0732. Решение задачки по химии

2. Calculate the emf of the cell Pt(t) | Hg(l)| Hg2Cl2(t)| HCl(aq)| H2(r)| Pt(s) with an activity of HCl(aq) equal to 1.256 and hydrogen fugacity of 1 atm, knowing the standard electrode potential of 0.2688 In the reaction Hg2Cl2(s) 2e– = 2Cl–(aq) 2Hg(l), at 25 °C.

If you do not open the file - install the RAR archiver. Inside the archive you will find a solution in Word format.