0732. Solving a chemistry problem
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1. Calculate the emf of the cell Pt(t) | Pb(t)| PbI2(s) | HI(aq)| H2(r)| Pt(s) with activity HI(aq) equal to 0.815 and hydrogen fugacity 1 atm, knowing the standard electrode potential –0.364 V reaction PbI2(s) 2e– = 2I–(aq) Pb(s), at 25 °C.
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