What is the emf of this cell when [fe2+]= 1.8 m , [fe3+]= 1.1×10−2 m, po2= 0.43 atm and the ph of the solution in the cathode compartment is 4.2?

Emf of the cell is 0.335 V

Let's write the reaction involved:

→

where n= number of electrons,

Since, oxidation takes place at cathode. Thus the reaction is:

  → , the value of reduction potential =1.229V

As, reduction takes place at anode. Thus the reaction will be:

→ , the value of reduction potential =0.771 V

 In this Reaction , Fe is oxidized from +2 to +3 state and oxygen is reduced to H₂O.

= 1.229 - 0.771 = 0.458 V

Insert the value in Nernst Equation

E = 0.458 V

n =4

It is given that:

pO₂=0.43 atm  

pH = 4.2  

pH = -log

= antilog(-4.2)

The value of K can be calculated as

On substituting the values,

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Emf of the cell is 0.335 V

Explanation:

Here ,

n = number of electrons transferred

Cathode : Reduction-Reaction

.......1.229 V

Anode :Oxidation Reaction  

.........0.771 V

The electron are balanced by multiplying Anode with 4(n= 4)

In this Reaction , Fe is oxidized from +2 to +3 state and oxygen is reduced to H2O

= +1.229 V (from standard reduction potential table)

= 0.771 V

= 1.229 - 0.771 = 0.458 V

Insert the value in Nernst Equation

E = 0.458 V

n =4

= 0.011 M

pH = 4.2

pH = -log[H+]

[H+] = antilog(-4.2)

b) mutualism

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