Electrochemistry Review - Cell Potential & Notation, Redox Half Reactions, Nernst Equation
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- เผยแพร่เมื่อ 30 ก.ย. 2024
- This electrochemistry review video tutorial provides a lot of notes, equations, and formulas that you need to pass your next chemistry test / exam. It’s a nice video that provides an introduction or overview of electrochemistry including the most important fundamental topics and concepts. It contains plenty of examples and practice problems.
Intro to Galvanic & Voltaic Cells:
• Introduction to Galvan...
How To Draw Galvanic Cells:
• How To Draw Galvanic C...
Standard Reduction Potentials:
• Standard Reduction Pot...
Cell Potential Problems:
• Cell Potential Problem...
Cell Notation Problems:
• Cell Notation Practice...
___________________________________
Concentration Cells:
• Concentration Cells & ...
Cell Potential & Gibbs Free Energy:
• Cell Potential & Gibbs...
Cell Potential & Equilibrium K:
• Equilibrium Constant K...
Nernst Equation:
• Nernst Equation Explai...
Electrolysis of Water:
• Electrolysis of Water ...
_____________________________________
Electrolysis of Sodium Chloride:
• Electrolysis of Sodium...
Electrolysis & Electroplating Problems:
• Electrolysis & Electro...
Electrochemistry Practice Problems:
• Electrochemistry Pract...
SAT Chemistry Subject Test Review:
• SAT Chemistry Subject ...
Carbon -14 Dating:
• Carbon 14 Dating Probl...
Beer Lambert's Law:
• Beer Lambert's Law, Ab...
______________________________________
Final Exams and Video Playlists:
www.video-tuto...
Full-Length Videos and Worksheets:
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Chemistry PDF Worksheets:
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Chemistry PDF Worksheets: www.video-tutor.net/chemistry-basic-introduction.html
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No
If you turn on captions by pressing on cc you’ll see it
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At 58:26, If you multiply the reaction by 2, wouldn't the half reaction then be 1.54V? And the new total would be 1.77V?
edit: 1:03:12 The answer is no.
If the reaction is multiplied by a coefficient the cell potential won't be affected or altered.
It is only altered when the reaction is reversed.
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58:14 it is 1 volt
yes I also got 1.0 V
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2:16 What if We plug in the battery to outlet and to my device at the same time..... it's (the battery) is receiving energy (electrolytic) and also supplying energy (galvanic).....how does this work ?
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At 59:13 you did their sum wrong its supposed to be 1 volt 0.23+0.77=1
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34:34, shouldn't it be PLUS RTlnQ
Yes it is plus
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At 34.26 minutes of video,
∆G = ∆G^o - RT lnQ
I have 2 textbook here of Gen Chemistry saying that
∆G = ∆G^o + RT lnQ
^o is to superscript, to set delta G in standard state
I am learning a lot from you but I like to clarify this formula..Thank you.
The textbooks are:
Fundamentals of Chemistry, 4th Ed. by Frank Brescia etc. page 311
Principles of General Chemistry, 3rd edition by Martin S. Silberberg, page 676
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Hello, thank you so much for the video. In response to question 3 45:57, should it be 1776.61 hours?
no
58:23, 1.0 V not 1.1 V
why when i watch this video it seems much simpler and easy but why the questions i always get complex question T.T
Remember "Red Cat An Ox" reduction, cathode, anode oxidation
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How does he know, the concentration for Zn^2+ and Cu^2+. That it is 0,01 and 10?? Please answer
COOL-om
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How is zinc the product and copper is the reactant if the copper is gaining mass in the cell?
Mark Keith, this reaction is not about product or reactant. It's electro chemical reaction pertaining to cell. We call them product/reactant only to simplify.
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anybody get a different answer for K at 20:40? mine keeps coming up 6.22 x 10^-38…
I tried it it’s correct. Notice there is a minus sign and delta G is minus
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how is the cathode less positive if its supposed to be more positive then the anode?
But isn't it ΔG=ΔG⁰+RTlog(Q) ? I am just asking because looking at my note, i am confused.
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Isnt the nernst equation +0.0591/n*.... And not -?
Oh didnt notice that it is + when it is [oxidator]/[reductor] and - when its [product] /[educt]
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At 26:16 you switch the 10 M and the .01M which makes your math wrong. Your explanation for the increase in E doesn't make any sense now since the Q was wrong.
Looking back I found that he did reactants over products for Q, so if you use the right numbers it makes sense
Exactly, i was lost there as well but now i understand how the values got "flipped" around; The liguids and solids does not inclunde in the q expression, so the values left over are in the expression
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13:20 I dont know wen it comes to electro chemistry whether or not my syllabus is different from what you are teaching here .I check my textbook and this what you're explaining ain't in there .
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Can any one help in how you come to 25080 if you 1.1 - 0.97 = 013 divided again by 0.591 does not make 25080. Clear this up for me please he not explaining well about 2 x 1.1 - 0.97 ÷ 0.591 my calculator is not coming up with any of his answers
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Mistake at 32:52, this is not a product-favoured system, Q is the initial concentration ratio. However, the positive standard cell potential is indicating that it is a product-favoured system since the reaction is spontaneous.
Am I correct?
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31:59 into the video how do I get the answer 25080. 1.1v - 0.97 ÷ 0.0591 can't get the answer as he not say if you multiple by 2 or 10 or just 1.1v - 0.97. Help
22:29 how did you know that CuSO4 have 10M
I think just made up the concentration values in order to plug them into the Q expression
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14:58 Why it's 212 Joules when you calculated -212267 J/mol?
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How does one figure out the electronegativity/cell potential of organic compounds, complexes ECT?
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