# What is the potential of the 1/2 cell Cu^(2+)+2erightleftharpoonsCu if [Cu^(2+)]=0.1M?  E^@=+0.34V.

Dec 1, 2016

$\textsf{E = + 0.31 \textcolor{w h i t e}{x} V}$

#### Explanation:

The Nernst Equation for the electrode potential of a 1/2 cell gives us:

sf(E=E^(@)-(RT)/(zF)ln([[Red]]/([Ox]))

In reality, activities should be used but molarities are acceptable at low concentrations as $\textsf{a \Rightarrow 1}$

In this case we have:

$\textsf{C {u}^{2 +} + 2 e r i g h t \le f t h a r p \infty n s C u}$

So $\textsf{z = 2}$

At $\textsf{{25}^{\circ} C}$ a convenient form of the equation is:

$\textsf{E = {E}^{\circ} + \frac{0.05916}{z} \log \left(\frac{\left[O x\right]}{\left[R e d\right]}\right)}$

Putting in the numbers:

$\textsf{E = + 0.34 + \frac{0.05916}{2} \log \left(0.1\right)}$

$\textsf{E = + 0.34 - 0.02958 \textcolor{w h i t e}{x} V}$

$\textsf{E = + 0.31 \textcolor{w h i t e}{x} V}$

This shows that reducing the concentration like this causes a drop in potential.