
Statement: During electrolysis of acidified water, at the cathode, ${{H}_{2}}$ is evolved and at the anode, ${{O}_{2}}$ is evolved.
Enter 1 if true else 0.
Answer
588k+ views
Hint: Water undergoes electrolysis and releases two gases, oxygen and hydrogen. The cation flows towards the cathode while anion flows towards the anode. So we need to see what is being formed as cation and what as anion in order to know which gas is being evolved at the respective electrode.
Complete answer:
-Electrolysis is a technique used to separate a compound in its different constituents with the help of DC current. It is used in non-spontaneous chemical reactions.
-The amount of voltage to be given depends on the type of the electrolyte used. The voltage is called decomposition potential.
-There are two electrodes, cathode and anode dipped in the electrolyte. They are connected to the battery. In acidified water, water is mixed with dilute sulfuric acid. This mixture acts as electrolyte.
-Cathode is the electrode where the cations are attracted. Reduction takes place at cathode. Anode is the electrode where the anions are attracted. Oxidation takes place at anode.
-The conductivity of normal water is very less. So addition of sulfuric acid increases its conductivity. Thus, the decomposition of acidified water takes place at a higher rate compared to normal water.
-Water decomposes into its respective ions ${{H}^{+}}$ and $O{{H}^{-}}$.
${{H}^{+}}$is a cation as it is electron deficient and hence moves towards the cathode. There, it reduces to form hydrogen gas.
$O{{H}^{-}}$is an anion as it has an excess of electrons and thus moves to the anode. There, it oxidised to form oxygen gas.
-The overall balanced equation for the electrolysis of acidified water can be shown as
\[2{{H}_{2}}O\left( l \right)\to 2{{H}_{2}}\left( g \right)+{{O}_{2}}\left( g \right)\]
-From the equation, it is clear that the volume of hydrogen gas formed is twice that of oxygen gas formed.
Therefore, the correct answer is 1.
During electrolysis of acidified water, at the cathode, ${{H}_{2}}$ is evolved and at the anode, ${{O}_{2}}$ is evolved.
Note:
If sulfuric acid is not added in water, the conductivity will be very less and the rate of electrolysis will be very low.
Steel and iron are used as electrolyte in the process. They are reduced at cathode and oxidised at anode.
Complete answer:
-Electrolysis is a technique used to separate a compound in its different constituents with the help of DC current. It is used in non-spontaneous chemical reactions.
-The amount of voltage to be given depends on the type of the electrolyte used. The voltage is called decomposition potential.
-There are two electrodes, cathode and anode dipped in the electrolyte. They are connected to the battery. In acidified water, water is mixed with dilute sulfuric acid. This mixture acts as electrolyte.
-Cathode is the electrode where the cations are attracted. Reduction takes place at cathode. Anode is the electrode where the anions are attracted. Oxidation takes place at anode.
-The conductivity of normal water is very less. So addition of sulfuric acid increases its conductivity. Thus, the decomposition of acidified water takes place at a higher rate compared to normal water.
-Water decomposes into its respective ions ${{H}^{+}}$ and $O{{H}^{-}}$.
${{H}^{+}}$is a cation as it is electron deficient and hence moves towards the cathode. There, it reduces to form hydrogen gas.
$O{{H}^{-}}$is an anion as it has an excess of electrons and thus moves to the anode. There, it oxidised to form oxygen gas.
-The overall balanced equation for the electrolysis of acidified water can be shown as
\[2{{H}_{2}}O\left( l \right)\to 2{{H}_{2}}\left( g \right)+{{O}_{2}}\left( g \right)\]
-From the equation, it is clear that the volume of hydrogen gas formed is twice that of oxygen gas formed.
Therefore, the correct answer is 1.
During electrolysis of acidified water, at the cathode, ${{H}_{2}}$ is evolved and at the anode, ${{O}_{2}}$ is evolved.
Note:
If sulfuric acid is not added in water, the conductivity will be very less and the rate of electrolysis will be very low.
Steel and iron are used as electrolyte in the process. They are reduced at cathode and oxidised at anode.
Recently Updated Pages
Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

There are 720 permutations of the digits 1 2 3 4 5 class 11 maths CBSE

Discuss the various forms of bacteria class 11 biology CBSE

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE

State the laws of reflection of light

Explain zero factorial class 11 maths CBSE

