electrolysis of copper(II) sulphate with copper electrode

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[Lim Tze Khai]

As we know, copper metal deposited at cathode and ionise at anode.

we all know about the theory behind it.


BUT if we conduct experiment using concentration 0.1 mol dm^-3, colourless gas produce at cathode. I didn't test for the gas but suspected is hydrogen gas.....

Any can comment on this?

For me, i can of the factor of concentration of [H⁺] and [Cu²⁺].

This is important bcos it confused our students. 

[Ting Tin Chung]

As we know, copper metal deposited at cathode and ionise at anode.

we all know about the theory behind it.


BUT if we conduct experiment using concentration 0.1 mol dm^-3, colourless gas produce at cathode. I didn't test for the gas but suspected is hydrogen gas.....

The copper metal only ionise at anode if we are using Cu as anode. (3rd factor that affect the product during electrolysis). I have conducted this experiment by using CuSO₄ 0.01 mol dm^-3 and carbon electrodes. What i have observed was, at the cathode, after half an hour, a brown solid was deposited on the electrode. At the anode, a colour gas bubbles was released (oxygen gas). Then i let the experiment going on about 2 hours and i hv observe that a colourless gas was formed at the cathode. i guess the gas must be hydrogen gas. because by theory, after sometimes discharging the copper ions, the concentration of copper ion decreases and the concentration of H⁺ ions increases..... so of course discharge the hydrogen ions.... am i right??  

[Goh Kaw Sze]

Yes, the second factor now play its role.
This is an good experiment that shown how the 3 factors affect the product of electrolysis

[Eric Chong]

mind if i comment a bit?
just to recall the knowledge.

electrolysis of copper(II) sulphate ( aqueos )
using Copper electrode.
Anode
CU(solid) - 2e -> Cu2+
( The electrode mass will decrease as time go on )

Cathode
Cu2+(aq) +2e -> Cu(solid)
( The deposition of brownish solid matter on the surface of cathode electrode)

In the aqueous solution consist of Cu2+, H+, OH-.
Normal electrolysis using carbon electrode, maybe the H+ will be deionised at cathode due to effect of concentration of ion.
But, now the situation is copper electrode. The Cu2+ ion concentration won finish as long as the copper anode electrode will exist. The anode electrode will continuoe to dissolve and replace the deposited Cu2+ in aqueos.
Please correct me if i made mistake. Sharing.

[Lim Tze Khai]

electrolysis of copper(II) sulphate ( aqueos )
using Copper electrode.
Anode
CU(solid) - 2e -> Cu2+

Just to share knowledge, when write half equation for donating electron @ oxidation,

electron must be written on the right hand side...

example,  Cu --> Cu²⁺ + 2e

TQ

[Eric Chong]

thank mr lim,
                 though i graduated from biochemistry, now i m teaching biology. Thanks for correcting me. i think i had made a big technical mistake in half equation writting.

Cheer.

Thanks for sharing

[Sim Thiam Huat]

Dear mr. lim,
     any reason why when we write a substance donating electron, we must write the electron on the right hand side like Mg ---- Mg2+ + 2e ? During my time, we were taught Mg - 2e ----Mg2+ and I always maintaini it is easier for students to understand and remember with the electron on the left hand side (because donate means give out, means losing, means minus !
     another thing about the periodic table, during our time we used group 1, 2, 3, 4, 5, 6, 7, 8 but now we are using group 1, 2, 13, 14. 15...... For me,it is so much easier to comprehend and more meaningful to use group 3 because  these elements have 3 valence electrons !
    These 2 are still bothering me and I am hoping someone can provide a satisfactory answer. tahnk you very much

[Jong Fui Joon]

i am facing the same problem with you all.

i conduct the electrolysis of copper(II) sulphate solution by using carbon electrodes, and the worse is that i obtain hydrogen gas(half the test tube) and deposition of copper metal at cathode!

and when i tried to test the presence of oxygen gas at anode with glowing splinter,it produces a small "pop" sound!  i am almost crazy with it.

[Jong Fui Joon]

Mr Sim,  as what i heard from Mr Jong Kah Yin very long ago, he said that now the format of marking the half equation is that everything is positive, we cannot write donate electron with -e.

So, i forced my students to change the position of -e to the right hand side to become +e.(just like maths)

[Jong Kah Yin]

Hi,
Answers to Mr. Sim T H's questions:
Electrons released is considered as a product, so rightfully, it should be written on the right hand side of the half eqn. Definitely, it will be easier for stds to understand by writing the other way round, but we have to train students to differentiate : electrons on the left hand side of the eqn means accepting electrons (reduction), whereas electrons on the right hand side means release electrons (oxidation)
Answers to Jong F J's answers:
Maybe the electrolysis process is left for too long a time and also both the electrodes are too close as such some gas bubbles accidentaly collected at anode. Concentration of solution may play a part here.
I welcome all comments / correction of facts, if any. Thanks

[Sim Thiam Huat]

Thanks to both the Mr. Jongs for the clarification.
another request please !
Could u please explain with examples the meaning of 'operational defination' . this term is quite foreign to me. T.Q.

[Jong Kah Yin]

Hi,
Operational Definition is a clearly defined sets of procedures for obtaining a measure of the construct of interest } to convert it from a concept level statement to a measurable and objective operation.
Example : Operational definition of an acid : substance / chemical that changes moist blue litmus paper to red } which is different from conceptual definition
Conceptual definition of an acid : Acid is a chemical substance which ionises in water to produce hydroxonium ions.

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