Great tutorial! Thx! But I have one question. In predicting reaction feasibility 3, E cell = 0.80 -2.00. But 0.80 is the potential of the "oxidised" half equation, right? How come it is not the other way round?
Same question because the cell potential is obtained by subtracting the reduced value from oxidised value. Here the reduced value is +2.00 and oxidised one is +0.80 if we subtract them both answer we get is +1.20
At 57:05 { predicting reaction feasibility 3 } , why is the E cell value negative ? As to the theory explained prior isn't (E cell = E reduced - E oxidised )
Depends on what they ask for. You are expected to balance a half equation using protons, water and electrons however sometimes they will provide you with a list of them if they want you to calculate E° values etc...
54:34 just clarifying that the Eo value for Fe2+ would not be doubled, right? Coz in the balancing of the equations they're being doubled but in the calculation of Eo cell it's not.
