Question
Sat October 27, 2012 By:

As excess of liquid mercury is added to an acidified solution of 1.0x10^-3M Fe^+3. Its is found that 5% of Fe+3 remains at equilibrium at 298K. Calculate E0 Hg2^+2/Hg assuming that the only reaction that occurs is 2Hg(l) + 2Fe^+3(aq) >>>>> Hg2^+2(aq) + 2Fe^+3(aq) E0 Fe+3/Fe+2 = 0.77 Volt

Expert Reply
Fri November 09, 2012
          2Hg +    2 Fe                    ------>  Hg2 2+            +                2 F2 2+
t0                   1.0 x 10-3                                                                  0
 
 
teq  excess   5/100   x 1.0 x 105 M        (1.0 x 10-3  - 5.0 x 10-5)/2    1.0 x 10-3  - 5.0 x 10-5
 
                      = 5.0 x 10-5                        = 4.75 x 10-4 M                       = 9.5 x 10-4  M
 
 
E cell = ER- EL = EFe3+, Fe 2+    -  EHg2+, Hg  ..............................(I)
 
Keq  = [Hg2 2+][Fe2+]2 / [Fe3+]2
 
= 4.75 x 10-4 M   x ( 9.5 x 10-4  M)2   / ( 5.0 x 10-5  ) 2
 
= 0.1715 M
 
 
E cel l= RT/ nF   x ln Keq, n = 2 for this reaction
 
=  (0.05915 V)/ 2   x log 0.1715 = -0.023V ...................(II)
 
From eq I and II,
EFe3+, Fe 2+    -  EHg2+, Hg  = -0.023 V
 
 
 EHg2+, Hg   = 0.023 + 0.77 v  = 0.793 V
 
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