Electrochemistry MCQ Quiz - Objective Question with Answer for Electrochemistry - Download Free PDF

CONCEPT:

Limiting Molar Conductivity

• Limiting molar conductivity (Λ_m) of an ion is the conductivity of that ion when it is present in a very dilute solution, where the ion's movement is not hindered by other ions.
• For ions in water at 298 K, the limiting molar conductivity depends on factors such as ion size and charge, with smaller and more highly charged ions typically exhibiting higher conductivity.
• The limiting molar conductivities of ions at 298 K are given as:
  • H⁺: 349.8 S cm² mol^-1
  • Na⁺: 50.11 S cm² mol^-1
  • K⁺: 73.52 S cm² mol^-1
  • Ca²⁺: 119 S cm² mol^-1
  • Mg²⁺: 106.12 S cm² mol^-1

EXPLANATION:

• The higher the molar conductivity, the better the ion conducts electricity in solution. Based on the given values, we can arrange the ions in decreasing order of limiting molar conductivity:
  • H⁺: 349.8 S cm² mol^-1
  • Ca²⁺: 119 S cm² mol^-1
  • Mg²⁺: 106.12 S cm² mol^-1
  • K⁺: 73.52 S cm² mol^-1
  • Na⁺: 50.11 S cm² mol^-1
• Therefore, the correct order of limiting molar conductivity for cations is: H⁺ > Ca²⁺ > Mg²⁺ > K⁺ > Na⁺.

Therefore, the correct answer is: H⁺ > Ca²⁺ > Mg²⁺ > K⁺ > Na⁺.

CONCEPT:

Molar Conductivity (Λ_m) and Degree of Dissociation (α)

• Molar conductivity (Λ_m) of an electrolyte is the conductivity of a solution containing one mole of electrolyte, divided by the molar concentration.
• For a weak electrolyte, the molar conductivity at any concentration (Λ_m) is related to the molar conductivity at infinite dilution (Λ_m⁰) by the degree of dissociation (α).
• The degree of dissociation (α) is calculated using the formula:

  α = Λ_m / Λ_m⁰

• For a monobasic weak acid, the molar conductivity at infinite dilution is given by:

  Λ_m⁰ = Λ₊⁰ + Λ_-⁰

EXPLANATION:

• Calculate Λ_m⁰:
  • Λ_m⁰ = Λ₊⁰ + Λ_-⁰
  • = 349.6 S cm² mol^-1 + 50.4 S cm² mol^-1
  • = 400 S cm² mol^-1
• Calculate α:
  • α = Λ_m / Λ_m⁰
  • = 90 S cm² mol^-1 / 400 S cm² mol^-1
  • = 0.225

Therefore, the degree of dissociation (α) is 0.225, and the correct answer is Option 3.

Correct answer: 3)

Concept:

• A fuel cell is a device that converts chemical potential energy (energy stored in molecular bonds) into electrical energy.
• A fuel cell is a lot like a battery.
• It has two electrodes where the reactions take place and an electrolyte that carries the charged particles from one electrode to the other.
• A fuel, such as hydrogen, is fed to the anode, and the air is fed to the cathode.

Explanation:

• In the H2−O2 fuel cell, hydrogen is oxidized at the anode, and oxygen is reduced at the cathode.
• The half-cell reactions are:
• At anode: 2OH−+H2→2H2O+2e−
• At cathode: 2H2O+O2+4e−→4OH−
• Overall reaction: 2H2(g) + O2(g) → 2H2O(l)
• Fuel cells provide clean energy and emit no pollution.
• Moreover, it also offers high efficiency and zero emissions.
• No carbon dioxide is produced while generating chemical energy from a fuel cell.

Conclusion:

Thus, the 2H2(g) + O2(g) → 2H2O(l) reaction is used to make a fuel cell.

CONCEPT:

Fuel Cell Electrolyte

• In a fuel cell, the electrolyte is a key component that allows the movement of ions between the anode and cathode while preventing direct mixing of the reactants (fuel and oxidant).
• The H₂–O₂ fuel cell uses hydrogen (H₂) as the fuel and oxygen (O₂) as the oxidant to generate electricity, with water as the byproduct.
• In the case of a typical H₂–O₂ fuel cell, the electrolyte is **potassium hydroxide (KOH)**, which facilitates the movement of hydroxide ions (OH⁻) during the electrochemical reactions.

EXPLANATION:

• Option A: H₂SO₄ – Sulfuric acid (H₂SO₄) is not used as an electrolyte in H₂–O₂ fuel cells.
• Option B: NaCl – Sodium chloride (NaCl) is not used as the electrolyte because it would result in unwanted side reactions.
• Option C: KOH – **Correct answer**. Potassium hydroxide (KOH) is commonly used as the electrolyte in alkaline H₂–O₂ fuel cells.
• Option D: NH₄Cl – Ammonium chloride (NH₄Cl) is not used in fuel cells as the electrolyte.

Therefore, the correct answer is C) KOH.

CONCEPT:

Lead Storage Battery (Secondary Cell)

• The lead storage battery is a secondary cell commonly used in vehicles.
• It can be recharged by passing an electric current through it, which reverses the discharge reactions.
• The reactions in a lead storage battery are reversible, making it possible to recharge and reuse the cell multiple times.

EXPLANATION:

• Assertion (A): The lead storage battery is a secondary cell. This is true because it can be recharged multiple times.
• Reason (R): It can be recharged by passing an electric current and reactions are reversible. This is also true because the chemical reactions that occur during discharging can be reversed during charging, allowing the battery to be reused.
• Since both the assertion and reason are true, and the reason correctly explains why the lead storage battery is a secondary cell, the correct answer is:

Therefore, the correct answer is A) Both A and R are true, and R is the correct explanation of A.

Bleaching powder by its nature is an Oxidising agent.

• Stable bleaching powder is widely used as a disinfectant in water purification, as well as in the textile and pulp and paper industries.
• "Bleaching powder" is made by the action of chloride gas on calcium hydroxide.
• The reaction being essentially:
• 2Ca (OH)₂ + 2Cl₂ → Ca(OCl)₂ + CaCl₂ + 2H₂O.
• In the production of bleaching powder, slaked lime spread on the floors of large rectangular chambers of lead or concrete is exposed to chlorine gas.
• Bleaching powder, a solid combination of chlorine and slaked lime, was introduced in 1799 by Scottish chemist Charles Tennant.

The correct answer is Anodising.

Key Points

• Anodising is the process of forming a thick oxide layer of aluminum.
  • The process is called anodizing because the part to be treated forms the anode electrode of an electrolytic cell.
  • This aluminum oxide coat makes it resistant to further corrosion. 
  • It is also useful in architectural finishing.

Additional Information

• Ductility is the ability of a material to be drawn or plastically deformed without fracture.
  • The ductility of steel varies depending on the types and levels of alloying elements present.
• Galvanisation or galvanization (or galvanizing as it is most commonly called) is the process of applying a protective zinc coating to iron or steel, to prevent rusting.
  • The most common method is hot-dip galvanizing, in which steel sections are submerged in a bath of molten zinc.
• Corrosion is a natural process that converts a refined metal into a more chemically stable form such as oxide, hydroxide, carbonate, or sulfide.
  • It is the gradual destruction of materials (usually a metal) by chemical or electrochemical reactions with their environment.

• Rusting of iron is an example of a Redox reaction. 
• Redox reaction= Oxidation-Reduction reaction.In this reaction, an oxidation number of a molecule, atom, or ion changes either by gaining or losing an electron. 
• The substance which gets reduced in a chemical reaction is known as the oxidizing agent and a substance that gets oxidized in a chemical reaction is known as the reducing agent.
• Rusting is the corrosion of iron. Iron forms red-brown hydrated metal oxide (rust) in the presence of water and air. 
• Iron is oxidized to Fe²⁺ and oxygen is reduced to water. Rust keeps on forming due to the subsequent oxidation of Fe2+ by atmospheric oxygen.
• Therefore we can say that rusting is a redox reaction because oxygen acts as an oxidising agent and iron acts as a reducing agent.

The correct answer is AlCl3 + 3H2O → Al(OH)3 + 3HCl.

Explanation:

• Redox reactions involve the oxidation and reduction of reacting species simultaneously. Thus, a change in oxidation state determines whether or not a reaction is a redox.
• An oxidizing agent (also oxidant) is the element or compound that accepts an electron from another species in an oxidation-reduction (redox) reaction. Because the oxidizing agent is gaining electrons (and thus is frequently referred to as an electron acceptor), it is said to be reduced.
• During a redox reaction, a reductant is a reactant that donates electrons to other reactants.

AlCl3 + 3H2O → Al(OH)3 + 3HCl is not an example of a redox reaction because in this reaction oxidation and reduction do not take place.

Additional Information

2NaH → 2Na + H₂ 
          oxidation-reduction  

4Fe + 3O₂ → 2Fe₂O₃
               oxidation reduction
     
CuSO₄ + Zn → Cu + ZnSO₄
                    reduction  oxidation

Thus, the above 3 reactions are an example of redox reactions because in these reactions oxidation and reduction take place simultaneously.

Concept:

Oxidation:

It is the process in which an atom, molecule, or ion loses one or more electrons.

Reduction:

It is the process in which an atom, molecule, or ion gains one or more electrons.

Reducing agent/ Reductant:

• It is one that donates electrons to a species and thereby brings about its reduction.
• The reducing agent is oxidized by having its electrons taken away.
• Eg: In the reaction \(Mn{O_2} + 4HCl \to MnC{l_2} + C{l_2} + 2{H_2}O\), the oxidizing agent is MnO2 and the reducing agent is HCl.

Oxidizing agent/ Oxidizer:

• It is defined as a substance that removes electrons from another reactant in a redox chemical reaction.
• The oxidizing agent is reduced by taking electrons onto itself.
• In the reaction \(2Mg + {O_2} \to 2MgO\), the oxidizing agent is O2 and the reducing agent is Mg.

Explanation:

• As the electronegativity of a species increases, its tendency to pull electrons also increases and eventually behaves as a stronger oxidizing agent.
• Greater the reduction potential, greater will be the tendency of oxidizing agent to get reduced easily and hence acts as a stronger oxidizing agent.

H₂O₂:

• Hydrogen peroxide acts as an oxidizing agent and reducing agent.
• Eg: of reaction in which H₂O₂ acts as an oxidizing agent is as follows: \({H_2}{O_2} + 2{H^ + }_{(aq)} + 2{e^ - } \to 2{H_2}{O_{(l)}};\,{E^0} = + 1.77V\)

O₃:

• Ozone is a strong oxidizing agent.
• It can easily lose nascent oxygen.
• An example of a reaction in which O₃ acting as an oxidizing agent is: \({O_3} + 2{H^ + }_{(aq)} + 2{e^ - } \to {O_2} + 2O{H^ - };\,{E^0}/SRP = + 2.07V\)

K₂Cr₂O₇:

• It acts as an oxidizing agent in an acidic medium and can oxidize reducing agents like ferrous sulfate, nitrite, sulfite, etc.
• In these reactions in acid solution, dichromate is reduced to green Cr³⁺ ions: \(C{r_2}O_7^{2 - } + 14{H^ + }_{(aq)} + 6{e^ - } \to 2C{r^{3 + }} + 7{H_2}O;\,{E^0}/SRP = + 1.33V\)

KMnO₄:​

• It dissolves in water to give an intensely purple solution.
• KMnO₄ is stronger than any other oxidizing agent because it contains Mn in its highest oxidation state +7.
• Elements become more electronegative as the oxidation state of their atoms is increased.
• Their reaction is as follows: \(MnO_4^ - + 8{H^ + }_{(aq)} + 5{e^ - } \to M{n^{2 + }} + 4{H_2}O;\,{E^0}/SRP = + 1.51V\)

The increasing order of reduction potential of the given oxidizing agent is: \({E^o}C{r_2}O_7^{2 - }/C{r^{3 + }}( + 1.33V) < {E^o}Mn{O_4}^ - /M{n^{2 + }}( + 1.51V) < {O_3}/{O_2}( + 1.77V) < {H_2}{O_2}/{H_2}O( + 2.07V)\)

Hence, the increasing order of power of the oxidizing agent is : \({K_2}C{r_2}{O_7} < KMn{O_4} < {H_2}{O_2} < {O_3}\)

Therefore, ozone is the strongest oxidizing agent.

Explanation:

Kohlrausch's law states that the equivalent conductivity of an electrolyte at infinite dilution is equal to the sum of the conductances of the anions and cations. If a salt is dissolved in water, the conductivity of the solution is the sum of the conductances of the anions and cations.

Calculation:

According to Kohlrausch’s law, the molar conductivity of HA at infinite dilution is given as,

\({\Lambda_m}^\circ \left( {{\rm{HA}}} \right) = \left[ {{\Lambda _m}^\circ \left( {{{\rm{H}}^ + }} \right) + {\Lambda _m}^\circ \left( {{\rm{C}}{{\rm{l}}^ - }} \right)} \right] + \left[ {{\Lambda _m}^\circ \left( {{\rm{N}}{{\rm{a}}^ + }} \right) + {\Lambda _m}^\circ \left( {{A^ - }} \right)} \right] - \left[ {{\Lambda _m}^\circ \left( {{\rm{N}}{{\rm{a}}^ + }} \right) + {\Lambda _m}^\circ \left( {{\rm{C}}{{\rm{l}}^ - }} \right)} \right.\) ]

= 425.9 + 100.5 – 126.4

= 400 S cm² mol^-1

Also, molar conductivity (Λm°) at given concentration is given as,

\(\Lambda_m = \frac{{1000 \times k}}{M}\)

Given, k = conductivity ⟹ 5 × 10^-5 S cm^-1

M = Molarity ⟹ 0.001 M

\({\therefore \Lambda_m} = \frac{{1000 \times 5 \times {{10}^{ - 5}}{\rm{\;Sc}}{{\rm{m}}^{ - 1}}}}{{{{10}^{ - 3}}{\rm{M}}}}\)

= 50 S cm² mol^-1

Therefore, degree of dissociation (α), of HA is,

\(\alpha = \frac{{{\Lambda_m}}}{{{\Lambda_m}^\circ {\rm{\;}}}} = \frac{{50\;S\;c{m^2}mo{l^{ - 1}}}}{{400\;S\;c{m^2}\;mo{l^{ - 1}}}} = 0.125\)

Option 2 is the correct answer: Decomposition reactions are always endothermic in nature.

• In a decomposition reaction, a chemical compound is broken into its constituent components.
• The process takes place through the breaking of bonds between the constituent atoms of the compound.
• The reaction in which heat or light energy is absorbed during the process, are called endothermic reactions.
• In decomposition reactions, the energy is required to break the chemical bonds. therefore they are aways endothermic in nature.
• Combustion reaction - Combustion means reaction with oxygen, hence combustion reactions are generally oxidation reactions (Exothermic).
• Displacement reaction - One constituent component is replaced by some other component (Spontaneous and exothermic).
• Combination reaction - Two or more elements or compounds combine. New bonds are formed and energy is released (Exothermic).

Concept:

• The conductivity is depending on the number of ions present in the unit volume of solution. The decreasing order of the electrical conductivity of the aqueous solution is based on the acid strength.
• The more the acid strength more will be the dissociation of acid into ion and more will be the conductivity.
• Acid strength refers to the tendency of an acid, symbolized by the chemical formula HA, to dissociate into a proton H⁺, and an anion, A⁻. The dissociation of a strong acid in solution is effectively complete, except in its most concentrated solutions.
• Electrical conductivity is the measure of the amount of electrical current a material can carry or its ability to carry a current. Electrical conductivity is also known as specific conductance.

The Order of acidic strength is HCOOH (formic acid) > C₆ H₅ COOH (benzoic acid) > CH₃COOH (Acetic acid).

The correct option is a Double Displacement reaction.

Key PointsDouble Displacement Reaction

• The chemical reaction is in which one component of both of the reacting molecules gets exchanged to form the products.
• In other words, the reaction in which two different atoms or groups of atoms are replaced by other atoms or groups of atoms.
• CuSO₄ + H₂S  → CuS ↓ + H₂SO₄
• In the above reaction on passing hydrogen sulphide gas through an aqueous solution of copper sulphate, a black precipitate of copper sulphide is formed.
• Downward Arrow(↓) indicates the formation of a precipitate.
• Two compounds exchange their ions and one of the products formed is insoluble which is precipitate.
• These reactions usually occur in ionic compounds when dissolved in water.
• These reactions are fast reactions and take place within a fraction of a second.

Additional Information Addition reaction

• An addition reaction is also known as a combination reaction.
• The reaction in which two or more substances (elements are compounds) simply combine to form a new substance.

Decomposition reaction

• A reaction in which a single compound breaks down to produce two or simpler substances is called a decomposition reaction.
• It is opposite to the combination reaction.
• A decomposition reaction is of three types:
• thermal decomposition, electrolytic decomposition and photochemical decomposition reaction.

Displacement reaction

• Those reactions in which more reactive elements can displace less reactive elements from a compound are called displacement reactions.

2e^- + 2H⁺ (aq) → H₂(g)

\(\mathrm{E}=\mathrm{E}^{\circ}-\frac{0.059}{\mathrm{n}} \log \frac{\mathrm{P}_{\mathrm{H}_2}}{\left[\mathrm{H}^{+}\right]^2} \)

\(0=0-\frac{0.059}{2} \log \frac{\mathrm{P}_{\mathrm{H}_2}}{\left(10^{-7}\right)^2} \)

\(\log \frac{\mathrm{P}_{\mathrm{H}_2}}{\left(10^{-7}\right)^2}=0 \)

\(\frac{\mathrm{P}_{\mathrm{H}_2}}{10^{-14}}=1 \)

\(\mathrm{P}_{\mathrm{H}_2}=10^{-14} \text { bar } \)