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(A) In the electrolysis of a concentrated aqueous solution of $NaCl$ (brine),the ions present are $Na^+$,$Cl^-$,$H^+$,and $OH^-$.At the anode,$Cl^-$ i

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$Cl_2$ and $H_2$

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(A) In the electrolysis of a concentrated aqueous solution of $NaCl$ (brine),the ions present are $Na^+$,$Cl^-$,$H^+$,and $OH^-$.At the anode,$Cl^-$ ions are preferentially oxidized over $OH^-$ ions due to their higher concentration,producing $Cl_2$ gas: $2Cl^- \rightarrow Cl_2 + 2e^-$.At the cathode,$H^+$ ions are preferentially reduced over $Na^+$ ions because the reduction potential of $H^+$ is higher than that of $Na^+$,producing $H_2$ gas: $2H^+ + 2e^- \rightarrow H_2$.Therefore,$Cl_2$ is produced at the anode and $H_2$ is produced at the cathode.

Electrolysis of a concentrated aqueous solution of $NaCl$ with graphite electrodes leads to what results at the anode and cathode respectively?

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