How to find Ecell for a chemical reaction. Here, I don't even tell you which direction the cell goes ... we'll figure it out along the way! Check me out: www.chemistnate...
The equation E cell = E cathode - E anode works like this: 0.34 - (- 0.44) = 0.78 , which is the same as what CHEMIST Nate did. He just flipped the sign to make it the 'oxidation potential'
To keep people from getting confused, you could also point out that if you left the Fe equation as is (instead of flipping it) you could have done 0.34-(-0.44) which would give the same number, because that's how I was taught it: Ereduction-Eoxidation
yes, you balance the reactions like you would any other balanced equation, and coefficient affects everything in the second equation. sorry this is a month late.
I think when voltage is negative the reaction is actually spontaneous and when the rxn is pos. It's non spontaneous. My notes say other wise but my exam study guide proves my point
+Omar Alami No, I think you are confusing this with Gibb's free energy, or deltaG. When deltaG is negative, your reaction is spontaneous, and vice versa.
Quick question; Are all galvanic cells spontaneous? Since we are basically guessing and if I chose something that is negative and make that my answer as the galvanic cell is not spontaneous; how can this be correct?
Just a student here, so someone correct me if I'm wrong, but.... I think the reason that you always choose the spontaneous one is that a non-spontaneous reaction, by definition, doesn't happen.* Say we mix all this iron and copper and they all decompose in the solution. Now we have a solution that has Fe(s), Cu(s), Fe2+ and Cu2+. Only the spontaneous reaction is going to happen. The non-spontaneous reaction is non-spontaneous, so it's not going to happen. That's why Mr. Nate uses the spontaneity test to figure out which way to write the reactions. *Of course, if we were adding extra energy to the system, the nonspontaneous reaction could occur, but we aren't doing that; we're just mixing stuff and seeing what happens. Bonus: I also noticed that you don't actually have to guess. When you look at your table of standard reduction potentials and find your two values, you just have to ask yourself, "Which one of these values do I need to change the sign so that when I add them I get a positive value?" Then you know which half-reaction to flip.
ok I got it now. He's doing it right but just using a different method. You can either look at the table and flip the signs according to what you have been given or you could just look at the table and see which value is more positive and use that as a E_red in your equation i described above without worrying about flipping the signs. For example, the cooper/zinc example he did, you could have just looked at the values and see which is more positive. Obviously it is +0.34 which is more positive than -0.44 right. SO just use 0.34 as a E-red and -0.44 as E-oxi and you will get the same results. If the values were, say, -0.34 and -0.44 which is more positive? obviously -0.44 and now use that as your E_red. That's it.
yea what you have is correct because from your example you would still get the same +0.10 as i did but in his example it was completely off because he flipped the sign then he added and so instead of getting a positive results he gets a negative one
That’s because here we are *adding two equations together* like Hess’ Law … we purposefully flipped one equation from Oxidation to Reduction, then added. You COULD subtract the two oxidation half-reaction voltages from each other, as you suggest.
I've had some kids tell me in 1-on-1 tutoring that their teacher required the reduction on the right as well. I'm not aware of an official rule for this, so I did it whatever way felt natural at the time. But you're right that it would have been better (i.e. more helpful for everyone) to have done it that way ...
I don't know what ages are your target group here(i mean if it is high school it could be fine),but your video is very confussing and utterly wrong considering the IUPAC directions.All (half)reactions conserning the Eo value are written as reductions in the international bibliography to avoid confusion and make clear that the value of E is just the voltage diffrence between the element in question and the SHE(Standard Hydrogen Electrode).My point is that by taking a ''practical'' approach of just flipping the half-reactions and changing your signs you are making more difficult for a student to understand the theory behind these excersises(e.g. the half-reactions are just a chemical equilibrium,the use of Eo = Ered - Eoc , etc) Anyway,this is not a hostile comment I am just trying to clear up some misunderstandings that took me a while to solve in the University exactly because of these kind of practices by my high school teachers.
This had confused me for a while as well, as EVERYWHERE I'd look online, I'd see the method above, whereas in my lectures we do Ered-EOx. I've been trying to understand why there's such different info everywhere, and I realize why now x_x It took a concept I THOUGHT I understood just fine, and made it more confusing
