4. Решение задачки по химии №4

5.63 Write a diagram of a galvanic cell and the reaction on the electrodes, and calculate the electrode potentials for the cell in which the reverse reaction occurs according to the equation:
2Fe3 H3AsO4 2H2O=2Fe2 H3AsO4 2H
The emf of a galvanic cell at a pressure of 1 bar and a temperature of 298 K is 0.338 V. The activity of the ions participating in the reaction:
aFe2 = 0.005; aFe3 = 0.01; aH3AsO4 = 0.2. aHAsO2 = 0.1. aH = 0.01 aH2O = 1 ε0Fe3 / Fe2 = 0.771 V

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