Kaplan electrochemistry test error?

This forum made possible through the generous support of SDN members, donors, and sponsors. Thank you.

HotHamH2O

Turn on the fire hose and open wide!
10+ Year Member
Joined
Jun 24, 2010
Messages
649
Reaction score
31
Capture.png


The explanation claims that at:
Cathode 1: copper is reduced
Cathode 2: OH- is oxidized

But, the reduction potential of water is higher than copper and the oxidation potential of sulfate is higher than OH-. So, given their reduction potentials, water should be reduced at cathode 1 and sulfate should be oxidized at cathode 2 right? Is this an error on their part or am I not seeing something?

Members don't see this ad.
 
Seriously, none of you geniuses on here can bring some insight to this?
 
This is an electrolytic cell, not a galvanic cell. Redox reactions are going to occur in a direction that opposes the natural flow of electrons
 
This is an electrolytic cell, not a galvanic cell. Redox reactions are going to occur in a direction that opposes the natural flow of electrons

Right, but in an electrolytic cells that occur in aqueous solutions, the other species still have to compete with the reduction potential of water. So, wherever a cathode is, the most favorable reduction will occur there. This is why if you have NaBraq in an electrolytic cell, water will be reduced and Br will be oxidized because Na+ has a reduction potential -2.71 and water is -0.83. Neither are spontaneous but water is more favorable, so it beats out sodium.

I seriously just think this is a bad problem in kaplan, because I have watched chad's vids on this and read a ton on it now and can't find anything that supports their reasoning here.
 

Similar threads

Top