## Ch 14 q 55

$E_{cell} = E_{cell}^{\circ}-\frac{RT}{nF}\ln Q$

Ariana de Souza 4C
Posts: 99
Joined: Wed Sep 21, 2016 2:56 pm

### Ch 14 q 55

We are supposed to come up with the anode and cathode reaction for 1 M NiSo4 solution electrolyzed.
When it comes to choosing the anode, the solution manual says to choose the oxidation reaction with the most negative reduction potential. so why did we choose
2H20 --> O2 + 4H+ +4e- with an standard reduction potential of +1.23
2H20 + 2e- --> H2 + 2OH- with a potential of -0.83?
its more negative, so why didn't we choose that?

breinhardt3G
Posts: 10
Joined: Wed Sep 21, 2016 2:56 pm

### Re: Ch 14 q 55

2H20 + 2e- --> H2 + 2OH- has a negative reduction potential because this chemical equation shows a reduction reaction (H+ from the H2O molecule is being reduced to H2 with and oxidation number of 0)
As a result, this reaction would not be a suitable candidate for the anode because it is not a strong oxidation reaction (this is also one of the reasons why it is considered, although not chosen, for the cathode rxn in part a.)

ntyshchenko
Posts: 21
Joined: Wed Sep 21, 2016 2:59 pm

### Re: Ch 14 q 55

In this same question, why can't you use SO4- as the cathode?

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