Question

The standard cell potential (E°cell) for the reaction below is +0.63 V. The cell potential for this reaction is __________V when the concentration of Zn2+ = 1.0 M and theconcentration of Pb2+ = 2.0 x 10-4 M.Pb2+(aq) + Zn(s) ---> Zn2+(aq) + Pb(s)Please include details of how to calculate the voltage...I need the steps. Thanks in advance.

Answer 1

Answer: The cell potential of the above reaction is 0.52 V

Explanation:

The given chemical equation follows:

$Pb^{2+}(aq.)+Zn(s)\rightarrow Zn^{2+}(aq.)+Pb(s)$

Oxidation half reaction:  $Zn(s)\rightarrow Zn^{2+}(aq.)+2e^-$

Reduction half reaction:  $Pb^{2+}(aq.)+2e^-\rightarrow Pb(s)$

To calculate the EMF of the cell, we use the Nernst equation, which is:

$E_{cell}=E^o_{cell}-\frac{0.059}{n}\log \frac{[Zn^{2+}]}{[Pb^{2+}]}$

where,

$E_{cell}$ = electrode potential of the cell = ? V

$E^o_{cell}$ = standard electrode potential of the cell = 0.63 V

n = number of electrons exchanged = 2

$[Zn^{2+}]=1.0M$

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

Putting values in above equation, we get:

$E_{cell}=0.63-\frac{0.059}{2}\times \log(\frac{1.0}{2.0\times 10^{-4}})\\\\E_{cell}=0.52V$

Hence, the cell potential of the above reaction is 0.52 V