# Glucose-Oxygen fuel cell conventional representation

Hi everyone,

I'm tutoring electrochemical cells to a student and I got a bit stumped on this question.

The question asked for the conventional representation for a glucose-oxygen fuel cell and the answer was (simplified for typing purposes):

C | Glucose , H+ | CO2 || O2 | H+ , H2O | Pt

Now for conventional representation you always put the species that has the highest oxidation state next to the salt bridge for each half cell. Surely that would be H+ (oxidation state of +1) as CO2 and O2 would have an oxidation state of 0.

Can someone explain why this isn't the case?
I think the current convention (I had to look it up) is that reading the cell from left to right will tell you the reaction that should happen. Glucose is being oxidised to CO2 ; O2 is being reduced to water. This means that glucose must be on the left and separated from the CO2 by a phase boundary. The O2 must be next to the bridge because it is becoming water. If we put H+ next to the bridge on the RHS then we would have to put the water with it, because they are both in the same phase. This would force O2 to be placed on the very far right, which would imply that the reaction is H2O becomes O2 which is the opposite of what we want. This is also the reason that the H+ must be placed to the left of the water - as we read the cell representation we understand that O2 and H+ are becoming water - this wouldn't work if the H+ and H2O were switched.
Original post by tony_dolby
I think the current convention (I had to look it up) is that reading the cell from left to right will tell you the reaction that should happen. Glucose is being oxidised to CO2 ; O2 is being reduced to water. This means that glucose must be on the left and separated from the CO2 by a phase boundary. The O2 must be next to the bridge because it is becoming water. If we put H+ next to the bridge on the RHS then we would have to put the water with it, because they are both in the same phase. This would force O2 to be placed on the very far right, which would imply that the reaction is H2O becomes O2 which is the opposite of what we want. This is also the reason that the H+ must be placed to the left of the water - as we read the cell representation we understand that O2 and H+ are becoming water - this wouldn't work if the H+ and H2O were switched.

That's different from what the textbook says though where it says it has to be the chemical with the higher oxidation state and there are definitely examples used by AQA that don't have it going in reaction order.
Original post by Munrot07
That's different from what the textbook says though where it says it has to be the chemical with the higher oxidation state and there are definitely examples used by AQA that don't have it going in reaction order.

I'm none the wiser after reading the IUPAC manual: "The value of E measured when the left-hand electrode is at virtual equilibrium, and hence acting as a reference electrode, may be called 'the potential of the (right-hand) electrode with respect to the (left- hand) reference electrode. This terminology should not be taken as implying that single electrode potentials are measurable."

Looking at it another way, the 'rule of thumb' is still partially correct. The average oxidation state of carbons in glucose is zero. In carbon dioxide, the oxidation state of carbon is +4. If applied strictly to the RHE, it wouldn't work because the H+ is obviously higher than zero in O2. I'm not convinced that H+, H2O |O2 | Pt is a viable solution for the RHE even though the species with the highest oxidation state is next to the salt bridge.

As you can see, I don't really understand this question either! It sounds like we need an expert!
(edited 1 year ago)