# electrochemistry question (chemistry q-pack passage #20)

#### theonlytycrane

5+ Year Member

From the description, reduction happens at the Ag electrode so this is the cathode.

The Cd electrode would be the anode (equation 3 shows Cd oxidation).

I was thinking about equations 1 & 2 for the cathode, though. The passage says the oxygen is getting reduced, but equation 2 shows the higher reduction potential. Would equation 1 get flipped?

#### bobeanie95

2+ Year Member
No equation 1 would remain the same. The passage states that oxygen gas is bubbled through and it is reduced. Since equation 1 shows the net gain of 4 electrons you would keep it as is. Also, if you look at the oxidation states oxygen gas has an oxidation of 0, while on the right side it goes to a -1 oxidation state, so its reduced.

OP

#### theonlytycrane

5+ Year Member
@bobeanie95 Would Ag then be oxidized at this electrode? It's weird because it has the higher reduction potential.

#### bobeanie95

2+ Year Member
I don't know the rest of the passage but if they mention that Ag is at the cathode like you said, then it would be reduced (Or is this something you reasoned out?). Reduction always occurs at the cathode regardless if its an electrochemical or electrolytic cell. This would also make sense with the voltage potentials since the 2nd equation would have the highest reduction potential.

#### laczlacylaci

2+ Year Member
@bobeanie95 Would Ag then be oxidized at this electrode? It's weird because it has the higher reduction potential.
I just came across this as well. @theonlytycrane, did you figure it out?

O2 is reduced at Ag electrode, so Ag must be the cathode. Making Cd the anode.

From my understanding:
The site of reduction=cathode=higher standard cell potential.
The site of oxidation=anode=lower standard cell potential.

Eq3 is already flipped. It's original potential is -0.8, making Cd an anode. Singe Ag has a more + potential, making it the cathode.

Between eq 1 and 2.
eq1 is reduced, but a site of oxidation (in a way) (or a oxidation agent)=lower standard cell potential
eq2 is the site of reduction=higher standard cell potential.

OP

#### theonlytycrane

5+ Year Member
I just came across this as well. @theonlytycrane, did you figure it out?

O2 is reduced at Ag electrode, so Ag must be the cathode. Making Cd the anode.

From my understanding:
The site of reduction=cathode=higher standard cell potential.
The site of oxidation=anode=lower standard cell potential.

Eq3 is already flipped. It's original potential is -0.8, making Cd an anode. Singe Ag has a more + potential, making it the cathode.

Between eq 1 and 2.
eq1 is reduced, but a site of oxidation (in a way) (or a oxidation agent)=lower standard cell potential
eq2 is the site of reduction=higher standard cell potential.
The presentation of equations 1 / 2 with the wording from the passage make it crappy. They say O2 is reduced at the Ag electrode so I guess we have to take their word for it. If O2 gets reduced, equation 2 gets flipped such that Ag gets oxidized.

Equation 3 is presented correctly as the anode.

If there was a question asking for Ecell, my best attempt would be +.40 - .80 + .81 for a positive and spontaneous voltage. The 3 equations make it confusing though.

laczlacylaci

#### BerkReviewTeach

##### Company Rep & Bad Singer
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10+ Year Member
This question is tricky by design. I'm really glad it was posted, because I feel sort of vindicated. I got an earful (more like a page full) about a tricky question in our books that involved something very much like this. It is because of AAMC questions like this that we make some of our questions as hard and tricky as we do. But enough of that.

The big thing to realize is that it is a silver ELECTRODE, not a silver reactant. Silver is already reduced, so you need not consider it. The MCAT test writers are notorious for giving you extraneous information you don't really need, such as Equation 2. The only reason it is there is to support the notion that Ag will NOT oxidize (given its emf), so you need only consider reactions 1 and 3 when determining the overall emf.

You need reaction 1 and 2 X reaction 3 to get the overall reaction of:

O2 + 2 Cd + 2 H2O <=> 2 Cd(OH)2

Be sure to recall that you need not multiple emf values by the stoichiometric number. The overall reaction has an emf of 1.21 V

OP

#### theonlytycrane

5+ Year Member
This question is tricky by design. I'm really glad it was posted, because I feel sort of vindicated. I got an earful (more like a page full) about a tricky question in our books that involved something very much like this. It is because of AAMC questions like this that we make some of our questions as hard and tricky as we do. But enough of that.

The big thing to realize is that it is a silver ELECTRODE, not a silver reactant. Silver is already reduced, so you need not consider it. The MCAT test writers are notorious for giving you extraneous information you don't really need, such as Equation 2. The only reason it is there is to support the notion that Ag will NOT oxidize (given its emf), so you need only consider reactions 1 and 3 when determining the overall emf.

You need reaction 1 and 2 X reaction 3 to get the overall reaction of:

O2 + 2 Cd + 2 H2O <=> 2 Cd(OH)2

Be sure to recall that you need not multiple emf values by the stoichiometric number. The overall reaction has an emf of 1.21 V
So the redox pair would be O2 (getting reduced) and Cd getting oxidized. The Cd2+ comes from the Cd electrode, but does the Ag electrode stay as Ag(s)?

#### betterfuture

2+ Year Member
If you say the Ag is the electrode and not the reactant, then how would we differentiate between Cd and Ag? Cd is also an electrode so. I am just wondering because it still doesn't seem clear.