Page 636 - Handbook of Battery Materials
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610  17 Liquid Nonaqueous Electrolytes

                    Table 17.19  Diffusion coefficients of lithium salts in organic liquids.

                                              ◦
                                                                  2
                                                                    −1
                    System      c           T ( C)  Method    D(cm ·s )      References
                         a
                                                                 +
                    LiTFSI in   1mol·L −1     –   pfg-NMR     D(Li ) =        [500]
                    GBL                                       1.65 × 10 −6
                                                                   −
                                                              D(TFSI ) =
                                                              2.26 × 10 −6
                                                                 +
                    LiTFSI in   n(solvent)/  30   pfg-NMR     D(Li ) =        [470]
                    GBL         n(salt) = 20                  1.6 × 10 −6
                                                                   −
                                                              D(TFSI ) =
                                                              2.6 × 10 −6
                                                                 +
                    LiTFSI in PC  1 mol·l −1  –   pfg-NMR     D(Li ) =        [500]
                                                              1.32 × 10 −6
                                                                   −
                                                              D(TFSI ) =
                                                              0.96 × 10 −6
                    LiTFSI in PC  1 mol·l −1  –   pfg-NMR     D(Li ) =        [460]
                                                                 +
                                                              0.77 × 10 −6
                                                                   −
                                                              D(TFSI ) =
                                                              0.93 × 10 −6
                    LiTFSI in EC  n(solvent)/  40  pfg-NMR    D(Li ) =        [470]
                                                                 +
                                n(salt) = 20                  2.1 × 10 −6
                                                                   −
                                                              D(TFSI ) =
                                                              3.1 × 10 −6
                    LiTFSI in   n(solvent)/  30   pfg-NMR     D(Li ) =        [470]
                                                                 +
                    DMC         n(salt) = 20                  5.8 × 10 −6
                                                                   −
                                                              D(TFSI ) =
                                                              6.0 × 10 −6
                    LiTFSI in   1mol·L −1     –   pfg-NMR     D(Li ) =        [460]
                                                                 +
                    THF                                       6.21 × 10 −6
                                                                   −
                                                              D(TFSI ) =
                                                              6.25 × 10 −6
                                                                 +
                    LiTFSI in   0.32 mol·kg −1  21  pfg-NMR   D(Li ) =        [509]
                         b
                    DEME TFSI                                 4.67 × 10 −8
                                                                   −
                                                              D(TFSI ) =
                                                              6.43 × 10 −8
                    LiBF 4 in   1mol·L −1     –   pfg-NMR     D(Li ) =        [455]
                                                                 +
                    EC/EMC                                    2.37 × 10 −6
                    (2/8)
                                                                   −
                                                              D(BF 4 ) =
                                                              2.62 × 10 −6
                                                                 +
                    LiBF 4 in GBL  1 mol·L −1  –  pfg-NMR     D(Li ) =        [500]
                                                              1.95 × 10 −6
                                                                   −
                                                              D(BF 4 ) =
                                                              4.0 × 10 −6
                                                                 +
                    LiBF 4 in PC  1 mol·L −1  –   pfg-NMR     D(Li ) =        [500]
                                                              0.94 × 10 −6
                                                                   −
                                                              D(BF 4 ) =
                                                              1.23 × 10 −6
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