Question

Calculate the cell potential for the following electrochemical cell:

Zn | [Zn²⁺]=4.2M || [Cu²⁺]=0.004M | Cu

 

Answer 1

Let us first write the reaction taking place

Zn    -------------> Zn⁺²   + 2e^-           ...........(1)

Cu²⁺ 2e^-      ---------->Cu               .............(2)

The first reaction is an oxidation reaction . The second reaction is an reduction reaction.

E^o1_ox for reaction (1) = + 0.76 V

E^o2_red for reaction (2) = +0.34 V

Overall reaction of the above electrochemical cell will be

Cu²⁺ +Zn ----------> Zn⁺² +Cu               ................(3)

E^o for reaction three will be summation of E^o of reaction (1) and reaction (2)

E^o =E^o1_ox + E^o2_red = 0.76+0.34 = 1.1 V

To find the cell potential we will make use of nernst equation,

Nernst equation is given by,

here number of electron tranfer (n) = 2

Zn+2 = 4.2 M

Cu+2 = 0.004 M

Substituting all the values in above equation ,

From above equation , E = 1.01057 V

CELL POTENTIAL FOR ABOVE REACTION IS 1.01057 V