Question

Question #3a73e

Answer 1

`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.