Question

The $E_{\text{cell}}^0$ for the given cell reaction is $-0.32\text{V}$ at ${25}^{\circ }\text{C}$.
$Fe\left(s\right)+Z{n}^{2+}\left(aq\right)\rightleftharpoons Zn\left(s\right)+F{e}^{2+}\left(aq\right)$
What will be the value of $\log \left[F{e}^{2+}\right]$ at equilibrium when a piece of iron is placed in a $1\text{M}$ of $Z{n}^{2+}$ solution?

Answer 1

We have the Nernst equation at equilibrium at ${25}^{\circ }\text{C}$

$E^0=\frac{0.0591}{n}\log K......\left(\text{Eqn.}1\right)$

Since $E_{\text{cell}}^0$ for the given reaction is negative, therefore, the reverse reaction is feasible for which $E_{\text{cell}}^0$ will be $+0.32\text{V}$.
Thus,

$Zn\left(s\right)+F{e}^{2+}\left(aq\right)\rightleftharpoons Fe\left(s\right)+Z{n}^{2+}\left(aq\right);E_{cell}^0=+0.32\text{V}$

By Equation (1), we get

$E^0=\frac{0.0591}{n}\log \frac{\left[Z{n}^{2+}\right]}{\left[F{e}^{2+}\right]}$

$\left[Z{n}^{2+}\right]=1\text{M}$

$\Rightarrow 0.32=\frac{0.0591}{2}\log \frac{1}{\left[F{e}^{2+}\right]}$

$\Rightarrow 0.32=-\frac{0.0591}{2}\log \left[F{e}^{2+}\right]$

$\Rightarrow \log \left[F{e}^{2+}\right]=-10.829\approx -10.83$