Question

The following electrochemical cell has been set-up :
$Pt(1)|F{e}^{3+},F{e}^{2+}(a=1)|C{e}^{4+},C{e}^{3+}(a=1)|Pt(2)$
$E^{\circ }(F{e}^{3+},F{e}^{2+})=0.77V$
and $E^{\circ }(C{e}^{4+},C{e}^{3+})=1.61V$
If an ammeter is connected between the two platinum electrodes, predict the direction of flow of current, will the current increases or decreases with time?

Answer 1

Since, activities of all the ions are unity, $E_{\text{cell }}=E_{\text{cell }}^{\circ }$ Also, left hand electrode is at lower reduction potential, it act as anode and
$E^{\circ }=E^{\circ }\left(C{e}^{4+},C{e}^{3+}\right)-E^{\circ }\left(F{e}^{3+},F{e}^{2+}\right)=0.84$
i.e. electrons will flow from left to right hand electrode and current from right hand electrode [Pt (2)] to left hand electrode $[Pt(1)]$
Also, $E=E^{\circ }-0.0592\log \frac{\left[F{e}^{3+}\right]\left[C{e}^{3+}\right]}{\left[F{e}^{2+}\right]\left[C{e}^{4+}\right]}$
As clectrolysis proceeds, $E$ will decrease and therefore, current.