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Question: Statement: During electrolysis of acidified water, at the cathode, \({{H}_{2}}\) is evolved and at t...

Statement: During electrolysis of acidified water, at the cathode, H2{{H}_{2}} is evolved and at the anode, O2{{O}_{2}} is evolved.
Enter 1 if true else 0.

Explanation

Solution

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+{{H}^{+}} and OH−O{{H}^{-}}.
H+{{H}^{+}}is a cation as it is electron deficient and hence moves towards the cathode. There, it reduces to form hydrogen gas.
OH−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
2H2O(l)→2H2(g)+O2(g)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, H2{{H}_{2}} is evolved and at the anode, O2{{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.