Page 199 - Basic physical chemistry for the atmospheric sciences
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Answers, hints,  and solutions to exercises    1 8 S
                     Solution: From Table 6.2 we see that the  standard potential
                     for  the  FeH (aq)  - Fe2 + ( aq)  couple  is  0.77 1  V  .   Since  the
                     seawater  has  a  redox  potential  of  0.600  V,  it  provides  a
                                                                     2
                     better reducin g  environment  than  the  FeH (aq)  - Fe + (aq)
                     couple.  Therefore ,  this  couple  will  be  reduced ;  that  is,  it
                     will be driven from left to right.
                                                        �ed =  0 . 7 7 1   V

                     The  seawater  will  be  involved  in  the  oxidation  half-cell
                     reduction  (i. e . ,  it acts as  the  reductant)  with  E0x  =  - (0.600
                     V) =  - 0.600  V  .   When  the  seawater  is  in  equilibrium  with
                     the  iron  system  Ecell = Ere d  + E0x = 0  or  Ered = 0 . 6 00  V.  For
                     nonstandard concentrations at 298K ,  the total cell potential
                     developed  by the iron  system when paired with the  hydro­
                     gen half-cell is given by Eq.  (6.26).  However,  the hydrogen
                     half-cell  generates  zero  electrode  potential .  Therefore,  the
                     electrode  potential  developed  by  the  iron  system  is,  from
                     Eq.  (6 . 26)
                                                          2
                                              0.059 1   [ F [Fe + (aq)]
                                    £ d = �e d  -  -  1 -  log  eH (aq)]
                                     re
                     or,
                                                            2
                                                         [Fe + (aq)]
                                   0 . 6 00 = 0 . 77 1 - 0.059 1  l o g
                                                         [ FeH (aq)]
                     Therefor ,
                             e
                                 2
                               Fe + ( aq)
                                         782
                               FeH (aq)
            6.28.     1  equiv.  of H2S(aq)  is  1 7   g,  and  1  equiv .  of HN03(aq)  is  1
                                                                         2
                                           6
                       .   Hint: From Exercise  . 1 1   we have
                     g
                                                      1  mole of the species
                       Oxidation number of a  species = 1   .         ·
                                                      eqmv.  o  t
                                                              f  h  e  species
                                 0
            6 . 2 9.   Normality is  . 5 7 N  .

                                       CHAPTER  7
            7 . 7 .    Ratios of energies are  I :   0 . 08:  0 . 8 :   0.004.
            7.8 .     4 x  1 0 5  .
                            J
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