The standard cell potential (e°cell) for the reaction below is +0.63 v. the cell potential for this reaction is v when [ zn2+] = 3.5 m and [pb2+] = 2.0⋅10−4 m.
pb2+ (aq) + zn (s) → zn2+ (aq) + pb (s)
a) 0.84
b) 0.76
c) 0.50
d) 0.63
e) 0.39

Answer : The cell potential for this reaction is 0.50 V

Explanation :

The given cell reactions is:

The half-cell reactions are:

Oxidation half reaction (anode):  

Reduction half reaction (cathode):  

First we have to calculate the cell potential for this reaction.

Using Nernest equation :

where,

F = Faraday constant = 96500 C

R = gas constant = 8.314 J/mol.K

T = room temperature =

n = number of electrons in oxidation-reduction reaction = 2

= standard electrode potential of the cell = +0.63 V

= cell potential for the reaction = ?

= 3.5 M

=

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

Therefore, the cell potential for this reaction is 0.50 V