The following electrochemical cell has been set up.
`Pt_((I))|Fe^(3+),Fe^(2+)(c=1)||Ce^(4+),Ce^(3+)(c=1)|Pt_((II))`
`E_(Fe^(3+)//Fe^(2+))=0.77V` and `E_(Ce^(4+)//Ce^(3+))^(@)=1.61V`
If an ammeter is connected between the two platinum electrode, predict the direction of flow of current. Will the current increase or decrease with time ?

Explanation

The e.m.f of given cell `=E_(OP_(Fr^(2+)//Fe^(3+)))+E_(RP_(Ce^(3+)//Ce^(3+)))` or
`E_("cell")^(@)=E_(OP_(Fe^(2+)//Fe^(3+)))^(@)-(0.059)/(1)"log"([Fe^(3+)])/([Fe^(2+)])+E_(RP_(Ce^(3+)//Ce^(4+)))^(@)+(0.059)/(1)"log"([Ce^(4+)])/([Ce^(3+)])`
`=E_(OP_(Fe^(2+)))^(@)+E_(OP_(Ce^(4+)))^(@)+(0.059)/(1)"log"([Ce^(4+)][Fe^(2+)])/([Ce^(3+)][Fe^(3+)])`
`=-0.77+1.61+(0.059)/(1)log 1 therefore E_("cell")=0.84V`
Thus, `Pt_((I))Fe^(3+)//Fe^(2+)` acts as anode and `Pt_((II))Ce^(4+)//Ce^(3+)` acts as cathode. The electrons flow from left right and thus current will flow from right to left. The current strength will decrease with time.