169.

The following electrochemical cell has been set up
Pt(1) | Fe³⁺, Fe²⁺ (a = 1) || Ce⁴⁺, Ce³⁺ (a = 1) | Pt (2)
E°_(Fe²⁺) = 0.77 V, and E°_{(Ce⁴⁺,Ce³⁺)} = 1.61 V
If an ammeter is connected between the two platinum electrodes, predict the direction of flow of current. Will the current increase or decrease with time?

For the electrochemical cell
Pt(1) | Fe³⁺, Fe²⁺ (a = 1) || Ce⁴⁺, Ce³⁺ (a = 1) | Pt (2)
the cell regions are

  Right-half cell: reduction
                     ${\mathrm{Ce}}^{4+}+{\mathrm{e}}^{-} \to {\mathrm{Ce}}^{3+}$
 
Left-half cell: oxidation
                      ${\mathrm{Fe}}^{2+} \to {\mathrm{Fe}}^{3+}+{\mathrm{e}}^{-}$
______________________________
Add ${\mathrm{Ce}}^{4+}+{\mathrm{Fe}}^{2+}\to {\mathrm{Ca}}^{3+}+{\mathrm{Fe}}^{3+}$

The net cell potential is
E° _Cell = E° _R – E° _L = 1.61 V – 0.77 V = 0.84 V.
Since E°_cell is positive, the cell reaction will be spontaneous.
The current in the external circuit will flow from Pt (1) (which serves as anode to Pt(2) which serves as cathode.
With the passage of time, E_cell will decrease and so is the current in the external circuit.