Question

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. 57 m, [h ] = 0. 090 m and ph2 = 1. 0 atm ? v

Answer 1

Answer:

The emf of the electrochemical cell has been calculated to be 3.364.

in order to calculate the emf we need to apply Nernst equation.

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= E⁰ cell - 0.059/n. log 1/concentration

Computation for the emf of the cell

  The given cell has the number of electrons transfer, n= 2

  The concentration of   Pb2= 0.57 M

  The concentration of   H= 0.090 M

The cell potential of the reaction has been= -0.126

Substituting the values for the emf of the cell:

                  emf= -0.126- 0.059/2 * log 1/ [0.57]x [0.090] ²

                  emf= 1.555* 2.337

                  emf=3.634 v

The emf of the electrochemical cell has been calculated to be 3.634 v.

Learn more about emf here-

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