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chemistry question

please can someone help me on this question?
thanks!

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Reply 1
here is the question
Original post by Ashirs
here is the question

It isn’t working for me.

Can you summarise what the question is asking and show me what you’ve done so far?
Reply 3
53 During electrolysis of an aqueous solution of sodium sulfate the half equations for the electrode
reactions are:
Anode (positive electrode): 2H2O(l) O2(g) + 4H+
(aq) + 4e–
Cathode (negative electrode): 2H2O(l) + 2e–
H2(g) + 2OH–
(aq)
Which of the following deductions, if any, can be made from these equations?

1 The ratio by moles of hydrogen to oxygen produced at the electrodes is 1:1.
2 The sodium sulfate solution will become more concentrated as the electrolysis
proceeds.
3 The whole solution will become acidic due to formation of H+
ions at the anode.
A none of them
B 1 only
C 2 only
D 3 only
E 1 and 2 only
F 1 and 3 only
G 2 and 3 only
H 1, 2 and 3
Reply 4
i haven't done much since it is not a calculation....
Reply 5
Original post by Ashirs
53 During electrolysis of an aqueous solution of sodium sulfate the half equations for the electrode
reactions are:
Anode (positive electrode): 2H2O(l) O2(g) + 4H+
(aq) + 4e–
Cathode (negative electrode): 2H2O(l) + 2e–
H2(g) + 2OH–
(aq)
Which of the following deductions, if any, can be made from these equations?

1 The ratio by moles of hydrogen to oxygen produced at the electrodes is 1:1.
2 The sodium sulfate solution will become more concentrated as the electrolysis
proceeds.
3 The whole solution will become acidic due to formation of H+
ions at the anode.
A none of them
B 1 only
C 2 only
D 3 only
E 1 and 2 only
F 1 and 3 only
G 2 and 3 only
H 1, 2 and 3

Are you in year 13?
Reply 6
Original post by Sakuna
Are you in year 13?


year 11 but could you explain it?
Reply 7
Original post by Ashirs
year 11 but could you explain it?

I'm year 12 n I don't really remember doing this in year 11 my bad
Original post by Ashirs
53 During electrolysis of an aqueous solution of sodium sulfate the half equations for the electrode
reactions are:
Anode (positive electrode): 2H2O(l) O2(g) + 4H+
(aq) + 4e–
Cathode (negative electrode): 2H2O(l) + 2e–
H2(g) + 2OH–
(aq)
Which of the following deductions, if any, can be made from these equations?

1 The ratio by moles of hydrogen to oxygen produced at the electrodes is 1:1.
2 The sodium sulfate solution will become more concentrated as the electrolysis
proceeds.
3 The whole solution will become acidic due to formation of H+
ions at the anode.
A none of them
B 1 only
C 2 only
D 3 only
E 1 and 2 only
F 1 and 3 only
G 2 and 3 only
H 1, 2 and 3


You’re in year 11? This sort of thing is year 12/13 content…. By any chance, are you looking at a medical admissions paper or something?

Ok, so you want to make the electrons cancel out. That involves scaling up your equations as appropriate (note that you may only need to scale up 1), then add them and cancel out anything that appears on both sides.

Anode (positive electrode): 2H2O O2 + 4H+ + 4e–
Cathode (negative electrode): 2H2O + 2e–
H2 + 2OH–

To cancel the electrons, you must scale the cathode equation up, so that the number of electrons consumed in the cathode reaction equals the number of electrons produced in the anode equation. How would you do that?
(edited 1 year ago)
Reply 9
Original post by TypicalNerd
You’re in year 11? This sort of thing is year 12/13 content…. By any chance, are you looking at a medical admissions paper or something?

Ok, so you want to make the electrons cancel out. That involves scaling up your equations as appropriate (note that you may only need to scale up 1), then add them and cancel out anything that appears on both sides.

Anode (positive electrode): 2H2O O2 + 4H+ + 4e–
Cathode (negative electrode): 2H2O + 2e–
H2 + 2OH–

To cancel the electrons, you must scale the cathode equation up, so that the number of electrons consumed in the cathode reaction equals the number of electrons produced in the anode equation. How would you do that?

yes, it was from a nsaa paper :biggrin:, i was able to understand most of the questions except this one
so you would multiply the cathode equation by 2?
how does that help answer the question?
thanks!
Original post by Ashirs
yes, it was from a nsaa paper :biggrin:, i was able to understand most of the questions except this one
so you would multiply the cathode equation by 2?
how does that help answer the question?
thanks!


Indeed you do.

Now add up the equations and cancel all the things that appear on both sides.

The result is an equation that summarises what happens in the electrolysis of sodium sulphate solution.
Reply 11
Original post by TypicalNerd
Indeed you do.

Now add up the equations and cancel all the things that appear on both sides.

The result is an equation that summarises what happens in the electrolysis of sodium sulphate solution.

the end equation is 6h20 --> o2 +4h+ + 4oh- +2h2
therefore the ratio hydrogen to oxygen is 2:1.
i am also unsure about the other 2 statements however the answer is F and statement 1 is wrong?
(edited 1 year ago)
Original post by Ashirs
the end equation is 6h20 --> o2 +4h+ + 4oh- +2h2
therefore the ratio hydrogen to oxygen is 2:1.
i am also unsure about the other 2 statements however the answer is F and statement 1 is wrong?


Let’s critically think about it.

The number of OH- ions and H+ ions produced is the same according to the equation. What does that tell you about the solution’s pH?

The only reactant is water. That means the amount of water in the solution is decreasing and the amount of sodium sulphate is unchanged.

Now what can you infer?
Reply 13
since the number of OH- ions and H+ ions are equal, the solution is neutral?
the concentration so sodium sulphate is greater than that of water?
however is the ratio of hydrogen to oxygen not 2:1
Original post by Ashirs
since the number of OH- ions and H+ ions are equal, the solution is neutral?
the concentration so sodium sulphate is greater than that of water?
however is the ratio of hydrogen to oxygen not 2:1


So you’ve worked out that 1 is wrong.

Statement 2 must be right, because concentration = mass (or moles) of solute / volume of solvent. Because water is the solvent and the amount of it decreases, the volume must decrease and the concentration of sodium sulphate must increase.

Statement 3 says the solution must get more acidic, but as you’ve said, it’s neutral. That means statement 3 is also false.

Which answer corresponds to the above?
(edited 1 year ago)
Reply 15
Original post by TypicalNerd
So you’ve worked out that 1 is wrong.

Statement 2 must be right, because concentration = mass (or moles) of solute / volume of solvent. Because water is the solvent and the amount of it decreases, the volume must decrease and the concentration of sodium sulphate must increase.

Statement 3 says the solution must get more acidic, but as you’ve said, it’s neutral. That means statement 3 is also false.

Which answer corresponds to the above?


the answer that corresponds is C however the answer in the markscheme is F.
I have attached the reasoning below but i don't get it?
Original post by Ashirs
the answer that corresponds is C however the answer in the markscheme is F.
I have attached the reasoning below but i don't get it?

Doesn’t seem to work, like the last link.

I’m intrigued by the answer being F, because it just doesn’t seem right.
Reply 17
Original post by TypicalNerd
Doesn’t seem to work, like the last link.

I’m intrigued by the answer being F, because it just doesn’t seem right.


Picture1.png

can you see the picture i have attached to this?
Reply 18
the reasoning is the exact same that you gave so i think that they accidentally put the answer wrong?
Their reasoning states that 1 is wrong, but they answered 1 as correct. Thus, it is a mistake on the mark scheme.:smile:

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