Question

solve only question no. 2,,,,, fasttttt

Answer 1

✸GOOD AFTERNOON✸

Let me give you a Summary of Electrolysis of Copper Sulfate

Here you go..✌

Whenever copper sulfate or CuSO4 is added to water, it gets dissolved in the water. As the CuSO4 is an electrolyte, it splits into Cu+ + (cation) and SO4 − − (anion) ions and move freely in the solution. electrolyte copper sulphate

Now we will immerse two copper electrodes in that solution. The Cu+ + ions (cation) will be attracted towards cathode i.e. the electrode connected to the negative terminal of the battery.

On reaching on the cathode, each Cu+ + ion will take electrons from it and becomes neutral copper atoms. Similarly the SO4 − − (anion) ions will be attracted by anode i.e. the electrode connected to the positive terminal of the battery.

So SO4 − − ions will move towards anode where they give up two electrons and become SO4 radical.

But since SO4 radical can not exist in the electrical neutral state, it will attack copper anode and will form copper sulfate. In the above process, after taking electrons the neutral copper atoms get deposited on the cathode. At the same time, SO4 reacts with copper anode and becomes CuSO4 but in water it can not exist as single molecules instead of that CuSO4 will split into Cu+ +, SO4 − − and dissolve in water. So it can be concluded that, during electrolysis of copper sulfate with copper electrodes, copper is deposited on cathode and same amount of copper is removed from anode.

I will give some hints also..Find out the answer from that...When you think and try to find the answer u will never forget

➠Some Hints:

★It's Cu+2 which is responsible for blue colour.

★Reduction always occur in Cathode...Oxidation will occur at Anode.

★Gaining of electron is called Reduction And Losing of electron is called Oxidation.

With these Hints and Summary Try to answer these questions.

IF ANY DOUBTS FEEL FREE TO ASK IN COMMENT BOX

Hope it Helpz

❀Naz❀

THANK YOU