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Hint: Electrolysis is the process by which ionic substances are decomposed into simpler substances when an electric current is passed through it.
Complete step by step answer:
When sodium chloride is dissolved in water, it ionizes as
$NaCl \rightleftharpoons N{a^ + } + C{l^ - }$
Water dissociates as ,
${H_2}O \rightleftharpoons {H^ + } + O{H^ - }$
During passing of electric current through this solution using platinum electrode , $N{a^ + }$ and ${H^ + }$ ions move towards the cathode. However, only ${H^ + }$ ions are discharged more readily than $N{a^ + }$ ions because of their low discharge potential since , in electromotive series hydrogen is lower sodium. These ${H^ + }$ ions gain electrons and change into neutral atoms.
At cathode,
${H^ + } + {e}^- \to H \\
H + H \to {H_2} \\$
$C{l^ - }$ and $O{H^ - }$ ions move towards anode.
$C{l^ - }$ ions lose electrons and change into neutral atoms.
At anode ,
$C{l^ - } + {e}^- \to Cl$
$Cl + Cl \to C{l_2}$
If mercury is used as cathode ${H^ + }$ ions are not discharged at mercury cathode because mercury has a high hydrogen overvoltage. $N{a^ + }$ ions are discharged at the cathode in preference to ${H^ + }$ ions, yielding sodium, which dissolves in mercury to form sodium amalgam.
At cathode $N{a^ + } + {e}^- \to Na$.
Thus, more voltage is required to reduce ${H^ + }$ at amalgam cathode than a platinum cathode.
Hence , the required option for the given question is b).
Note: During passing of electric current through this solution using platinum electrode, $N{a^ + }$ and ${H^ + }$ ions move towards the cathode. However, only ${H^ + }$ ions are discharged more readily than $N{a^ + }$ ions because of their low discharge potential since, in electromotive series hydrogen is lower sodium. These ${H^ + }$ ions gain electrons and change into neutral atoms.
Complete step by step answer:
When sodium chloride is dissolved in water, it ionizes as
$NaCl \rightleftharpoons N{a^ + } + C{l^ - }$
Water dissociates as ,
${H_2}O \rightleftharpoons {H^ + } + O{H^ - }$
During passing of electric current through this solution using platinum electrode , $N{a^ + }$ and ${H^ + }$ ions move towards the cathode. However, only ${H^ + }$ ions are discharged more readily than $N{a^ + }$ ions because of their low discharge potential since , in electromotive series hydrogen is lower sodium. These ${H^ + }$ ions gain electrons and change into neutral atoms.
At cathode,
${H^ + } + {e}^- \to H \\
H + H \to {H_2} \\$
$C{l^ - }$ and $O{H^ - }$ ions move towards anode.
$C{l^ - }$ ions lose electrons and change into neutral atoms.
At anode ,
$C{l^ - } + {e}^- \to Cl$
$Cl + Cl \to C{l_2}$
If mercury is used as cathode ${H^ + }$ ions are not discharged at mercury cathode because mercury has a high hydrogen overvoltage. $N{a^ + }$ ions are discharged at the cathode in preference to ${H^ + }$ ions, yielding sodium, which dissolves in mercury to form sodium amalgam.
At cathode $N{a^ + } + {e}^- \to Na$.
Thus, more voltage is required to reduce ${H^ + }$ at amalgam cathode than a platinum cathode.
Hence , the required option for the given question is b).
Note: During passing of electric current through this solution using platinum electrode, $N{a^ + }$ and ${H^ + }$ ions move towards the cathode. However, only ${H^ + }$ ions are discharged more readily than $N{a^ + }$ ions because of their low discharge potential since, in electromotive series hydrogen is lower sodium. These ${H^ + }$ ions gain electrons and change into neutral atoms.
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