COMPOUNDS OF OXIDANT AND REDUCTANT 41 



E^ = ^" + -^^+0.0601 log \| ^ + 0.0601 log ^^ + "^^^ 

 2 ^Kr Ko + VKoKr 



= ^l±Il + 0.0601 log ^^^'^o+KoVK^ 

 2 KoVKk+KrVK'o 



2 

 From equations 28 and 40, by addition: 



2 E, - {E', + E,) = 0.0601 logi^fJ-l^ 



iogMiIM = o 



[RB,] [R] 



and 



[05^] [0] = [RB,] [/?] (a) 



but from equations 31 and 32: 



[OB,] [fiB,] 

 Hence : 



[0] [R] = [OB^] [RB,] (b) 



From equations a and h it follows that: 



[0] = [RB2] and [R] = [OB2] 

 and from equations 34, 35 and 42: 



[OB2] + [^^2] - 0.5 S 



i.e., half the metalloporphyrin is combined with base. Then from equations 

 43 and 44, y = l/.r, and from equation 48, Z = I. Hence: 



[B] = (Sb - ZS) = Sb- S 



It will be observed that, so long as q = r, the point at which [7^]^'" = KqKr 

 satisfies the above conditions, irrespective of the value of r, except that 



Z = r/2 and [B] = Sb - r/2 S. 



{3) When [B]^ = Ko: 



E, = E',+ 0.06 log^ + 0.06 log^^+^ 

 Kr 2Ko 



= Eo - 0.06 log ^^ ^ ^ from equation 39 

 Kb + Ko 



but since Kr is much smaller than Ko: 



El - Eh = 0.06 log 2 = 0.0181 v. 



