Question

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

Answer 1

after reaction $\left( Fe ^{+2}\right)=1+.8=1.8$
$Ca ^{+2}=1-.8=.2$
$E _{\text {cell }} = E _{\text {cell }}^{\circ}-\frac{.0591}{2} \log \frac{\left[ Fe ^{+2}\right]}{\left[ Cd ^{+2}\right]}$
$=.04-\frac{.0591}{2} \log \frac{.0591}{2}$
$=.0118\, V$