401. In electrolysis of molten \(MgCl_2\), the products formed at the electrodes are
ⓐ. magnesium at the cathode and chlorine at the anode
ⓑ. magnesium at the anode and chlorine at the cathode
ⓒ. hydrogen at the cathode and chlorine at the anode
ⓓ. magnesium oxide at the cathode and chlorine at the anode
Correct Answer: magnesium at the cathode and chlorine at the anode
Explanation: In molten \(MgCl_2\), the only mobile ions are \(Mg^{2+}\) and \(Cl^-\). At the cathode, reduction occurs:
\[Mg^{2+} + 2e^- \rightarrow Mg\]
At the anode, oxidation occurs:
\[2Cl^- \rightarrow Cl_2 + 2e^-\]
Therefore magnesium is formed at the cathode and chlorine at the anode.
402. During silver electroplating of an article, the mass of the silver anode decreases by \(1.08\,\text{g}\). If the process is ideal, the mass of silver deposited on the cathode is
ⓐ. \(0.54\,\text{g}\)
ⓑ. \(2.16\,\text{g}\)
ⓒ. \(0.27\,\text{g}\)
ⓓ. \(1.08\,\text{g}\)
Correct Answer: \(1.08\,\text{g}\)
Explanation: In silver electroplating, the silver anode dissolves to form \(Ag^+\) ions, and the same ions are reduced at the cathode to deposit silver. For an ideal process, the amount of silver lost from the anode equals the amount gained at the cathode. Hence the deposited mass is \(1.08\,\text{g}\).
403. What volume of chlorine gas at STP is liberated at the anode when a charge of \(19300\,\text{C}\) is passed through molten sodium chloride?
(Use \(F = 96500\,\text{C mol}^{-1}\) and molar gas volume at STP \(= 22.4\,\text{L mol}^{-1}\))
ⓐ. \(1.12\,\text{L}\)
ⓑ. \(2.24\,\text{L}\)
ⓒ. \(4.48\,\text{L}\)
ⓓ. \(5.60\,\text{L}\)
Correct Answer: \(2.24\,\text{L}\)
Explanation: \(\textbf{Given:}\)
\[Q = 19300\,\text{C}\]
\[F = 96500\,\text{C mol}^{-1}\]
Molar gas volume at STP \(= 22.4\,\text{L mol}^{-1}\)
\(\textbf{Required:}\)
Volume of \(Cl_2\) liberated
\(\textbf{Relevant anode reaction:}\)
\[2Cl^- \rightarrow Cl_2 + 2e^-\]
\(\textbf{Why this reaction applies:}\)
In molten sodium chloride, chloride ions are oxidized at the anode.
\(\textbf{Find moles of electrons:}\)
\[\text{moles of electrons} = \frac{Q}{F} = \frac{19300}{96500} = 0.20\]
\(\textbf{Use stoichiometry:}\)
\(2\) moles of electrons produce \(1\) mole of \(Cl_2\)
So,
\[\text{moles of }Cl_2 = \frac{0.20}{2} = 0.10\]
\(\textbf{Convert to volume at STP:}\)
\[V = 0.10 \times 22.4\]
\(\textbf{Final simplification:}\)
\[V = 2.24\,\text{L}\]
\(\textbf{Final Answer:}\)
\[2.24\,\text{L}\]
404. During electrorefining of copper, which statement about the impurities is correct?
ⓐ. All impurities dissolve and then deposit on the cathode.
ⓑ. Every impurity remains fixed inside the anode and does not separate.
ⓒ. Reactive impurities dissolve; noble impurities form anode mud.
ⓓ. All impurities are oxidized to gases and leave the cell.
Correct Answer: Reactive impurities dissolve; noble impurities form anode mud.
Explanation: In copper electrorefining, impure copper at the anode dissolves as \(Cu^{2+}\). Impurities more reactive than copper can also enter the solution as ions. Less reactive impurities such as silver and gold do not dissolve easily and settle below the anode as anode mud.
405. How much charge is required to deposit \(5.4\,\text{g}\) of silver from \(Ag^+\) ions?
(Atomic mass of Ag \(= 108\); \(F = 96500\,\text{C mol}^{-1}\))
ⓐ. \(24125\,\text{C}\)
ⓑ. \(4825\,\text{C}\)
ⓒ. \(38600\,\text{C}\)
ⓓ. \(48250\,\text{C}\)
Correct Answer: \(4825\,\text{C}\)
Explanation: \(\textbf{Given:}\)
Mass of Ag deposited, \(m = 5.4\,\text{g}\)
Atomic mass of Ag \(= 108\)
\(\textbf{Electrode reaction:}\)
\[Ag^+ + e^- \rightarrow Ag\]
\[F = 96500\,\text{C mol}^{-1}\]
\(\textbf{Required:}\)
Charge required, \(Q\)
\(\textbf{Relevant principle:}\)
First find moles of silver, then use the electron stoichiometry.
\(\textbf{Why this principle applies:}\)
Each silver ion requires one electron for deposition.
\(\textbf{Find moles of Ag:}\)
\[\text{moles of Ag} = \frac{5.4}{108} = 0.050\]
\(\textbf{Use stoichiometry:}\)
\[\text{moles of electrons} = 0.050\]
\(\textbf{Convert to charge:}\)
\[Q = 0.050 \times 96500\]
\(\textbf{Final simplification:}\)
\[Q = 4825\,\text{C}\]
The computed value is \(4825\,\text{C}\), so that is the physically correct result.
\(\textbf{Final Answer:}\)
\[4825\,\text{C}\]
406. A current of \(2.0\,\text{A}\) is passed through acidified water for \(2\,\text{h},40\,\text{min}\). What volume of oxygen is liberated at STP?
(Use \(F = 96500\,\text{C mol}^{-1}\) and molar gas volume at STP \(= 22.4\,\text{L mol}^{-1}\))
ⓐ. \(2.24\,\text{L}\)
ⓑ. \(1.12\,\text{L}\)
ⓒ. \(4.48\,\text{L}\)
ⓓ. \(5.60\,\text{L}\)
Correct Answer: \(1.12\,\text{L}\)
Explanation: \(\textbf{Given:}\)
\[I = 2.0\,\text{A}\]
\[t = 2\,\text{h},40\,\text{min}\]
\[F = 96500\,\text{C mol}^{-1}\]
Molar gas volume at STP \(= 22.4\,\text{L mol}^{-1}\)
\(\textbf{Required:}\)
Volume of \(O_2\)
\(\textbf{Relevant anode reaction:}\)
\[2H_2O \rightarrow O_2 + 4H^+ + 4e^-\]
\(\textbf{Why this reaction applies:}\)
At the anode, water is oxidized to oxygen.
\(\textbf{Convert time into seconds:}\)
\[2\,\text{h},40\,\text{min} = 160\,\text{min} = 9600\,\text{s}\]
\(\textbf{Find charge:}\)
\[Q = It = 2.0 \times 9600 = 19200\,\text{C}\]
\(\textbf{Find moles of electrons:}\)
\[\text{moles of electrons} = \frac{19200}{96500} \approx 0.199 \approx 0.20\]
\(\textbf{Use stoichiometry:}\)
\(4\) moles of electrons produce \(1\) mole of \(O_2\)
So,
\[\text{moles of }O_2 \approx \frac{0.20}{4} = 0.050\]
\(\textbf{Convert to volume:}\)
\[V = 0.050 \times 22.4\]
\(\textbf{Final simplification:}\)
\[V = 1.12\,\text{L}\]
The computed value is \(1.12\,\text{L}\), so that is the physically correct result.
\(\textbf{Final Answer:}\)
\[1.12\,\text{L}\]
407. During electrolysis of aqueous \(CuSO_4\) using inert electrodes, which observation is expected?
ⓐ. Copper is deposited at the anode and oxygen at the cathode.
ⓑ. Hydrogen is liberated at the cathode and chlorine at the anode.
ⓒ. Copper deposits at the cathode and oxygen evolves at the anode.
ⓓ. Copper dissolves at both electrodes.
Correct Answer: Copper deposits at the cathode and oxygen evolves at the anode.
Explanation: In aqueous \(CuSO_4\) with inert electrodes, \(Cu^{2+}\) ions are reduced at the cathode:
\[Cu^{2+} + 2e^- \rightarrow Cu\]
At the anode, water is oxidized to oxygen. Hence copper is deposited at the cathode and oxygen is evolved at the anode.
408. A charge of \(0.30F\) is passed through a solution containing \(Al^{3+}\) ions. What mass of aluminium is deposited?
(Atomic mass of Al \(= 27\))
ⓐ. \(0.90\,\text{g}\)
ⓑ. \(2.70\,\text{g}\)
ⓒ. \(8.10\,\text{g}\)
ⓓ. \(1.80\,\text{g}\)
Correct Answer: \(2.70\,\text{g}\)
Explanation: \(\textbf{Given:}\)
Charge passed \(= 0.30F\)
\(\textbf{For aluminium:}\)
\[Al^{3+} + 3e^- \rightarrow Al\]
Atomic mass of Al \(= 27\)
\(\textbf{Required:}\)
Mass of aluminium deposited
\(\textbf{Relevant principle:}\)
\(1F\) corresponds to \(1\) mole of electrons.
\(\textbf{Why this principle applies:}\)
Three moles of electrons are required to deposit one mole of aluminium.
\(\textbf{Find moles of electrons:}\)
\[\text{moles of electrons} = 0.30\]
\(\textbf{Use stoichiometry:}\)
\[\text{moles of Al} = \frac{0.30}{3} = 0.10\]
\(\textbf{Convert to mass:}\)
\[m = 0.10 \times 27\]
\(\textbf{Final simplification:}\)
\[m = 2.70\,\text{g}\]
\(\textbf{Final Answer:}\)
\[2.70\,\text{g}\]
409. To electroplate a steel spoon with nickel, which arrangement is correct?
ⓐ. Steel spoon as anode, nickel strip as cathode, dilute hydrochloric acid as electrolyte
ⓑ. Steel spoon as cathode, nickel strip as anode, solution containing a soluble nickel salt as electrolyte
ⓒ. Steel spoon as cathode, platinum as anode, distilled water as electrolyte
ⓓ. Steel spoon as anode, nickel strip as cathode, solution containing a soluble nickel salt as electrolyte
Correct Answer: Steel spoon as cathode, nickel strip as anode, solution containing a soluble nickel salt as electrolyte
Explanation: In electroplating, the article to be coated is made the cathode so that metal ions deposit on its surface. The plating metal is made the anode so that it can dissolve and replenish metal ions in the solution. Therefore the correct setup uses the steel spoon as cathode, nickel as anode, and a soluble nickel salt as electrolyte.
410. In ideal electrorefining of copper, if \(31.75\,\text{g}\) of copper is deposited at the cathode, the mass lost by the copper anode is
ⓐ. \(15.875\,\text{g}\)
ⓑ. \(63.5\,\text{g}\)
ⓒ. \(47.625\,\text{g}\)
ⓓ. \(31.75\,\text{g}\)
Correct Answer: \(31.75\,\text{g}\)
Explanation: In copper electrorefining, copper dissolves from the anode as \(Cu^{2+}\) and an equal amount of \(Cu^{2+}\) is reduced at the cathode to copper metal. Under ideal conditions, the mass of copper lost by the anode is equal to the mass deposited at the cathode. Hence the anode loses \(31.75\,\text{g}\).
411. During electrolysis of concentrated aqueous sodium chloride using inert electrodes, the solution near the cathode becomes
ⓐ. alkaline because water reduction produces \(OH^-\) ions
ⓑ. acidic because chlorine dissolves near the cathode
ⓒ. neutral because sodium metal is formed there
ⓓ. acidic because \(H^+\) ions are released at the cathode
Correct Answer: alkaline because water reduction produces \(OH^-\) ions
Explanation: At the cathode in aqueous sodium chloride, water is reduced:
\[2H_2O + 2e^- \rightarrow H_2 + 2OH^-\]
This produces hydroxide ions, so the solution near the cathode becomes alkaline.
412. What mass of magnesium is produced when a charge of \(193000\,\text{C}\) is passed through molten \(MgCl_2\)?
(Atomic mass of Mg \(= 24\); \(F = 96500\,\text{C mol}^{-1}\))
ⓐ. \(12\,\text{g}\)
ⓑ. \(48\,\text{g}\)
ⓒ. \(24\,\text{g}\)
ⓓ. \(6\,\text{g}\)
Correct Answer: \(24\,\text{g}\)
Explanation: \(\textbf{Given:}\)
\[Q = 193000\,\text{C}\]
\[F = 96500\,\text{C mol}^{-1}\]
\(\textbf{For magnesium deposition:}\)
\[Mg^{2+} + 2e^- \rightarrow Mg\]
Atomic mass of Mg \(= 24\)
\(\textbf{Required:}\)
Mass of Mg produced
\(\textbf{Relevant formula / principle:}\)
First find moles of electrons from charge, then use reaction stoichiometry.
\(\textbf{Why this formula applies:}\)
Two moles of electrons are needed to deposit one mole of magnesium.
\(\textbf{Find moles of electrons:}\)
\[\text{moles of electrons} = \frac{Q}{F} = \frac{193000}{96500} = 2.0\]
\(\textbf{Use stoichiometry:}\)
\(2\) moles of electrons deposit \(1\) mole of Mg
So,
\[\text{moles of Mg} = \frac{2.0}{2} = 1.0\]
\(\textbf{Convert to mass:}\)
\[m = 1.0 \times 24\]
\(\textbf{Final simplification:}\)
\[m = 24\,\text{g}\]
\(\textbf{Final Answer:}\)
\[24\,\text{g}\]
413. The same current is passed for the same time through solutions of \(AgNO_3\) and \(CuSO_4\) connected in series. If \(10.8\,\text{g}\) of silver is deposited, the mass of copper deposited is
(Atomic mass of Ag \(= 108\); atomic mass of Cu \(= 63.5\))
ⓐ. \(6.35\,\text{g}\)
ⓑ. \(10.8\,\text{g}\)
ⓒ. \(1.587\,\text{g}\)
ⓓ. \(3.175\,\text{g}\)
Correct Answer: \(3.175\,\text{g}\)
Explanation: \(\textbf{Given:}\)
Mass of Ag deposited \(= 10.8\,\text{g}\)
\(\textbf{For silver:}\)
\[Ag^+ + e^- \rightarrow Ag\]
\(\textbf{For copper:}\)
\[Cu^{2+} + 2e^- \rightarrow Cu\]
\(\textbf{Atomic masses:}\)
Ag \(= 108\), Cu \(= 63.5\)
\(\textbf{Required:}\)
Mass of Cu deposited
\(\textbf{Relevant law:}\)
For the same quantity of electricity, deposited masses are proportional to equivalent masses.
\(\textbf{Why this law applies:}\)
Cells in series receive the same charge.
\(\textbf{Find equivalent masses:}\)
Equivalent mass of Ag \(= \frac{108}{1} = 108\)
Equivalent mass of Cu \(= \frac{63.5}{2} = 31.75\)
\(\textbf{Write proportional relation:}\)
\[\frac{m_{Cu}}{m_{Ag}} = \frac{31.75}{108}\]
\(\textbf{Substitution:}\)
\[m_{Cu} = 10.8 \times \frac{31.75}{108}\]
\(\textbf{Intermediate simplification:}\)
\[\frac{10.8}{108} = 0.10\]
So,
\[m_{Cu} = 0.10 \times 31.75\]
\(\textbf{Final simplification:}\)
\[m_{Cu} = 3.175\,\text{g}\]
\(\textbf{Final Answer:}\)
\[3.175\,\text{g}\]
414. Which pair of electrode reactions correctly represents copper electrorefining?
ⓐ. Anode: \(Cu \rightarrow Cu^{2+} + 2e^-\) ; Cathode: \(Cu^{2+} + 2e^- \rightarrow Cu\)
ⓑ. Anode: \(Cu^{2+} + 2e^- \rightarrow Cu\) ; Cathode: \(Cu \rightarrow Cu^{2+} + 2e^-\)
ⓒ. Anode: \(2H_2O \rightarrow O_2 + 4H^+ + 4e^-\) ; Cathode: \(Cu^{2+} + 2e^- \rightarrow Cu\)
ⓓ. Anode: \(Cu \rightarrow Cu^+ + e^-\) ; Cathode: \(Cu^+ + e^- \rightarrow Cu\)
Correct Answer: Anode: \(Cu \rightarrow Cu^{2+} + 2e^-\) ; Cathode: \(Cu^{2+} + 2e^- \rightarrow Cu\)
Explanation: In electrorefining, impure copper at the anode is oxidized to \(Cu^{2+}\). These copper ions move through the electrolyte and are reduced at the cathode to pure copper metal. Thus the correct pair is:
Anode: \(Cu \rightarrow Cu^{2+} + 2e^-\)
Cathode: \(Cu^{2+} + 2e^- \rightarrow Cu\).
415. A current of \(2.0\,\text{A}\) is passed through a solution of \(Cu^{2+}\) ions. How long will it take to deposit \(6.35\,\text{g}\) of copper?
(Atomic mass of Cu \(= 63.5\); \(F = 96500\,\text{C mol}^{-1}\))
ⓐ. \(32\,\text{min},10\,\text{s}\)
ⓑ. \(80\,\text{min},25\,\text{s}\)
ⓒ. \(160\,\text{min},50\,\text{s}\)
ⓓ. \(16\,\text{min},5\,\text{s}\)
Correct Answer: \(160\,\text{min},50\,\text{s}\)
Explanation: \(\textbf{Given:}\)
\[I = 2.0\,\text{A}\]
Mass of Cu deposited, \(m = 6.35\,\text{g}\)
Atomic mass of Cu \(= 63.5\)
\(\textbf{Reaction:}\)
\[Cu^{2+} + 2e^- \rightarrow Cu\]
\[F = 96500\,\text{C mol}^{-1}\]
\(\textbf{Required:}\)
Time, \(t\)
\(\textbf{Relevant formula / principle:}\)
First find moles of Cu, then required charge, then use \(Q = It\).
\(\textbf{Why this formula applies:}\)
Each mole of copper requires two moles of electrons.
\(\textbf{Find moles of Cu:}\)
\[\text{moles of Cu} = \frac{6.35}{63.5} = 0.10\]
\(\textbf{Use stoichiometry:}\)
\[\text{moles of electrons} = 2 \times 0.10 = 0.20\]
\(\textbf{Find required charge:}\)
\[Q = 0.20 \times 96500 = 19300\,\text{C}\]
Use \(Q = It\):
\[t = \frac{Q}{I} = \frac{19300}{2.0} = 9650\,\text{s}\]
\(\textbf{Convert to minutes and seconds:}\)
\[9650\,\text{s} = 160\,\text{min},50\,\text{s}\]
\(\textbf{Final Answer:}\)
\[160\,\text{min},50\,\text{s}\]
416. In electrolysis of aqueous \(AgNO_3\) using silver electrodes, which statement is correct?
ⓐ. Silver is deposited at the anode and oxygen is evolved at the cathode.
ⓑ. Silver dissolves at the anode and deposits at the cathode.
ⓒ. Hydrogen is evolved at the cathode and oxygen at the anode, so no silver is transferred.
ⓓ. Nitrate ions are reduced at the cathode and silver dissolves at both electrodes.
Correct Answer: Silver dissolves at the anode and deposits at the cathode.
Explanation: With silver electrodes in aqueous \(AgNO_3\), the anode reaction is
\[Ag \rightarrow Ag^+ + e^-\]
and the cathode reaction is
\[Ag^+ + e^- \rightarrow Ag\].
Thus silver is transferred from anode to cathode, and the concentration of \(Ag^+\) in solution remains nearly unchanged.
417. In copper electrorefining, which electrode gains mass during the process?
ⓐ. Impure copper anode
ⓑ. Electrolyte
ⓒ. Pure copper cathode
ⓓ. Anode mud
Correct Answer: Pure copper cathode
Explanation: In electrorefining, oxidation occurs at the anode and reduction occurs at the cathode. Impure copper at the anode dissolves as \(Cu^{2+}\) ions, while these ions gain electrons at the cathode and deposit as pure copper. Therefore the cathode gains mass.
418. A charge of \(2F\) is passed through acidified water. What volumes of \(H_2\) and \(O_2\) are liberated at STP?
ⓐ. \(22.4\,\text{L}\) of \(H_2\) and \(11.2\,\text{L}\) of \(O_2\)
ⓑ. \(11.2\,\text{L}\) of \(H_2\) and \(22.4\,\text{L}\) of \(O_2\)
ⓒ. \(22.4\,\text{L}\) of \(H_2\) and \(22.4\,\text{L}\) of \(O_2\)
ⓓ. \(11.2\,\text{L}\) of \(H_2\) and \(11.2\,\text{L}\) of \(O_2\)
Correct Answer: \(22.4\,\text{L}\) of \(H_2\) and \(11.2\,\text{L}\) of \(O_2\)
Explanation: \(\textbf{Given:}\)
Charge passed \(= 2F\)
\(\textbf{Required:}\)
Volumes of \(H_2\) and \(O_2\) at STP
\(\textbf{Relevant electrode reactions:}\)
\(\textbf{Cathode:}\)
\[2H^+ + 2e^- \rightarrow H_2\]
\(\textbf{Anode:}\)
\[2H_2O \rightarrow O_2 + 4H^+ + 4e^-\]
\(\textbf{Why these reactions apply:}\)
Acidified water gives hydrogen at the cathode and oxygen at the anode.
\(\textbf{For hydrogen:}\)
\(2\) moles of electrons produce \(1\) mole of \(H_2\)
Since \(2F\) means \(2\) moles of electrons,
\[\text{moles of }H_2 = 1\]
Volume of \(H_2\) at STP:
\[1 \times 22.4 = 22.4\,\text{L}\]
\(\textbf{For oxygen:}\)
\(4\) moles of electrons produce \(1\) mole of \(O_2\)
So for \(2\) moles of electrons,
\[\text{moles of }O_2 = \frac{2}{4} = 0.5\]
Volume of \(O_2\) at STP:
\[0.5 \times 22.4 = 11.2\,\text{L}\]
\(\textbf{Final Answer:}\)
\(22.4\,\text{L}\) of \(H_2\) and \(11.2\,\text{L}\) of \(O_2\)
419. During electrolysis of aqueous \(CuSO_4\) using inert electrodes, \(0.10\) mole of copper is deposited at the cathode. The amount of oxygen liberated at the anode is
ⓐ. \(0.20\) mole
ⓑ. \(0.10\) mole
ⓒ. \(0.025\) mole
ⓓ. \(0.050\) mole
Correct Answer: \(0.050\) mole
Explanation: \(\textbf{Given:}\)
Copper deposited \(= 0.10\) mole
\(\textbf{Required:}\)
Moles of \(O_2\) liberated
\(\textbf{Relevant half-reactions:}\)
\(\textbf{Cathode:}\)
\[Cu^{2+} + 2e^- \rightarrow Cu\]
\(\textbf{Anode:}\)
\[2H_2O \rightarrow O_2 + 4H^+ + 4e^-\]
\(\textbf{Why these reactions apply:}\)
In aqueous \(CuSO_4\) with inert electrodes, copper is deposited at the cathode and water is oxidized at the anode.
\(\textbf{Find moles of electrons used for copper deposition:}\)
\(1\) mole of Cu requires \(2\) moles of electrons
So,
\(0.10\) mole of Cu requires \(0.20\) mole of electrons
\(\textbf{Use anode stoichiometry:}\)
\(4\) moles of electrons produce \(1\) mole of \(O_2\)
Therefore,
\[\text{moles of }O_2 = \frac{0.20}{4} = 0.050\]
\(\textbf{Final Answer:}\)
\(0.050\) mole
420. Which statement best explains the formation of sodium hydroxide during electrolysis of concentrated aqueous sodium chloride using inert electrodes?
ⓐ. Sodium metal is first formed at the cathode and then reacts with water.
ⓑ. Water is reduced at the cathode, while \(Na^+\) remains in solution.
ⓒ. Chloride ions are reduced to \(OH^-\) ions directly at the cathode.
ⓓ. Sulfate ions present in water combine with sodium ions to form sodium hydroxide.
Correct Answer: Water is reduced at the cathode, while \(Na^+\) remains in solution.
Explanation: In brine electrolysis, sodium ions are not discharged at the cathode. Instead, water is reduced to hydrogen and hydroxide ions:
\[2H_2O + 2e^- \rightarrow H_2 + 2OH^-\]
The \(Na^+\) ions already present in solution remain there and combine with \(OH^-\) in the solution, giving sodium hydroxide.
