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

Key Points

• During the electrolysis of brine (aqueous NaCl solution), NaOH, H₂, and Cl₂ are produced.
• Chlorine gas (Cl₂) is liberated at the anode (oxidation reaction).
• Hydrogen gas (H₂) is liberated at the cathode (reduction reaction).
• Sodium hydroxide (NaOH) is formed in the solution as a by-product.

Important Points

• Electrolysis of brine: Brine is a concentrated solution of sodium chloride (NaCl) in water. When subjected to electrolysis, it undergoes decomposition into its components.
• Chemical equation: 2NaCl(aq) + 2H₂O(l) → 2NaOH(aq) + H₂(g) + Cl₂(g).
• The process is widely used in the industrial production of chlorine, hydrogen, and sodium hydroxide.

The correct answer is Copper.

Key Points

• Copper has the highest electrical conductivity among the given elements at 298.15 K, with a conductivity value of approximately 5.96 × 10⁷ S/m.
• Copper is widely used in electrical wiring, electronics, and power transmission due to its excellent conductivity.
• The high conductivity of copper is attributed to its free electron density and minimal resistance to electron flow.
• Copper is more conductive than gold, iron, and sodium, making it a preferred choice for applications requiring efficient energy transfer.
• It is also highly durable and resistant to corrosion, enhancing its reliability in electrical systems.

Additional Information

• Electrical Conductivity
  • Electrical conductivity is the measure of a material's ability to allow the flow of electric current.
  • It is expressed in units of siemens per meter (S/m).
  • Materials with high conductivity, like copper, are classified as conductors, while materials with low conductivity are called insulators.
• Copper's Properties
  • Copper has a low resistivity and high thermal conductivity, making it ideal for both electrical and heat transfer applications.
  • It is a ductile and malleable metal, allowing for easy shaping and processing.
  • Copper is widely used in alloys like bronze and brass for various industrial applications.
• Comparison with Gold
  • Although gold is highly conductive, it is less conductive than copper and significantly more expensive.
  • Gold is often used in specialized applications where corrosion resistance is critical, such as high-end electronics.
• Other Metals
  • Iron has a lower electrical conductivity compared to copper and is primarily used in structural and magnetic applications.
  • Sodium is a good conductor but is rarely used in electrical systems due to its chemical reactivity and instability.

CONCEPT:

Limiting Molar Conductivity

• Limiting molar conductivity (Λ_m⁰) is the molar conductivity of an electrolyte at infinite dilution, where the concentration of the electrolyte approaches zero.
• It represents the maximum conductivity that an electrolyte can achieve in solution, as there are no inter-ionic interactions affecting the movement of ions.
• Limiting molar conductivity is important for understanding the intrinsic conductivity properties of ions in a given solvent.

EXPLANATION:

• As the concentration of an electrolyte decreases, ions have more freedom to move without interaction with other ions.
• At infinite dilution (concentration approaching zero), the molar conductivity reaches a limiting value known as limiting molar conductivity (Λ_m⁰).
• This value is specific to each electrolyte and is used to compare the conductive properties of different electrolytes.
• For example, the limiting molar conductivity of sodium chloride (NaCl) in water can be determined experimentally and is used as a reference for other electrolytes.

Therefore, the term used when the concentration of electrolyte approaches zero is the limiting molar conductivity.

Oxidation Half - Cell Reaction:

Ag(s) → Ag⁺ (aq) + e^-

Reduction Half - Cell Reaction:

AgCl(s) + e^- → Ag(s) + Cl^- (aq)

Net Cell Reaction:

AgCl(s) → Ag⁺ (aq) + Cl^- (aq)

At Equilibrium:

E_cell = 0 ⇒ E^o_cell = ^RT⁄_nF ln K_sp

And E^o_cell = E^o_Cr/AgCl/Ag - E^o_Ag⁺/Ag

= -0.57 V

K_sp = 10^-10 × √10

= 3.16 × 10^-10

Concept:

• For weak acid acetic acid, the dissociation is weak 
• \(CH_{3}COOH\rightleftharpoons CH_{{3}}COO^{-}\, +\, H^{+} \)
• If C is the initial concentration of acetic acid and \(\alpha \) is the degree of dissociation, the equilibrium constant for the reaction can be expressed as;
• \(K\, =\, \frac{C\alpha ^{2}}{1-\alpha } \)
• \(\alpha \) can be expressed as \(\alpha =\frac{\lambda }{\lambda ^{0}} \)
• Where, \(\lambda\) and \(\lambda^{0} \) are the conductances at concentration C and zero.
• Putting the values of  \(\alpha \) in equilibrium constant we get,
• \(K=\frac{C\lambda ^{2}}{\lambda ^{0}\left ( \lambda ^{0}-\lambda \right )}\)

Explanation:

• We know,  \(K=\frac{C\lambda ^{2}}{\lambda ^{0}\left ( \lambda ^{0}-\lambda \right )} \)
• Rearranging this we can write
• \(\frac{C}{K\lambda ^{0}}=\frac{\lambda ^{0}-\lambda }{\lambda ^{2}} \)
• Multiplying \(\frac{\lambda }{\lambda ^{0}}\) on both sides we get
• \(\frac{C\lambda }{K\lambda ^{0^{2}}}=\frac{\lambda ^{0}-\lambda }{\lambda\, \lambda ^{0}}=\frac{1}{\lambda }-\frac{1}{\lambda ^{0}} \)
• Further rearranging this we can get 
• \(\frac{1}{\lambda}=\frac{1}{\lambda^\circ}+\frac{C\lambda}{K\lambda^{\circ^2}} \)

Conclusion: -

For a weak electrolyte such as acetic acid, the relation among conductance (λ), equilibrium constant (K) and concentration (C) can be expressed as  \(\frac{1}{\lambda}=\frac{1}{\lambda^\circ}+\frac{C\lambda}{K\lambda^{\circ^2}} \)

Concept:

According to Gibb's Helmohltz equation,

 \(\Delta G\, =\, \Delta H\, -\, T\Delta S \)

\(\Delta G\, =\, -nFE \)

\(\Delta S\, =\, -\frac{d}{dT}\, \left ( \Delta G/\Delta T \right )\, =\, nF\, \frac{dE}{dT} \)

\(\frac{dE}{dT}\, =\, \frac{E_{2}-E_{1}}{T_{2}-T_{1}}\)= temperature coefficient of a cell

Where,

\(\Delta G,\, \Delta H,\, \Delta S \) are the change in free energy, change in enthalpy and change in entropy.

E₁, E₂ are the emf of the cell at temperature T₁ and T₂.

n is the number of electrons in the cell reaction, F is the Faraday i.e. 96485 C. mol^-1

Explanation:

Given,

E⁰ = 0.675 V,  \(\frac{dE^{0}}{dT}\)= -6.5\(\times\)10^-4 V K^-1 , T = 27 ⁰C = 300 K

\(\Delta G^{0}\, =\, -nFE^{0} \)

= -2 \(\times\) 96485 \(\times\) 0.675

= -130.25 kJ

For the reaction Cd + 2AgCl → 2Ag + CdCl2         n=2  

\(\Delta S\, =\, nF\, \frac{dE}{dT}\)

= 2 \(\times\) 96485 \(\times\) (-6.5\(\times\)10^-4)

= -0.125 kJ

\(\Delta H\, = \Delta G\, +\, T\Delta S \)

= -130.25 + 300 (-0.125)

= -167.75 kJ 

Conclusion: -

The ΔH (kJ mol-1) value for the reaction Cd + 2AgCl → 2Ag + CdCl2 is closest to -168. 

Concept:

• The vapourisation of a liquid occurs from the surface of the liquid.
• When we add non-volatile solute molecules to the solvent, the solute molecules occupy space on the surface.
• This gives lesser space for solvent molecules on the surface and thus a lower number of molecules go to the vapour phase.
• Thus, the vapour pressure also becomes less on the addition of non-volatile solute.
• The lowering of vapour pressure was given by Raoult's law, which states that for a solution of volatile liquids, each component's partial vapour pressure is directly proportional to its mole fraction present in the solution.
• The lowering of vapour pressure is derived from Raoult's law which is calculated by the formula:
  ​\(Δ p = p_0 × x_2\), where x2 = mole fraction of solute and p0 = vapour pressure of the pure solution.

Calculation:

Given:

• Vapor pressure of toluene = p⁰_toluene = 0.13 atm
• Weight of solute (hydrocarbon) = 6g
• Weight of solvent (toluene) = 92g
• Vapour pressure of the solution = .12 atm. 
• Change in vapour pressure = Δ p = .13 - .12 = .01atm
• Molecular mass of toluene = 92.14g
• Hence, the number of moles of toluene = 92/92.14 = 1 mol
• Let the Molecular Mass of the Hydrocarbon be 'M'.
• Number of moles of Hydrocarbon = 6/M
• Total number of moles = 1 + 6/M
• Mole fraction of solute
  \(x_2 = {6/M \over 1+6/M}\), 
• The change in vapour pressure is given by:
  ​\(\frac{1}{13}\;=\;\frac{6/M}{1+6/M}\)
• \(\frac{78}{M}\;=\;1+\frac{6}{M}\)
• \(\frac{78}{M}-\frac{6}{M}\;=1\)
• M = 72 \(\frac{g}{mol}\)
  Hence, the molar mass of the hydrocarbon is 72.

Concept:

Electric current:

• The flow of electric charges through a conductor constitutes an electric current.
• Quantitatively, electric current in a conductor across an area held perpendicular to the direction of flow of charge is defined as the amount of charge flowing across that area per unit time i.e.,

\({\rm{Electric\;curernt\;}}\left( {\rm{I}} \right) = \frac{{{\rm{Electric\;charge}}\;\left( {\rm{Q}} \right)}}{{{\rm{Time\;}}\left( {\rm{t}} \right)}}\)

• SI unit of current is ampere and it is denoted by the letter A.

• SI unite of current is Ampere.

Calculation:

Given:

• The voltage of the Battery = 5V
• Current Passed = 1.5A
• Time for which current is passed = 2 hours = 2 × 3600 = 7200secs
• So, the total charge =

​ Q = I × t = 1.5 × 7200 = 10800 Coulombs.

Hence, the total charge that has passed through the circuit is 10800 Coulombs.

Concept:

The effective concentration of an ion or a molecule in a given solution is known as the activity.

\(a_{salt} = m_{+} \gamma_{+}.m_{-} \gamma_{-} \)

whereas the mean activity coefficient  ( \({\gamma _ \pm }\) ) mainly depends on the number of ions that a molecule gives on dissociation. The activity coefficients can't be measured independently, it is because the solution must be electrically neutral. 

consider general electrolytic compound dissolving in water.

\({A^x}\, + \,{B^y}\, \to \,x{A^{z + }}\, + \,y{B^{z - }}\)

If the molality of the solution is m, then the ion concentration in the given solution is given by

\(\left[ A \right]\, = \,xm\)  and   \(\left[ B \right]\, = \,ym\)

Therefore, the activity of the solution becomes 

\( \begin{array}{c} a_A^x\,a_B^y\, &= \,\,\left( {\,{\gamma _A}\,{m_A}} \right){\,^x}\,\left( {\,{\gamma _B}\,{m_B}} \right){\,^y}\,\\ &= \,\,\left( {\,{\gamma _A}\,xm} \right){\,^x}\,\left( {\,{\gamma _B}\,ym} \right){\,^y}\,\\ &= \,\left( {\,{x^x}\,{y^y}} \right)\,m{\,^{\left( {x + y} \right)}}\,\left( {\,\gamma _A^{\,x}\,\gamma \,_B^y} \right) \end{array}\)

Mean activity coefficient can be defined as, 

\({\gamma _ \pm }\, = \,\,\left( {\mathop \gamma \nolimits_A^x \,\,\mathop \gamma \nolimits_B^y } \right){\,^{\frac{1}{{\left( {x + y} \right)}}}}\)

​therefore, we can rewrite the equation as, 

\(\mathop a\nolimits_A^x \,\mathop a\nolimits_B^y \, = \,\left( {\mathop x\nolimits^x \,\mathop y\nolimits^y } \right)\,{m^{\left( {x + y} \right)}}\,\mathop \gamma \nolimits_ \pm ^{\left( {x + y} \right)} \)

Explanation:

Given ‘m’ molal CuSO4 solution, 

\(CuS{O_4}\,\left( s \right)\,\, \mathbin{\lower.3ex\hbox{$\buildrel\textstyle\rightarrow\over {\smash{\leftarrow}\vphantom{_{\vbox to.5ex{\vss}}}}$}} \,C{u^{2 + }}\,\left( {aq} \right)\, + SO_4^{2 - }\,\left( {aq} \right)\)

here, x = 1 , y = 1 

Therefore the activity of the solution in terms of mean activity coefficient ( \({\gamma _ \pm }\)) is given by, 

\(\mathop a\nolimits_A^x \,\mathop a\nolimits_B^y \, = \,\left( {\mathop x\nolimits^x \,\mathop y\nolimits^y } \right)\,{m^{\left( {x + y} \right)}}\,\mathop \gamma \nolimits_ \pm ^{\left( {x + y} \right)} \)

therefore,

\(\mathop a\nolimits_A^x \,\mathop a\nolimits_B^y \, = \,{m^2}\,\mathop \gamma \nolimits_ \pm ^2 \)

Concept:

→ In a common glass electrode, the sensitivity of the electrode potential to changes in pH (known as the slope) is influenced by the composition of the electrolyte solution in which it is immersed.

→ One common problem with glass electrodes is the alkaline error, which occurs when the measured potential deviates from the theoretical value at high pH values (above pH 10).

→ To minimize the alkaline error, a high concentration of potassium chloride (KCl) is commonly used as the electrolyte solution. This is because the presence of high concentrations of potassium ions (K⁺) suppresses the ion exchange between the glass membrane and the solution, reducing the effect of the alkaline error.

Explanation:

→ 0.01 M NaCl: This is a relatively low concentration of salt, which may not be sufficient to suppress the alkaline error. In addition, the presence of sodium ions (Na⁺) may not be as effective as potassium ions (K⁺) in reducing the ion exchange between the glass membrane and the solution, leading to a higher alkaline error.

→ 1.0 M NaCl: This is a high concentration of salt, but the presence of sodium ions (Na⁺) may not be as effective as potassium ions (K⁺) in reducing the ion exchange between the glass membrane and the solution, leading to a higher alkaline error.

→ 1.0 M LiCl: Lithium ions (Li⁺) are smaller than potassium or sodium ions, and their smaller size may allow for more effective ion exchange between the glass membrane and the solution, leading to a higher alkaline error.

→1.0 M KCl: A solution of 1.0 M KCl is often used because it is a relatively high concentration that effectively minimizes the alkaline error while still maintaining a reasonable electrical conductivity. Lower concentrations of KCl may not be effective in suppressing the alkaline error, while higher concentrations can lead to increased electrical resistance and polarization effects.

Conclusion:
The correct answer is 1.0 M KCl.

Explanation:-

• Redox reaction- The reaction which involves reduction and oxidation simultaneously is called a redox reaction. 
• Reduction Half - The half-reaction in the electrolytic process which explains reduction.
• Oxidation Half- The half-reaction in the electrolytic process which shows the oxidation.
• Nernst Equation- The mathematical representation to explain the cell potential with respect to concentrations of reactants and products.

 E_Mⁿ⁺/M =⁰E_Mⁿ⁺/M - RT/nF ln [M] / [Mⁿ⁺]

Calculation:- 

E⁰_cell = E_cathode - E_anode  ⇒ -0.12V -(0.36V) ⇒ +0.24 V. 

ΔrH⁰ = nF [ T \(\left(\frac{\partial E^0_{cell}}{\partial T}\right)_P\) _p - E] ⇒

 ΔrH0 = 2 x 96500 C mol^-1 [298 x 1.5 × 10-4 VK^-1 - 0.24V] ⇒ 

 Δr H⁰ =  -37692.9 CV mol^-1 ⇒ 

Δr H0 = -37692.9 J mol^-1  ⇒ 

Δr H0 = -37.69 KJ mol-1 

Concept:-

• Nerst equation gives a relationship between the electrode potential and ionic concentration of the electrolyte solution.
• For a reduction occurring at an electrode Mn+|M

Mn+ + ne- → M (s)

E = Eº - \(\frac{RT}{nF}ln\frac{[M]}{[M^{n+}]}\)

E = Eº + \(\frac{RT}{nF}ln{[M^{n+}]}\) (molar concentration of pure solid and liquid is taken as unity)

E = Eº + \(\frac{0.0591}{n}log{[M^{n+}]}\) at 25ºC

where, n is the number of electrons exchanged in the reaction.

• The relation between the activity of the individual ions and the concentration of the solute is,

\(a_+ = \gamma _+(C_m)_+\).............(ii)​

where a+ is the activity of the cation,

\(\gamma _+\) is the activity coefficient of the cation,

\((C_m)_+\) is the concentration of the cation.

Explanation:-

• The given electrochemical cell is,

Zn(s)|ZnBr2(aq, 0.20 mol/kg) ||AgBr(s)|Ag(s)|Cu

• The corresponding cell reaction is

2 AgBr(s) + Zn(s) → 2Ag(s) + ZnBr2 (aq)

​Reaction at Anode:

Zn → Zn+2 + 2e-

​Reaction at Cathode:​

AgBr(s) + e- → Ag(s) + Br-

• For the cell,

\(E^{o}\) = \(E^0_{AgBr/Ag}\) - \(E^0_{zn^{+2}/zn}\)

=  +0.730V - ( -0.762V)

= 1.492 V

• The EMF of the cell will be,

E = \(E^{o}\) - \(\frac{RT}{nF}ln\frac{a_{Ag}\times a_{ZnBr_2}}{a_{AgBr}^2 \times a_{Ag}}\)

E = 1.492 V - \(\frac{0.059}{2}ln\frac{a_{ZnBr_2}}{a_{AgBr}^2}\)

E = 1.492 V - (-0.074) (γ± of ZnBr2 solution = 0.462)

or, E = 1.566 V

Conclusion:-

Hence, the cell potential (in V) at 298 K and 1 bar for the following cell is 1.566

Explanation:-

Standard Reduction Potential:-

• The standard reduction potential of a chemical species is a measurement of its tendency to be reduced.
• The standard oxidation potential is a measurement of a chemical species' tendency to be oxidized rather than reduced.
• The gas fluorine is strongly electronegative and seeks electrons to complete its outer shell.
• Fluorine has a comparatively high reduction potential of +2.87V as a result of this, indicating that it is more likely to acquire electrons and become reduced.

Given,

 \(E^0_{Fe^{+3_{/Fe}}}\) = -0.04 V; \(E^0_{Fe^{+2_{/Fe}}}\) = 0.44 V

Fe⁺³ (aq) + 3e- → Fe(s) ; E⊖ = -0.04 V . . . (i)

Also,

Fe+2 (aq) + 2e- → Fe(s); E⊖ = - 0.44 V 

Fe(s) → Fe+2 (aq) + 2e- ; E⊖ = 0.44 V . . . (ii)

Now, from equation (i) and (ii) we got,

\(E^0_{Fe^{+3_{/Fe}}}=\frac{n_1\times E^0_{Fe^{+3_{/Fe^{+2}}}}+n_2\times E^0_{Fe^{+2_{/Fe}}}}{n}\)

\(E^0_{Fe^{+3_{/Fe}}}=\frac{1\times E^0_{Fe^{+3_{/Fe^{+2}}}}+2\times -0.44}{1+2}\)

\(-0.04=\frac{1\times E^0_{Fe^{+3_{/Fe^{+2}}}} -0.88}{3}\)

\( E^0_{Fe^{+3_{/Fe^{+2}}}}=-0.04\times3+0.88\)

\( E^0_{Fe^{+3_{/Fe^{+2}}}}=0.76V\)

Conclusion:-

Hence, the value of \(E^0_{Fe^{+3_{/Fe}2+}}\) is 0.76 V.

Concept:

The Debye length is a measure of the range of electrostatic forces in an electrolyte solution. It is inversely proportional to the square root of the ionic strength of the solution. The ionic strength is a measure of the concentration of ions in the solution.

The ionic strength of a solution is calculated as follows:

\(I = 1/2 Σ z^2 c\)

where: I is the ionic strength, z is the charge of the ion and c is the concentration of the ion in moles per liter.

Explanation: 

The Debye length is calculated as follows:

\(\lambda_D = 1/\kappa\)

where: \(\lambda_D\) is the Debye length and \(\kappa\) is the Debye-Huckel constant.

The Debye-Huckel constant is a temperature-dependent constant that is different for different solvents. For water at 298 K, the Debye-Huckel constant is 0.334 nm^-1.

Using the above equations, we can calculate the Debye length for each of the solutions.

The Debye length for 0.01 M LaCl₃ is the smallest. Therefore, the electrolyte solution that has the smallest Debye-length at 298 K is 0.01 M LaCl₃

Conclusion:-

The electrolyte solution that has the smallest Debye-length at 298 K is 0.01 M LaCl3