Electrolysis and Electrodes Reactions

What is Electrolyte?

When we dissolve some specific compounds in water, the compounds split up in positive and negative ions. Due to the presence of ions these solutions can carry electric current. We call these water diluted solutions as electrolytes.


To understand the concept of electrolysis let us take an example of a common electrolyte like copper sulfate (CuSO4). When we dissolve copper sulfate in water it breaks up in Cu++ and SO42- ions. Now, if we immersed to copper rods in the copper sulfate solution. Also, we apply a potential difference between these two copper rods. As a result, the rod connected with higher polarity becomes positive. As the same time, the rod connected with lower polarity becomes negative. Then we refer these two rods as electrodes. We call the positive electrode as anode.
On the other hand, the negative electrode is the cathode. Due to the electrostatic attraction of cathode positive Cu++ ions will to it. On the side negative SO42- ions will come closer to the anode. Since positive ions go to cathode we call these as cations. Alternatively, the ions which go to the anode are anions. Then at cathode copper cations will receive electrons and become neutral atoms.

Therefore copper is deposited on the cathode. On the other side anion or SO42- ions donate their electrons to the anode and react with the copper of the anode itself. As a result, the reaction produces copper sulfate. But copper sulfate itself is an electrolyte compound. So it can not remain in its molecular form. Therefore it again split up in Cu++ and SO42- ions. Hence, it dissolves in existing solutions.

Outcomes of Electrolysis of Copper Sulfate with Copper Electrodes

In this way, during this process, the weight of the cathode increases. At the same time the anode losses the same amount of weight. When cation (Cu++) goes to the cathode, these carry an electric charge. Because each of the cations has a positive charge. Again when SO42- ions go to the anode these carry the negative charge. Because each of the anions has a negative charge.

Actually, the movement of these ions causes electric currents in the electrolyte has two components. One is for movement of cations and other is for movement anions. The total current through the electrolyte is the sum of these two components. However, the current through the external circuit is due to the movement of electrons only.

Electrolytes of Copper Sulfate with Platinum Electrodes

Platinum does not react with SO4. So, at the platinum anode, SO4 will combine with the hydrogen of water. As a result, H2SO4 (Sulfuric Acid) forms in addition to oxygen.


Electrolysis of Silver Nitrate (AgNO3)

Silver nitrate is a common electrolyte. Soon as we dip to silver electrodes in the diluted silver nitrate, electrolysis takes place. Here, cations Ag+ come to the cathode. Then these cations give up their positive charge to the cathode. As a result, they become neutral silver atoms and get deposited on the cathode itself. At the anode, NO3 ions leave their charge and create silver nitrate (AgNO3).

At the same time, AgNO3 splits up in Ag+ and NO3 ions and gets dissolve in the existing solution. Finally, the cathode gains weight. At the same time anode losses the same weight.

Electrolysis of Sulfuric Acid with Platinum Electrodes

Diluted sulfuric acid remains as to H+ and SO42- ions. When we place platinum electrodes in the diluted sulfuric acid, the electrolysis process starts. Here, H+ ions come to the cathode and give up their positive charge. That means H+ ions receive electrons from the cathode and become neutral H2 gas. Then it liberates in the form of hydrogen gas bubbles. Along the platinum cathode rod.

Then SO42- ions come to the platinum anode. There they give up their negative charge ( electrons). But these can not react with platinum. Instead, they combine with hydrogen water and form sulfuric acid with gaseous oxygen.


Then H2SO4 again get dissolved in the existing solution as H+ and SO42- ions. Finally, gaseous oxygen will come out from the solution along the anode in the form of bubbles.

Again we know that the oxygen is bivalent and hydrogen is monovalent. Therefore for a given period of electrolysis the hydrogen liberated is twice of that of oxygen.

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