401. A fuel-cell vehicle produces water at the electrochemical stack but is not automatically environmentally impact-free because:
ⓐ. water formation guarantees that hydrogen production has no upstream emissions
ⓑ. oxygen reduction at the cathode removes all pollutants associated with the vehicle
ⓒ. the catalyst eliminates the energy and resources needed to produce and handle hydrogen
ⓓ. hydrogen production, storage, and transport can consume energy and resources
Correct Answer: hydrogen production, storage, and transport can consume energy and resources
Explanation: Water is the principal direct product of the hydrogen-oxygen cell reaction. This gives the cell an important local-emission advantage. The overall environmental effect also depends on how the hydrogen is produced. Compression, liquefaction, storage, distribution, and catalyst materials can require substantial energy or cost. Evaluating the whole system therefore requires more than inspecting the stack exhaust alone.
402. A hydrogen-oxygen fuel cell receives \(4.0\,mol\) of \(H_2\) and \(1.5\,mol\) of \(O_2\). Assuming complete reaction of the limiting reactant, the amount of water formed and the excess hydrogen remaining are:
ⓐ. \(3.0\,mol\) of \(H_2O\) and \(1.0\,mol\) of \(H_2\)
ⓑ. \(4.0\,mol\) of \(H_2O\) and \(0.5\,mol\) of \(H_2\)
ⓒ. \(1.5\,mol\) of \(H_2O\) and \(2.5\,mol\) of \(H_2\)
ⓓ. \(3.0\,mol\) of \(H_2O\) and no hydrogen remaining
Correct Answer: \(3.0\,mol\) of \(H_2O\) and \(1.0\,mol\) of \(H_2\)
Explanation: Use the overall fuel-cell reaction:
\[
2H_2+O_2\rightarrow2H_2O
\]
For \(1.5\,mol\) of oxygen, the hydrogen required is:
\[
n(H_2)=2(1.5)
\]
\[
n(H_2)=3.0\,mol
\]
Since \(4.0\,mol\) of hydrogen is supplied, oxygen is the limiting reactant.
The excess hydrogen is:
\[
n(H_2,\text{ excess})=4.0-3.0
\]
\[
n(H_2,\text{ excess})=1.0\,mol
\]
The reaction forms two moles of water per mole of oxygen:
\[
n(H_2O)=2(1.5)
\]
\[
n(H_2O)=3.0\,mol
\]
The limiting-reactant check prevents the incorrect assumption that all supplied hydrogen reacts.
403. A hydrogen-oxygen fuel cell operates at \(9.65\,A\) for \(1.00\times10^4\,s\). The volume of oxygen consumed at STP is approximately:
ⓐ. \(22.4\,L\)
ⓑ. \(11.2\,L\)
ⓒ. \(5.6\,L\)
ⓓ. \(2.8\,L\)
Correct Answer: \(5.6\,L\)
Explanation: Calculate the charge:
\[
Q=It
\]
\[
Q=(9.65)(1.00\times10^4)
\]
\[
Q=9.65\times10^4\,C
\]
This equals one faraday:
\[
n(e^-)=1.00\,mol
\]
The cathodic reaction is:
\[
O_2+2H_2O+4e^-\rightarrow4OH^-
\]
Therefore:
\[
4\,mol\ e^-:1\,mol\ O_2
\]
\[
n(O_2)=\frac{1.00}{4}=0.250\,mol
\]
At STP:
\[
V=0.250(22.4\,L)
\]
\[
V=5.6\,L
\]
404. Corrosion of a metal is best understood as:
ⓐ. a physical loss of metal caused by abrasion and surface wear in the environment
ⓑ. a spontaneous redox process that oxidises the metal through environmental reaction
ⓒ. an electrolytic metal-deposition process driven by an external power source
ⓓ. a redox process in which the corroding metal always acts as the cathode
Correct Answer: a spontaneous redox process that oxidises the metal through environmental reaction
Explanation: Corrosion involves conversion of a metal into more oxidised chemical forms such as oxides, hydroxides, or salts. The metal atoms lose electrons and therefore undergo oxidation. Environmental substances such as dissolved oxygen or hydrogen ions accept those electrons in a cathodic process. The coupled reactions can occur spontaneously without an external power source. Corrosion is therefore an electrochemical redox phenomenon rather than only mechanical damage.
405. Select the row that correctly identifies the charge-transport paths in electrochemical corrosion.
| Row | Through the metal | Through the surface moisture |
|---|
| P | Electrons | Ions |
| Q | Cations only | Electrons |
| R | Anions only | Free metal atoms |
| S | Electrons and ions equally | Electrons only |
ⓐ. Row P
ⓑ. Row Q
ⓒ. Row R
ⓓ. Row S
Correct Answer: Row P
Explanation: The metallic body provides an electronic conducting path. Electrons released at anodic regions travel through the metal toward cathodic regions. The thin moisture layer behaves as an electrolyte and conducts through ionic movement. Electrons do not normally pass through the aqueous layer as the main charge carriers. Both paths are needed to complete the local electrochemical circuit.
406. A clean metal surface remains in dry air but begins corroding rapidly after a conducting moisture film forms. The moisture film mainly:
ⓐ. prevents environmental oxidants from reaching the surface
ⓑ. acts as an electronic conductor between the local anodic and cathodic regions
ⓒ. provides an ionic path between local anodic and cathodic sites
ⓓ. carries electrons externally from the cathode to the anode
Correct Answer: provides an ionic path between local anodic and cathodic sites
Explanation: Local anodic and cathodic reactions must be connected by both electronic and ionic paths. The metal itself carries electrons between the reaction sites. A moisture film containing dissolved ions permits charge balance through ionic migration. It can also dissolve oxygen or other species involved in cathodic reduction. Without a suitable electrolyte layer, the electrochemical corrosion circuit is greatly hindered.
407. Assertion: Corrosion can occur even when the anodic and cathodic regions are parts of the same continuous metal object.
Reason: Differences in composition, stress, oxygen access, or surface condition can create local differences in electrode behaviour.
ⓐ. Both Assertion and Reason are true, but Reason does not explain Assertion
ⓑ. Assertion is true, but Reason is false
ⓒ. Assertion is false, but Reason is true
ⓓ. Both Assertion and Reason are true, and Reason explains Assertion
Correct Answer: Both Assertion and Reason are true, and Reason explains Assertion
Explanation: A corrosion cell does not require two separately manufactured electrodes. Different microscopic areas of one metal object can develop different oxidation or reduction tendencies. Local environmental or structural differences create these electrochemical inequalities. One region then supplies electrons while another consumes them. The Reason explains how a single metal surface can contain both anodic and cathodic zones.
408. For a metal \(M\) forming ions \(M^{n+}\), the general anodic corrosion reaction is:
ⓐ. \(M(s)\rightarrow M^{n+}(aq)+ne^-\)
ⓑ. \(M^{n+}(aq)+ne^-\rightarrow M(s)\)
ⓒ. \(M(s)+ne^-\rightarrow M^{n-}(aq)\)
ⓓ. \(M^{n+}(aq)\rightarrow M(s)+ne^-\)
Correct Answer: \(M(s)\rightarrow M^{n+}(aq)+ne^-\)
Explanation: Corrosion begins with oxidation of metal atoms at anodic sites. Oxidation places electrons on the product side of the half-reaction. The metal enters the electrolyte as positively charged ions. The number of released electrons equals the positive charge developed by each ion. Those electrons become available for the accompanying cathodic process.
409. In neutral, aerated moisture, a common cathodic corrosion reaction is:
ⓐ. \(4OH^-(aq)\rightarrow O_2(g)+2H_2O(l)+4e^-\)
ⓑ. \(O_2(g)+2H_2O(l)+4e^-\rightarrow4OH^-(aq)\)
ⓒ. \(O_2(g)\rightarrow2O^{2-}(aq)+4e^-\)
ⓓ. \(2H_2O(l)+2e^-\rightarrow H_2(g)+2OH^-(aq)\)
Correct Answer: \(O_2(g)+2H_2O(l)+4e^-\rightarrow4OH^-(aq)\)
Explanation: Dissolved oxygen commonly acts as the electron acceptor in neutral aerated water. It gains four electrons and reacts with water to produce hydroxide ions. Because electrons are consumed, the reaction is cathodic reduction. The reverse equation would represent oxidation rather than the normal cathodic corrosion process. The hydroxide ions formed can later react with metal ions produced at anodic regions.
410. A metal object is partly covered by a thin electrolyte film. Oxidation occurs at one location, but the cathodic reaction is experimentally blocked everywhere else. The corrosion current falls sharply because:
ⓐ. anodic oxidation proceeds independently once the metal surface becomes wet
ⓑ. blocking the cathodic process prevents electrons from moving through the metal
ⓒ. the corrosion current is controlled only by ion mobility in the electrolyte film
ⓓ. anodic oxidation requires a simultaneous reduction that consumes its electrons
Correct Answer: anodic oxidation requires a simultaneous reduction that consumes its electrons
Explanation: Oxidation and reduction must occur together in a complete redox process. Anodic metal dissolution releases electrons into the metal. If no cathodic reaction consumes them, excess negative charge develops and further oxidation is inhibited. Blocking the cathodic process therefore suppresses the overall corrosion current. The two half-reactions are chemically coupled even when they occur at different surface locations.
411. The anodic reaction responsible for iron dissolution during rusting is:
ⓐ. \(Fe(s)\rightarrow Fe^{2+}(aq)+2e^-\)
ⓑ. \(Fe^{2+}(aq)+2e^-\rightarrow Fe(s)\)
ⓒ. \(Fe(s)+3e^-\rightarrow Fe^{3-}(aq)\)
ⓓ. \(2Fe(s)+O_2(g)\rightarrow2FeO(s)+4e^-\)
Correct Answer: \(Fe(s)\rightarrow Fe^{2+}(aq)+2e^-\)
Explanation: Iron atoms lose two electrons at anodic regions of the surface. The immediate dissolved product is commonly represented as \(Fe^{2+}\). Electron loss confirms that the reaction is oxidation. The released electrons move through the iron toward cathodic regions. Subsequent reactions convert the iron ions into hydroxides and hydrated oxide products.
412. At a cathodic region on rusting iron exposed to neutral aerated water, the principal reaction is:
ⓐ. \(Fe^{2+}(aq)\rightarrow Fe^{3+}(aq)+e^-\)
ⓑ. \(2H^+(aq)+2e^-\rightarrow H_2(g)\)
ⓒ. \(O_2(g)+2H_2O(l)+4e^-\rightarrow4OH^-(aq)\)
ⓓ. \(4OH^-(aq)\rightarrow O_2(g)+2H_2O(l)+4e^-\)
Correct Answer: \(O_2(g)+2H_2O(l)+4e^-\rightarrow4OH^-(aq)\)
Explanation: Dissolved oxygen reaches the cathodic parts of the wet iron surface. It accepts electrons arriving through the metal. In neutral water, the balanced reduction product is hydroxide ion. Four electrons are required for one oxygen molecule in this equation. The resulting hydroxide later combines with iron ions generated at the anode.
413. When the iron-oxidation reaction is combined with oxygen reduction in neutral water, the initial electrochemical stage can be represented by:
ⓐ. \(Fe(s)+O_2(g)\rightarrow FeO_2(s)\)
ⓑ. \(2Fe(s)+O_2(g)+2H_2O(l)\rightarrow2Fe^{2+}(aq)+4OH^-(aq)\)
ⓒ. \(4Fe^{2+}(aq)+O_2(g)\rightarrow4Fe^{3+}(aq)+2O^{2-}(aq)\)
ⓓ. \(2Fe(s)+4OH^-(aq)\rightarrow2Fe^{2+}(aq)+2H_2O(l)+O_2(g)\)
Correct Answer: \(2Fe(s)+O_2(g)+2H_2O(l)\rightarrow2Fe^{2+}(aq)+4OH^-(aq)\)
Explanation: The anodic half-reaction is:
\[
Fe\rightarrow Fe^{2+}+2e^-
\]
Multiply it by two:
\[
2Fe\rightarrow2Fe^{2+}+4e^-
\]
The cathodic half-reaction is:
\[
O_2+2H_2O+4e^-\rightarrow4OH^-
\]
Adding the half-reactions cancels the four electrons:
\[
2Fe+O_2+2H_2O\rightarrow2Fe^{2+}+4OH^-
\]
The equation describes the coupled production of \(Fe^{2+}\) and \(OH^-\) before hydroxide precipitation occurs.
414. The first hydroxide precipitate commonly formed when \(Fe^{2+}\) meets \(OH^-\) during rusting is produced by:
ⓐ. \(Fe^{2+}+OH^-\rightarrow FeOH^+\)
ⓑ. \(Fe^{3+}+3OH^-\rightarrow Fe(OH)_3\)
ⓒ. \(2Fe^{2+}+O_2\rightarrow2FeO_2\)
ⓓ. \(Fe^{2+}+2OH^-\rightarrow Fe(OH)_2\)
Correct Answer: \(Fe^{2+}+2OH^-\rightarrow Fe(OH)_2\)
Explanation: Each \(Fe^{2+}\) ion carries a charge of \(+2\). Two hydroxide ions are therefore needed to form a neutral hydroxide precipitate. The product is iron(II) hydroxide, \(Fe(OH)_2\). This compound is not the final rust material. It undergoes further oxidation in the presence of oxygen and water.
415. A simplified equation for further oxidation of iron(II) hydroxide during rust formation is:
ⓐ. \(4Fe(OH)_2+O_2+2H_2O\rightarrow4Fe(OH)_3\)
ⓑ. \(2Fe(OH)_3\rightarrow2Fe(OH)_2+H_2O+O_2\)
ⓒ. \(Fe(OH)_2+2e^-\rightarrow Fe+2OH^-\)
ⓓ. \(4Fe(OH)_3+O_2\rightarrow4FeO_2+6H_2O\)
Correct Answer: \(4Fe(OH)_2+O_2+2H_2O\rightarrow4Fe(OH)_3\)
Explanation: Iron(II) hydroxide contains iron in the \(+2\) oxidation state. Oxygen can oxidise it further toward iron(III) hydroxide. The displayed equation balances iron, hydrogen, and oxygen atoms. Iron(III) hydroxide and related hydrated products can then undergo changes leading to rust. Rusting is therefore a sequence of reactions rather than a single direct combination of iron and oxygen.
416. Rust is commonly represented approximately as ______.
ⓐ. \(FeO\cdot xH_2O\)
ⓑ. \(Fe_3O_4\cdot xH_2O\)
ⓒ. \(Fe(OH)_2\cdot xH_2O\)
ⓓ. \(Fe_2O_3\cdot xH_2O\)
Correct Answer: \(Fe_2O_3\cdot xH_2O\)
Explanation: Rust is not usually a single pure compound with one fixed composition. It consists largely of hydrated iron(III) oxide together with related iron oxyhydroxide materials. The notation \(Fe_2O_3\cdot xH_2O\) represents its variable hydration. The value of \(x\) need not be a fixed whole number. This approximate formula is more realistic than treating rust as pure anhydrous \(Fe_2O_3\).
417. A water droplet rests on an iron surface. The centre beneath the droplet is relatively oxygen-poor, while the edge is oxygen-rich. The most likely assignment is:
ⓐ. centre cathodic and edge anodic
ⓑ. centre anodic and edge cathodic
ⓒ. both centre and edge anodic
ⓓ. both centre and edge cathodic
Correct Answer: centre anodic and edge cathodic
Explanation: Oxygen reduction is favoured where dissolved oxygen is more readily available. The oxygen-rich edge therefore acts mainly as the cathodic region. The oxygen-poor centre becomes relatively anodic and iron dissolves there. This separation is called differential-aeration corrosion. A difference in oxygen concentration can thus create a corrosion cell on an otherwise continuous iron surface.
418. In the differential-aeration cell formed beneath a water droplet on iron, electrons move:
ⓐ. through the water film from the oxygen-rich edge toward the centre
ⓑ. through the water film from the oxygen-poor centre toward the edge
ⓒ. through the iron from the oxygen-rich edge toward the oxygen-poor centre
ⓓ. through the iron from the oxygen-poor centre toward the oxygen-rich edge
Correct Answer: through the iron from the oxygen-poor centre toward the oxygen-rich edge
Explanation: The oxygen-poor centre is the anodic zone where iron releases electrons. The oxygen-rich edge is the cathodic zone where oxygen consumes those electrons. Electronic conduction occurs through the iron itself. The moisture layer carries ions rather than serving as the main electron path. The electron direction is therefore from the centre toward the edge.
419. Arrange the principal stages of rust formation in the most reasonable order.
Stage P: \(Fe^{2+}\) combines with \(OH^-\) to form \(Fe(OH)_2\).
Stage Q: Iron is oxidised to \(Fe^{2+}\).
Stage R: Hydrated iron(III) oxide products develop.
Stage S: Oxygen is reduced to produce \(OH^-\).
ⓐ. Q and S occur together, followed by P and then R
ⓑ. R occurs first, followed by Q, P, and S
ⓒ. P must occur before either electron-transfer reaction
ⓓ. S occurs only after all \(Fe^{2+}\) has become rust
Correct Answer: Q and S occur together, followed by P and then R
Explanation: Rusting begins with simultaneous anodic and cathodic reactions at different surface regions. Iron oxidation produces \(Fe^{2+}\), while oxygen reduction produces \(OH^-\). These ions then meet and form \(Fe(OH)_2\). Further oxidation produces iron(III) hydroxide and related hydrated compounds. The final rust material develops only after the initial electrochemical and precipitation stages.
420. Iron corrodes more rapidly in salt water than in distilled water mainly because dissolved salt:
ⓐ. removes dissolved oxygen and suppresses the cathodic reaction
ⓑ. converts iron into a more noble metal through alloy formation
ⓒ. raises ionic conductivity and allows a larger corrosion current
ⓓ. prevents separate anodic and cathodic regions from developing
Correct Answer: raises ionic conductivity and allows a larger corrosion current
Explanation: Electrochemical corrosion requires ionic conduction through the moisture layer. Dissolved salt supplies mobile ions and lowers the electrical resistance of that layer. Charge can therefore move more readily between local anodic and cathodic regions. The coupled oxidation and reduction reactions proceed at a greater rate. Salt does not protect the iron or eliminate dissolved oxygen.
