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  • (a) The cell in which the following reaction occurs :  <img alt="2Fe^{3+}\left ( aq \right )+2I^{-}\left ( aq \right )\rightarrow 2Fe^{2+}+I_{2}\left ( s \right )" src="https://entrancecorner.codecogs.com/gif.latex?2Fe%5E%7B3&plus;%7D%5

(a) The cell in which the following reaction occurs :
 2Fe^{3+}\left ( aq \right )+2I^{-}\left ( aq \right )\rightarrow 2Fe^{2+}+I_{2}\left ( s \right ) has E^{0}_{cell}= 0\cdot 236 \, V\; at\; 298\, K.
Calculate the standard Gibbs energy of the cell reaction.(Given : 1 F = 96,500 C mol-1)
(b) How many electrons flow through a metallic wire if a current of 0·5 A is passed for 2 hours ? (Given : 1 F = 96,500 C mol-1)

 

 

 

 
 
 
 
 

Answers (1)

(a) 2Fe^{3+}\left ( aq \right )+2I^{-}\left ( aq \right )\rightarrow 2Fe^{2+}+I_{2}\left ( s \right )
     E^{0}_{cell}= 0\cdot 236 \, V , T= 298\, K, IF= 96500\, Cmol^{-1}
   \Delta _{r}G^{0}= ?                 n = 2
   \Delta _{r}G^{0}= -nFE^{0}_{cell}
               = -2\times 96500\times 0\cdot 236
              = -45548\, J\, mol^{-1}


(b) I= 0\cdot 5A
     t= 2\, hours\, = 7200\, s
     Q=It
     Q=0\cdot 5\times 7200
         =3600\, c
Number of e^{-}= \frac{6\cdot 023\times 10^{23}\times 3600}{96500}
                       = 0\cdot 22469\times 10^{23}
                       = 2\cdot 25\times 10^{23}

Posted by

Sumit Saini

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