Question #3a73e

1 Answer
Jan 18, 2017

Answer:

#sf((A))#

Explanation:

The expression for half - cell potential is:

#sf(E=E^@-(RT)/(zF)ln""[[red]]/(["ox"]))#

For a metal/metal ion system at 298 K this can be simplified to:

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

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

#:.##sf(0.24=0.34+0.0591/(2)log[Cu^(2+)])#

#:.##sf(log[Cu^(2+)]=-3.3841)#

From which:

#sf([Cu^(2+)]=4.13xx10^(-4)color(white)(x)"mol/l")#

So (A) is the best result.