The following electrochemical cell has been set up
`Pt_(1)|Fe^(3+)|Fe^(2+)1M||Ce^(4+)|Ce^(3+)(1M)Pt_(2)`
`E_(Fe^(3+)//Fe^(2+))^(@)=0.77 V and E_(Ce^(4+)//Ce^(3+))^(@)=1.61 V`
If an ammeter is conncted between between the two platinum electrodes predict the direction of flow of current wil the current increse or decrease with time?

Explanation

Since, activities of allthe ions are unity ,`E_(cell)=E_(cell)^(@)`.Also left hand electrode is at lower reduction potential,it act as anode and `E=E^(@)(Ce^(4+),Ce^(3+))-E^(@)(Fe^(3+),Fe^(2+))=0.84` i.e.electones 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^(@)-0.0592log. ([Fe^(3+)][Ce^(2+)])/([Fe^(2+)][Ce^(4+)])`
As electrolysis proceds, E will decrease and therefore ,current will also decrease.