Question

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

• A. $-10.83$
• B. $-5.42$
• C. $+10.83$
• D. $+5.42$

Answer 1

The correct option is A $-10.83$

We have the Nernst equation at equilibrium at ${25}^{o}C$

$E^0=\frac{0.0591}{n}logK......(Eqn.1)$

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

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

By Equation (1), we get

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

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

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

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

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