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Question

The standard reduction potential for Cu2+|Cu is +0.34V. Calculate the reduction potential at pH =14 for the above couple if Ksp of Cu(OH)2 is 1.0×1019.


A
0.2205 V
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B
+0.2205 V
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C
0.11 V
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D
+0.11 V
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Solution

The correct option is A 0.2205 V
pH = 14 means pOH = 0 or OH=1M.

Cu(OH)2Cu2++2OH

Also, Ksp=Cu2+OH2.

Substituting values in the above equation.
1×1019=Cu2+12Cu2+=1×1019

The expression for the reduction potential of the electrode is:

Ecell=E0cell+0.05922logCu2+

=0.34V+0.05922log1×1019

=0.2205V.

Hence, option A is correct.

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