For anyone wondering why at 7:40 the answer is -2.20V it's because electrolytic cells can have negative cell potentials which is the difference between galvonic cells. When electrolytic cells have negative cell potential values they need energy to run. It's like when your smartphone is dead; you need to recharge it first before it can run again
another neat way to determine which cell potential is "flipped" is which is more negative. The more negative a cell potential is, the more it wishes to be oxidized. And when we oxidize reactions in problems similar to these we will flip the reaction, which also can flip the sign to its corresponding reduction potential
Professor Organic Chemistry Tutor, thank you for showing How to analyze and calculate cell potential of a galvanic cell/ electrolytic cell in AP/General Chemistry. In this video, the examples are simple to follow and understand from start to finish. This is an error free video/lecture on TH-cam TV with the Organic Chemistry Tutor.
Keep in mind when he says that it has to be positive, he means the overall potential for the entire cell, and it only applies to galvanic cells. Electrolytic cells can be negative because they are powered by an external source. You can also use the value of the cell potential to determine if the reaction will be in equilibrium (Ecell = 0) or if it’s spontaneous (Ecell = positive) or non-spontaneous (Ecell = negative). Hopefully you knew this already but I’m adding it just in case :)
@@AverageMED He's wrong. An electrolytic cell, by definition, has a negative Ecell. Always negative. Never positive. First time I've seen him make a big mistake in a video.
Thank you so much for this video, it is exactly what I was looking for! One question: for cases like Ag/Mg reaction, why does multiplying the Ag equation by 2 not change the cell potential at all? I keep thinking that if I have to move twice as many electrons, the potential should double.
hi dude the basic thing of this chapter is electrode potential do not change with how many moles , pressure , temperature you add the ep donot change . i think my answer would clear your doubt .
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Can anyone explain equation 3 ? Thank you. In contrast to a galvanic cell,the most positive standard reduction half reaction within an electrolytic cell is the one that experiences oxidation. With that in mind, Why did he reverse iron half reaction and not the Br half reaction (where Br would be our anode and Fe be our cathode? Ered of cathode - E red of anode.
Br is the one receiving the electron therefore it is a reduction and Fe is losing electron therefore it’s a oxidation. Knowing that reduction is cathode and oxidation is anode. Using the formula Ered-Eox or Ecathode - Eanode = 1.09-(0.77) which is 0.32
yes it does. for calculating cell potential in standard conditions (1M) use the method in this video. for concentrations other than 1M use nernst equation
In an electrolytic cell, you have to put in energy to do the electrolysis, so Gibbs free energy is positive here, which means cell potential needs to be negative, not positive.
Question- why the volts are added together inside of subtracting it? Wouldnt you be incorporating based on the following emf formula E cell= E cathode - E anode instead?
On question 3) (7:35) you don't multiply Fe^2+ potential by two. So the overall result should be -0.45v not +0.32v. If this is the answer why galvanic cell result negative?
Electrode potential is an intensive property and non additive in nature so we can't directly add it the way you did. Instead what we have to do is add all the reactions and equate it's Gibbs free energy to the final free energy. Could you pls solve this confusion for me?
First you say the cell potential has to be positive, then you introduce a nonspontaneous reaction (were the potential is negative) without explaining. Im confused. So i wouldve had the wrong answer if i flipped the Standard for aluminum so i could do 1,66 - 0,535?
I just wanted to know if the balancing of the electrons changed the cell potential for my homework. Thankfully I have found somewhere where it says that it does not change. Where else can I find the answer to this question? Like actually I was surfing this channels videos to find this answer.
If a galvanic cell, invert the reaction that will result in a positive E. If an electrolytic cell, it depends what they are looking for. The "E" can be negative, positive, or zero. Thus it doesn't matter which you invert.
Dont you think when we subtract one from other with opposite charge the net result with sign of those which is larger term...as you did in case of secnd e.g ...suggest me please
@@sajidmehmood6976 If Ni is used as oxidation it's wrong cause with half - reactions, one half - reaction must be reduced and the other oxidized . Both of them can't be oxidized. Fe is already oxidized
hmm I'm not quite understanding how he got +0.21V for Q2. After reversing the iron's rxn, it becomes an oxidation (anode) with E= +0.44. So should it be: E cell = E cat - E ano = -0.23 - 0.44 which is
@@sandeepsammy9480 Sure thing. For this question, it'd be helpful to arrange the half-rxns in the decreasing order of standard cell potential values - we will have Ni (-0.23V) then Fe (-0,44V) so Ni+2 rxn is reduction (cathode) and Fe+2 is oxidation (anode). The formula (Cathode - Anode) works according to the standard potential values of reduction half-rxn. Thus, since anode = -0.44, Cathode - Anode = -0.23 - (-0.44) = 0.23 + 0.44. Hope this helps :)
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For anyone wondering why at 7:40 the answer is -2.20V it's because electrolytic cells can have negative cell potentials which is the difference between galvonic cells.
When electrolytic cells have negative cell potential values they need energy to run. It's like when your smartphone is dead; you need to recharge it first before it can run again
thanks!
How did u get -2.20V at problem 3?
THANKS!
God bless your soul for pointing this out!!!
another neat way to determine which cell potential is "flipped" is which is more negative. The more negative a cell potential is, the more it wishes to be oxidized.
And when we oxidize reactions in problems similar to these we will flip the reaction, which also can flip the sign to its corresponding reduction potential
Thanks!!!
Tnx
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Professor Organic Chemistry Tutor, thank you for showing How to analyze and calculate cell potential of a galvanic cell/ electrolytic cell in AP/General Chemistry. In this video, the examples are simple to follow and understand from start to finish. This is an error free video/lecture on TH-cam TV with the Organic Chemistry Tutor.
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"A standard cell potential has to be positive" can't believe my professor never mentioned that in lecture man haha
Keep in mind when he says that it has to be positive, he means the overall potential for the entire cell, and it only applies to galvanic cells. Electrolytic cells can be negative because they are powered by an external source. You can also use the value of the cell potential to determine if the reaction will be in equilibrium (Ecell = 0) or if it’s spontaneous (Ecell = positive) or non-spontaneous (Ecell = negative). Hopefully you knew this already but I’m adding it just in case :)
@@avationmusic can an electrolytic cell be positive?? I am having a hard time wondering why they ever would be
@@AverageMED He's wrong. An electrolytic cell, by definition, has a negative Ecell. Always negative. Never positive. First time I've seen him make a big mistake in a video.
@@user-pr5py8cw9x yeah i was gonna say by definition it has to be negative
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Thank you so much for this video, it is exactly what I was looking for! One question: for cases like Ag/Mg reaction, why does multiplying the Ag equation by 2 not change the cell potential at all? I keep thinking that if I have to move twice as many electrons, the potential should double.
Cell potential is an intensive property, so it will not change even if the number of moles vary.
hi dude the basic thing of this chapter is electrode potential do not change with how many moles , pressure , temperature you add the ep donot change . i think my answer would clear your doubt .
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what do you do when you're not told if it's electrolytic or galvanic?
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Great.please make video how to affect using active and inert electrode on the product of electrolysis
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In Standard cell potential don't we subtract the cell potential of cathode from anode as per formula?
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Mohammad nasir you do, but he reversed one of the reactions, that’s why he still got the correct answer despite adding them.
@@shubhamsinghhh99 I think you should check yourself before accusing him.
yeap it should, that's how is shown on the books
His half rxn are standard hence he can choose to reverse the other one
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Can anyone explain equation 3 ? Thank you.
In contrast to a galvanic cell,the most positive standard reduction half reaction within an electrolytic cell is the one that experiences oxidation.
With that in mind, Why did he reverse iron half reaction and not the Br half reaction (where Br would be our anode and Fe be our cathode? Ered of cathode - E red of anode.
Br is the one receiving the electron therefore it is a reduction and Fe is losing electron therefore it’s a oxidation. Knowing that reduction is cathode and oxidation is anode. Using the formula Ered-Eox or Ecathode - Eanode = 1.09-(0.77) which is 0.32
Hi question, does molarity of the substances affect the cell potential?
yes it does. for calculating cell potential in standard conditions (1M) use the method in this video. for concentrations other than 1M use nernst equation
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should we multiply the E°cell value when equation is being multiplied?
No since the standard potential is in volts and not in volts/mole.
Interesting. It’s a bit confusing when it comes to the redox flow battery. Can you do it for iron redox flow battery reactions.
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How did u get -2.20 ..I got +2.20 in the last question
In an electrolytic cell, you have to put in energy to do the electrolysis, so Gibbs free energy is positive here, which means cell potential needs to be negative, not positive.
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Question- why the volts are added together inside of subtracting it? Wouldnt you be incorporating based on the following emf formula E cell= E cathode - E anode instead?
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On question 3) (7:35) you don't multiply Fe^2+ potential by two. So the overall result should be -0.45v not +0.32v. If this is the answer why galvanic cell result negative?
because its asking for an electrolytic cell.
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Please tell me why souldn't we need to balance the last example?
how did you come up with the reduction potentials?
Electrode potential is an intensive property and non additive in nature so we can't directly add it the way you did. Instead what we have to do is add all the reactions and equate it's Gibbs free energy to the final free energy. Could you pls solve this confusion for me?
Yea thats what im thinking
thanks master 🙇
Forgalvanic cell cell potential hs to be positive. For electrolytic cell it can be negative positive or zero.
I learned in class the formula for ecell was to subtract ecathode and eanode? Can you explain why you are adding them?
This is the same problem I'm facing 😭
Thank you po
First you say the cell potential has to be positive, then you introduce a nonspontaneous reaction (were the potential is negative) without explaining.
Im confused.
So i wouldve had the wrong answer if i flipped the Standard for aluminum so i could do 1,66 - 0,535?
same here! XP
for galvanic cell, that's the rule that it must be positive. in electrolytic cell it can be either
Thank you sir
Thanks man!
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I just wanted to know if the balancing of the electrons changed the cell potential for my homework. Thankfully I have found somewhere where it says that it does not change. Where else can I find the answer to this question? Like actually I was surfing this channels videos to find this answer.
Where did they get the value of Voltage?
Thank you
Will the cell potential value for each reaction be given ?
Good explain
5:51 how did you get those cell potentials
Why cell potential of lead acid battery is 2volt if lead and sulpher potential is much lower ????
when the electrons are present on the same side then which reaction is to be inversed and on what basis?
If a galvanic cell, invert the reaction that will result in a positive E.
If an electrolytic cell, it depends what they are looking for. The "E" can be negative, positive, or zero. Thus it doesn't matter which you invert.
I will do my best
Do we have to remember the data as you have shown in example 3..
Dont you think when we subtract one from other with opposite charge the net result with sign of those which is larger term...as you did in case of secnd e.g ...suggest me please
As in Ni and Fe eg. The net sign is negatve...
I have read this eg in another book with Ni used as oxidation
@@sajidmehmood6976 If Ni is used as oxidation it's wrong cause with half - reactions, one half - reaction must be reduced and the other oxidized . Both of them can't be oxidized. Fe is already oxidized
🩷🩷🩷💯Thankx you so much
hmm I'm not quite understanding how he got +0.21V for Q2. After reversing the iron's rxn, it becomes an oxidation (anode) with E= +0.44. So should it be: E cell = E cat - E ano = -0.23 - 0.44 which is
Even I have the same doudt and going through the comments...so that anyone can reply me .
If your doudt has clarified .. can you please tell ?
@@sandeepsammy9480 Sure thing. For this question, it'd be helpful to arrange the half-rxns in the decreasing order of standard cell potential values - we will have Ni (-0.23V) then Fe (-0,44V) so Ni+2 rxn is reduction (cathode) and Fe+2 is oxidation (anode). The formula (Cathode - Anode) works according to the standard potential values of reduction half-rxn. Thus, since anode = -0.44, Cathode - Anode = -0.23 - (-0.44) = 0.23 + 0.44. Hope this helps :)
@@sandeepsammy9480 Also it's a galvanic cell meaning Cathode - Anode always > 0
@@venusz6134 thankyou ...I understood ..thanks for the clarification
Thank you soo much