Consider the cell described below at 287 k: pb | pb2+ (1.07 m) || fe3+ (2.13 m) | fe given the standard reduction potentials found on the sheet attached to the exam, calculate the cell potential after the reaction has operated long enough for the [fe3+] to have changed by 1.052 m.

Answer : The cell potential is, 0.090 V

Explanation :

The values of standard reduction electrode potential of the cell are:

From the given cell representation we conclude that, the lead (Pb) undergoes oxidation by loss of electrons and thus act as anode. Iron (Fe) undergoes reduction by gain of electrons and thus act as cathode.

The half reaction will be:

Reaction at anode (oxidation) :    

Reaction at cathode (reduction) :    

The balanced cell reaction will be,  

First we have to calculate the standard electrode potential of the cell.

Now we have to calculate the cell potential when the concentration of changed by 1.052 M.

Initial concentration of = 2.13 M

New concentration of = 2.13 - 2x

As we are given that, 2x = 1.052

So, x = 0.526

New concentration of = 2.13 - 2x = 2.13 - 1.052 = 1.078 M

Initial concentration of = 1.07 M

New concentration of = 1.07 + 3x = 1.07 + 3(0.526) = 2.648 M

Using Nernest equation :

n = number of electrons in oxidation-reduction reaction = 6

= emf of the cell = ?

Now put all the given values in the above equation, we get:

Therefore, the cell potential is, 0.090 V

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