aVO2+ 2aV3+ aV2+ (2-19) Membrane Bulk electrolyte (2-19) Figure 2-2: Self-discharge reactions due to the vanadium crossover in the negative half-cell V2+ ions are forced to leave the negative half-cell, because in the positive half-cell, no ions in this oxidation state are present and thus a concentration gradient exists. We can calculate the resulting ionic flux using Eq. (2-15), yielding Eq. (2-23). J D c2C c2C A 2C 2Mem Mem Mem (2-23) Now we are interested in the flux of VO2+ and VO2+ ions penetrating the negative half- cell through the membrane from the positive half-cell. These fluxes are given by the Eqs. (2-24) and (2-25). J J D D c4C c4C A Mem Mem c5C c5C A Mem Mem (2-24) (2-25) 4C 4Mem 5Mem 5C 22PDF Image | Model-based Design Vanadium Redox Flow Batteries
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