electrochem - sign error
I'm getting -5.2 mV. Why is it positive? I considered the cu2+ ions of cathode as reactant.
12 Replies
@Dexter
Note for OP
+solved @user1 @user2... to close the thread when your doubt is solved. Mention the users who helped you solve the doubt. This will be added to their stats.can u show what work u did?
e = -0.0591log(cu2+ anode/ cu2+ cathode)
Okay so my logic on this, you said reactant are Cu2+ molecules on cathode side, that's well and good for electrolysis to occur normally
However what is asked here is "back potential" generated due to ion concentration imbalance
Cu2+ concentration is greater at the anode than cathode, and copper reduction potential is positive, whereas only oxidation occurs at anode
So copper ions must migrate to the cathode, in order to get reduced
you can think of this
Cathode: Cu---->Cu2+
Anode: Cu2+----> Cu
Cuz conc of Cu2+ is reduced at anode (not using this rxn but to get an idea)
I'm sure this is opposite to that of what you had considered
(I hope it's not too long)
Also note that reduction also doesn't happen at anode, that's just a purely fictional thing I've written, only the effect of back potential is generated
These 2 equations actually don't occur, only to help you understand that your assumption maybe was wrong
This back potential is overcome anyways by the battery
And the process continues
or maybe i just blabbered something this can be fully wrong
watch a vid on concentration cells for better description
and which one did u take as anode?
because inverse log in negative of actual
ah okay
that makes sense
thanks btw. appreciate it
Np anytime
Inka bhi opinion dekh lo I'm not so sure about my ans
+solved @coolguy.
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