Question

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 1

Answer : The cell potential for this reaction is 0.50 V

Explanation :

The given cell reactions is:

`Pb^(2+)(aq)+Zn(s)\rightarrow Zn^(2+)(aq)+Pb(s)`

The half-cell reactions are:

Oxidation half reaction (anode):
`Zn\rightarrow Zn^(2+)+2e^-`

Reduction half reaction (cathode):
`Pb^(2+)+2e^-\rightarrow Pb`

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

Using Nernest equation :

`E_(cell)=E^o_(cell)-(2.303RT)/(nF)\log ([Zn^(2+)])/([Pb^(2+)])`

where,

F = Faraday constant = 96500 C

R = gas constant = 8.314 J/mol.K

T = room temperature =
`25^oC=273+25=298K`

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

`E^o_(cell)` = standard electrode potential of the cell = +0.63 V

`E_(cell)` = cell potential for the reaction = ?

`[Zn^(2+)]` = 3.5 M

`[Pb^(2+)]` =
`2.0* 10^(-4)M`

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

`E_(cell)=(+0.63)-(2.303* (8.314)* (298))/(2* 96500)\log (3.5)/(2.0* 10^(-4))`

`E_(cell)=0.50V`

Therefore, the cell potential for this reaction is 0.50 V