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Using the standard electrode potentials given in the Table 8.1 predict if the reaction between the following is

8.26 Using the standard electrode potentials given in the Table 8.1, predict if the reaction between the following is feasible

(a) Fe^{3+}(aq) and I^{-}(aq)
(b) Ag^{+}(aq) and Cu(s)
(c) Fe^{3+}(aq) and Cu(s)
(d) Ag(s) and Fe^{3+} (aq)
(e) Br_{2}(aq) and  Fe^{2+}(aq).

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Answer- 

If E^{0} for the overall reaction is positive \rightarrow feasible

                                                 negative\rightarrownot feasible

(a)                        [Fe^{3+}+ e^{-}\rightarrow Fe^{2+} ]*2: E^{0} = 0.77V

                                               2I^{-}\rightarrow I_{2}+2e^{-}: E^{0} = -0.54V

                      ---------------------------------------------------------------------------------------

                             2Fe^{3+}+2I^{-}\rightarrow Fe^{2+}+I_{2}: E^{0} = +0.23V

 

(b)                           Ag^{+}+e^{-}\rightarrow Ag(s)]*2: E^{0} = +0.80V

                               Cu\rightarrow Cu^{2+}+2e^{-}:E^{0} =-0.34V

                    ---------------------------------------------------------------------------------------

                   2Ag^{+}+Cu\rightarrow 2Ag+Cu^{2+}: E^{0}=+0.46V

 

(c)     Fe^{3+}+e^{-}\rightarrow Fe^{2+}]*2 :E^{0}=+0.77V

        2Br^{-}\rightarrow Br+2e^{- }:E^{0}=-1.09V

-------------------------------------------------------------------------

2Fe^{3+}+2Br^{-}\rightarrow 2Fe^{3+}+Br_{2}:E^{0}=-0.32

 

(d)     Ag\rightarrow Ag^{+}+e^{-}:E^{0}=-0.80V

         Fe^{3+}+e^{-}\rightarrow Fe^{2+}:E^{0}==0.77V

------------------------------------------------------------------------------

   Ag+Fe^{3+}\rightarrow Ag^{+}+Fe^{2+}:E^{0}=-0.03V

 

(e)   Br_{2}+2e^{-}\rightarrow 2Br^{-}:E^{0}=+1.09V

       Fe^{2+}\rightarrow Fe^{3+}+e^{-}]*2:E^{0}=-0.77V

-------------------------------------------------------------------------------

Br_{2}+2Fe^{2+}\rightarrow 2Br^{-}+2Fe^{3+}:E^{0}=+0.32V

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