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Rowanblu
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#1
you are told that if you double the current in an electrolysis experiment then you will get twice as much product produced in the same time.
you are provided with aqueous copper(II) sulphate solution, graphite electrodes and any other standard laboratory equipment you may need. outline an experiment that you could do to investigate the statement.
you should refer in your answers to:
the main points of the method you would use
what you would measure
how you would make your experience a 'fair test'
how you would use the results of the experiment to draw a conclusion
i'm not really sure where to get started with this question?? could someone please help??
you are provided with aqueous copper(II) sulphate solution, graphite electrodes and any other standard laboratory equipment you may need. outline an experiment that you could do to investigate the statement.
you should refer in your answers to:
the main points of the method you would use
what you would measure
how you would make your experience a 'fair test'
how you would use the results of the experiment to draw a conclusion
i'm not really sure where to get started with this question?? could someone please help??
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y.u.mad.bro?
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#2
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#2
Have you studied electrolysis? Do you know what is attracted to the cathode and the anode? The copper(II) sulphate solution contains 2 different ions which will each go to either anode/cathode.
Think of these questions and try to answer. Reply with what you know already and then we can help you more.
Think of these questions and try to answer. Reply with what you know already and then we can help you more.
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Rowanblu
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#3
(Original post by y.u.mad.bro?)
Have you studied electrolysis? Do you know what is attracted to the cathode and the anode? The copper(II) sulphate solution contains 2 different ions which will each go to either anode/cathode.
Think of these questions and try to answer. Reply with what you know already and then we can help you more.
Have you studied electrolysis? Do you know what is attracted to the cathode and the anode? The copper(II) sulphate solution contains 2 different ions which will each go to either anode/cathode.
Think of these questions and try to answer. Reply with what you know already and then we can help you more.
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y.u.mad.bro?
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#4
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#4
(Original post by Rowanblu)
I have studied electrolysis but I've never fully understood it and I've always struggled. I know that the anode is the positive electrode and non-metals are attracted to it. I know that the cathode is the negative electrode and metals are attracted to it. I know that you use graphite electrodes because graphite has valency electrons meaning it can conduct electricity easily.
I have studied electrolysis but I've never fully understood it and I've always struggled. I know that the anode is the positive electrode and non-metals are attracted to it. I know that the cathode is the negative electrode and metals are attracted to it. I know that you use graphite electrodes because graphite has valency electrons meaning it can conduct electricity easily.
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Rowanblu
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#5
(Original post by y.u.mad.bro?)
Good. Now can you tell me what two ions the CuSO4 solution contains?
Good. Now can you tell me what two ions the CuSO4 solution contains?
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17IAfzaa
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#6
y.u.mad.bro?
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#7
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#7
(Original post by Rowanblu)
oh dear ummm...copper and oxygen? i feel like thats totally wrong
oh dear ummm...copper and oxygen? i feel like thats totally wrong
(Original post by 17IAfzaa)
x
x
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Rowanblu
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#8
(Original post by y.u.mad.bro?)
You have CuSO4. Copper(II) and Sulphate. Do you know what ions they form? Keep in mind the (II) represents the charge on the copper.
Would prefer if you don't just give the answer. Let OP work for it
You have CuSO4. Copper(II) and Sulphate. Do you know what ions they form? Keep in mind the (II) represents the charge on the copper.
Would prefer if you don't just give the answer. Let OP work for it
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y.u.mad.bro?
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#9
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#9
(Original post by Rowanblu)
oh right the charge! so copper is positive while so is negative so would it be cu+2 and so4-2? i've probably gotten it wrong again
oh right the charge! so copper is positive while so is negative so would it be cu+2 and so4-2? i've probably gotten it wrong again
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Rowanblu
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#10
(Original post by y.u.mad.bro?)
No thats correct. Good job. Now you know the ions and the electrodes. you also know that opposite charges attract and CuSO4 is aqueous, how can you seperate them using what you have? Heres a hint: think of electrolysis... electro means electricity.
No thats correct. Good job. Now you know the ions and the electrodes. you also know that opposite charges attract and CuSO4 is aqueous, how can you seperate them using what you have? Heres a hint: think of electrolysis... electro means electricity.
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#11
(Original post by Rowanblu)
so you use the electrical currents in the graphite electrodes to attract the different ions. copper is attracted to the negative electrode - cathode. but does sulphur or oxygen get attracted to the positive electrode - anode?
so you use the electrical currents in the graphite electrodes to attract the different ions. copper is attracted to the negative electrode - cathode. but does sulphur or oxygen get attracted to the positive electrode - anode?
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Rowanblu
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#12
(Original post by y.u.mad.bro?)
Its SO4 as a whole not oxygen so sulphur goes to the anode. Now using all this info can you answers the questions?
Its SO4 as a whole not oxygen so sulphur goes to the anode. Now using all this info can you answers the questions?
by electrolysing copper(II) sulphate solution, you are producing copper ions at the cathode and sulphur ions at the anode. first, measure out (HOW MUCH SOLUTION??) of your copper(II) sulphate solution using a burette and place it into a beaker with graphite electrodes. next, record the amount of product produced in half an hour. next, repeat the same steps using the same materials and apparatus. however, this time double the current and record the amount of product produced in half an hour.
IM NOT SURE HOW TO WRITE THE CONCLUSION? WOULD DOUBLiNG THE CURRENT ACTUALLY MEAN TWICE THE PRODUCT?
i also know that if you put too much current then the products wont be able to electrolyse...dont know if i can fit that in idk if its important
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y.u.mad.bro?
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#13
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#13
(Original post by Rowanblu)
how about this?
by electrolysing copper(II) sulphate solution, you are producing copper ions at the cathode and sulphur ions at the anode. first, measure out (HOW MUCH SOLUTION??) of your copper(II) sulphate solution using a burette and place it into a beaker with graphite electrodes. next, record the amount of product produced in half an hour. next, repeat the same steps using the same materials and apparatus. however, this time double the current and record the amount of product produced in half an hour.
IM NOT SURE HOW TO WRITE THE CONCLUSION? WOULD DOUBLiNG THE CURRENT ACTUALLY MEAN TWICE THE PRODUCT?
i also know that if you put too much current then the products wont be able to electrolyse...dont know if i can fit that in idk if its important
how about this?
by electrolysing copper(II) sulphate solution, you are producing copper ions at the cathode and sulphur ions at the anode. first, measure out (HOW MUCH SOLUTION??) of your copper(II) sulphate solution using a burette and place it into a beaker with graphite electrodes. next, record the amount of product produced in half an hour. next, repeat the same steps using the same materials and apparatus. however, this time double the current and record the amount of product produced in half an hour.
IM NOT SURE HOW TO WRITE THE CONCLUSION? WOULD DOUBLiNG THE CURRENT ACTUALLY MEAN TWICE THE PRODUCT?
i also know that if you put too much current then the products wont be able to electrolyse...dont know if i can fit that in idk if its important
Have a read at the following links and see if you understand what happens in more detail.
http://www.rsc.org/learn-chemistry/r...id=CMP00005019
http://www.docbrown.info/page01/ExIn...hemistry04.htm
https://www.onlinemathlearning.com/e...-sulphate.html
Once you have read the following, redesign your plan and reply with it. Then we can start to refine your answer.
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Rowanblu
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#14
(Original post by y.u.mad.bro?)
You are sort of on the right track. Might need to refine it a bit.
Have a read at the following links and see if you understand what happens in more detail.
http://www.rsc.org/learn-chemistry/r...id=CMP00005019
http://www.docbrown.info/page01/ExIn...hemistry04.htm
https://www.onlinemathlearning.com/e...-sulphate.html
Once you have read the following, redesign your plan and reply with it. Then we can start to refine your answer.
You are sort of on the right track. Might need to refine it a bit.
Have a read at the following links and see if you understand what happens in more detail.
http://www.rsc.org/learn-chemistry/r...id=CMP00005019
http://www.docbrown.info/page01/ExIn...hemistry04.htm
https://www.onlinemathlearning.com/e...-sulphate.html
Once you have read the following, redesign your plan and reply with it. Then we can start to refine your answer.
okay so I learned that copper forms at the cathode and the reaction will be Cu+2 (aq) + 2e- --> Cu (s)
then oxygen bubbles off at the anode and the reaction will be 2H20 (l) --> 02 (g) + 4H+ (aq) + 4e-
then the blue colour of the solution will fade as more copper is deposited.
However, I still don't know if doubling the current will actually make it go faster as it could, but if you put too much electricity into the solution it won't actually electrolyse
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Rowanblu
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#15
(Original post by y.u.mad.bro?)
You are sort of on the right track. Might need to refine it a bit.
Have a read at the following links and see if you understand what happens in more detail.
http://www.rsc.org/learn-chemistry/r...id=CMP00005019
http://www.docbrown.info/page01/ExIn...hemistry04.htm
https://www.onlinemathlearning.com/e...-sulphate.html
Once you have read the following, redesign your plan and reply with it. Then we can start to refine your answer.
You are sort of on the right track. Might need to refine it a bit.
Have a read at the following links and see if you understand what happens in more detail.
http://www.rsc.org/learn-chemistry/r...id=CMP00005019
http://www.docbrown.info/page01/ExIn...hemistry04.htm
https://www.onlinemathlearning.com/e...-sulphate.html
Once you have read the following, redesign your plan and reply with it. Then we can start to refine your answer.
the electrolysis of copper(II) sulphate solution leads to the deposition of copper at the cathode and the reaction will be Cu+2 (aq) + 2e- --> Cu (s). oxygen will then bubble off at the anode, and the reaction will be 2H2O (l) --> )2 (g) + 4H+ (aq) + 4e-. First, measure out 200 ml of your copper(II) sulphate solution using a burette and lace into a beaker with graphite electrodes secured by a clamp. Next, record the amount of product produced in half an hour. The blue colour of your solution will fade as more copper is deposited. Next, repeat the same steps using the same materials and apparatus, making sure the conditions are the same. This time, double the current used to electrolyse and record the amount of product produced in half an hour. Has the amount of product doubled by using twice the amount of current in the same time?
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y.u.mad.bro?
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#16
(Original post by Rowanblu)
okay so hows this:
the electrolysis of copper(II) sulphate solution leads to the deposition of copper at the cathode and the reaction will be Cu+2 (aq) + 2e- --> Cu (s). oxygen will then bubble off at the anode, and the reaction will be 2H2O (l) --> )2 (g) + 4H+ (aq) + 4e-. First, measure out 200 ml of your copper(II) sulphate solution using a burette and lace into a beaker with graphite electrodes secured by a clamp. Next, record the amount of product produced in half an hour. The blue colour of your solution will fade as more copper is deposited. Next, repeat the same steps using the same materials and apparatus, making sure the conditions are the same. This time, double the current used to electrolyse and record the amount of product produced in half an hour. Has the amount of product doubled by using twice the amount of current in the same time?
okay so hows this:
the electrolysis of copper(II) sulphate solution leads to the deposition of copper at the cathode and the reaction will be Cu+2 (aq) + 2e- --> Cu (s). oxygen will then bubble off at the anode, and the reaction will be 2H2O (l) --> )2 (g) + 4H+ (aq) + 4e-. First, measure out 200 ml of your copper(II) sulphate solution using a burette and lace into a beaker with graphite electrodes secured by a clamp. Next, record the amount of product produced in half an hour. The blue colour of your solution will fade as more copper is deposited. Next, repeat the same steps using the same materials and apparatus, making sure the conditions are the same. This time, double the current used to electrolyse and record the amount of product produced in half an hour. Has the amount of product doubled by using twice the amount of current in the same time?
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Rowanblu
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#17
(Original post by y.u.mad.bro?)
Yeah but rather than measuring the amount formed after 30 minutes, measure the amount formed every 2 minutes or 5 minutes. This will help you when you compare the prodcution with increased current.
Yeah but rather than measuring the amount formed after 30 minutes, measure the amount formed every 2 minutes or 5 minutes. This will help you when you compare the prodcution with increased current.
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y.u.mad.bro?
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#18
(Original post by Rowanblu)
so i should say measure the amount of product produced every 5 minutes so in the end they can come up with a ratio?
so i should say measure the amount of product produced every 5 minutes so in the end they can come up with a ratio?
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Rowanblu
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#19
(Original post by y.u.mad.bro?)
Yup.
Yup.
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y.u.mad.bro?
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#20
(Original post by Rowanblu)
thank you so much you've been so much help I can't believe you actually went through all the trouble of helping me even though I'm so stupid and you actually looked for links and resources. You've been so much help and I really can't thank you enough. Do you mind if I ask you what you want to be when you're older and what subjects you're studying for your a levels? You don't have to answer if you don't want to, I was just curious at how you were so good at chemistry and teaching.
thank you so much you've been so much help I can't believe you actually went through all the trouble of helping me even though I'm so stupid and you actually looked for links and resources. You've been so much help and I really can't thank you enough. Do you mind if I ask you what you want to be when you're older and what subjects you're studying for your a levels? You don't have to answer if you don't want to, I was just curious at how you were so good at chemistry and teaching.
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