# Question 3a73e

Jan 18, 2017

$\textsf{\left(A\right)}$

#### 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:

$\textsf{E = {E}^{\circ} + \frac{0.0591}{z} \log \left[i o n\right]}$

Where $\textsf{z}$ is the number of moles of electrons transferred which, in this case, is 2.

$\therefore$$\textsf{0.24 = 0.34 + \frac{0.0591}{2} \log \left[C {u}^{2 +}\right]}$

$\therefore$$\textsf{\log \left[C {u}^{2 +}\right] = - 3.3841}$

From which:

$\textsf{\left[C {u}^{2 +}\right] = 4.13 \times {10}^{- 4} \textcolor{w h i t e}{x} \text{mol/l}}$

So (A) is the best result.