Question

Calculate the potential of the standard Iron-Cadmium cell after the reaction has proceeded to $80\%$ completion. Initially $1M$ of each taken and $E^{\circ }$ for cell $=0.04V$. $Fe(s)\left|F{e}^{+2}\right|\left|C{d}^{+2}\right|Cd$

Answer 1

Answer: 0.01

after reaction $\left(F{e}^{+2}\right)=1+.8=1.8$
$C{a}^{+2}=1-.8=.2$
$E_{\text{cell }}=E_{\text{cell }}^{\circ }-\frac{.0591}{2}\log \frac{\left[F{e}^{+2}\right]}{\left[C{d}^{+2}\right]}$
$=.04-\frac{.0591}{2}\log \frac{.0591}{2}$
$=.0118V$