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 activity of each ion inboth the half-cells is unity so potential of each half-cell will be standard potential.
Oxidation of LHE:
`Fe^(2+) to Fe^(3+)+e`
Reduction at RHE:
`Ce^(4+) +e to Ce^(3+)`
Cell reaction: `F_((a=1))^(2+)+Ce_((a=1))^(4+) to F_((a=1))^(3+) +Ce_((a=1))^(3+)`
`E_("cell")^(0)=E_(Ce^(4+)//Ce^(3+))^(0)-E_(Fe^(3+)//Fe^(2+))^(0)`
=1.61-0.77=0.84V
`E_("cell")^(0)` is positive, so `triangleG^(@)` is negative and hence the above cell reaction is spontaneous (feasible).
Thus, left and electrode (LHE) i.e, ferrous ferric redox electrode will behave as anode and the RHE i.e, cerric-cerrium redox electrode will behave as cathode. Electrons will flow from left to right via external circuit since the direction of current (conventional) is taken to be opposite to the direction of flow of electrons so current will flow from Pt(2) to Pt(1). The curcent.will decrease with time due to decrease in the concentration of reactants.