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

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Answer 1 · 2018-03-15T17:58:01

+0.38 V

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.

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

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