Answer
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Hint: Electrolysis is a process by which electricity is passed through an electrolyte for a chemical reaction to happen at the electrode. Here, electricity is used to drive a nonspontaneous reaction. Electrolysis takes place in an electrolytic cell. It contains two electrodes- cathode and anode. The conducting chemical substance (usually a solution) that contains free ions and charge carriers for the reaction to happen is called electrolyte.
Complete answer:
From the hint, we understood what electrolysis is. Now, let’s see the anodic reaction of the electrolysis of brine solution (NaCl). The possible half-cell reactions at anode are:
$C{l^ - }(aq)\, \to \,\dfrac{1}{2}C{l_2}(g)\, + \,\mathop e\limits^\_ \,\,\,\,\,\,\,\,;E{^\circ _{cell}} = 1.36\,V$
$2{H_2}O(l)\, \to \,{O_2}(g) + 4{H^ + }(aq) + 4\mathop e\limits^\_ \,\,\,\,\,;E{^\circ _{cell}} = 1.23\,V$
At anode, the reaction with lower $E{^\circ _{cell}}$ is preferred. Therefore, oxidation of water is the preferred anodic reaction. But, chlorine gas is liberated at anode. This is due to the phenomenon called overvoltage. It is the additional voltage that is required for the actual electrolytic reaction to occur than from the calculated standard electrode potential. The over voltage required for formation of chlorine gas is lesser than that of oxygen. So, chlorine gas is liberated at anode.
Therefore, the correct option is (A) Both assertion and reason are correct and reason is the correct explanation for assertion.
Note:
The electrodes in which oxidation and reduction takes place are called anode and cathode respectively.
The standard electrode potential of a cell $(E{^\circ _{cell}})$ is the potential of a half cell reaction which is measured against the Standard Hydrogen Electrode (SHE) when the concentration of all species involved is unity at standard conditions.
Complete answer:
From the hint, we understood what electrolysis is. Now, let’s see the anodic reaction of the electrolysis of brine solution (NaCl). The possible half-cell reactions at anode are:
$C{l^ - }(aq)\, \to \,\dfrac{1}{2}C{l_2}(g)\, + \,\mathop e\limits^\_ \,\,\,\,\,\,\,\,;E{^\circ _{cell}} = 1.36\,V$
$2{H_2}O(l)\, \to \,{O_2}(g) + 4{H^ + }(aq) + 4\mathop e\limits^\_ \,\,\,\,\,;E{^\circ _{cell}} = 1.23\,V$
At anode, the reaction with lower $E{^\circ _{cell}}$ is preferred. Therefore, oxidation of water is the preferred anodic reaction. But, chlorine gas is liberated at anode. This is due to the phenomenon called overvoltage. It is the additional voltage that is required for the actual electrolytic reaction to occur than from the calculated standard electrode potential. The over voltage required for formation of chlorine gas is lesser than that of oxygen. So, chlorine gas is liberated at anode.
Therefore, the correct option is (A) Both assertion and reason are correct and reason is the correct explanation for assertion.
Note:
The electrodes in which oxidation and reduction takes place are called anode and cathode respectively.
The standard electrode potential of a cell $(E{^\circ _{cell}})$ is the potential of a half cell reaction which is measured against the Standard Hydrogen Electrode (SHE) when the concentration of all species involved is unity at standard conditions.
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