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An electrochemical cell is constructed as: Pt(I) | Fe3+, Fe2+ (a = 1) || Ce4+, Ce3+ (a = 1) | Pt(II), with E°(Fe3+/Fe2+) = 0.77 V and E°(Ce4+/Ce3+) = 1.61 V. If an ammeter is connected across the two platinum electrodes, what is the direction of conventional current flow in the external circuit, and how does the current change with time?

1. Current flows in the external circuit from Pt(II) to Pt(I), and the current decreases with time.
2. Current flows in the external circuit from Pt(I) to Pt(II), and the current decreases with time.
3. Current flows in the external circuit from Pt(II) to Pt(I), and the current increases with time.
4. Current flows in the external circuit from Pt(I) to Pt(II), and the current increases with time.

Correct answer: Current flows in the external circuit from Pt(II) to Pt(I), and the current decreases with time.

Solution

Ce4+/Ce3+ has the higher E° (1.61 V) so the cerium electrode (Pt II) is the cathode where Ce4+ is reduced; the iron electrode (Pt I) is the anode where Fe2+ is oxidized. Conventional current in the external circuit flows from the cathode (+) terminal Pt(II) to the anode Pt(I). The cell EMF = 1.61 - 0.77 = 0.84 V drives the reaction toward equilibrium; as Fe2+ and Ce4+ deplete the EMF decreases, so the current decreases with time until equilibrium (zero current).

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