Answer
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Hint Brine solution is aqueous solution of common salt or solution of $NaCl$ in water. In electrolysis two electrodes are dipped in solution and electric current is passed through the solution. These electrodes are cathode and anode where the reactions occur between the ions present in solution due to flow of current.
Complete step by step solution
On missing $NaCl$ we get $N{a^ + }$, $C{l^ - }$, ${H^ + }$ and $O{H^ - }$ ions in solution.
Reactions occurs at the negative Cathode electrode in brine solution during electrolysis:
As cathode is negative then it attracts sodium $N{a^ + }$ ions and hydrogen ${H^ + }$ ions. Only the hydrogens ions are discharged at cathode and release hydrogen gas. The hydrogen ions are reduced by gain electrons from cathode $2{H^ + }(aq) + 2{e^ - } \to {H_2}(g)$ or $2{H_2}O + 2{e^ - } \to {H_2}(q) + O{H^ - }(aq)$.
Reactions occurs at the positive Anode electrode in brine solution during electrolysis:
The positive anode attracts the hydroxide $O{H^ - }$ ions and chloride $C{l^ - }$ ions as they are negatively charged. Only the chloride ions are discharged in an appreciable quantity. The chloride ions are oxidised by losing electron to electron deficient anode and gives us the chloride molecule or chloride gas
$2C{l^ - }(aq) \to C{l_2} + 2{e^ - }$
From solution hydrogen ions and chloride are removed using electrolysis and sodium ions and hydroxide ions are left in solution and form $NaOH$.
$N{a^ + } + O{H^ - } \to NaOH$
The final product occurs are ${H_2}$, $C{l_2}$ and $NaOH$.
Hence the correct answer is option D.
Note A cathode hydrogen ion is discharged because it has more tendency to electron compared to sodium ions that why sodium ions don't get discharged at cathode. Similarly, at anode chloride ions have more tendency to lose an electron and it gets discharged. Here we notice that hydrogen gas is released which is inflammable and burnt with blast then this process is done slowly for safety.
Complete step by step solution
On missing $NaCl$ we get $N{a^ + }$, $C{l^ - }$, ${H^ + }$ and $O{H^ - }$ ions in solution.
Reactions occurs at the negative Cathode electrode in brine solution during electrolysis:
As cathode is negative then it attracts sodium $N{a^ + }$ ions and hydrogen ${H^ + }$ ions. Only the hydrogens ions are discharged at cathode and release hydrogen gas. The hydrogen ions are reduced by gain electrons from cathode $2{H^ + }(aq) + 2{e^ - } \to {H_2}(g)$ or $2{H_2}O + 2{e^ - } \to {H_2}(q) + O{H^ - }(aq)$.
Reactions occurs at the positive Anode electrode in brine solution during electrolysis:
The positive anode attracts the hydroxide $O{H^ - }$ ions and chloride $C{l^ - }$ ions as they are negatively charged. Only the chloride ions are discharged in an appreciable quantity. The chloride ions are oxidised by losing electron to electron deficient anode and gives us the chloride molecule or chloride gas
$2C{l^ - }(aq) \to C{l_2} + 2{e^ - }$
From solution hydrogen ions and chloride are removed using electrolysis and sodium ions and hydroxide ions are left in solution and form $NaOH$.
$N{a^ + } + O{H^ - } \to NaOH$
The final product occurs are ${H_2}$, $C{l_2}$ and $NaOH$.
Hence the correct answer is option D.
Note A cathode hydrogen ion is discharged because it has more tendency to electron compared to sodium ions that why sodium ions don't get discharged at cathode. Similarly, at anode chloride ions have more tendency to lose an electron and it gets discharged. Here we notice that hydrogen gas is released which is inflammable and burnt with blast then this process is done slowly for safety.
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