Unit 4 questions thread

[soccerboi]

For part b) , I put O₂ + 2H₂O + 4e^- ---> 4OH^- , i just grabbed it of the Electrochemical series
But why have they written it as O₂ + 4e^- ---> 2O^2- ? What's the thought process to getting this equation?

Thanks

[nisha]

For part b) , I put O₂ + 2H₂O + 4e^- ---> 4OH^- , i just grabbed it of the Electrochemical series
But why have they written it as O₂ + 4e^- ---> 2O^2- ? What's the thought process to getting this equation?

Thanks

Hey, thats what I did too when I did the question. But you have to look a bit more closely than that. You know the anode reaction, and you also know that water has to be the ONLY PRODUCT. Therefore, if we combine the anode ration given and the equation for moist air (the one you thought it was) we are going to have OH- as a product (which we do not need) and H20 as a reactant (which we do not need as it says, "hydrogen and oxygen are the only reactants").
So work backwards: If the anode reaction is: H2 + O2- ->H20 + 2e-
The Cathode reaction must start with O2 (as it is a reactant), must cancel out the O2- because that is not a product/reactant that we need (but make sure we have the right number of oxygens...hence a 2 in the front). To do this, we must put it on the products side, and then balance by using electrons.
Therefore:
O2 + 4e- ---> 2O2-
Then we multiply the anode reaction by 2 to get the overall!
I hope that all makes sense.

[soccerboi]

Thanks Nisha for the explanation

[soccerboi]

When writing an equilibrium expression, why don't we need to include H2O if its in liquid state? If we do include it, would we lose the mark?

[thushan]

When writing an equilibrium expression, why don't we need to include H2O if its in liquid state? If we do include it, would we lose the mark?

H2O concentration is constant. If you include it, you will lose the mark.

[soccerboi]

Thanks Thushan

Will the % ionisation of a strong acid always be 100%?

[Jenny_2108]

Thanks Thushan

Will the % ionisation of a strong acid always be 100%?

Strong acids ~ or = 100% ionisation

[charmanderp]

The exception being amphiprotic acids, I guess, where ionisations after the first one (which occurs almost to completion) are relatively weak.

[thushan]

Thanks Thushan

Will the % ionisation of a strong acid always be 100%?

By definition, yes. Well, effectively 100% (like 99.9999999999999999999999999%)

As for derp's comment, I'd agree, although I like to think of say H2SO4 as a strong acid, and HSO4- as a weak acid as a separate species altogether.

[soccerboi]

VCCA 2010 exam, Q 4 B)i)

Why did they not accept "to increase the reaction rate" as an answer? Is it because since the vessel is sealed, it is not necessary to increase the rate as benzoic acid would eventually all react?.

Also, why is temp change lower in a calorimeter, when insulation is absent?

[nisha]

In a calorimeter, it matters that complete combustion has occurred (and that the maximum temp change has spread into the solution) rather than the rate of reaction. Rate of reaction implies that time is a factor and is dependent on changing the results. Whilst this can change the energy supplied, it doesn't really have any relation to the context in the question. You may have mistakes it for a disguised equilibrium question where adding such a substance would change the position, but it is not this in this case.


Temp change is lower because there is less heat finally as some of the heat is lost to the surroundings. Eg: With insulation Tf-27 degrees
Ti-20 degrees. Therefore temp change is =7 degrees
Without insulation, Tf-=23 degrees (as heat is lost), whilst Ti=20 degrees. Hence temp change is =3 degrees

Correct me, if i'm wrong though.

[charmanderp]

Yeah as Nisha said, rate of reaction is mostly irrelevant. What you've said isn't incorrect but in chemistry you have to give the 'best answer' and consider the background to which the question is written. In this case the purpose of the experiment is calibration which is all about ensuring that you get a 'complete' energy transfer from what you're combusting to heat. Rate of reaction doesn't play a role in the extent of a reaction, whereas making sure that there is sufficient oxygen to react all of the methanoic acid or whatever is was is a key factor, otherwise your calculations would be all off as you'd have the correct amount in mol of acid but a lower value for ΔH and ΔT.

[soccerboi]

Can someone clarify if my understanding is correct:

In galvanic cells, the strongest oxidant will always react with the strongest reductant.
In electrolytic cells, the weakest oxidant will always react with the weakest reductant.
Also, in electrolysis, the oxidation eqn is the higher eqn of the two in consideration and occurs in reverse on the electrochemical series.

Sorry if i haven't worded the last bit clearly enough and thanks for any help

[charmanderp]

The strongest oxidant will always react with the strongest reductant. The distinction between galvanic and electrolytic cells is that with the former you can have a 'positive gradient' if you draw a line linking the oxidant and the reductant ie. the oxidation reaction is higher than the reduction reaction, whereas with galvanic cells you need to have a 'negative gradient' in order to have a spontaneous reaction ie. the reduction reaction must be higher than the oxidation reaction.

[Tonychet2]

chemderp is right , the strongest oxidant , going up on the left side of the electrochemical series , always reacts with the strongest reductant, going down on the rights ide of the electrochemical series

just circle all the species present in a solution and circle the highest one on the left and the lowest on the right , dont forget h20, and lookout for OH- or H+ if it says acidic/basic environment !