chemistry question
hi, please could I have help on question v on page 9 of this paper:
https://www.physicsandmathstutor.com/pdf-pages/?pdf=https%3A%2F%2Fpmt.physicsandmathstutor.com%2Fdownload%2FChemistry%2FA-level%2FPast-Papers%2FCAIE%2FPaper-4%2FQP%2FMarch%202020%20(v2)%20QP.pdf
I've figured why E3 is greater than E2 and that's all the ms wants but I don't get how that is related to why aqua regia can dissolve gold?
Thank you.
https://www.physicsandmathstutor.com/pdf-pages/?pdf=https%3A%2F%2Fpmt.physicsandmathstutor.com%2Fdownload%2FChemistry%2FA-level%2FPast-Papers%2FCAIE%2FPaper-4%2FQP%2FMarch%202020%20(v2)%20QP.pdf
I've figured why E3 is greater than E2 and that's all the ms wants but I don't get how that is related to why aqua regia can dissolve gold?
Thank you.
Reply 1
Original post by anonymous56754
hi, please could I have help on question v on page 9 of this paper:
https://www.physicsandmathstutor.com/pdf-pages/?pdf=https%3A%2F%2Fpmt.physicsandmathstutor.com%2Fdownload%2FChemistry%2FA-level%2FPast-Papers%2FCAIE%2FPaper-4%2FQP%2FMarch%202020%20(v2)%20QP.pdf
I've figured why E3 is greater than E2 and that's all the ms wants but I don't get how that is related to why aqua regia can dissolve gold?
Thank you.
https://www.physicsandmathstutor.com/pdf-pages/?pdf=https%3A%2F%2Fpmt.physicsandmathstutor.com%2Fdownload%2FChemistry%2FA-level%2FPast-Papers%2FCAIE%2FPaper-4%2FQP%2FMarch%202020%20(v2)%20QP.pdf
I've figured why E3 is greater than E2 and that's all the ms wants but I don't get how that is related to why aqua regia can dissolve gold?
Thank you.
if E3 is more positive than E2 then NO3-/NO shifts towards NO and [AuCl4]-/Au,Cl- shifts towards [AuCl4]-...
so if you had a system with NO3- (aq) and Cl- (aq) and you added Au (s) into this system, [AuCl4]- (aq) would form, ie Au has dissolved.
(eg NaCl (s) dissolving is just the formation of Na+ (aq) and Cl- (aq))
Reply 2
Original post by anonymous56754
hi, please could I have help on question v on page 9 of this paper:
https://www.physicsandmathstutor.com/pdf-pages/?pdf=https%3A%2F%2Fpmt.physicsandmathstutor.com%2Fdownload%2FChemistry%2FA-level%2FPast-Papers%2FCAIE%2FPaper-4%2FQP%2FMarch%202020%20(v2)%20QP.pdf
I've figured why E3 is greater than E2 and that's all the ms wants but I don't get how that is related to why aqua regia can dissolve gold?
Thank you.
https://www.physicsandmathstutor.com/pdf-pages/?pdf=https%3A%2F%2Fpmt.physicsandmathstutor.com%2Fdownload%2FChemistry%2FA-level%2FPast-Papers%2FCAIE%2FPaper-4%2FQP%2FMarch%202020%20(v2)%20QP.pdf
I've figured why E3 is greater than E2 and that's all the ms wants but I don't get how that is related to why aqua regia can dissolve gold?
Thank you.
Aqua regia can dissolve gold because it's a mix of concentrated hydrochloric acid and concentrated nitric acid. When combined, these acids create a strong oxidizer that breaks gold down into its ions.
To see how this works, let's look at some half-equations. These equations show the chance of a substance to gain or lose electrons, which helps us understand what's happening.
In the second half-equation, we see:
[AuCl4]^-(aq) + 3e- ==> Au(s) + 4Cl^-(aq)......................Eo/V = +1.00
This tells us that turning [AuCl4]^- into gold is pretty easy and happens on its own, without needing extra energy. The +1.00 Eo/V shows that this reaction likes to happen.
When you add hydrochloric acid, it raises the level of Cl^- ions, which can interact with AuCl4^- to make a more stable form, [AuCl4]^2-. This change boosts the E value, making it even more likely for [AuCl4]^- to turn into gold.
Now, looking at the third half-equation:
NO3^-(aq) + 4H^+(aq) + 3e- ==> NO(g) + 2H2O(l)...........Eo/V = +0.96
This reaction, where nitrate turns into nitrogen monoxide, also happens on its own but isn't as favorable as the one with gold. The +0.96 Eo/V indicates it’s less likely to occur.
Adding nitric acid raises the level of NO3^- ions, which can create a stable form with gold, [Au(NO3)4]^-. This change also raises its E value, making that reaction more favorable.
So, when you put the two acids together, they form a strong oxidizer that breaks down gold into ions.
My 2 cents!
Bye,
Sandro
(edited 1 month ago)
Reply 3
Original post by goaway1239846
if E3 is more positive than E2 then NO3-/NO shifts towards NO and [AuCl4]-/Au,Cl- shifts towards [AuCl4]-...
so if you had a system with NO3- (aq) and Cl- (aq) and you added Au (s) into this system, [AuCl4]- (aq) would form, ie Au has dissolved.
(eg NaCl (s) dissolving is just the formation of Na+ (aq) and Cl- (aq))
so if you had a system with NO3- (aq) and Cl- (aq) and you added Au (s) into this system, [AuCl4]- (aq) would form, ie Au has dissolved.
(eg NaCl (s) dissolving is just the formation of Na+ (aq) and Cl- (aq))
ohhh that makes sense, but what is the purpose of the no3- then?
Reply 4
Original post by anonymous56754
ohhh that makes sense, but what is the purpose of the no3- then?
Please look at my Reply 2 above.
Bye,
Sandro
Reply 5
I suppose you could consider the impact on the electrode potentials for systems 2 and 3 using the following logic.
[AuCl4]^- (aq) + 3e^- <=> Au (s) + 4Cl^- (aq) (E° = +1.00 V)
NO3^- (aq) + 4H^+ (aq) + 3e^- <=> NO (g) + 2H2O (l) (E° = +0.96V)
E°(cell) = (+0.96 V) - (+1.00 V) = -0.04 V
The electrode potential values given in the table are only correct if they are measured under standard conditions. As such, if the conditions were standard, a mixture of HCl and HNO3 would fail to dissolve gold.
However, under standard conditions, E°(cell) is only ever so slightly more negative than zero and so adjusting the conditions could make E(cell) positive and thus the reaction can be made spontaneous.
They’ve told you that the concentrations of each acid are much greater than the standard concentration of (circa) 1.00 mol dm^-3, so the conditions are most certainly non-standard. Let’s now consider each equilibrium separately:
[AuCl4]^- (aq) + 3e^- <=> Au (s) + 4Cl^- (aq) (E° = +1.00 V)
By using a large [HCl], you ensure a large [Cl^-] and so drive this equilibrium left. This means that E in practice is less -ve than the standard value given by some amount x because the equilibrium goes to the side that produces electrons. Let’s say that E = (+1.00 - x) V.
NO3^- (aq) + 4H^+ (aq) + 3e^- <=> NO (g) + 2H2O (l) (E° = +0.96V)
By using a large [HNO3], you ensure a large [NO3^-] and a large [H^+]. This drives this equilibrium right and so E increases by some amount y. Thus let’s say E = (+0.96 + y) V.
E(cell) = (+0.96 + y) V - (+1.00 - x) V = (x + y - 0.04) V, which is positive
So Au can dissolve in aqua regia in practice
[AuCl4]^- (aq) + 3e^- <=> Au (s) + 4Cl^- (aq) (E° = +1.00 V)
NO3^- (aq) + 4H^+ (aq) + 3e^- <=> NO (g) + 2H2O (l) (E° = +0.96V)
E°(cell) = (+0.96 V) - (+1.00 V) = -0.04 V
The electrode potential values given in the table are only correct if they are measured under standard conditions. As such, if the conditions were standard, a mixture of HCl and HNO3 would fail to dissolve gold.
However, under standard conditions, E°(cell) is only ever so slightly more negative than zero and so adjusting the conditions could make E(cell) positive and thus the reaction can be made spontaneous.
They’ve told you that the concentrations of each acid are much greater than the standard concentration of (circa) 1.00 mol dm^-3, so the conditions are most certainly non-standard. Let’s now consider each equilibrium separately:
[AuCl4]^- (aq) + 3e^- <=> Au (s) + 4Cl^- (aq) (E° = +1.00 V)
By using a large [HCl], you ensure a large [Cl^-] and so drive this equilibrium left. This means that E in practice is less -ve than the standard value given by some amount x because the equilibrium goes to the side that produces electrons. Let’s say that E = (+1.00 - x) V.
NO3^- (aq) + 4H^+ (aq) + 3e^- <=> NO (g) + 2H2O (l) (E° = +0.96V)
By using a large [HNO3], you ensure a large [NO3^-] and a large [H^+]. This drives this equilibrium right and so E increases by some amount y. Thus let’s say E = (+0.96 + y) V.
E(cell) = (+0.96 + y) V - (+1.00 - x) V = (x + y - 0.04) V, which is positive
So Au can dissolve in aqua regia in practice
Reply 6
Original post by anonymous56754
ohhh that makes sense, but what is the purpose of the no3- then?
see TypicalNerd's reply: the extra NO3- AND the extra Cl- both tilt their respective eqms.
[extra Cl- ⇒ E2 more negative (good)... extra NO3- ⇒ E3 more positive (also good)]
Half equations/reactions do not just happen, they happen in conjunction with another half eqn/rxn; the reason E2 shifts left and E3 shifts right is because E3 is more positive than E2.
Reply 7
Original post by TypicalNerd
I suppose you could consider the impact on the electrode potentials for systems 2 and 3 using the following logic.
[AuCl4]^- (aq) + 3e^- <=> Au (s) + 4Cl^- (aq) (E° = +1.00 V)
NO3^- (aq) + 4H^+ (aq) + 3e^- <=> NO (g) + 2H2O (l) (E° = +0.96V)
E°(cell) = (+0.96 V) - (+1.00 V) = -0.04 V
The electrode potential values given in the table are only correct if they are measured under standard conditions. As such, if the conditions were standard, a mixture of HCl and HNO3 would fail to dissolve gold.
However, under standard conditions, E°(cell) is only ever so slightly more negative than zero and so adjusting the conditions could make E(cell) positive and thus the reaction can be made spontaneous.
They’ve told you that the concentrations of each acid are much greater than the standard concentration of (circa) 1.00 mol dm^-3, so the conditions are most certainly non-standard. Let’s now consider each equilibrium separately:
[AuCl4]^- (aq) + 3e^- <=> Au (s) + 4Cl^- (aq) (E° = +1.00 V)
By using a large [HCl], you ensure a large [Cl^-] and so drive this equilibrium left. This means that E in practice is less -ve than the standard value given by some amount x because the equilibrium goes to the side that produces electrons. Let’s say that E = (+1.00 - x) V.
NO3^- (aq) + 4H^+ (aq) + 3e^- <=> NO (g) + 2H2O (l) (E° = +0.96V)
By using a large [HNO3], you ensure a large [NO3^-] and a large [H^+]. This drives this equilibrium right and so E increases by some amount y. Thus let’s say E = (+0.96 + y) V.
E(cell) = (+0.96 + y) V - (+1.00 - x) V = (x + y - 0.04) V, which is positive
So Au can dissolve in aqua regia in practice
[AuCl4]^- (aq) + 3e^- <=> Au (s) + 4Cl^- (aq) (E° = +1.00 V)
NO3^- (aq) + 4H^+ (aq) + 3e^- <=> NO (g) + 2H2O (l) (E° = +0.96V)
E°(cell) = (+0.96 V) - (+1.00 V) = -0.04 V
The electrode potential values given in the table are only correct if they are measured under standard conditions. As such, if the conditions were standard, a mixture of HCl and HNO3 would fail to dissolve gold.
However, under standard conditions, E°(cell) is only ever so slightly more negative than zero and so adjusting the conditions could make E(cell) positive and thus the reaction can be made spontaneous.
They’ve told you that the concentrations of each acid are much greater than the standard concentration of (circa) 1.00 mol dm^-3, so the conditions are most certainly non-standard. Let’s now consider each equilibrium separately:
[AuCl4]^- (aq) + 3e^- <=> Au (s) + 4Cl^- (aq) (E° = +1.00 V)
By using a large [HCl], you ensure a large [Cl^-] and so drive this equilibrium left. This means that E in practice is less -ve than the standard value given by some amount x because the equilibrium goes to the side that produces electrons. Let’s say that E = (+1.00 - x) V.
NO3^- (aq) + 4H^+ (aq) + 3e^- <=> NO (g) + 2H2O (l) (E° = +0.96V)
By using a large [HNO3], you ensure a large [NO3^-] and a large [H^+]. This drives this equilibrium right and so E increases by some amount y. Thus let’s say E = (+0.96 + y) V.
E(cell) = (+0.96 + y) V - (+1.00 - x) V = (x + y - 0.04) V, which is positive
So Au can dissolve in aqua regia in practice
that makes sense, thank you!
Reply 8
Original post by goaway1239846
see TypicalNerd's reply: the extra NO3- AND the extra Cl- both tilt their respective eqms.
[extra Cl- ⇒ E2 more negative (good)... extra NO3- ⇒ E3 more positive (also good)]
Half equations/reactions do not just happen, they happen in conjunction with another half eqn/rxn; the reason E2 shifts left and E3 shifts right is because E3 is more positive than E2.
[extra Cl- ⇒ E2 more negative (good)... extra NO3- ⇒ E3 more positive (also good)]
Half equations/reactions do not just happen, they happen in conjunction with another half eqn/rxn; the reason E2 shifts left and E3 shifts right is because E3 is more positive than E2.
Yep, but to clarify, doesn't the shifts cause E3 to become more positive than E2?
Reply 9
Original post by anonymous56754
Yep, but to clarify, doesn't the shifts cause E3 to become more positive than E2?
Yes, E3 > E2 under the conditions used.
The shifts cause E3 to increase by some amount (which I called y V) and E2 to decrease by some amount (which I called x V)
E(cell) = (+0.96 + y) V - (+1.00 - x) V = (x + y -0.04) V
Where x + y > 0.04
Really and truly, the best way to justify that E(cell) is greater than 0 would be to use the Nernst equation and use the fact that aqua regia is typically a 3:1 mix of conc HCl and conc HNO3 (e.g [HCl] = 9 mol dm^-3 and [HNO3] = 4 mol dm^-3, so [H^+] = 13 mol dm^-3, [Cl^-] = 9 mol dm^-3 and [NO3^-] = 4 mol dm^-3), though neither bit of information is given in the question, unfortunately. Admittedly the Nernst equation would likely break down as the concentrations of reagents are huge and so don’t approximate the activities of each species well, but that’s a massive digression and a very much post-A level consideration.
(edited 1 month ago)
Reply 10
Original post by TypicalNerd
Yes, E3 > E2 under the conditions used.
The shifts cause E3 to increase by some amount (which I called y V) and E2 to decrease by some amount (which I called x V)
E(cell) = (+0.96 + y) V - (+1.00 - x) V = (x + y -0.04) V
Where x + y > 0.04
Really and truly, the best way to justify that E(cell) is greater than 0 would be to use the Nernst equation and use the fact that aqua regia is typically a 3:1 mix of conc HCl and conc HNO3 (e.g [HCl] = 9 mol dm^-3 and [HNO3] = 4 mol dm^-3, so [H^+] = 13 mol dm^-3, [Cl^-] = 9 mol dm^-3 and [NO3^-] = 4 mol dm^-3), though neither bit of information is given in the question, unfortunately. Admittedly the Nernst equation would likely break down as the concentrations of reagents are huge and so don’t approximate the activities of each species well, but that’s a massive digression and a very much post-A level consideration.
The shifts cause E3 to increase by some amount (which I called y V) and E2 to decrease by some amount (which I called x V)
E(cell) = (+0.96 + y) V - (+1.00 - x) V = (x + y -0.04) V
Where x + y > 0.04
Really and truly, the best way to justify that E(cell) is greater than 0 would be to use the Nernst equation and use the fact that aqua regia is typically a 3:1 mix of conc HCl and conc HNO3 (e.g [HCl] = 9 mol dm^-3 and [HNO3] = 4 mol dm^-3, so [H^+] = 13 mol dm^-3, [Cl^-] = 9 mol dm^-3 and [NO3^-] = 4 mol dm^-3), though neither bit of information is given in the question, unfortunately. Admittedly the Nernst equation would likely break down as the concentrations of reagents are huge and so don’t approximate the activities of each species well, but that’s a massive digression and a very much post-A level consideration.
(edited 1 month ago)
Reply 11
Original post by Nitrotoluene
Aqua regia can dissolve gold because it's a mix of concentrated hydrochloric acid and concentrated nitric acid. When combined, these acids create a strong oxidizer that breaks gold down into its ions.
To see how this works, let's look at some half-equations. These equations show the chance of a substance to gain or lose electrons, which helps us understand what's happening.
In the second half-equation, we see:
This tells us that turning [AuCl4]^- into gold is pretty easy and happens on its own, without needing extra energy. The +1.00 Eo/V shows that this reaction likes to happen.
When you add hydrochloric acid, it raises the level of Cl^- ions, which can interact with AuCl4^- to make a more stable form, [AuCl4]^2-. This change boosts the E value, making it even more likely for [AuCl4]^- to turn into gold.
Now, looking at the third half-equation:
This reaction, where nitrate turns into nitrogen monoxide, also happens on its own but isn't as favorable as the one with gold. The +0.96 Eo/V indicates it’s less likely to occur.
Adding nitric acid raises the level of NO3^- ions, which can create a stable form with gold, [Au(NO3)4]^-. This change also raises its E value, making that reaction more favorable.
So, when you put the two acids together, they form a strong oxidizer that breaks down gold into ions.
My 2 cents!
Bye,
Sandro
To see how this works, let's look at some half-equations. These equations show the chance of a substance to gain or lose electrons, which helps us understand what's happening.
In the second half-equation, we see:
[AuCl4]^-(aq) + 3e- ==> Au(s) + 4Cl^-(aq)......................Eo/V = +1.00
This tells us that turning [AuCl4]^- into gold is pretty easy and happens on its own, without needing extra energy. The +1.00 Eo/V shows that this reaction likes to happen.
When you add hydrochloric acid, it raises the level of Cl^- ions, which can interact with AuCl4^- to make a more stable form, [AuCl4]^2-. This change boosts the E value, making it even more likely for [AuCl4]^- to turn into gold.
Now, looking at the third half-equation:
NO3^-(aq) + 4H^+(aq) + 3e- ==> NO(g) + 2H2O(l)...........Eo/V = +0.96
This reaction, where nitrate turns into nitrogen monoxide, also happens on its own but isn't as favorable as the one with gold. The +0.96 Eo/V indicates it’s less likely to occur.
Adding nitric acid raises the level of NO3^- ions, which can create a stable form with gold, [Au(NO3)4]^-. This change also raises its E value, making that reaction more favorable.
So, when you put the two acids together, they form a strong oxidizer that breaks down gold into ions.
My 2 cents!
Bye,
Sandro
Thank you
Reply 12
Original post by anonymous56754
Yep, but to clarify, doesn't the shifts cause E3 to become more positive than E2?
I won’t use the word shift to avoid confusion…
E is effectively a measure of the “inherent” equilibrium constant of these reduction equilibria. The tendency of the thing on the LHS to be reduced and form the RHS.
The additional Cl- and NO3- would change the positions of equilibrium as per Le Chatelier: the eqm now lies further to the left for E2 (which is a more -ve E than before/standard) and further to the right for E3 (which is a more +ve E than before/standard), when the two half equations are combined with something else (perhaps the standard hydrogen electrode, so we can put a number on it)
—
But what is the overall reaction? As I said, half equations need to be combined with another one. A simplified way of thinking about it is this, [but see the aforementioned nernst equation for more details - all we are really saying with an “overall reaction” is which side of the overall eqm is favoured with K > 1, usually >> 1]
If E3 is more +ve than E2, this means effectively the equilbrium lies further to the right for E3 than it does for E2. So the overall reaction is the forward reaction for E3 and the backward reaction for E2.
—
There could be a situation (eg our solutions of HCl and HNO3 are not concentrated enough) where we’ve made E2 more -ve than standard (further to the LHS than it would be in standard conditions) and E3 more +ve (further to the RHS) than standard, but not by enough.
In this case, E3 is still less positive than E2, and so “the eqm lies further to the RHS for E2 than it does for E3 overall”**
The ‘overall reaction’ in this case would be the combination of the backward reaction for E3 and the forward reaction for E2, ie the Au not dissolving.
—
** this, and the equivalent previous statement, is a nonsense. Something like: “the LHS of E2 has a greater tendency to be reduced and form the RHS than the LHS of E3 does” would be more accurate.
(edited 1 month ago)
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