201. The cathodic half-reaction during electrolysis of molten \(NaCl\) is:
ⓐ. \(Na^+(l)+e^-\rightarrow Na(l)\)
ⓑ. \(Na(l)\rightarrow Na^+(l)+e^-\)
ⓒ. \(2Cl^-(l)\rightarrow Cl_2(g)+2e^-\)
ⓓ. \(Cl_2(g)+2e^-\rightarrow2Cl^-(l)\)
Correct Answer: \(Na^+(l)+e^-\rightarrow Na(l)\)
Explanation: The cathode is the site of reduction. Each sodium ion accepts one electron and becomes a neutral sodium atom. Electrons must therefore appear on the reactant side of the equation. The resulting sodium is collected in the liquid state under the hot operating conditions. An equation showing sodium releasing an electron would represent anodic oxidation rather than cathodic reduction.
202. The balanced anodic half-reaction for molten sodium chloride is:
ⓐ. \(Cl^-(l)+e^-\rightarrow Cl(l)\)
ⓑ. \(Cl_2(g)\rightarrow2Cl^-(l)+2e^-\)
ⓒ. \(2Cl^-(l)+2e^-\rightarrow Cl_2(g)\)
ⓓ. \(2Cl^-(l)\rightarrow Cl_2(g)+2e^-\)
Correct Answer: \(2Cl^-(l)\rightarrow Cl_2(g)+2e^-\)
Explanation: Chloride ions undergo oxidation at the anode. Two chloride ions each lose one electron and combine to form one chlorine molecule. The equation therefore produces two electrons. Both chlorine atoms and total charge are balanced. Writing electrons on the reactant side would incorrectly describe reduction of chlorine gas.
203. Adding the balanced electrode reactions for electrolysis of molten \(NaCl\) gives:
ⓐ. \(NaCl(l)\rightarrow Na^+(l)+Cl^-(l)\)
ⓑ. \(2NaCl(l)\rightarrow2Na(l)+Cl_2(g)\)
ⓒ. \(2Na(l)+Cl_2(g)\rightarrow2NaCl(l)\)
ⓓ. \(NaCl(l)+H_2O(l)\rightarrow NaOH(aq)+HCl(aq)\)
Correct Answer: \(2NaCl(l)\rightarrow2Na(l)+Cl_2(g)\)
Explanation: Two sodium ions require two electrons to form two sodium atoms. Oxidation of two chloride ions supplies exactly those two electrons and forms one chlorine molecule. The electrons cancel when the half-reactions are added. Combining the remaining ions gives the decomposition of two formula units of sodium chloride. Water does not appear because the electrolyte is molten rather than aqueous.
204. Consider the following statements about molten \(NaCl\) electrolysis.
Statement I: \(Na^+\) migrates toward the cathode and undergoes reduction.
Statement II: \(Cl^-\) migrates toward the anode and undergoes oxidation.
Statement III: Water competes with \(Na^+\) for reduction at the cathode.
ⓐ. Statements I and III only
ⓑ. Statements II and III only
ⓒ. Statements I and II only
ⓓ. Statements I, II and III
Correct Answer: Statements I and II only
Explanation: Cations move toward the negatively charged cathode, where they may gain electrons. Sodium ions are therefore reduced to sodium metal. Anions move toward the positive anode, where chloride ions lose electrons and form chlorine gas. Molten sodium chloride contains no water. Statement III incorrectly imports aqueous-electrolysis competition into a molten electrolyte.
205. Use the arrangement described below. Electrode P is connected to the negative terminal of a direct-current source, while electrode Q is connected to the positive terminal. Both electrodes are immersed in molten \(NaCl\). The correct interpretation is:
ⓐ. P is the anode forming \(Cl_2\), and Q is the cathode forming \(Na\)
ⓑ. P is the cathode forming \(Cl_2\), and Q is the anode forming \(Na\)
ⓒ. P is the anode forming \(Na\), and Q is the cathode forming \(Cl_2\)
ⓓ. P is the cathode forming \(Na\), and Q is the anode forming \(Cl_2\)
Correct Answer: P is the cathode forming \(Na\), and Q is the anode forming \(Cl_2\)
Explanation: The negative source terminal supplies electrons to electrode P. Reduction therefore occurs at P, making it the cathode. Sodium ions gain electrons there and form sodium metal. Electrode Q is connected to the positive terminal and functions as the anode. Chloride ions release electrons at Q and form chlorine gas.
206. During electrolysis of molten \(NaCl\), \(0.40\,mol\) of sodium is produced. The amount of chlorine gas formed simultaneously is:
ⓐ. \(0.10\,mol\)
ⓑ. \(0.20\,mol\)
ⓒ. \(0.40\,mol\)
ⓓ. \(0.80\,mol\)
Correct Answer: \(0.20\,mol\)
Explanation: \( \textbf{Overall reaction:} \)
\[
2NaCl(l)\rightarrow2Na(l)+Cl_2(g)
\]
The equation gives the product mole ratio:
\[
n(Na):n(Cl_2)=2:1
\]
\( \textbf{Given sodium amount:} \)
\[
n(Na)=0.40\,mol
\]
\( \textbf{Chlorine amount:} \)
\[
n(Cl_2)=0.40\,mol\times\frac{1\,mol\ Cl_2}{2\,mol\ Na}
\]
\[
n(Cl_2)=0.20\,mol
\]
The same result follows because two moles of electrons form two moles of sodium and one mole of chlorine.
\( \textbf{Final answer:} \) The amount of chlorine gas is \(0.20\,mol\).
207. Electrolysis of acidified water using inert electrodes produces:
ⓐ. \(H_2(g)\) at the cathode and \(O_2(g)\) at the anode
ⓑ. \(O_2(g)\) at the cathode and \(H_2(g)\) at the anode
ⓒ. hydrogen and oxygen together at the cathode only
ⓓ. metallic hydrogen at the cathode and oxide ions at the anode
Correct Answer: \(H_2(g)\) at the cathode and \(O_2(g)\) at the anode
Explanation: Reduction occurs at the cathode, where hydrogen ions gain electrons and form hydrogen gas. Oxidation occurs at the anode, where water molecules release electrons and form oxygen gas. Inert electrodes provide conducting surfaces without appearing in the overall reaction. The acid supplies mobile ions and improves conductivity. The net chemical change is decomposition of water into hydrogen and oxygen.
208. In acidic representation, the cathodic half-reaction during electrolysis of acidified water is:
ⓐ. \(H_2(g)\rightarrow2H^+(aq)+2e^-\)
ⓑ. \(2H_2O(l)+2e^-\rightarrow H_2(g)+2OH^-(aq)\)
ⓒ. \(2H^+(aq)\rightarrow H_2(g)+2e^-\)
ⓓ. \(2H^+(aq)+2e^-\rightarrow H_2(g)\)
Correct Answer: \(2H^+(aq)+2e^-\rightarrow H_2(g)\)
Explanation: The cathode supplies electrons for reduction. Two hydrogen ions each accept one electron and combine to form one hydrogen molecule. The equation conserves both hydrogen atoms and electrical charge. The acidic representation uses \(H^+\) rather than \(OH^-\). Reversing the equation would describe oxidation of hydrogen gas.
209. The balanced anodic half-reaction for acidified-water electrolysis is:
ⓐ. \(O_2(g)+4H^+(aq)+4e^-\rightarrow2H_2O(l)\)
ⓑ. \(2H_2O(l)+2e^-\rightarrow H_2(g)+2OH^-(aq)\)
ⓒ. \(2H_2O(l)\rightarrow O_2(g)+4H^+(aq)+4e^-\)
ⓓ. \(2H_2O(l)\rightarrow H_2(g)+O_2(g)\)
Correct Answer: \(2H_2O(l)\rightarrow O_2(g)+4H^+(aq)+4e^-\)
Explanation: Oxygen formation at the anode is an oxidation process. Two water molecules produce one oxygen molecule, four hydrogen ions, and four electrons. The equation contains four hydrogen atoms and two oxygen atoms on each side. Its total charge is also balanced at zero. The reverse equation would represent oxygen reduction rather than anodic oxygen evolution.
210. The overall reaction obtained by combining the acidified-water electrode reactions is:
ⓐ. \(H_2(g)+O_2(g)\rightarrow H_2O_2(l)\)
ⓑ. \(2H_2O(l)\rightarrow2H_2(g)+O_2(g)\)
ⓒ. \(2H^+(aq)+2OH^-(aq)\rightarrow H_2(g)+O_2(g)\)
ⓓ. \(H_2O(l)\rightarrow H_2(g)+O_2(g)\)
Correct Answer: \(2H_2O(l)\rightarrow2H_2(g)+O_2(g)\)
Explanation: The cathode reaction must be multiplied by two so that it consumes four electrons. Adding it to the anode reaction cancels the electrons. Four hydrogen ions produced at the anode also cancel with four hydrogen ions consumed at the cathode. The remaining equation shows two water molecules forming two hydrogen molecules and one oxygen molecule. This balanced equation determines the gas-production ratio.
211. Assertion: Dilute \(H_2SO_4\) is added to water before electrolysis mainly to improve electrical conductivity.
Reason: Pure water contains very few ions, whereas the added acid supplies mobile ions that can carry charge.
ⓐ. Both Assertion and Reason are true, but Reason does not explain Assertion
ⓑ. Assertion is true, but Reason is false
ⓒ. Both Assertion and Reason are true, and Reason explains Assertion
ⓓ. Assertion is false, but Reason is true
Correct Answer: Both Assertion and Reason are true, and Reason explains Assertion
Explanation: Pure water is a very poor electrical conductor because its ion concentration is extremely low. Adding a small amount of dilute sulphuric acid supplies mobile ions. These ions allow charge to pass through the solution more effectively. The principal net products remain hydrogen and oxygen when inert electrodes and suitable conditions are used. The Reason therefore explains why acidification is needed for practical electrolysis.
212. At constant current, the volumes of \(H_2\) and \(O_2\) produced during acidified-water electrolysis are plotted separately against time. The correct graph description is:
ⓐ. both are straight lines through the origin with equal slopes
ⓑ. both pass through the origin, with the \(O_2\) slope twice the \(H_2\) slope
ⓒ. both pass through the origin, with the \(H_2\) slope twice the \(O_2\) slope
ⓓ. the \(H_2\) line slopes downward while the \(O_2\) line slopes upward from the origin
Correct Answer: both pass through the origin, with the \(H_2\) slope twice the \(O_2\) slope
Explanation: At constant current, equal amounts of charge pass through the cell in equal time intervals. Gas production is therefore proportional to time while the current efficiency remains constant. The balanced reaction produces hydrogen and oxygen in a \(2:1\) mole ratio. Under identical conditions, their volume-production rates have the same ratio. The hydrogen graph consequently has twice the slope of the oxygen graph.
213. A total of \(90\,mL\) of gas is collected from both electrodes during electrolysis of acidified water. The gases are measured under identical conditions and remain separate until measurement. Their individual volumes are:
ⓐ. \(45\,mL\) of \(H_2\) and \(45\,mL\) of \(O_2\)
ⓑ. \(30\,mL\) of \(H_2\) and \(60\,mL\) of \(O_2\)
ⓒ. \(75\,mL\) of \(H_2\) and \(15\,mL\) of \(O_2\)
ⓓ. \(60\,mL\) of \(H_2\) and \(30\,mL\) of \(O_2\)
Correct Answer: \(60\,mL\) of \(H_2\) and \(30\,mL\) of \(O_2\)
Explanation: \( \textbf{Required gas ratio:} \)
\[
V(H_2):V(O_2)=2:1
\]
The total ratio contains:
\[
2+1=3\ \text{parts}
\]
\( \textbf{Volume of one part:} \)
\[
\frac{90\,mL}{3}=30\,mL
\]
\( \textbf{Hydrogen volume:} \)
\[
V(H_2)=2\times30\,mL=60\,mL
\]
\( \textbf{Oxygen volume:} \)
\[
V(O_2)=1\times30\,mL=30\,mL
\]
The two volumes add to the measured total of \(90\,mL\).
\( \textbf{Final answer:} \) The gases are \(60\,mL\) of hydrogen and \(30\,mL\) of oxygen.
214. The principal anodic half-reaction during electrolysis of concentrated brine with inert electrodes is:
ⓐ. \(2H_2O(l)+2e^-\rightarrow H_2(g)+2OH^-(aq)\)
ⓑ. \(Cl_2(g)+2e^-\rightarrow2Cl^-(aq)\)
ⓒ. \(2H_2O(l)\rightarrow O_2(g)+4H^+(aq)+4e^-\)
ⓓ. \(2Cl^-(aq)\rightarrow Cl_2(g)+2e^-\)
Correct Answer: \(2Cl^-(aq)\rightarrow Cl_2(g)+2e^-\)
Explanation: Oxidation occurs at the anode. Each chloride ion loses one electron, and two chlorine atoms combine to form one \(Cl_2\) molecule. The released electrons appear on the product side of the balanced half-reaction. In concentrated brine, chloride oxidation is favoured over oxygen evolution under the usual industrial conditions. The high chloride concentration and kinetic effects at the anode both contribute to this product choice.
215. The overall reaction for electrolysis of concentrated aqueous sodium chloride is:
ⓐ. \(2NaCl(aq)+2H_2O(l)\rightarrow2NaOH(aq)+H_2(g)+Cl_2(g)\)
ⓑ. \(2NaCl(aq)+2H_2O(l)\rightarrow2Na(s)+Cl_2(g)+2H_2(g)+O_2(g)\)
ⓒ. \(2NaCl(aq)+2H_2O(l)\rightarrow2NaOH(aq)+2HCl(aq)\)
ⓓ. \(2NaOH(aq)+H_2(g)+Cl_2(g)\rightarrow2NaCl(aq)+2H_2O(l)\)
Correct Answer: \(2NaCl(aq)+2H_2O(l)\rightarrow2NaOH(aq)+H_2(g)+Cl_2(g)\)
Explanation: At the cathode, water is reduced to hydrogen and hydroxide ions. At the anode, chloride ions are oxidised to chlorine gas. The electrons cancel when the two electrode equations are combined. Sodium ions remain in solution and accompany the hydroxide ions, producing sodium hydroxide solution. The balanced equation gives the product ratio \(NaOH:H_2:Cl_2=2:1:1\).
216. Consider the following statements about electrolysis of concentrated brine.
Statement I: Water is reduced at the cathode.
Statement II: Chloride ions are oxidised at the anode.
Statement III: Sodium ions remain mainly in solution and accompany the hydroxide ions formed.
ⓐ. Statements I and II only
ⓑ. Statements I, II and III
ⓒ. Statements II and III only
ⓓ. Statement III only
Correct Answer: Statements I, II and III
Explanation: Cathodic reduction of water forms \(H_2\) and \(OH^-\). At the anode, chloride ions lose electrons and form \(Cl_2\). Sodium ions are not readily reduced in the aqueous medium. They remain in solution and balance the negative charge of the hydroxide ions. All three statements describe complementary parts of the brine-electrolysis process.
217. In an industrial brine electrolyser, a membrane or separator is mainly used to:
ⓐ. conduct ions while limiting contact between chlorine and sodium hydroxide
ⓑ. carry electrons internally while ions remain in separate compartments
ⓒ. convert sodium ions to sodium metal while chloride remains in solution
ⓓ. mix chlorine with sodium hydroxide while maintaining ionic conduction
Correct Answer: conduct ions while limiting contact between chlorine and sodium hydroxide
Explanation: The electrolytic circuit requires ionic movement between the electrode regions. A suitable membrane permits selected ions to migrate and helps maintain electrical continuity. It also restricts direct mixing of chlorine with the alkaline cathode product. Such mixing would contaminate the products and could lead to secondary chemical reactions. Electrons remain confined to the external metallic circuit and power source.
218. A concentrated aqueous \(NaCl\) solution and a very dilute aqueous \(NaCl\) solution are electrolysed using inert electrodes. The most reasonable anodic comparison is:
ⓐ. Chlorine forms at the anode in both solutions, independent of concentration
ⓑ. Oxygen forms only from concentrated brine, while chlorine forms only from dilute solution
ⓒ. Concentrated brine favours chlorine, while dilution makes oxygen more competitive
ⓓ. The products must be identical because both solutions contain \(Na^+\) and \(Cl^-\)
Correct Answer: Concentrated brine favours chlorine, while dilution makes oxygen more competitive
Explanation: Aqueous electrolysis involves competition between chloride oxidation and water or hydroxide oxidation. High chloride concentration favours chlorine production under chlor-alkali conditions. At very low chloride concentration, oxygen evolution becomes more competitive. Electrode material and overvoltage also influence the observed product. Product prediction must therefore include concentration rather than using the ion list alone.
219. Select the row that describes the expected local changes during concentrated-brine electrolysis with inert electrodes.
| Row | Near the cathode | Near the anode |
|---|
| P | \(OH^-\) decreases | \(Cl^-\) increases |
| Q | \(Na^+\) is deposited | \(H_2\) is formed |
| R | \(OH^-\) increases | \(Cl^-\) is consumed |
| S | \(Cl_2\) is reduced | \(Na^+\) is oxidised |
ⓐ. Row P
ⓑ. Row Q
ⓒ. Row R
ⓓ. Row S
Correct Answer: Row R
Explanation: Cathodic reduction of water produces hydroxide ions, so the region near the cathode becomes more alkaline. At the anode, chloride ions are consumed as chlorine gas forms. Sodium ions are not deposited from the aqueous solution. Hydrogen forms at the cathode rather than at the anode. Row R correctly represents both local composition changes.
220. During electrolysis of concentrated brine, \(0.25\,mol\) of \(Cl_2\) is produced. According to the overall reaction, the amount of \(NaOH\) formed is:
ⓐ. \(0.125\,mol\)
ⓑ. \(0.50\,mol\)
ⓒ. \(0.25\,mol\)
ⓓ. \(1.00\,mol\)
Correct Answer: \(0.50\,mol\)
Explanation: \( \textbf{Overall reaction:} \)
\[
2NaCl(aq)+2H_2O(l)\rightarrow2NaOH(aq)+H_2(g)+Cl_2(g)
\]
The stoichiometric relation is:
\[
n(NaOH):n(Cl_2)=2:1
\]
\( \textbf{Given chlorine amount:} \)
\[
n(Cl_2)=0.25\,mol
\]
\( \textbf{Calculate the sodium hydroxide amount:} \)
\[
n(NaOH)=0.25\,mol\times\frac{2\,mol\ NaOH}{1\,mol\ Cl_2}
\]
\[
n(NaOH)=0.50\,mol
\]
The hydrogen amount formed simultaneously would be \(0.25\,mol\).
\( \textbf{Final answer:} \) The process forms \(0.50\,mol\) of \(NaOH\).
