Post by: soccerboi on July 13, 2012, 11:53:21 am
Would the equilibrium expression still be the same if the equation was reversed?
e.g wA+xB <--> yC
k=[C]y/([A]wX[B ]x)
but if the eqn was reversed i.e yC <--> wA + xB
Would the equilibrium expression still be k=[C]y/([A]wX[B ]x)?
Post by: FlorianK on July 13, 2012, 12:38:40 pm
Kn=K^-1
And you multiply [A]^q with ^x not add them
Post by: Graphite on July 13, 2012, 01:27:12 pm
Which shows how the new K is related to the initial K
Post by: soccerboi on July 13, 2012, 02:18:31 pm
Post by: soccerboi on August 04, 2012, 10:27:06 pm
From neap study guide
Q) Write the equation occurring at the cathode while the cell is recharging.
A) PbSO4(s)+H++2e---->Pb(s)+HSO4-
From checkpoints
Q) Give the equation of the reaction that occurs at the negative electrode when the accumulator is being recharged.
A) PbSO4(s)+2e---->Pb(s)+SO42-
The first question asks for at the cathode and the second question asks for the negative electrode(anode) so why are the equations the same? Which one is correct?
Post by: soccerboi on August 05, 2012, 05:04:29 pm
Ag2O(s) + Zn(s) ---> 2Ag(s) + Zn(OH2)(s)
Give the eqn for the reaction occurring at the anode and cathode.
Could someone please show me how to split it into the two separate eqns. I don't know how to do it correctly. :(
I got:
Zn + 2H2O ---> Zn(OH)2 +2H++2e-
Ag2O +2H+ +2e- ---> 2Ag + H2O
but answers have it as:
Zn + 2OH- --->Zn(OH)2 +2e-
Ag2O +H2O +2e- ---> 2Ag +2OH-
Post by: soccerboi on August 05, 2012, 08:09:19 pm
Post by: Jenny_2108 on August 05, 2012, 09:49:09 pm
Post by: soccerboi on August 05, 2012, 10:09:36 pm
Why does your school study so fast? Mine has just finished AOS1Nah we're not fast, we skipped ch 23,24,25 and will come back to it. They wanted to get the topics which require more conceptual understanding out of the way first.
Post by: golden on August 07, 2012, 04:39:26 pm
In galvanic cells, why must the oxidant and reductant not come into contact? What happens if they do contact?
They must be kept separate so that they do not react directly - that would use up all your energy source at once, and potentially generate a lot of heat, which all together is a waste (because they are usually used as an energy source for a long time).
Post by: soccerboi on August 28, 2012, 05:45:09 pm
Why can't Al be deposited from aqeous solutions of Al3+ ?
Thanks
Post by: jaydee on August 28, 2012, 09:01:08 pm
Post by: soccerboi on August 28, 2012, 09:09:43 pm
Al can't be deposited because if you look at your electrochemical series, water (at -0.83V) is a stronger oxidant than Al3+ and will reduce in preference to Al3+. the word 'solution' refers to involving water.Oh ok thank you, but how do you know which water eqn to look at? I can see 4 on the electrochemical series that contain water.
Post by: Jenny_2108 on August 28, 2012, 09:13:47 pm
Al can't be deposited because if you look at your electrochemical series, water (at -0.83V) is a stronger oxidant than Al3+ and will reduce in preference to Al3+. the word 'solution' refers to involving water.Oh ok thank you, but how do you know which water eqn to look at? I can see 4 on the electrochemical series that contain water.
Depending on the question
Post by: jaydee on August 28, 2012, 09:20:13 pm
Post by: soccerboi on September 10, 2012, 06:24:07 pm
Thanks in advance
Post by: Lasercookie on September 10, 2012, 07:08:01 pm
The solutions are all 0.10M. We can see that the pH's vary, that is some donate their
That's the lowest pH we could get out of a monoprotic acid with 0.10M of acid.
So in this case, if we were to have a pH less than 1, that would mean we'd need a
Hence, Acid III, as that has a pH of 0.7.
Interestingly, if we solve in general:
So when pH is less than one, then
Post by: soccerboi on September 13, 2012, 08:35:19 pm
Question 16
When comparing the electrolysis of molten NaF and that of a 1.0 M aqueous solution of NaF, which one of the
following statements is correct?
A. The product at the anodes is the same in both cells and the product at the cathodes is the same in both
cells.
B. The product at the anodes is the same in both cells but the products at the cathodes are different.
C. The product at the cathodes is the same in both cells but the products at the anodes are different.
D. The products at the cathodes of the cells are different and also the products at the anodes are different.
Can someone please explain this question to me, my teacher couldn't explain it =.=
Post by: soccerboi on September 13, 2012, 08:53:14 pm
Post by: soccerboi on September 17, 2012, 07:21:04 am
Post by: joseph95 on September 17, 2012, 09:34:13 pm
VCAA 2008
Question 16
When comparing the electrolysis of molten NaF and that of a 1.0 M aqueous solution of NaF, which one of the following statements is correct?
A. The product at the anodes is the same in both cells and the product at the cathodes is the same in both
cells.
B. The product at the anodes is the same in both cells but the products at the cathodes are different.
C. The product at the cathodes is the same in both cells but the products at the anodes are different.
D. The products at the cathodes of the cells are different and also the products at the anodes are different.
Can someone please explain this question to me, my teacher couldn't explain it =.=
For this question, you have to determine for each electrolysis what can possibly act as a reductant or an oxidant.
For molten NaF, the only oxidant present is Na+, and the only reductant present is F-.
Thus at the anode, F- ions will be oxidised to F2, and at the cathode Na+ ions will be reduced to Na.
Anode: F-(aq) --> F2(g) + 2e
Cathode: Na+ + e --> Na(s)
For an aqueous solution of NaF, Na+, F- and H2O are present.
H2O is a stronger oxidant than Na+, so water instead will be reduced at the cathode. H2O is also a stronger reductant than F-, and thus will be oxidised at the anode.
Anode: 2H2O(l) -->O2(g) + 4H+(aq) + 4e
Cathode: 2H2O(l) + 2e --> H2(g) + 2OH-(aq)
Thus the answer is D: the products at the cathodes and anodes are different.
Post by: charmanderp on September 17, 2012, 09:56:30 pm
Which terminal of a rechargeable galvanic cell connects to the negative terminal of the power supply when its recharging? The positive terminal of the galvanic cell?In recharging a cell, the negative terminal of the power supply is connected to the cathode. This is where reduction occurs. Since reduction is here the reverse process of oxidation, that means the cathode was previously the anode, which was the negative terminal of the galvanic cell.
Post by: soccerboi on September 20, 2012, 07:03:23 pm
Post by: charmanderp on September 20, 2012, 07:16:19 pm
Post by: Lasercookie on September 20, 2012, 07:42:04 pm
why isn't natural gas considered to be renewable? Can't we or cows just fart to make more? LOLI like this question. I'm not too sure, but I'll have a go at answering. Be aware that there might be inaccuracies here.
Natural gas is the mixture of hydrocarbon gases, it's extracted from natural gas reserves / oil reserves. "Oil and natural gas are produced by the same geological process according fossil fuel suggestion: anaerobic decay of organic matter deep under the Earth's surface". I believe when we say 'natural gas', I think that's what we refer to.
http://en.wikipedia.org/wiki/List_of_natural_gas_fields
http://en.wikipedia.org/wiki/Natural_gas_field
"In common usage, deposits rich in oil are known as oil fields, and deposits rich in natural gas are called natural gas fields."
http://en.wikipedia.org/wiki/Natural_gas#Sources
In fear of ruling out on what might be a revolutionary energy source that will fix the world's problems, I would be inclined to think that farts are not an efficient power source to harness. I'm not sure how you would go about it. http://en.wikipedia.org/wiki/Flatulence#Composition
Recall that "non-renewable" means that the rate of depletion is much greater than the rate of production.
I'm not entirely sure on this but stuff like Petroleum (http://en.wikipedia.org/wiki/Petroleum#Formation) is still being formed, just at a very very very slow rate (centuries, millions of years seem to be the figures thrown around from quick googling) If that implies that fossil fuels supply may one day - one day meaning a very very long time away - replenish themselves, I'm not sure. Either way, that won't be something that would happen in the next few centuries.
Also I just realised why fossil fuels are called fossil fuels. Now what was I talking about?
Oh yes, so there's Biogas, which is formed from from decaying "organic matter". That's a renewable source as you know.
There's also this http://en.wikipedia.org/wiki/Renewable_natural_gas
http://en.wikipedia.org/wiki/Substitute_natural_gas
Post by: soccerboi on September 25, 2012, 09:32:19 pm
The two relevant half-equations for the discharge reaction are
Cd(s) + 2OH–(aq) → Cd(OH)2(s) + 2e–
NiOOH(s) + H2O(l) + e– → Ni(OH)2(s) + OH–(aq)
a. Will the pH in the region immediately surrounding the anode increase or decrease as the cell discharges? Give a reason for your answer.
Solution
Decrease because hydroxide ions are being consumed, meaning that the [OH–] is decreasing.
Isnt OH- also being produced in the Cd(OH)2(s)?
Post by: nerfsdacier on September 25, 2012, 10:09:33 pm
Post by: soccerboi on September 26, 2012, 08:03:13 am
a. Write half-equations for the oxidation and reduction reactions.
i. oxidation reaction
ii. reduction reaction
Solution
i. 2H2O(l) → O2(g) + 4H+(aq) + 4e–
ii. Cu2+(aq) + 2e– → Cu(s)
i'm still confused about which water eqn to use? Why have they used that water eqn and not say the one at E value of 0.00?
Post by: charmanderp on September 26, 2012, 05:24:09 pm
A beaker containing 250 mL of 1.00 M CuSO4(aq) is electrolysed using carbon rods.The equation at 0.00 does not involve water in any way...It's the reduction of H+ ions to form hydrogen gas (or oxidation of hydrogen gas yaddah yaddah).
a. Write half-equations for the oxidation and reduction reactions.
i. oxidation reaction
ii. reduction reaction
Solution
i. 2H2O(l) → O2(g) + 4H+(aq) + 4e–
ii. Cu2+(aq) + 2e– → Cu(s)
i'm still confused about which water eqn to use? Why have they used that water eqn and not say the one at E value of 0.00?
Since you have Cu2+ acting as an oxidant and this is electrolysis, clearly you need water to act as a reductant. The occasion where water acts as reductant with the strongest reduction potential is the equation at +1.23 which the solution has used.
Post by: soccerboi on September 26, 2012, 05:45:15 pm
Post by: charmanderp on September 26, 2012, 05:50:27 pm
Post by: VCEstudentguy on September 28, 2012, 07:12:40 pm
Post by: aznxD on September 28, 2012, 08:36:12 pm
Post by: atom on September 29, 2012, 08:54:24 pm
Post by: charmanderp on September 29, 2012, 09:06:15 pm
Post by: soccerboi on October 02, 2012, 03:35:45 pm
b. i. Circle the best solution below for the student to choose for the electrolyte in the cell.
deionised water
sodium sulfate
copper sulfate
sodium iodide
Solution
sodium sulfate
ii. Give an explanation for your choice.
Solution
Sodium sulfate will provide ions that can carry charge in the solution but which will not react
in preference to the H2O(l).
I dont understand it, can someone elaborate please
Cheers
Post by: Nobby on October 02, 2012, 04:34:07 pm
A student wishes to electrolyse some water by setting up an electrolytic cell.
b. i. Circle the best solution below for the student to choose for the electrolyte in the cell.
deionised water
sodium sulfate
copper sulfate
sodium iodide
Solution
sodium sulfate
ii. Give an explanation for your choice.
Solution
Sodium sulfate will provide ions that can carry charge in the solution but which will not react
in preference to the H2O(l).
I dont understand it, can someone elaborate please
Cheers
So there are two main things you're looking for in your electrolyte - it must conduct, and it can't react. So you can't use deionised water, because it won't conduct electricity. You can't use copper sulfate, because once if you run a current through the solution, copper is going to deposit at the cathode instead of H2 gas, just look at your Eo table. You can't use sodium iodide, because as you can glean from the Eo table I2 gas will bubble off the anode instead of O2. And finally, you can use sodium sulfate, because sodium is so far down the electrochemical series it's not going to reduce in preference to water.
Post by: charmanderp on October 02, 2012, 05:33:21 pm
And finally, you can use sodium sulfate, because sodium is so far down the electrochemical series it's not going to reduce in preference to water.
Also Nobby, it's referring to Na+ here which is an oxidant, not a reductant. So it wouldn't be reduced in preference to water rather than it not reducing in preference to water. Sorry for being pedantic but I just thought I'd point that out (:
Post by: Nobby on October 02, 2012, 05:54:24 pm
Also Nobby, it's referring to Na+ here which is an oxidant, not a reductant. So it wouldn't be reduced in preference to water rather than it not reducing in preference to water. Sorry for being pedantic but I just thought I'd point that out (:
Ah yeah probably should have said 'sodium ions' and 'be reduced'.
Post by: charmanderp on October 02, 2012, 06:00:32 pm
Also Nobby, it's referring to Na+ here which is an oxidant, not a reductant. So it wouldn't be reduced in preference to water rather than it not reducing in preference to water. Sorry for being pedantic but I just thought I'd point that out (:
Ah yeah probably should have said 'sodium ions' and 'be reduced'.
Yeah I figured you probably knew that that was the case but others might have read it and taken it at its word.
Post by: soccerboi on October 04, 2012, 10:43:29 pm
Thanks a lot for the explanation but could you reexplain why sodium iodide is not a suitable electrolyte? I always have trouble with knowing which half eqns to look at on the E.SA student wishes to electrolyse some water by setting up an electrolytic cell.
b. i. Circle the best solution below for the student to choose for the electrolyte in the cell.
deionised water
sodium sulfate
copper sulfate
sodium iodide
Solution
sodium sulfate
ii. Give an explanation for your choice.
Solution
Sodium sulfate will provide ions that can carry charge in the solution but which will not react
in preference to the H2O(l).
I dont understand it, can someone elaborate please
Cheers
So there are two main things you're looking for in your electrolyte - it must conduct, and it can't react. So you can't use deionised water, because it won't conduct electricity. You can't use copper sulfate, because once if you run a current through the solution, copper is going to deposit at the cathode instead of H2 gas, just look at your Eo table. You can't use sodium iodide, because as you can glean from the Eo table I2 gas will bubble off the anode instead of O2. And finally, you can use sodium sulfate, because sodium is so far down the electrochemical series it's not going to reduce in preference to water.
Also, you refered to I2 gas in your explanation, but on the E.S it solid, so do we till refer to that half eqn?
Post by: thushan on October 04, 2012, 10:48:08 pm
Al can't be deposited because if you look at your electrochemical series, water (at -0.83V) is a stronger oxidant than Al3+ and will reduce in preference to Al3+. the word 'solution' refers to involving water.Oh ok thank you, but how do you know which water eqn to look at? I can see 4 on the electrochemical series that contain water.
You see the one that uses water as a REACTANT.
Post by: charmanderp on October 04, 2012, 10:54:27 pm
I- is a stronger reductant (left hand side of the serieS) than H2O, so you'd have to apply a current for a certain amount of time before this I- supply was exhausted (oxidised to I2) and you started producing O2(g) and electrolysing H2O at the anode.Thanks a lot for the explanation but could you reexplain why sodium iodide is not a suitable electrolyte? I always have trouble with knowing which half eqns to look at on the E.SA student wishes to electrolyse some water by setting up an electrolytic cell.
b. i. Circle the best solution below for the student to choose for the electrolyte in the cell.
deionised water
sodium sulfate
copper sulfate
sodium iodide
Solution
sodium sulfate
ii. Give an explanation for your choice.
Solution
Sodium sulfate will provide ions that can carry charge in the solution but which will not react
in preference to the H2O(l).
I dont understand it, can someone elaborate please
Cheers
So there are two main things you're looking for in your electrolyte - it must conduct, and it can't react. So you can't use deionised water, because it won't conduct electricity. You can't use copper sulfate, because once if you run a current through the solution, copper is going to deposit at the cathode instead of H2 gas, just look at your Eo table. You can't use sodium iodide, because as you can glean from the Eo table I2 gas will bubble off the anode instead of O2. And finally, you can use sodium sulfate, because sodium is so far down the electrochemical series it's not going to reduce in preference to water.
Post by: soccerboi on October 04, 2012, 11:04:48 pm
Post by: charmanderp on October 04, 2012, 11:19:55 pm
Post by: soccerboi on October 06, 2012, 12:08:50 pm
But why have they written it as O2 + 4e- ---> 2O2- ? What's the thought process to getting this equation?
Thanks
Post by: nisha on October 06, 2012, 01:07:12 pm
For part b) , I put O2 + 2H2O + 4e- ---> 4OH- , i just grabbed it of the Electrochemical seriesHey, 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").
But why have they written it as O2 + 4e- ---> 2O2- ? What's the thought process to getting this equation?
Thanks
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.
Post by: soccerboi on October 06, 2012, 01:50:18 pm
Post by: soccerboi on October 11, 2012, 05:41:08 pm
Post by: thushan on October 11, 2012, 05:41:39 pm
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.
Post by: soccerboi on October 11, 2012, 07:23:22 pm
Will the % ionisation of a strong acid always be 100%?
Post by: Jenny_2108 on October 11, 2012, 07:43:31 pm
Thanks Thushan
Will the % ionisation of a strong acid always be 100%?
Strong acids ~ or = 100% ionisation
Post by: charmanderp on October 11, 2012, 08:22:04 pm
Post by: thushan on October 11, 2012, 11:21:39 pm
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.
Post by: soccerboi on October 12, 2012, 05:58:44 pm
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?
Post by: nisha on October 12, 2012, 06:17:01 pm
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.
Post by: charmanderp on October 12, 2012, 06:43:24 pm
Post by: soccerboi on October 28, 2012, 06:37:55 pm
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 :)
Post by: charmanderp on October 28, 2012, 06:42:19 pm
Post by: Tonychet2 on October 29, 2012, 11:40:11 pm
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 !
Post by: charmanderp on October 29, 2012, 11:41:08 pm
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 seriesYeah this is the first thing do whenever I see an electrolysis question. Write down what species are present.
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 !
Post by: joseph95 on October 29, 2012, 11:44:45 pm
whereas with galvanic cells you need to have a 'negative gradient' in order to have a spontaneous reaction ie. the oxidation reaction must be higher than the reduction reaction.
Isn't it the other way around? Yes there's a negative gradient, but the reduction reaction is always above the oxidation reaction in galvanic cells I thought.
Post by: charmanderp on October 29, 2012, 11:50:30 pm
Right you are!whereas with galvanic cells you need to have a 'negative gradient' in order to have a spontaneous reaction ie. the oxidation reaction must be higher than the reduction reaction.
Isn't it the other way around? Yes there's a negative gradient, but the reduction reaction is always above the oxidation reaction in galvanic cells I thought.
Post by: joseph95 on October 29, 2012, 11:52:40 pm
Post by: mals22 on November 02, 2012, 01:49:41 pm
Eg. Question 4
Dinitrogen tetroxide, N2O4(g), dissociates to form nitrogen dioxide, NO2(g), according to the equation
N2O4(g) 2NO2(g)
0.45 mol of N2O4 gas is placed in an empty 1.00 L vessel at 100°C. When the system reaches equilibrium, there
is 0.36 mol of NO2 gas present in the vessel.
a. Calculate the numerical value of the equilibrium constant for this reaction at 100°C.
How do you figure out how much is remaining.. :/
Thanks for your help in advance.
Post by: thushan on November 02, 2012, 01:54:49 pm
Post by: soccerboi on November 02, 2012, 02:25:53 pm
Thank you for any assistance :)
Post by: soccerboi on November 10, 2012, 08:23:06 am
Thanks