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?

Answer

Flow of current from cathode to anode and the current will decrease with time.

Explanation

For electrochemical cell
`Pt_((1))|Fe^(3+),Fe^(2+)(a=1)|Ce^(4+),Ce^(3+)(a=1)|Pt_((2))`
`E^(c-)._((Fe^(3+)|Fe^(2+)))=0.77V`
`E^(c-)._(Ce^(4+)|Ce^(3+))=1.61V`
On the basis of standard reduction potential `(E^(c-))`,
`E^(c-)` of `FegtE^(c-)` of `Ce`. So in such cell `Fe` acts as anode while `Ce` acts as cathode.
`E^(c-)._(cell)=(E^(c-)._(cathode)-E^(c-)._(anode ))_(red)`
`=1.61-0.77=0.84V`
Hence , the flow of current from cathode to anode `(` right to left `),` and current will decrease with time.