# Does Le Chatelier's principle apply to electrochemical cells?

Apr 24, 2015

Le Chatelier's Principle is very important when applied to electrochemical cells.

Consider the Daniel Cell:

The 2 half cells are:

$Z {n}^{2 +} + 2 e r i g h t \le f t h a r p \infty n s Z n$ ${E}^{o} = - 0.76 \text{V}$

$C {u}^{2 +} + 2 e r i g h t \le f t h a r p \infty n s C u$ ${E}^{0} = + 0.34 \text{V}$

To find ${E}_{c e l l}^{0}$ you subtract the least positive value from the most positive value:

${E}_{c e l l}^{0} = + 0.34 - \left(- 0.76\right) = + 1.1 \text{V}$

And the cell reaction is therefore:

$Z n + C {u}^{2 +} \rightarrow Z {n}^{2 +} + C u$

These refer to standard conditions i.e unit concentrations, 298K and pH zero.

What would happen if we were to reduce the concentration of $C {u}_{\left(a q\right)}^{2 +}$ from $1 \text{mol/l}$ to $0.1 \text{mol/l}$ ?

Looking at the Cu2+/Cu half - cell Le Chatelier's Principle would predict that would cause the position of equilibrium to shift to the left to produce more Cu2+.

This would cause the ${E}^{0}$ value to become less positive. This would therefore reduce the potential difference between the 2 half - cells and therefore the value of ${E}_{c e l l}$.

We can calculate the new value using the Nernst Equation which in this case is:

${E}_{c e l l} = {E}_{c e l l}^{0} - \frac{0.0591}{2} \log \left(\frac{\left[Z {n}^{2 +}\right]}{\left[C {u}^{2 +}\right]}\right)$

${E}_{c e l l} = 1.1 - \frac{0.0591}{2} \times 1$

${E}_{c e l l} = + 1.07 \text{V}$

So as predicted by Le Chatelier,. the value of ${E}_{c e l l}$ has been reduced.