JEE Advance - Chemistry (2000 - No. 2)
The following electrochemical cell has been set up.
Pt(1) | Fe3+, Fe2+ (a = 1) | Ce4+, Ce3+ (a=1) | Pt(2)
Eo (Fe3+, Fe2+) = 0.77 V; Eo (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?
Pt(1) | Fe3+, Fe2+ (a = 1) | Ce4+, Ce3+ (a=1) | Pt(2)
Eo (Fe3+, Fe2+) = 0.77 V; Eo (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?
Iron electrode to cerium electrode, increase
Iron electrode to cerium electrode, decrease
Cerium electrode to iron electrode, remains constant
Cerium electrode to iron electrode, increase
Cerium electrode to iron electrode, decrease
Uitleg
For electrochemical cell,
$$Pt(1)|F{e^{3 + }}$$ , $$Fe(a = 1)||C{e^{4 + }}$$ , $$C{e^{3 + }}(a = 1)|Pt(2)$$
$$E_{F{e^{3 + }}|F{e^{2 + }}}^0 = 0.77\,V$$, $$E_{C{e^{4 + }}|C{e^{3 + }}}^0 = 1.61\,V$$
$$\therefore$$ $$E_{cell}^0 = E_{cathode}^0 - E_{anode}^0 = 1.61 - 0.77 = 0.84\,V$$
Since, $$E_{cell}^0$$ is +ve, the cell reaction will occur spontaneously.
Hence, the flow of current will be from cathode to anode (right to left), and the current will decrease with time.