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Cu+ is not stable and undergoes disproportion. E for cu+ disproportionation. (E ° cu2+/cu =0.153V , E ° cu+/cu=0.53V). (a) +0.683 V (b) -0.367 V (c) +0.3415 V (d) +0.367 V

 Cu+ is not stable and undergoes disproportion. E for cu+ disproportionation.


(E°  cu2+/cu =0.153V , E°  cu+/cu=0.53V).


(a)  +0.683 V


(b)  -0.367 V


(c) +0.3415 V


(d) +0.367 V

Grade:12

2 Answers

Ramesh V
70 Points
13 years ago

I want th make a small correction to question

Its cu2+/cu+ =0.153V  i.e.,                     cu2+ + e-   ------>  cu+            E0 = 0.153 V

Now going into problem

Cu+ solutions, tend to give a disproportion reaction:
2Cu+ ----- >   Cu + Cu2+

                       
                     cu+ + e-    ------>  cu                                E0 = 0.53 V        so,  nE0 = 0.53 V    

                     cu+            ------>  cu2+ + e-               E0 = - 0.153 V       so,  n E0 = - 0.153 V 

on adding above rxns we get   
 2Cu+ ----- >   Cu + Cu2+                  with  E0 = +0.377 V   

 

Option is D

ankit singh
askIITians Faculty 614 Points
2 years ago
The relationship between standard gibbs energy change for a reaction and its cell potential is ΔGo=nFEcello.
 
For copper electrodes,
 
ΔGCu2+Cuo=ΔGCu2+Cu+o+ΔGCu+Cuo
 
nFECu2+Cuo=nFECu2+Cu+onFECu+Cuo
 
2×ECu2+Cuo=1×ECu2+Cu+o+1×ECu+Cuo
 
Substitute values in the above equation.
 
2×ECu2+Cuo=1×ECu2+Cu+o+1×ECu+Cuo
 
                        =0.153V+0.53V
 
ECu2+Cu=0.3415 V

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