## Textbook Problem 6O #3

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

Rami_Z_AbuQubo_2K
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### Textbook Problem 6O #3

Can someone explain to me the reasoning behind this problem? I do not understand what specifically causes water to be reduced over metal and vise versa.

Emmaraf 1K
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Joined: Sat Oct 20, 2018 12:16 am

### Re: Textbook Problem 6O #3

Basically during electrolysis, one species is always reduced and another is always oxidized but in contrast to a galvanic cell, Ecell is negative, meaning it requires energy to power the reactions to happen. Therefore, when looking at the reduction potential of water and the reduction potential of a metal, you have to recognize that the species with the highest, most positive reduction potential is actually oxidized so that the Ecell is negative. This is because when you reverse the sign of the Eo with the highest, most positive reduction potential as you oxidize it, your Ecell comes out to be negative.

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