Batteries, Fuel Cells and Corrosion Dry cell and lead accumulator; Fuel cells.: NEET MCQs with Answers

15 free questions on Batteries, Fuel Cells and Corrosion Dry cell and lead accumulator; Fuel cells. for NEET (UG), starting with previous-year NEET and AIPMT questions. Try each one, then tap Show answer for the answer and a short explanation.

Practise all 43 Batteries, Fuel Cells and Corrosion Dry cell and lead accumulator; Fuel cells. questions free Take a free timed test

Question 1

Previous-year question
A device that converts the energy of combustion of fuels like hydrogen and methane directly into electrical energy is known as [AIPMT 2015]
  1. electrolytic cell
  2. dynamo
  3. Ni–Cd cell
  4. fuel cell
Show answer
Answer: (D) fuel cell
Explanation: This is the definition of a fuel cell.

Question 2

For the rusting step 2Fe(s) + O₂(g) + 4H⁺(aq) → 2Fe²⁺(aq) + 2H₂O(l), what is the standard Gibbs energy change? (E° of O₂,H⁺/H₂O = +1.23 V, E° of Fe²⁺/Fe = −0.44 V, F = 96500 C mol⁻¹)
  1. −161.2 kJ
  2. −322.3 kJ
  3. −1289.2 kJ
  4. −644.6 kJ
Show answer
Answer: (D) −644.6 kJ
Explanation: E°cell = 1.23 − (−0.44) = 1.67 V. Two Fe atoms each lose 2 electrons, so n = 4 for the equation as written. ΔG° = −nFE° = −4 × 96500 × 1.67 = −644620 J, or about −644.6 kJ. The large negative value shows the process is spontaneous.

Question 3

Iron may be protected by (i) a coat of paint, (ii) a covering of a chemical such as bisphenol, (iii) a block of magnesium joined to it, (iv) a block of zinc joined to it. Which methods work only by keeping the atmosphere away from the surface?
  1. (i) and (ii) only
  2. (i) and (iii) only
  3. (ii) and (iv) only
  4. (iii) and (iv) only
Show answer
Answer: (A) (i) and (ii) only
Explanation: Paint and chemicals such as bisphenol simply prevent the metal surface from coming in contact with the atmosphere. Magnesium or zinc joined to iron acts as a sacrificial electrode. This method is electrochemical, because the more active metal corrodes itself and saves the iron.

Question 4

Rusting can be pictured in two stages: 2Fe(s) + O₂(g) + 4H⁺(aq) → 2Fe²⁺(aq) + 2H₂O(l), followed by 4Fe²⁺(aq) + O₂(g) + 4H₂O(l) → 2Fe₂O₃(s) + 8H⁺(aq). Combining the stages, what is the net result when 4 mol of Fe finally ends up as Fe₂O₃?
  1. 3 mol of O₂ is used and 4 mol of H⁺ is produced.
  2. 3 mol of O₂ is used and there is no net change in the amount of H⁺.
  3. 2 mol of O₂ is used and 8 mol of H⁺ is consumed.
  4. 1 mol of O₂ is used and 8 mol of H⁺ is produced.
Show answer
Answer: (B) 3 mol of O₂ is used and there is no net change in the amount of H⁺.
Explanation: The first stage must be doubled for 4 mol Fe, which uses 2 mol O₂ and 8 mol H⁺ and forms 4 mol Fe²⁺ and 4 mol H₂O. The second stage uses 1 more mol of O₂ and 4 mol H₂O and gives back 8 mol H⁺. Adding them gives 4Fe + 3O₂ → 2Fe₂O₃, so H⁺ is regenerated and not used up overall.

Question 5

When an iron object rusts under a film of moisture, oxidation occurs at one spot and reduction at another spot on the same object. Which statement correctly describes the electrons released at the anodic spot?
  1. They move through the metal to the cathodic spot, where they reduce O₂.
  2. They move through the metal to the cathodic spot, where they reduce Fe²⁺ to iron.
  3. They move through the water film to the cathodic spot, where they reduce O₂.
  4. They remain at the anodic spot, where they reduce H₂O to H₂.
Show answer
Answer: (A) They move through the metal to the cathodic spot, where they reduce O₂.
Explanation: Iron is a metallic conductor, so the electrons released by Fe → Fe²⁺ + 2e⁻ travel through the metal itself. At another spot, which behaves as the cathode, they reduce O₂ in the presence of H⁺ to water. The object therefore acts like a small electrochemical cell.

Question 6

Iron railings in an industrial town whose air contains much SO₂ rust faster than identical railings in clean air under the same rainfall. Which explanation fits the electrochemical picture of rusting?
  1. SO₂ takes the place of O₂ as the oxidant throughout the rusting process.
  2. Dissolved SO₂ makes the standard electrode potential of Fe²⁺/Fe more positive.
  3. SO₂ reduces Fe²⁺ back to iron, which exposes fresh metal to the air.
  4. Acidic oxides dissolving in the water film supply more H⁺, which helps the reduction of O₂ at cathodic spots.
Show answer
Answer: (D) Acidic oxides dissolving in the water film supply more H⁺, which helps the reduction of O₂ at cathodic spots.
Explanation: At the cathodic spot O₂ is reduced in the presence of H⁺: O₂ + 4H⁺ + 4e⁻ → 2H₂O. In ordinary air this H⁺ is believed to come from H₂CO₃ formed when CO₂ dissolves in water. Other acidic oxides from the atmosphere also dissolve and add H⁺, so rusting becomes faster.

Question 7

In one design of hydrogen–oxygen fuel cell, the gases are bubbled through porous carbon electrodes into concentrated aqueous NaOH, and finely divided platinum or palladium is filled into the electrodes. What is the purpose of this metal?
  1. To increase the rate of the electrode reactions
  2. To act as the fuel that is oxidised at the anode
  3. To raise the standard potential of the cell
  4. To supply the OH⁻ ions used at the anode
Show answer
Answer: (A) To increase the rate of the electrode reactions
Explanation: Finely divided platinum or palladium acts as a catalyst for the electrode reactions, so they proceed faster. A catalyst does not change the overall reaction or its standard cell potential. The metal is not consumed, because hydrogen is the fuel and the OH⁻ ions come from the electrolyte.
Doing well? Create a free account to take timed tests on this topic and see your weak areas.

Question 8

Assertion: Electricity from a hydrogen–oxygen fuel cell is obtained at an efficiency of about 70%, against about 40% for a thermal power plant. Reason: A fuel cell converts the energy of combustion of the fuel directly into electrical energy instead of first releasing it as heat.
  1. Both Assertion and Reason are true, but Reason is not the correct explanation of Assertion.
  2. Assertion is false but Reason is true.
  3. Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
  4. Assertion is true but Reason is false.
Show answer
Answer: (C) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
Explanation: Fuel cells are galvanic cells designed to turn the energy of combustion of a fuel directly into electrical energy, at an efficiency of about 70%. A thermal plant first burns the fuel to produce heat, and its efficiency is only about 40%. The direct conversion accounts for the higher efficiency.

Question 9

A hydrogen–oxygen fuel cell delivers a steady current of 9.65 A for 1000 s. What masses of the two gases are consumed in this time? (F = 96500 C mol⁻¹, H = 1, O = 16)
  1. 0.05 g H₂ and 0.40 g O₂
  2. 0.20 g H₂ and 1.60 g O₂
  3. 0.10 g H₂ and 0.80 g O₂
  4. 0.10 g H₂ and 1.60 g O₂
Show answer
Answer: (C) 0.10 g H₂ and 0.80 g O₂
Explanation: The charge passed is 9.65 × 1000 = 9650 C, which is 0.10 mol of electrons. Each H₂ gives 2 electrons, so 0.05 mol H₂ (0.10 g) is used. Each O₂ takes 4 electrons, so 0.025 mol O₂ (0.80 g) is used.

Question 10

In the hydrogen–oxygen fuel cell with concentrated aqueous NaOH as electrolyte, the electrode reactions are 2H₂(g) + 4OH⁻(aq) → 4H₂O(l) + 4e⁻ and O₂(g) + 2H₂O(l) + 4e⁻ → 4OH⁻(aq). What is the net effect on the electrolyte while the cell runs?
  1. The OH⁻ used at the anode is replaced at the cathode, and there is a net gain of water.
  2. The total amount of OH⁻ rises steadily, and water is used up.
  3. The total amount of OH⁻ falls steadily, and the amount of water stays the same.
  4. The OH⁻ used at the anode is replaced at the cathode, and the amount of water stays the same.
Show answer
Answer: (A) The OH⁻ used at the anode is replaced at the cathode, and there is a net gain of water.
Explanation: Four OH⁻ ions are used at the anode and four are formed at the cathode, so the total amount of OH⁻ does not change. The anode forms 4 H₂O while the cathode uses only 2 H₂O. The net result is 2H₂ + O₂ → 2H₂O, a gain of water.

Question 11

The mercury cell used in hearing aids and watches gives a nearly constant voltage of about 1.35 V throughout its life. The best reason is that
  1. the cell is recharged continuously by the current it delivers
  2. its electrodes are inert and take no part in the cell reaction
  3. its overall reaction, Zn(Hg) + HgO(s) → ZnO(s) + Hg(l), involves no ion whose concentration changes in solution
  4. the zinc ion concentration around the anode rises steadily and offsets the loss of voltage
Show answer
Answer: (C) its overall reaction, Zn(Hg) + HgO(s) → ZnO(s) + Hg(l), involves no ion whose concentration changes in solution
Explanation: By the Nernst equation, cell potential changes only if the concentrations in the reaction quotient change. The overall reaction here involves only solids, a liquid and an amalgam, and the KOH electrolyte is not consumed. The reaction quotient therefore stays fixed as the cell discharges. In the dry cell, by contrast, ions are consumed and the voltage falls with use.

Question 12

Statement I: When the coating is scratched through, iron rusts at the exposed spot on a tin-plated sheet but remains protected on a galvanised sheet. Statement II: E° of Sn²⁺/Sn is higher than that of Fe²⁺/Fe, whereas E° of Zn²⁺/Zn is lower than that of Fe²⁺/Fe. Choose the correct option.
  1. Statement I is false, but Statement II is true
  2. Both statements are true, but Statement II does not explain Statement I
  3. Both statements are true, and Statement II correctly explains Statement I
  4. Statement I is true, but Statement II is false
Show answer
Answer: (C) Both statements are true, and Statement II correctly explains Statement I
Explanation: Tin only acts as a barrier: once it is broken, iron has the lower potential and becomes the anode of the local cell, so it rusts. Zinc has a lower potential than iron, so at a scratch zinc is oxidised instead and iron stays cathodic. The difference in electrode potentials is therefore the reason for the observed behaviour.

Question 13

A buried iron pipeline is to be protected by connecting it to a block of another metal that corrodes in its place. Given E° (V): Mg²⁺/Mg = −2.37, Al³⁺/Al = −1.66, Zn²⁺/Zn = −0.76, Fe²⁺/Fe = −0.44, Sn²⁺/Sn = −0.14, which metal cannot serve as this sacrificial anode?
  1. Aluminium
  2. Zinc
  3. Magnesium
  4. Tin
Show answer
Answer: (D) Tin
Explanation: A sacrificial anode must be oxidised more easily than iron, so its reduction potential must be lower than −0.44 V. Magnesium, aluminium and zinc all satisfy this. Tin has the higher potential, so in contact with tin the iron would become the anode and corrode faster.

Question 14

Three clean iron nails are kept for a week: nail X in freshly boiled water sealed under a layer of oil, nail Y in a closed tube of air dried by anhydrous CaCl₂, and nail Z partly dipped in ordinary water in an open tube. Only nail Z rusts. Which statement best explains this result?
  1. Rusting is an electrochemical process that needs oxygen for the cathodic reduction and a water film as the medium for ion movement
  2. Rusting is a direct reaction of iron with water alone, and oxygen plays no part in it
  3. Rusting occurs only when iron is first converted to Fe³⁺ by the action of oil-free air without any electron transfer
  4. Rusting needs only oxygen, and water merely washes away the oxide that forms
Show answer
Answer: (A) Rusting is an electrochemical process that needs oxygen for the cathodic reduction and a water film as the medium for ion movement
Explanation: Boiling expels dissolved oxygen and the oil layer stops more from entering, so nail X has no oxygen to reduce at the cathodic sites. In dry air, nail Y has no water film to carry ions between anodic and cathodic sites. Only nail Z has both, so the iron can be oxidised to Fe²⁺ while O₂ is reduced, and the product later forms hydrated ferric oxide.

Question 15

In the corrosion of iron, the overall change is 2Fe(s) + O₂(g) + 4H⁺(aq) → 2Fe²⁺(aq) + 2H₂O(l). Given E° for O₂/H₂O (acidic) = 1.23 V, E° for Fe²⁺/Fe = −0.44 V and F = 96500 C mol⁻¹, what is ΔG° for this change per mole of O₂ reduced?
  1. −644.6 kJ
  2. −152.5 kJ
  3. −322.3 kJ
  4. −1289.2 kJ
Show answer
Answer: (A) −644.6 kJ
Explanation: The cell potential is E° = 1.23 − (−0.44) = 1.67 V. Reduction of one mole of O₂ involves 4 moles of electrons. So ΔG° = −nFE° = −4 × 96500 × 1.67 J = −644620 J, or about −644.6 kJ.

Want more Batteries, Fuel Cells and Corrosion Dry cell and lead accumulator; Fuel cells. questions?

E-GYAN NIDHI has 43 questions on this topic and 63,000+ in total, all free: timed CBT tests, NEET marking and instant analysis.

Register free Free Sunday Mock Test

More topics in REDOX REACTIONS AND ELECTROCHEMISTRY

↑