# Two AAMC Test Questions (I won't put the real prob on here)

#### Mastac741

10+ Year Member
On AAMC MCAT Practice Test 5R #31, which asks about the concentration on the polymer. How does it find the answer to be (1/n)[HPO42-]. This one I have no idea.

ON AAMC MCAT Practice Test 7 Prob #24. I thought this was an easy one, but I guess I need some clarification. Since the Cu 2+ is getting reduced to Cu(s) that means it's the cathode, and since Oxygen is getting oxidized, it's the anode. So I plugged the electrode potentials in = E(cell) = E(cath) - E(anode) = .34 - -1.23 = 1.57V which is wrong.

The answer says you just add the two potentials. I know there's the equation E(cell) = E(red) + E(oxydation), but I thought this was the same as the equation above. Anyway, if someone could point out the flaw in my thought process, it'd be appreciated

#### WilliamsF1

10+ Year Member
You have to add the two potentials. You first have to orient the two half reactions so that they're facing the right direction. If you flip one the other way around, you have to change the sign of the potential. Then add the two together. I don't know the question to your first one, though.

Cu++ + 2e- -> Cu(s) .34 V
O2(g) + 4H+ + 4e- -> H2O 1.23 V

Since these are REDUCTION potentials, only the half reaction that is getting reduced stays in the same orientation. The oxidized half reaction must be flipped and the sign must change. Since O2 is getting oxidized you have to flip the 2nd half reaction. You'll see if you flip it, O2 gets oxidized because the O in H20 is at -2 and O2 is 0, it lost electroncs (oxidized). If you kept the 1/2 rxn oriented the same way, you'll see that oxygen was being reduced, which is incorrect.

So since Cu++ is being reduced, you keep the .34v
Since O2 is being oxidized, you must flip that 1/2 reaction and the sign, so -1.23
Add the two and you get -.89V, which should make it an electrolytic cell. It's electrolytic because the E is negative meaning an outside source is providing the power. If E was positive, it means the system is making energy and the rxn is spontaneous and a galvanic cell (delta G is negative).

OP
M

#### Mastac741

10+ Year Member
Thanks for your response. That makes sense. I am still confused when to use which equation. E= E(red) + E(oxydation) making sure to switch the oxidation sign.

And the E= E(cathode) - E(anode). This one you both use reduction potentials, which is a little more straight forward which is why I like it. However, I'm confused why it usually works and does not work here. This equation really is just a rewrite of the other equation using reduction potentials instead of the oxidation potential. *Shrug* Maybe the first equation only works on galvanic cells or something.

WilliamsF1 said:
You have to add the two potentials. You first have to orient the two half reactions so that they're facing the right direction. If you flip one the other way around, you have to change the sign of the potential. Then add the two together. I don't know the question to your first one, though.

Cu++ + 2e- -> Cu(s) .34 V
O2(g) + 4H+ + 4e- -> H2O 1.23 V

Since these are REDUCTION potentials, only the half reaction that is getting reduced stays in the same orientation. The oxidized half reaction must be flipped and the sign must change. Since O2 is getting oxidized you have to flip the 2nd half reaction. You'll see if you flip it, O2 gets oxidized because the O in H20 is at -2 and O2 is 0, it lost electroncs (oxidized). If you kept the 1/2 rxn oriented the same way, you'll see that oxygen was being reduced, which is incorrect.

So since Cu++ is being reduced, you keep the .34v
Since O2 is being oxidized, you must flip that 1/2 reaction and the sign, so -1.23
Add the two and you get -.89V, which should make it an electrolytic cell. It's electrolytic because the E is negative meaning an outside source is providing the power. If E was positive, it means the system is making energy and the rxn is spontaneous and a galvanic cell (delta G is negative).