For the reaction :1) B(OH) 3 + 3e- ----> B + 3OH - E o = -0.890V G 1 o = -nFE o

Question

For the reaction :1) B(OH)3 + 3e- ----> B + 3OH- Eo = -0.890V

G1o = -nFEo = -3F * (-0.89) = 2.67F kJ/mol

2) 12 B + 14e- +12H2O ----> B12H122- +12OH-   Eo=0.08V

here each B changes its oxidation state from 0 to -7/6 , total e- change per B = 0- (-7/6) =7/6

Total gain of e- =7/6*12 =14e-

G2o = - nFEo = -14F* 0.08 = -1.12F kJ/mol

To find potential Eo of

B(OH)3 ----> B12H122-

multiply eqn 1 by 12 and add it into eqn 2 we get

12B(OH)3 + 50e- +12H2O---->B12H122- + 48OH-

For this reaction:

Go = Go1 +Go2

50F * Eo =12* 2.67F + (-1.12F)

50Eo = 32.04 - 1.12

Eo =0.62V

Explanation / Answer

Inorg. Chem. CH450 Exam 4 (version 2) Fri. 10:00-12.00 Noon, December 15,2017 Problem 3 continued (b) Calculate the potential for the conversion of B(OH)s to Bi2Hi?. (10 pts) 036 B(OH) -0.890, B 0.08 B2H1 2 0.32 B

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