Can anyone explain a) i & ii?
Posted May 5, 2018, 8:10 PM UTC
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[deleted]
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How do you know btw? Since current is flowing it is electrolytic?
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[deleted]
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Apparently this is an electrolytic cell. The anode will be positive. Either water (the second reaction) or fluorine will be oxidized at the anode. Since water has a lower reduction potential than fluorine, it will be the one that will experience oxidation. Now, either water (the fourth reaction) or nickel will be reduced at the cathode (it is negative and attracts cations). It's clear that nickel will be reduced, since it has a higher reduction potential.
So, the two reactions are: Oxidation: 2H2O --> O2 + 4H+ + 4e Reduction: Ni2+ + 2e --> Ni
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Are fluorine and the second water reaction at the anode because of their positive potentials?
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Remember: Red Cat (reduction happens at cathode) and An Ox (oxidation happens at anode)
Edit: I realize that this doesn’t answer your question so ima do it now. The fluorine,H2O and O2 is being reduced while the Ni is being oxizided
My book says that Ni is being reduced though?
I think I have an explanation.
Basically this is going to be an electrolytic cell so the standard cell potential has to be negative, since it is non spontaneous. Since your reaction contains Fe and Ni look for the standard reduction potential combo that results in a negative E value. When you do the potential for Ni - Fe (Reduction - Oxidation) it is negative.
Might have a completely wrong explanation for this so if anyone can confirm this that would be great
Shoot I misread the problem. Since H2 is being produced at one end (the one with the bubbles) we can assume that the second and last half rxn on the chart are being reduced cause both are producing bubbles. The F and Ni are being oxides. If we add up the Ecells at each redox sites, the absolute voltage should be the same (one side is neg and the other is positive) which confirms it.
I believe you meant that reduction happens at the cathode.
Yeah that’s what I meant my b