Ye, delta G is in J/mol. But in your equation you give electrons mole unit, and cancel it out with R (J/mol) mole unit. So according to your equation delta G is in J. 3:06 in clip.
Oh man, I had to teach this and forgot and I thought I had a video on it and I didn't and now the kids are looking at me like I am an idiot - thanks - you saved me buddy!
I think that number of electrons is without unit. In an equation for reaction speed - (Kb*T/h)exp(-delta G/(RT)) - you are left with moles as unit, when it should be unitless. Kb is Boltzman constant and h being Planck constant.
So correct me if I am wrong... But E(cell)= E(Reduction)-E(Oxidation) If the standard reduction potential for Fe2+ (-0.44V) and reduction potential for Cu is 0.34V Then E(cell)= 0.34V -(-0.44)=0.78 So this is a spontaneous reaction.
Asclepius The reaction between Cu and Fe2+ is actually non-spontaneous. First step is to determine the reductant and oxidant (note that oxidant oxidises another compound and itself undergoes reduction reductant reduces another compound and itself undergoes oxidation). Cu solid is a reductant, meaning itself will be oxidised (oxidation). Fe2+ is a tricky one in some situations since it can be both a reductant and an oxidant. In this case it will be a oxidant with E=-0.44V as you have stated, meaning itself will be reduced (reduction). Now there is 2 ways of calculating E(cell) 1. E(cell)= E(Oxidant)-E(Reductant), E(cell)=-0.44-0.34= -0.78, Non-spontaneous reaction 2. E(cell)=E(Oxidation) + E(Reduction) write down reduction and oxidation half equation of the system. The oxidation of Cu solid has potential of -0.34V, we flipped the equation of Cu2+ reduction and meaning E value becomes negative accordingly. E value for Fe2+ reduction remain the same as we want E for reduction of Fe2+. now add the two E values, we get E(cell)=-0.34+(-0.44)= -0.78, non-spontaneous reaction. Both methods are pretty much the same. Personally I find method 2 easier to understand and remember when there are more formulas to remember. Either way its your personally preference which one to use. Best of luck to you with your studies.
@@billmao1109 It is a spontaneous reaction. Copper's reduction is positive on a reduction table. It is +0.34V. www5.csudh.edu/oliver/chemdata/data-e.htm
YES! I discovered this yesterday. In fact, you can use the E* values to get E(cell) for a solubility equilibrium, and use it to find K(sp) for the reaction. It rocked my world to learn that yesterday. Will make a video about it soon.
I have a problem with the units. I also thought that the final units for G would just be J but some of my professor's lecture slides have G in J/mol or in kj/mol. I thought the mol was supposed to cancel out bcz n is in mol and F also has mol. Is it possible that n is supposed to be unitless? That's the only way G could end up as J/mol or kj/mol. I'm having my final soon so a quick answer would be greatly appreciated.
Question: In the previous example it stated that Ecell for this reaction was -.78 V . Therefore shouldn't the calculated delta G be positive and the overall reaction be nonspontaneous as mentioned in the beginning of the video?
That previous example was for a purposefully constructed nonspontaneous reaction. In the video prior to that he showed the spontaneous version. Look at your class's or an online reduction potential table. :) The nonspontaneous version does not exist in reality without being in electrolysis which requires an outside electrical source to make it happen. In reality, copper (II) is reduced when in a voltaic cell with iron (II).
Great thanks al ot . please keep sharing this kind of simple videos to refresh our knowledge ........... we are following your channel waiting for more and more videos
I'm not sure if anyone has mentioned this but you made a small mistake in the multiplication where you inserted a positive Ecell into the equation, rather than the negative value. A negative Delta G indicates a spontaneous reaction, but your Ecell, which is negative, indicates a non-spontaneous reaction. The idea is that Ecell and Delta G for the same reaction will have opposite signs. Having the same sign for both indicates a mistake has been made.
The ecell is positive? It is positive 0.78V. So the delta G would be negative and spontaneous. A positive Ecell = negative delta G. I haven't any idea why you though the Ecell was negative originally. It's positive. Look at a reduction values table. :)
This video is misleading in multiple ways. He states that delta G = -nFE. The equation should be delta G standard (superscript degree symbol that signifies under standard state conditions) = -nFEcell standard. Delta G and delta G standard are two different things. The CORRECT equation for plain delta G: delta G = delta G (standard)+RTlnQ Where R is the universal gas constant (8.314 for measuring delta G in J/mol or .008314 for KJ/mol), T is temperature in Kelvins and Q is the reaction quotient (ratio of concentrations of products over reactants). Also, standard state conditions implies 1 atm, 1M and 298K.
+Nick Hammond Strictly speaking, standard state has nothing to do with concentration, but rather activity existing at unity. ALT + 0176 is ° handy, for reference.
In the case of this specific equation, they're interchangeable because you are using E naught cell. So it is implied that you are under standard conditions to be able to use this. If you were then to use E cell and solve for it firstly, you'd hopefully know the difference and know what you are doing. This is one easier example to get the ball rolling.
Because people seem so utterly confused by Nate showing a purposefully constructed nonspontaneous version in his last video which you should NOT do because you should be able to use a standard reductions table properly like in his prior videos to that. A positive value and higher up on the table means it is more strongly reduced! A negative reduction value means it is more likely to be oxidized. Copper (II) ion IS reduced in a voltaic cell with iron (II) ion. Cu =+0.34V, Fe=-0.41V, because Fe is being oxidized (as shown by their values and location on the table) you can flip the equation and sign so it becomes +0.41V. +0.34V+ (+0.41V) = + 0.75V. As he stated in THIS video his value was +0.78V because copper being reduced is the spontaneous reaction that would occur in this voltaic/galvanic cell. IF you were to use the nonspontaneous version where iron is reduced, you would use an electrolytic cell which requires energy to be PUT INTO the cell to cause the reaction to occur. Nature wants to reduce copper.
Also, it is MOLES OF ELECTRONS FOR THIS. chem.libretexts.org/Bookshelves/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Electrochemistry/Electrochemistry_and_Thermodynamics Just look it up. Unlike NERNST which shows it both ways depending on the source.
2:55 take a look at what just happened, yo
+polydipsiac Jesse Pinkman moment
He does chemistry too
You are so good. You just explain what I didn't understand in the class CLEARLY.
THANK YOU SO MUCHHHHH
I can tell he's Canadian by the way he pronounces the word "Out" like "Aut".
And yes, It's from 2013!
Thanks!
I think for "n", it is "1" instead of "1 mole", so the final answer should be -151 KJ/mole
I think you're right! I think he may have been confusing "n" (molar quantity: think of PV = nRT) with "z" (dimensionless quantity).
Actually happy I found this because I was thinking the exact same thing. Thanks for pointing out Lan Lan
it is 2 moles of electrons.
delta G naught is kJ/mole for thermo, it is a bit different when it is moles of electrons.
I love how happy you are! Excellent video - am off to do some practice with my calculatrice!
Ye, delta G is in J/mol. But in your equation you give electrons mole unit, and cancel it out with R (J/mol) mole unit. So according to your equation delta G is in J. 3:06 in clip.
Chemist Nate, Thank god for your Bonus Video! I am your big fan, you are amazing, and inspire me to learn more
yo dude thank you so much, i appreciate you going out of your way to help students out.
I disagree. Can you explain?
delta G is in J/mol
R is in J/mol.K
T is in K
so the units in the exponent cancel out...from my perspective.
Oh man, I had to teach this and forgot and I thought I had a video on it and I didn't and now the kids are looking at me like I am an idiot - thanks - you saved me buddy!
I thought faraday constant is 96500 but you used 96495 why?
I think that number of electrons is without unit. In an equation for reaction speed - (Kb*T/h)exp(-delta G/(RT)) - you are left with moles as unit, when it should be unitless. Kb is Boltzman constant and h being Planck constant.
It is moles of electrons. That is irrelevant in the case of Nernst Equation.
Y'all need to use google before you comment more.
So correct me if I am wrong... But E(cell)= E(Reduction)-E(Oxidation) If the standard reduction potential for Fe2+ (-0.44V) and reduction potential for Cu is 0.34V Then E(cell)= 0.34V -(-0.44)=0.78 So this is a spontaneous reaction.
Asclepius The reaction between Cu and Fe2+ is actually non-spontaneous. First step is to determine the reductant and oxidant
(note that oxidant oxidises another compound and itself undergoes reduction
reductant reduces another compound and itself undergoes oxidation).
Cu solid is a reductant, meaning itself will be oxidised (oxidation). Fe2+ is a tricky one in some situations since it can be both a reductant and an oxidant. In this case it will be a oxidant with E=-0.44V as you have stated, meaning itself will be reduced (reduction).
Now there is 2 ways of calculating E(cell)
1. E(cell)= E(Oxidant)-E(Reductant), E(cell)=-0.44-0.34= -0.78, Non-spontaneous reaction
2. E(cell)=E(Oxidation) + E(Reduction)
write down reduction and oxidation half equation of the system. The oxidation of Cu solid has potential of -0.34V, we flipped the equation of Cu2+ reduction and meaning E value becomes negative accordingly. E value for Fe2+ reduction remain the same as we want E for reduction of Fe2+.
now add the two E values, we get E(cell)=-0.34+(-0.44)= -0.78, non-spontaneous reaction.
Both methods are pretty much the same. Personally I find method 2 easier to understand and remember when there are more formulas to remember. Either way its your personally preference which one to use. Best of luck to you with your studies.
Correct. It is positive, as he showed? If you look at a standard reduction table, cu is reduced.
@@billmao1109 You're entirely wrong.
@@billmao1109 It is a spontaneous reaction. Copper's reduction is positive on a reduction table. It is +0.34V. www5.csudh.edu/oliver/chemdata/data-e.htm
YES! I discovered this yesterday.
In fact, you can use the E* values to get E(cell) for a solubility equilibrium, and use it to find K(sp) for the reaction.
It rocked my world to learn that yesterday. Will make a video about it soon.
could yo please paste the link of E cell video
and keep up the good work
I have a problem with the units. I also thought that the final units for G would just be J but some of my professor's lecture slides have G in J/mol or in kj/mol. I thought the mol was supposed to cancel out bcz n is in mol and F also has mol. Is it possible that n is supposed to be unitless? That's the only way G could end up as J/mol or kj/mol. I'm having my final soon so a quick answer would be greatly appreciated.
n is moles of electrons.
Can you make a video series on the Kohlrausch's Law?
Question: In the previous example it stated that Ecell for this reaction was -.78 V . Therefore shouldn't the calculated delta G be positive and the overall reaction be nonspontaneous as mentioned in the beginning of the video?
That previous example was for a purposefully constructed nonspontaneous reaction. In the video prior to that he showed the spontaneous version. Look at your class's or an online reduction potential table. :) The nonspontaneous version does not exist in reality without being in electrolysis which requires an outside electrical source to make it happen. In reality, copper (II) is reduced when in a voltaic cell with iron (II).
OMGosh thank you so much for this video :D
going to be having a chemistry test worth 12.5% of that course soon and this is part of it - now I get it!
If dead pool taught chemistry it would be this guy "Take a look at what just hap-end Yo" cracked up
Fabulous explanation
Sir why 151 KJ it will be 15 KJ.....
This really helped me aut! ;D Thanks
Great thanks al ot . please keep sharing this kind of simple videos to refresh our knowledge ........... we are following your channel waiting for more and more videos
what if i multiplied the whole reaction by 1/2 so the value of n will become 1
Why bother though?
Saving my IB, praise you!
Thank you so much for posting this video its really helped me! nice and simple to digest!!! keep making more videos and saving us! thank you
LOVE YOU! thank you!
but shouldn't it be kJ / mol ??
Yes, dG can have units in j/mol but not here. The units cancel out
Thank You, Boss!!
Beautiful
Good shit man
Thank you
I'm not sure if anyone has mentioned this but you made a small mistake in the multiplication where you inserted a positive Ecell into the equation, rather than the negative value.
A negative Delta G indicates a spontaneous reaction, but your Ecell, which is negative, indicates a non-spontaneous reaction. The idea is that Ecell and Delta G for the same reaction will have opposite signs. Having the same sign for both indicates a mistake has been made.
The ecell is positive? It is positive 0.78V. So the delta G would be negative and spontaneous. A positive Ecell = negative delta G. I haven't any idea why you though the Ecell was negative originally. It's positive. Look at a reduction values table. :)
Dude you are an absolute legend! And i hate you because this must be easy as piss for you! :P Keep it up bro
This video is misleading in multiple ways. He states that delta G = -nFE. The equation should be delta G standard (superscript degree symbol that signifies under standard state conditions) = -nFEcell standard. Delta G and delta G standard are two different things.
The CORRECT equation for plain delta G: delta G = delta G (standard)+RTlnQ
Where R is the universal gas constant (8.314 for measuring delta G in J/mol or .008314 for KJ/mol), T is temperature in Kelvins and Q is the reaction quotient (ratio of concentrations of products over reactants).
Also, standard state conditions implies 1 atm, 1M and 298K.
+Nick Hammond Strictly speaking, standard state has nothing to do with concentration, but rather activity existing at unity. ALT + 0176 is ° handy, for reference.
In the case of this specific equation, they're interchangeable because you are using E naught cell. So it is implied that you are under standard conditions to be able to use this. If you were then to use E cell and solve for it firstly, you'd hopefully know the difference and know what you are doing. This is one easier example to get the ball rolling.
Thankyou
Very useful info, thanks alot
Thank you!
Awesome! You rock dude!
Thanks for your video , it helps me a lot !!! :-)
thx a bunch
Thanks dude
Nice Vid
brazilian here!!
Canadian Eh?
you're so Canadian...I bet you listen to Robin Sparkles!
Vietnam hello
Minnesota accent %100
I think it's an Atlantic Canada accent
Because people seem so utterly confused by Nate showing a purposefully constructed nonspontaneous version in his last video which you should NOT do because you should be able to use a standard reductions table properly like in his prior videos to that. A positive value and higher up on the table means it is more strongly reduced! A negative reduction value means it is more likely to be oxidized. Copper (II) ion IS reduced in a voltaic cell with iron (II) ion. Cu =+0.34V, Fe=-0.41V, because Fe is being oxidized (as shown by their values and location on the table) you can flip the equation and sign so it becomes +0.41V. +0.34V+ (+0.41V) = + 0.75V. As he stated in THIS video his value was +0.78V because copper being reduced is the spontaneous reaction that would occur in this voltaic/galvanic cell. IF you were to use the nonspontaneous version where iron is reduced, you would use an electrolytic cell which requires energy to be PUT INTO the cell to cause the reaction to occur. Nature wants to reduce copper.
Also, it is MOLES OF ELECTRONS FOR THIS. chem.libretexts.org/Bookshelves/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Electrochemistry/Electrochemistry_and_Thermodynamics Just look it up. Unlike NERNST which shows it both ways depending on the source.