If electrode potential of 0.01M Cu^2+/Cu is 0.32V then what will be its E°_RP?

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Michael Share
Mar 15, 2018

Answer:

+0.38 V

Explanation:

The half cell is:

#sf(Cu^(2+)+2erightleftharpoonsCu)#

The potential of a metal / metal ion half cell is given by:

#sf(E=E^@-(RT)/(zF)lncolor(white)(x)(["reduced form"])/(["oxidised form"]))#

At #sf(25^@C)# this simplifies to:

#sf(E=E^@+0.0591/(z)log[Cu^(2+)])#

#sf(z)# is the number of moles of electrons transferred which, in this case = 2.

#:.##sf(+0.32=E^(@)+0.0591/(2)xx-2)#

#sf(E^@=+0.38color(white)(x)V)#

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