Electrolytes and Electrolysis Questions in English

(D) $1$. In an aqueous solution of $NaCl$, the ions present are $Na^+$, $Cl^-$, $H^+$, and $OH^-$.
$2$. At the cathode, the reduction reaction competes between $Na^+ + e^- \rightarrow Na(s)$ $(E^o = -2.71 \text{ V})$ and $2H_2O(l) + 2e^- \rightarrow H_{2(g)} + 2OH^-(aq)$ $(E^o = -0.83 \text{ V})$.
$3$. Since the reduction potential of water is higher than that of $Na^+$, water is reduced preferentially.
$4$. Therefore, $H_{2(g)}$ is produced at the cathode.

(C) In a dry cell (Leclanché cell), the electrolyte is a moist paste of $NH_4Cl$ (ammonium chloride) and $ZnCl_2$ (zinc chloride). The $MnO_2$ (manganese dioxide) acts as the depolarizer, while the zinc container acts as the anode and a graphite rod acts as the cathode.

(D) $1$. During the electrolysis of molten $NaCl$, the dissociation reaction is: $2NaCl(l) \rightarrow 2Na^+(l) + 2Cl^-(l)$.
$2$. At the anode (oxidation): $2Cl^-(l) \rightarrow Cl_2(g) + 2e^-$. Pale green $Cl_2$ gas is evolved.
$3$. At the cathode (reduction): $2Na^+(l) + 2e^- \rightarrow 2Na(l)$. Molten silvery-white sodium metal is deposited.
$4$. Since both statements $A$ and $B$ are correct, the correct option is $D$.

(D) $1$. During the electrolysis of molten $NaCl$, the electrolyte dissociates into $Na^+$ and $Cl^-$ ions.
$2$. The anode is the positively charged electrode, which attracts negatively charged ions (anions).
$3$. The chloride ions $(Cl^-)$ migrate to the anode.
$4$. At the anode, oxidation occurs, where $Cl^-$ ions lose electrons to form chlorine gas $(Cl_2)$: $2Cl^-(l) \rightarrow Cl_2(g) + 2e^-$.
$5$. Therefore, option $D$ is the correct reaction.

(D) $1$. Aqueous $NaCl$ contains $Na^+$, $Cl^-$, $H^+$, and $OH^-$ ions.
$2$. At the cathode, the reduction reaction with the higher standard reduction potential occurs.
$3$. The reduction potential of $H^+ + e^- \rightarrow \frac{1}{2} H_{2(g)}$ $(E^0 = 0.00 \text{ V})$ is higher than that of $Na^+ + e^- \rightarrow Na_{(s)}$ $(E^0 = -2.71 \text{ V})$.
$4$. Therefore, $H^+$ ions are reduced to $H_{2(g)}$ gas at the cathode.