
What will be the products of electrolysis of $AgN{{O}_{3}}$ solution in water with platinum electrodes?
(A) Ag is liberated at cathode and Ag is deposited in anode.
(B) Ag is liberated at cathode and ${{O}_{2}}$ is liberated at anode.
(C) Ag is liberated at anode and water is liberated at cathode.
(D) Ag is liberated at cathode and silver oxide is liberated at anode.
Answer
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Hint: Write the reactions which take place during electrolysis of silver nitrate solution. Identify the oxidation and reduction reactions. Determine which reaction will take place at cathode and anode accordingly.
Complete step by step solution:
-Electrolysis is the process in which cleavage of bond takes place due to the action of electricity.
-During electrolysis of $AgN{{O}_{3}}$ solution in water with platinum electrodes, the following reactions will take place.
$AgN{{O}_{3}}\to A{{g}^{+}}+NO_{3}^{-}$
${{H}_{2}}O\to {{H}^{+}}+O{{H}^{-}}$
Now let’s see which reactions occur at anode and cathode.
At cathode (reduction), silver ions donate one electron to form Ag metal and get deposited on the cathode.
$A{{g}^{+}}+{{e}^{-}}\to Ag$
At anode (oxidation), since platinum electrodes are present, self-ionization of water takes place which leads to liberation of oxygen gas.
${{H}_{2}}O\to 2{{H}^{+}}+{}^{1}/{}_{2}{{O}_{2}}+2{{e}^{-}}$
Therefore, from the above reactions, we can tell that silver metal gets deposited at cathode and oxygen gas is liberated at anode. So, option (B) is the correct answer.
Note: Remember LOAN (left oxidation anode). Oxidation will take place at anode and reduction will take place at the cathode. Or you can also remember CAR (cathode reduction). One of the important applications of this reaction is electroplating of silver in silver spoons, medals, cooking utensils, etc. Electroplating is a technique in which a thin layer of metal is coated on an object to get the desired finish or outcome. In electroplating, the object to be electroplated is the cathode and is completely dipped inside the electrolyte, $AgN{{O}_{3}}$. So, silver will get uniformly deposited on the object and the object becomes silver-plated.
Complete step by step solution:
-Electrolysis is the process in which cleavage of bond takes place due to the action of electricity.
-During electrolysis of $AgN{{O}_{3}}$ solution in water with platinum electrodes, the following reactions will take place.
$AgN{{O}_{3}}\to A{{g}^{+}}+NO_{3}^{-}$
${{H}_{2}}O\to {{H}^{+}}+O{{H}^{-}}$
Now let’s see which reactions occur at anode and cathode.
At cathode (reduction), silver ions donate one electron to form Ag metal and get deposited on the cathode.
$A{{g}^{+}}+{{e}^{-}}\to Ag$
At anode (oxidation), since platinum electrodes are present, self-ionization of water takes place which leads to liberation of oxygen gas.
${{H}_{2}}O\to 2{{H}^{+}}+{}^{1}/{}_{2}{{O}_{2}}+2{{e}^{-}}$
Therefore, from the above reactions, we can tell that silver metal gets deposited at cathode and oxygen gas is liberated at anode. So, option (B) is the correct answer.
Note: Remember LOAN (left oxidation anode). Oxidation will take place at anode and reduction will take place at the cathode. Or you can also remember CAR (cathode reduction). One of the important applications of this reaction is electroplating of silver in silver spoons, medals, cooking utensils, etc. Electroplating is a technique in which a thin layer of metal is coated on an object to get the desired finish or outcome. In electroplating, the object to be electroplated is the cathode and is completely dipped inside the electrolyte, $AgN{{O}_{3}}$. So, silver will get uniformly deposited on the object and the object becomes silver-plated.
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