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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 $Fe^{2+}(aq)$ .

Answers (1)

best_answer

Answer-

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

negative $\rightarrow$ not 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) $Br_{2}+2e^{-}\rightarrow 2Br^{-}:E^{0}=+1.09V$

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

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

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

Posted by

Gautam harsolia

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