Question

The following electrochemical cell has been set-up :
$ Pt(1)|Fe^{3+},Fe^{2+} (a = 1)|Ce^{4+},Ce^{3+}(a=1)|Pt(2)$
$ E^\circ (Fe^{3+},Fe^{2+})=0.77\, V$
and $ E^\circ (Ce^{4+},Ce^{3+})=1.61\, V$
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
$E^{\circ}=E^{\circ}\left( Ce ^{4+}, Ce ^{3+}\right)-E^{\circ}\left( Fe ^{3+}, Fe ^{2+}\right)=0.84$
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[ Fe ^{3+}\right]\left[ Ce ^{3+}\right]}{\left[ Fe ^{2+}\right]\left[ Ce ^{4+}\right]}$
As clectrolysis proceeds, $E$ will decrease and therefore, current.