For a galvanic cell using Fe | Fe2+(0.25 M) and Pb | Pb2+(0.25 M) half-cells, wh

Question

For a galvanic cell using Fe | Fe2+(0.25 M) and Pb | Pb2+(0.25 M) half-cells, which of the following statements is correct? Fe2+(aq) + 2e– Fe(s); E° = –0.41 V Pb2+(aq) + 2e– Pb(s); E° = –0.13 V

a. The iron electrode is the cathode

b. When the cell has completely discharged, the concentration of Pb2+ is zero

c. The mass of the iron electrode increases during discharge

d. The concentration of Pb2+ decreases during discharge

e. Electrons leave the lead electrode to pass through the external circuit during discharge

Explanation / Answer

The first option is incorrect, because Iron has the lowest E° which means that is reducting the lead, and to do this, it needs to be in the anode, not cathode.

The concentration of lead cannot be zero when the cell is discharged, cause the Q will not match in the expression of nerst.

Iron is oxidating so it mass would be decreasing instead of increasing

Lead is reducting, is gaining electrons, so it cannot leave the lead electrode.

So the correct option is d). and this is true cause is reducting.

Hope this helps

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