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Electrochemistry

Question
CBSEENCH12006107

The standard reaction potential for Cu2+/Cu is + 0.34 V. Calculate the reduction potential at pH = 14 for the above couple. Ksp of Cu(OH)2 is 1.0 x 10–19.

Solution

reduction potential is given by:

E =E° - RT2Fln 1Cu2+/M
From the given data PH =14 and KsP (Cu(OH2) =1.0 x 10-19 


we get 
H+ =10-14 M thus,

[OH]- =OH- =Kw[H+] =10-14M10-14M =1M[Cu2+]  =KspOH- =1.0×10-191 = 1.0 x 10-19ME =0.34 -0.0592log11.0 x10-190.34-0.059×192 = 0.34 -0.56 =-0.22
ans -0.22V