Question

Question 7 options: The cell potential of an electrochemical cell made of an Fe, Fe2 half-cell and a Pb, Pb2 half-cell is _____ V. Enter your answer to the hundredths place and do not leave off the leading zero, if needed.

Answer 1

Answer: Thus the cell potential of an electrochemical cell is +0.28 V

Explanation:

The calculation of cell potential is done by :

$E^0=E^0_{cathode}- E^0_{anode}$

Where both $E^0$ are standard reduction potentials.

$E^0_{[Fe^{2+}/Fe]}= -0.41V$

$E^0_{[Pb^{2+}/Pb]}=-0.13V$

As Reduction takes place easily if the standard reduction potential is higher(positive) and oxidation takes place easily if the standard reduction potential is less(more negative). Thus iron acts as anode and lead acts as cathode.

$E^0=E^0_{[Pb^{2+}/Pb]}- E^0_{[Fe^{2+}/Fe]}$

$E^0=-0.13- (-0.41V)=0.28V$

Thus the cell potential of an electrochemical cell is +0.28 V