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Electrochemistry - 2 - Free MCQ Practice Test with solutions, Chemistry


MCQ Practice Test & Solutions: Test: Electrochemistry - 2 (30 Questions)

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Test Highlights:

  • - Format: Multiple Choice Questions (MCQ)
  • - Duration: 90 minutes
  • - Number of Questions: 30

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Test: Electrochemistry - 2 - Question 1

A 19th century iron bridge is protected from corrosion by connecting it to a block of metal (sacrificial anode), which is replaced annually. The corrosion of iron, represented by the chemical equation:

Which of the following metals is best suited as sacrificial anode:

Test: Electrochemistry - 2 - Question 2

Consider the following electrochemical cell, from which current is drawn through an external resistor of 10 Ohms. During this process, the concentration of CuSO4 in the left and the right halfcells were measured, and the value of was calculated. From the init ial value of K = 10, predict the value of K after a very long time when the cell stopped giving any current.

Test: Electrochemistry - 2 - Question 3

The velocity of Li+ ion in water is 2 × 10–2 cm/sec when 100 V is applied between two electrodes separated by 2 cm. The mobility of Li+ ion in water is,

Detailed Solution: Question 3

Test: Electrochemistry - 2 - Question 4

At 20°C, the standard EMF of a certain cell is +0.2699 V, and at 30°C it is + 0.2669 V. What can you say about the standard entropy of this reaction? Assume that the standard ΔH° and ΔS° are independent of temperature:

Detailed Solution: Question 4

As per temperature coefficient of cell, dS= nF[dE/dT] at constant pressure Here we can se that as temp. Increase E is decrease it means if we take T as positive then E will be negative. So, here [dE/dT] this ratio will be negative which is directly proportional to the entropy change. Hence, we can say that dS= (-ve)

Test: Electrochemistry - 2 - Question 5

The cell potential for the following electrochemical system at 25°C is:

Al(s) | Al3+ (0.01 M) || Fe2+ (0.1 M) | Fe (s)

Given: Standard reduction potential of Al3+ + 3e → Al is –1.66 V at 25°C
Standard reduction potential of Fe2+ + 2e → Fe is –0.44 V at 25°C

Detailed Solution: Question 5

    Test: Electrochemistry - 2 - Question 6

    FAD is a redox active molecule which takes part in many important biological reactions. the redox potential of FAD at pH 7.0 is given below:

    Calculate the redox potential when the media is acidified to pH0

    Test: Electrochemistry - 2 - Question 7

    According to the Nernst equation, the potential of an electrode changes by 59.2 mV whenever the ratio of the oxidized and the reduced species changed by a factor of 10 at 25°C. What would be the corresponding change in the electrode potential if the experiment is carried out at 30°C:

    Detailed Solution: Question 7

    Test: Electrochemistry - 2 - Question 8

    Chemical oxidizing of water to produce O2 gas is an energy demanding reaction, done routinely by plants using the process called photosynthesis. By how many eV will it be uphill if the water oxidation reaction be carried out at pH = 0 versus at pH = 7.0:

    Test: Electrochemistry - 2 - Question 9

    The concentration of K+ ion inside a biological cell is 20 times higher than outside. The magnitude of potential difference between the two sides is [Given: 2.303 RT/F = 59 mV]

    Detailed Solution: Question 9

    Given: Concentration of K⁺ inside the cell is 20 times higher than outside:

    [K⁺]_inside / [K⁺]_outside = 20

    2.303 RT / F = 59 mV (at room temperature)

    Objective: Find the magnitude of potential difference (E) across the cell membrane due to this ion concentration difference.

    Relevant formula: Nernst Equation For a monovalent ion like K⁺:

    E = (2.303 RT / zF) log₁₀ ([K⁺]_outside / [K⁺]_inside)

    where:

    • E = membrane potential (in volts or millivolts)
    • z = +1 (charge of K⁺)
    • R = gas constant
    • T = absolute temperature
    • F = Faraday constant
    • [K⁺]_outside, [K⁺]_inside = concentrations outside and inside respectively

    Step 1: Substitute values E = 59 mV × log₁₀ (1/20) = 59 × log₁₀ (0.05)

    Step 2: Calculate logarithm log₁₀ (0.05) = log₁₀ (1/20) = -log₁₀ (20) ≈ -1.3010

    Step 3: Calculate E E = 59 × (-1.3010) = -76.76 mV

    Step 4: Interpret result The negative sign indicates that the inside is negative relative to the outside (typical for K⁺ potential).

    The magnitude of potential difference is:

    |E| = 76.76 mV

    Final answer: 76.8 mV (approx)

    Test: Electrochemistry - 2 - Question 10

    If the transport number of Na+ is 0.463 (dilute solution of NaCl in methanol), the transport number of H+ (dilute solut ion of HCl in methanol) is:

    (Given, (NaCl in methanol) = 96.9 Ω-1 cm2 mol-1 and (HCl in methanol) = 192 Ω-1 cm2 mol-1 

    Detailed Solution: Question 10

    The transport number of an ion in a solution indicates the fraction of the total current it carries. Here, we need to find the transport number of H+ in a dilute solution of HCl in methanol.

    • The transport number of Na+ in NaCl (methanol) is 0.463.
    • The molar conductivity of NaCl in methanol is 96.9 Ω-1 cm2 mol-1.
    • The molar conductivity of HCl in methanol is 192 Ω-1 cm2 mol-1.
    • To find the transport number of H+, use the ratio of their conductivities:
      • Ratio = 192 / 96.9 = approximately 1.98.
    • Since HCl is almost twice as conductive, the transport number of H+ will be higher than that of Na+.
    • The transport number of H+ is calculated to be approximately 0.73.

    Test: Electrochemistry - 2 - Question 11

    The molality of (NH4)2SO4 solution that has the same ionic strength as 1 mol kg-1 solution of KCl is:

    Detailed Solution: Question 11

    Correct option: A.

    The ionic strength is given by I = 1/2 Σ c_i z_i2, where c_i is the molal concentration of ion i and z_i its charge.

    For 1 mol kg-1 KCl, concentrations are [K+] = 1 mol kg-1 and [Cl-] = 1 mol kg-1. Thus

    I = 1/2 (1·12 + 1·12) = 1 mol kg-1.

    Let the molality of (NH4)2SO4 be m. On dissociation, [NH4+] = 2m and [SO42-] = m.

    Calculate ionic strength:

    I = 1/2 (2m·12 + m·22) = 1/2 (2m + 4m) = 3m.

    Equate ionic strengths: 3m = 1m = 1/3 mol kg-1.

    Therefore the required molality is 1/3 mol kg-1, option A.

    Test: Electrochemistry - 2 - Question 12

    A concentration cell with two hydrogen electrodes at two different pressure is depicted as

    The potential (Ecell) of the cell is

    Test: Electrochemistry - 2 - Question 13

    The molar conductivity of 0.009 M aqueous solution of a weak acid (HA) is 0.005 S m2 mol-1 and the limiting molar conductivity of HA is 0.05 S m2 mol-1 at 298 K. Assuming activity coefficients to be unity, the acid dissociation constant (Ka) of HA at this temperature is

    Detailed Solution: Question 13

    Correct answer: C

    Degree of dissociation (α) is given by α = Λ_c / Λ_∞.

    α = 0.005 / 0.05 = 0.1

    For the weak acid HA, at equilibrium [H+] = αc, [A-] = αc and [HA] = c(1 - α). Hence

    K_a = (α^2 · c) / (1 - α)

    Calculate the numerator: α^2 · c = (0.1)^2 × 0.009 = 9×10-5

    Now divide by (1 - α) = 0.9 to get

    K_a = (9×10-5)/(0.9) = 1.0×10-4

    Therefore, K_a = 1.0×10-4, so option C is correct.

    Test: Electrochemistry - 2 - Question 14

    The efficiency of a cell is 60%. Its cell reaction is:

    The standard electrode potential of cell is:

    Test: Electrochemistry - 2 - Question 15

    The temperature coefficient  for a given cell is 1.5 × 10–4 JK–1 at 300 K. The change in entropy during cell reaction:

    Detailed Solution: Question 15

    Test: Electrochemistry - 2 - Question 16

    For the following electrochemical cell at 298 K, 

    Pt(s) | H2 (g, 1 bar) | H+ (aq, 1 M) || M4+ (aq, M2 (aq)| Pt (s)

    The value of x is:

    Test: Electrochemistry - 2 - Question 17

    Which solution will conduct the electricity:

    Detailed Solution: Question 17

    In water magnesium chloride ionizes to form ions and these ions conduct electricity. Sugar and iodine are non electrolytes. They do not dissociate into ions in aqueous solution. Hence, they do not conduct electricity.

    Test: Electrochemistry - 2 - Question 18

    Which statement is not correct:

    Test: Electrochemistry - 2 - Question 19

    At infinite dilut ion, eachion of an electrolyte contributes a characteristic ionic conductance towards equivalent conductance of electrolyte which is independent of the nature of other ion present in solution. This statement was given by:

    Test: Electrochemistry - 2 - Question 20

    Which one of the following solutions has lowest conducting power:

    Detailed Solution: Question 20

    Test: Electrochemistry - 2 - Question 21

    The fraction of the total current carried by an ion is known as:

    Detailed Solution: Question 21

    The fraction of the total current carried by each ion is called the Transport number.

    It is also called the transference number or Hittorf number.

    Test: Electrochemistry - 2 - Question 22

    At pH = 2, E°Quinhydrone = 1.30 V, EQuinhydrone will be:

    Detailed Solution: Question 22

    Test: Electrochemistry - 2 - Question 23

    The time required to coat a metal surface of 80 cm2 with 5 × 10–3 cm thick layer of silver (density 1.05 g cm–3) with the passage of 3A current through a silver nitrate solution is:

    Detailed Solution: Question 23

    • First, calculate the volume of the silver layer: Volume = Area x Thickness = 80 cm² x (5 x 10⁻³ cm) = 0.4 cm³.
    • Next, determine the mass using the density: Mass = Volume x Density = 0.4 cm³ x 1.05 g/cm³ = 0.42 g.
    • Convert mass to moles by dividing by the atomic mass of silver (approximately 108 g/mol): Moles = 0.42 g / 108 g/mol ≈ 0.00389 mol.
    • According to Faraday's laws, the total charge required is given by Charge = Moles x Faraday's Constant (≈ 96500 coulombs/mol) = 0.00389 mol x 96500 C/mol ≈ 375 C.
    • Finally, the time required is Time = Charge / Current = 375 C / 3 A = 125 sec.

    Test: Electrochemistry - 2 - Question 24

    A hydrogen electrode placed in a buffer solution of CH3COONa and acetic acid in the ratio’s x : y and y : x has electrode potential values E1 volts and E2 volts respectively at 25°C. The pKa values of acetic acid is (E1 and E2 are oxidation potential)

    Test: Electrochemistry - 2 - Question 25

    Salts of A (atomic weight 7), B (atomic weight 27) and C (atomic weight 48) were electrolyzed under identical condition using the same quantity of electricity. It was found that when 2.1 g of A was deposited, the weights of B and C deposited were 2.7g and 7.2 g. The valencies of A, B and C respectively:

    Detailed Solution: Question 25

    According to faraday's law:
    W = ZQ = EQ/96500
    For A:
    2.1=(7/x)Q/96500
    For B:
    2.7=(27/y)Q/96500
    For C:
    7.2=(48/z)Q/96500
    x: y:z ::(7/2.1):(27/2.7):(48/7.2)

    = 3.33:10:6.66

    =1:3:2
    by solving these equations,
    x=1, y=3, z=2

    Test: Electrochemistry - 2 - Question 26

    The conductance of a solution of an electrolyte is same as that of its conductivity. The cell used can be said to have cell constant equal to:

    Test: Electrochemistry - 2 - Question 27

    The value of molar conductance of HCl is greater that of NaCl at particular temperature and dilution because:

    Test: Electrochemistry - 2 - Question 28

    The factors, which influence the conductance of solution are:

    Detailed Solution: Question 28

    Test: Electrochemistry - 2 - Question 29

    Which ion has exceptionally higher λ values:

    Test: Electrochemistry - 2 - Question 30

    The reduction potential of a half-reaction is E=+0.34 V. If the concentration of the reducing agent is increased by a factor of 10, what is the new reduction potential at 25°C?

    Detailed Solution: Question 30

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