Inorganic and Applied Chemistry

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Inorganic and Applied Chemistry

Example 6- E:

Calculation of cell potential and spontaneity

We are informed that a galvanic cell consists of the following two half cell reactions:

Ag+(aq) + e-  Ag(s)

Fe3+(aq) + e-  Fe2+(aq)

From a table we have the following reduction potentials:

Half cell reaction 1: Ag+(aq) + e-  Ag(s) ,^0 = 0.80 volt

Half cell reaction 2: Fe3+(aq)  Fe2+(aq) + e- ,    ^0 = 0.77 volt

We wish to determine which one of the following two reactions that runs spontaneously and what the

corresponding standard cell potential is:

Reaction 1: Ag+ + Fe2+ Ag + Fe3+

Reaction 2: Ag + Fe3+ Ag+ + Fe2+

In order to settle this question G^0 for the two half cell reactioner must be determined using equation

(6-2):

Half cell reaction 1: G 10 

1 F 0. 80 V  0. 80 V F

Half cell reaction 2: G 20 

1 F 0. 77 V  0. 77 V F

NowG^0 for the reaction 1 and 2 may be determined at appropriate combination of G 10 and G 20

corresponding to the half cell reactions:

Reaction 1: G^0 G 10 ^ G 20 ^ ^0.^80 V^ F ^0.^77 V^ F^0.^03 V^ F

Reaction 1: G^0 G 10 G 20   0. 80 V F  

0. 77 V F  0. 03 V F

A spontaneous reaction runs only when G^0 is less than 0 where reaction 1 is the reaction that runs in the

galvanic cell as shown in figure 6-4. The standard potential of the cell is calculated on the basis of the

determined value of G^0 (equation (6-2)):

Ag+ + Fe2+ Ag + Fe3+

 

V

F

V F

z F

G

G z F 0. 03

1

0 0 0 0 0.^03



 + +

Electrochemistry