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 :

$E_{cell}=E^o_{cell}-\frac{2.303RT}{nF}\log \frac{[Zn^{2+}]}{[Pb^{2+}]}$

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

$[Pb^{2+}]$ = $2.0\times 10^{-4}M$

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

$E_{cell}=(+0.63)-\frac{2.303\times (8.314)\times (298)}{2\times 96500}\log \frac{3.5}{2.0\times 10^{-4}}$

$E_{cell}=0.50V$

Therefore, the cell potential for this reaction is 0.50 V