Next Video: ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-iWqUWwyPkSs.html Chemistry PDF Worksheets: www.video-tutor.net/chemistry-basic-introduction.html Final Exams and Video Playlists: www.video-tutor.net/
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
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
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.
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
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.
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.
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 .
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
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 :)
Aye bro I don't know who you are, but Jesus the Christ, the Only Son of the Living God loves you sith all of His heart and cares about you. Jezus dragged me oit of mzsturbation, lust, pride amnd anger into His kindness, love amd peCe. He is different from the world. Don't followjp this temporary world, i tried it out , JESUS LOVES YOU MPRE THAN YOUR SIN! ayye, this world failed to satisfy me even for a little while, it didn't work, this world fails. Jeaus loves you much more than the people or the ,oney of thisworld, follow Him, He is the Way, The Pnly Way to Heaven, He is the Truth (truly, give Him a try), and He is the Life. His love spans this life, and the life after this one. Follow His love, and not your pleasures of this world, which will, in the end, dissapoint you, and lead you to Hell. Please, Jesus desperately lives you ( in a God to creation way) and He is truly a being of love and faithfulness. He will never hurt you, kill you or backstab you. ❤
Do they often give equations along side😢.Here at my school you don't dare expect any cell equation💔. Please do a video for assigning which one should be oxidized or reduced when two elements are under consideration. Or send me a link if it is already available . That is just my challenge presently😢
Where the hell comes india man.. can you see anybody commenting usa france Germany etc.. don't make india look bad because of your stupidity. . I am indian also but for god sake don't write everywhere i am from india love from india
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?
Changing the direction if a reaction does not change its sign. This is the most common mistake in electrochemistry that continues to mess students up. Stop, please
Gary Guzman if you subtract oxidation from reduction, don’t change the sign when you flip equations. But if you add them instead of subtract, then you need to change the sign when you flip an equation. You’ll get the same answer either way
when we calculate the cell potential we have to do subtraction. The formula is cathode - anode. In the first example you added the values and now i am confused.
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?
It is super helpful to look at the charges of each elements first! (for example, Fe^2+ in reactant side but on product side, it has 3+ charge, then it is getting bigger, which is oxidation= Anode). However, if the charge was Br2 itself and with a zero charge in reactant side and in product side, it is -1 charge, then it is getting less, which is reduction= cathode. It is super helpful to think of an number line! Going to left (decreasing charge= Cathode) and going to right (got bigger number of charges = Anode). But don't get confused, reduction means gaining ELECTRONS and oxidation means losing ELECTRONS, not charges!!
Question 3 is wrong. It states that its an electrolytic cell but has a positive value for standard potential value. That will result in negative value for delta G meaning its spontaneous. Electrolytic cells are supposed to be non spontaneous. Maybe that question should I have stated that is a voltaic/galvanic cell question.
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.
Sir , please I'm confused..... Is E cell= Ereduction - E oxidation How come did u just add together instead of subtracting please explain cos I'm kinda confused
wait, isn't getting the cell potential of a galvanic cell done in a way in that, E(oxidation)-E(reduction)? And it seems like we do not gave to change the signs when flipping the equations around....
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
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?
