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Enter your answer in the provided box. what is the emf of a cell consisting of a pb2 / pb half-cell and a pt / h / h2 half-cell if [pb2 ] = 0. 11 m, [h ] = 0. 087 m and ph2 = 1. 0 atm ? v

Respuesta :

The answer is-  EMF of cell = 0.096 V.

Nernst equation: EMF of a cell is the cell potential and Nernst equation is used to calculate the EMF of a cell.

Explain Nernst equation.

  • According to Nernst equation:

[tex]\[{{\rm{E}}_{{\rm{cell}}}}{\rm{\; = }}{{\rm{E}}^{\rm{0}}}{\rm{\;- }}\left[ {\frac{{0.0591}}{{\rm{n}}}} \right]{\rm{ log Q}}\][/tex]  

Where,

Ecell = EMF of the cell

E0 = Standard EMF under standard conditions

n = number of electrons transferred in the redox reaction

Q = reaction quotient

  • Now, the given reaction is-

[tex]Pb + 2H^+[/tex]→[tex]Pb^2^+ + H_2[/tex]

  • The Nernst equation for the given reaction is-[tex]Ecell = E_0 + \frac{0.0591}{n} log \frac{pH_2 * [Pb^2^+]}{[H^+]^2}[/tex]
  • Putting all the data, emf of the cell is-

E0 = 0.13 V

[tex]Ecell = 0.13 V - [\frac{0.0591}{2}] log \frac{(1.0 *0.11) }{(0.087)^2} \\ = 0.13V - 0.034= 0.096\ V[/tex]

  • Hence, the EMF of the given cell 0.096 V.

To learn more about EMF and Nernst equation, visit:

https://brainly.com/question/21601855

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