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PBCHEM
Following is correct representation of electrochemical cell of interest
Pb/[tex] PB^{2+} [/tex] // [tex] H_{2} [/tex]/ [tex] H_{2} [/tex]

The standard reduction potential of [tex] PB^{2+} [/tex]/ Pb and [tex] H_{2} [/tex]/[tex] H_{2} [/tex] is -0.126 v and 0.0 v respectively.

Now, in present cell using Nernst Eq. we have
Ecell = [tex] E^{0}cell - \frac{0.059}{n}log \frac{1}{[Pb^2^+]X[H^+]^2} [/tex]
where n = number of eletrons = 2 (in present case) 
Also, for present cell, [tex] E^{0}cell = 0.126 v

∴Ecell = 0.126 - \frac{0.059}{2}log \frac{1}{[0.83]X[0.064]^2} [/tex]
           = 0.0532 v

The emf of the electrochemical cell has been calculated to be 0.0532 V.

The emf has been the potential of the cell in the reaction with the change in the electrons in the reaction. The emf of the cell has been given by the Nernst equation as;

[tex]emf=E^\circ _{cell}-\dfrac{0.059}{n}\;log\;\dfrac{1}{\rm concentration} [/tex]

Computation for the emf of the cell

The given cell has the number of electrons transfer, [tex]n=2[/tex]

The concentration of [tex]\rm Pb^2^+=0.83\;M[/tex]

The concentration of [tex]\rm H^+=0.064\;M[/tex]

The cell potential of the reaction has been, [tex]E^\circ =-0.126\;\text V[/tex]

Substituting the values for the emf of the cell:

[tex]emf=-0.126\;-\dfrac{0.059}{2}\;\times\;log\;\dfrac{1}{[0.83]\;\times\;[0.064]^2}\\ emf=0.0532\;\text V [/tex]

The emf of the electrochemical cell has been calculated to be 0.0532 V.

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