Question

Select the correct answer from the list given in brackets: When dilute sodium chloride is electrolysed using graphite electrodes, the cation is discharged at the cathode most readily.
[$N{a^ + }$, $O{H^ - }$,${H^ + }$,$C{l^ - }$]

Answer 1

Hint: As we know, the cathode is an electrode that is connected with the negative terminal of the battery. Hence, positive ions will migrate towards the cathode and convert to neutral atoms by gaining electrons.

Complete step-by-step answer:
Electrolysis is defined as a technique that uses electric current to instigate non-spontaneous chemical reactions. Electrolysis is commercially significant as it is the step when the separation of elements from naturally occurring sources such as ores takes place. Electrolysis takes place through an electrolytic cell.
Electrolysis occurs in electrolytic cells by the establishment of a power supply, which generates the energy and thus forces electrons to flow in the non-spontaneous direction. Since electrolysis is carried out in solutions, which contain enough ions such that current can flow. When the solution consists of only one material, such as the electrolysis of molten sodium chloride, it is a simple matter to determine what is oxidized and what is reduced. On the other hand, like the electrolysis of aqueous sodium chloride, more than one species can be oxidized or reduced and the standard reduction potentials are used to determine the most likely oxidation (the half-reaction with the largest [most positive] standard reduction potential) and reduction (the half-reaction with the smallest [least positive] standard reduction potential).
Now writing down the equation of electrolysis of dilute sodium chloride (${H_2}O$ is present)
$NaCl\xrightarrow{{aq}}N{a^ + } + C{l^ - }$
${H_2}O\overset {} \leftrightarrows {H^ + } + O{H^ - }$
As we know the negatively charged electrode is called the cathode, therefore the positive ions (${H^ + }$and $N{a^ + }$ ) will move towards the cathode.
At cathode ${H^ + }$ ions are discharged in preference to $N{a^ + }$ions as the discharge potential of ${H^ + }$ions is lesser than $N{a^ + }$ions.

$${H^ + } + {e^ - } \to H \\\ H + H \to {H_2}(g) \\\$$

Hence the answer is $N{a^ + }$

Note: Preferential discharge theory explains that when more than one type of ions are pulled towards an electrode, then the discharged ion is the one that requires the least energy.