To be frank about it I can't remember all of that stuff about which is the anode/cathode and I'm not sure it matters in this case. You're looking for Ecell and my philosophy is to calculate Ecell and not worry about the anode and cathode. But I do remember ONE definition for the anode; the anode is where oxidation occurs (and the Ag^+ ==> Ag +e is the anode because that's where oxidation is.
Look at the equation.
I see Ag^+ on the right and Ag on the left which means your first equation must be reversed with E = -0.80v. Then I see O2 and H2O on the left and OH on the right which is the same way as your second equation with E = 0.40
Add the oxidation half and the reduction half to get -0.80v + 0.40v = -0.40v and that negative sign means the cell will not operate spontaneously the way it is set up but would operate in the reverse direction.
Calculate Ecell for reaction for the given info below..
O2(g)+2H2O(l)+4Ag(s)--> 4OH-(aq)+4Ag+(aq
Ag+(aq)+ 1e ---> Ag(s) Ered= 0.80v
O2(g)+2H2O(l) + 4e ---> 4OH- Ered= 0.40v
My confusion is I have done problems similar to this when the Ered was negatives, but two positives, does the same "the one that's the lowest is the anode and highest cathode"?
OR! do u have to rewrite the equation into 2 sepetate cells you have? and if you have to, why do you have to for this situation?
ANS: -0.40v
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