# Question #2eac3

Feb 6, 2016

$\left(a\right)$

${E}_{c e l l}^{\circ} = + 0.32 \text{V}$

$\left(b\right)$

${E}_{c e l l} = + 0.37 \text{V}$

#### Explanation:

(a)

List the standard electrode potentials:

$\textcolor{w h i t e}{\times \times \times \times \times \times \times \times \times x} {E}^{\circ} \text{V}$

$\stackrel{\textcolor{b l u e}{\leftarrow}}{\textcolor{w h i t e}{\times \times \times \times \times \times \times}}$

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

$F {e}^{2 +} + 2 e r i g h t \le f t h a r p \infty n s F e \text{ "-0.44}$

$\stackrel{\textcolor{red}{\rightarrow}}{\textcolor{w h i t e}{\times \times \times \times \times \times \times}}$

Overall reaction:

$Z {n}_{\left(s\right)} + F {e}_{\left(a q\right)}^{2 +} \rightarrow Z {n}_{\left(a q\right)}^{2 +} + F {e}_{\left(s\right)}$

To find ${E}_{c e l l}^{\circ}$ subtract the least +ve ${E}^{\circ}$ value from the most +ve:

${E}_{c e l l}^{\circ} = - 0.44 - \left(- 0.76\right) = + 0.32 \text{V}$

(b)

Now that the conditions are non - standard we need to use the Nernst Equation:

${E}_{c e l l} = {E}_{c e l l}^{\circ} - \frac{R T}{n F} \ln Q$

Where $Q$ is the reaction quotient.

$n$ is the number of moles of electrons transferred.

At 298K this simplifies down to:

${E}_{c e l l} = {E}_{c e l l}^{\circ} - \frac{0.05916}{2} \log Q$

$\therefore {E}_{c e l l} = {E}_{c e l l}^{\circ} - \frac{0.05916}{2} \log \left(\frac{\left[Z {n}^{2 +}\right]}{\left[F {e}^{2 +}\right]}\right)$

$\therefore {E}_{c e l l} = + 0.32 - \frac{0.05916}{2} \log \left(\frac{0.01}{0.5}\right)$

${E}_{c e l l} = + 0.32 - \frac{0.5916}{2} \times \left(- 1.698\right)$

${E}_{c e l l} = 0.32 + 0.05 = + 0.37 \text{V}$

You can see that this has increased the value of ${E}_{c e l l}$ indicating that the reaction has been driven to the right.

This is in accordance with Le Chatelier as $\left[Z {n}^{2 +}\right]$ has been decreased relative to $\left[F {e}^{2 +}\right]$ so shifting the position of equilibrium to the right.