Chemistry, HSSC-KPK First Year. This chapter is split into 36 topics and has
1,013 practice questions. The topics covered are listed below, followed by
10 free sample questions from the chapter.
Topics in this chapter
Chemistry of oxidizing and reducing agents30 questions
Corrosion, electroplating and galvanizing30 questions
Electrodes, electrode potential and SHE30 questions
Oxidation number: rules and calculation29 questions
Recognizing redox reactions, oxidizing and reducing agents30 questions
Conductors, electrolytes and electrolysis basics30 questions
Balancing redox equations: oxidation number method30 questions
Electrolytic cells and industrial electrolysis30 questions
A. Determining the boiling point of a pure liquid(correct answer)
B. Batteries using oxidation and reduction reactions to produce direct current
C. Production of hydrogen gas by electrolysis of water
D. Manufacture of heavy water, D2O
Answer: A.Determining the boiling point of a pure liquid is not an application of electrolysis; the STS box lists batteries, rocket propellant reactions, metal separation from ores, hydrogen production, heavy water manufacture, metal extraction (Na, K, Mg, Al), compound synthesis, electroplating, and metal purification.
Q2. Why is it impossible to measure the absolute potential difference between an electrode and its surrounding solution using a simple voltmeter?
A. Because electrodes have no measurable properties at all
B. Because the second probe dipped into the solution creates its own half-cell with its own electrode potential, so the voltmeter reads the difference between the two electrodes rather than the absolute potential of just one(correct answer)
C. Because solutions never have any potential difference
D. Because voltmeters cannot measure any potential differences
Answer: B.When a voltmeter probe is dipped into the solution, it acts as another half-cell with its own electrode potential; thus the voltmeter reads the difference in potential between the two electrodes rather than the absolute potential of the electrode under investigation.
Q3. According to the example given, Zn displaces Cu from CuSO4 because:
A. Zn and Cu have identical reduction potentials
B. Zn has a higher reduction potential than Cu
C. Zn (E°=-0.76V) is placed higher in the series with lower reduction potential, while Cu (E°=+0.34V) is placed lower with higher reduction potential(correct answer)
D. Cu is placed higher in the series than Zn
Answer: C.Zn displaces Cu from CuSO4 because Zn (E°=-0.76V) is placed higher in the series and has a lower reduction potential, while Cu (E°=+0.34V) is placed lower with a higher reduction potential; the element higher in the series can give electrons to the element below it.
Q4. According to the exercise short question, during electrolysis of fused (molten) NaCl, sodium metal is collected at the cathode and not at the anode because:
A. Sodium metal always forms at the anode in any electrolysis
B. There is no chemical basis for this observation
C. Chloride ions are attracted to the cathode instead
D. Na+ (cation) migrates to the cathode where it is reduced to Na metal, while Cl⁻ (anion) migrates to the anode where it is oxidized to Cl2(correct answer)
Answer: D.In electrolysis of fused NaCl, Na+ (the cation) migrates to the cathode where it gains electrons and is reduced to sodium metal, while Cl⁻ (the anion) migrates to the anode where it is oxidized to Cl2 gas — this is a direct consequence of cations moving to the cathode and anions to the anode.
Q5. Which of the following is NOT among the six methods for preventing rust formation mentioned in the text?
A. Alloying the metal — steel formation
B. Submerging the metal permanently in water(correct answer)
C. Coating the metal surface with paint, grease etc.
D. Cathodic protection
Answer: B.Submerging the metal permanently in water is NOT one of the six listed methods; the actual methods are: coating with paint/grease, dipping in a phosphate bath, alloying, electroplating with nickel/chromium, cathodic protection, and coating with a sacrificial metal.
Q6. Why is cathodic protection an effective method for preventing corrosion of a buried iron pipeline?
A. It coats the pipe with a layer of rust deliberately
B. It works by heating the pipeline to a very high temperature
C. By connecting the pipeline to a more easily oxidized (more reactive) metal, that sacrificial metal corrodes preferentially instead of the iron, protecting the iron(correct answer)
D. It has no actual protective effect
Answer: C.Cathodic protection works by connecting the iron structure to a more reactive (more easily oxidized) sacrificial metal, such as zinc or magnesium; since the sacrificial metal is more readily oxidized, it corrodes preferentially, protecting the iron pipeline from corrosion.
Q7. Given that H2S reduces both halogens (Cl2, Br2) and ferric salts (Fe³⁺), what is the common feature of H2S's behaviour in all these reactions?
A. H2S is reduced in every one of these reactions
B. H2S only reacts with halogens, not with metal salts
C. H2S never changes oxidation state in any of these reactions
D. In each case, S in H2S (oxidation number -2) is oxidized to elemental S (0), losing 2 electrons, which are transferred to the oxidizing agent(correct answer)
Answer: D.In all these reactions, S in H2S (oxidation number -2) is oxidized to elemental sulfur (0), losing 2 electrons per S atom; these electrons are transferred to whatever oxidizing agent (halogen or Fe³⁺) is present, consistent with H2S's role as a reducing agent.
Q8. According to the exercise descriptive question 4(i), explaining the terms oxidation and reduction with an example is best done using which framework covered in the chapter?
A. Only the classical (oxygen addition/removal) concept, since electron transfer is not covered
B. The electron transfer concept (loss of electrons = oxidation; gain of electrons = reduction), illustrated with a specific reaction(correct answer)
C. Only the concept of pH change
D. Neither concept, since oxidation and reduction have no formal definition
Answer: B.The exercise descriptive question on explaining oxidation and reduction with an example is best answered using the electron transfer concept covered in 12.1.1(B), where oxidation is loss of electrons and reduction is gain of electrons, illustrated with a specific worked reaction.
Q9. The oxidation number method is particularly convenient for balancing redox equations because it allows the chemist to:
A. Avoid identifying which species are oxidized or reduced
B. Only work with gaseous reactants
C. Directly compare the total increase in oxidation numbers (oxidation) to the total decrease (reduction), without needing to write separate ionic half-reactions(correct answer)
D. Skip balancing atoms entirely
Answer: C.The oxidation number method allows a chemist to directly compare the total increase in oxidation number (oxidation) to the total decrease (reduction) across the whole equation, without needing to write out separate ionic half-reactions as in the ion-electron method.
Q10. The classical concept and the electron transfer concept of oxidation-reduction are related in that:
A. They are completely unrelated and give contradictory classifications
B. The classical concept has completely replaced the electron transfer concept
C. Only one of the two concepts is ever valid
D. Addition of oxygen (classical oxidation) is generally accompanied by an increase in oxidation number (electron transfer oxidation), so both concepts agree for many common reactions(correct answer)
Answer: D.The classical concept (addition of oxygen = oxidation) and the electron transfer concept (increase in oxidation number = oxidation) are generally consistent for common reactions, since adding oxygen (an electronegative element) typically increases the oxidation number of the other atom involved, so both concepts agree in most everyday cases.