Help with difficult electrode GC question

[SonhosDaVida]

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So looking at question 18, I know it is referring to the oxidation or reduction of cations zinc and copper and trying to find a certain voltage.

But what if we were looking at anions such as the reduction of Br + 1e ---> Br- and oxidation of Cl- --> Cl +1e

Then would the equation be

Eobserved = E°cell - (0.059/n)(log(cathode/anode))

Or is Eobserved = E°cell - (0.059/n)(log(anode/cathode)) a set equation for finding of voltages of all kind, cations and anions? Thanks

 

[Doogie.Howser]

Before reading on and just to get you thinking, why should the cathode and anode be flipped?

I ask because you might run into a similar problem on test day that makes you question what you know. If you can't control your nerves, then you'll psych yourself out. When you come across a question that makes you think like this, ask yourself why would I do it differently this time? You may have to change what you're used to, but justify it.

The cathode is the cathode and the anode the anode. The equation doesn't change. The contents of the anode are still being oxidized, regardless if it's a cation or anion.

 

[SonhosDaVida]

@Doogie.Howser so the
Eobserved = E°cell - (0.059/n)(log(cathode/anode)) is an actual equation to be memorized? Can you explain to me why the cathode's reduction is on top while the anode's oxidation is on the bottom?

I thought cations will migrate to the cathode after being oxidized at the anode and anions migrate to the anode after being reduced at the cathode? There doesn't seem to be a clear product or reactant in this.....

 

[Doogie.Howser]

@Doogie.Howser so the
Eobserved = E°cell - (0.059/n)(log(cathode/anode)) is an actual equation to be memorized? Can you explain to me why the cathode's reduction is on top while the anode's oxidation is on the bottom?

I thought cations will migrate to the cathode after being oxidized at the anode and anions migrate to the anode after being reduced at the cathode? There doesn't seem to be a clear product or reactant in this.....

1) It's anode/cathode
2) I wouldn't memorize the equation. Something like this should be provided in the passage.
3) The original equation has the logQ. If you write out the equation, you'll see that prod/reac can be simplified to anode/cathode.

For the question, the equation comes out to Cu2+ (aq) + Zn (s) --> Cu (s) + Zn2+ (aq). Q = [Zn2+] / [Cu2+] whether it's prod/reac or anode/cathode.

Zn is oxidized and is the anode. So C and D are eliminated. B would give a >1 value for the log making it positive and an overall less than 1.15V value. Answer A would give a <1 value for the log making it negative and an overall greater than 1.15V value.

If you want to make up numbers for Cl and Br, go ahead. It'll end up being the same.