Chemistry: What is the emf of a cell consisting of a pb2+ / pb half-cell and a pt / h+ / h2 half-cell if [pb2+] = 0.83 m, [h+] = 0.064 m and ph2 = 1.0 atm ?

What is the emf of a cell consisting of a pb2+ - id21147840, milkshakegrande101

Ответ:

calebwoodall6477

14.03.2022

Following is correct representation of electrochemical cell of interest
Pb/ $PB^{2+}$ // $H_{2}$ / $H_{2}$

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

Now, in present cell using Nernst Eq. we have
Ecell = $E^{0}cell - \frac{0.059}{n}log \frac{1}{[Pb^2^+]X[H^+]^2}$
where n = number of eletrons = 2 (in present case)
Also, for present cell, $E^{0}cell = 0.126 v\\∴Ecell = 0.126 - \frac{0.059}{2}log \frac{1}{[0.83]X[0.064]^2}$
= 0.0532 v

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Ответ:

dinecree

14.03.2022

Chemistry

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;

$emf=E^\circ _{cell}-\dfrac{0.059}{n}\;log\;\dfrac{1}{\rm concentration}$