Q: Given the standard electrode potentials Fe2+/Fe = −0.44 V and Fe3+/Fe2+ = +0.77 V, if a block of iron is placed in contact with Fe3+ ions, what is expected to happen?

The amount of Fe3+ falls. Iron metal reduces Fe3+: Fe + 2Fe3+ -> 3Fe2+ (E_cell = 0.77 - (-0.44) > 0, spontaneous). Therefore the amount of Fe3+ falls while Fe2+ rises.

Question 1

During electrolysis of an aqueous solution of dipotassium succinate, which gas is evolved at the electrodes?

Accepted Answer

Ethene. During the electrolysis of dipotassium succinate, the organic compound undergoes oxidation, leading to the formation of ethene as a byproduct at the anode due to the elimination of hydrogen ions and electrons.

Question 2

Which expression gives the equivalent conductance at infinite dilution of Al2(SO4)3, if Λ°Al3+ and Λ°SO4²− denote the equivalent conductances at infinite dilution of the corresponding ions?

Accepted Answer

Λ°Al3+ + Λ°SO4²−. The correct option is right because the equivalent conductance of Al2(SO4)3 at infinite dilution is the sum of the conductances of its ions, Al3+ and SO4²−. Since the compound dissociates into 2 Al3+ ions and 3 SO4²− ions, the total conductance is represented by the sum of their contributions, which simplifies to Λ°Al3+ + Λ°SO4²−.

Question 3

The standard emf of a Daniell cell is 1.10 V. What is the greatest amount of electrical work that can be extracted from this cell?

Accepted Answer

212.3 kJ. Max electrical work = -dG = nFE = 2 * 96500 * 1.10 = 212300 J = 212.3 kJ.

Question 4

During the recharging of a lead storage battery, which reaction takes place at the cathode?

Accepted Answer

Pb2+ + 2e− → Pb. During recharging (electrolysis), reduction occurs at the cathode: PbSO4 + 2e- -> Pb + SO4^2-, i.e. Pb2+ + 2e- -> Pb. The answer is the reduction of Pb2+ to Pb.

Question 5

The limiting molar ionic conductivities of the ions X2+ and Y2− are 57 and 73, respectively. What will be the molar conductivity of the electrolyte formed by these ions?

Accepted Answer

130 S cm² mol−1. The molar conductivity of an electrolyte is the sum of the limiting molar ionic conductivities of its constituent ions. In this case, adding the conductivities of X2+ (57 S cm² mol⁻¹) and Y2− (73 S cm² mol⁻¹) gives a total of 130 S cm² mol⁻¹.

Question 6

According to Kohlrausch’s law, at what condition does each ion contribute a fixed amount to the conductance of an electrolyte, irrespective of the other ion present?

Accepted Answer

At infinite dilution, each ion contributes a definite amount to the equivalent conductance of an electrolyte, regardless of the other ion present.. Kohlrausch's law states that at infinite dilution each ion contributes a definite amount to the EQUIVALENT conductance of the electrolyte, independent of (regardless of) the other ion present. Option (d) is the precise statement; option (c) wrongly says 'conductance' instead of 'equivalent conductance.'

Question 7

At 298 K, the standard free energies of formation (kJ mol⁻¹) of H2O(l), CO2(g), and pentane(g) are −237.2, −394.4, and −8.2, respectively. What is the standard cell potential, E°cell, for a pentane–oxygen fuel cell?

Accepted Answer

1.0968 V. Combustion: C5H12 + 8 O2 -> 5 CO2 + 6 H2O. delta G = [5(-394.4) + 6(-237.2)] - (-8.2) = -3387 kJ. Each O2 accepts 4 e-, so n = 32. E = 3387000/(32*96485) = 1.097 V.

Question 8

For a reaction with a negative standard cell potential (E°cell), which statement correctly describes the signs of ΔG° and the equilibrium constant Keq?

Accepted Answer

ΔG° is positive and Keq is less than 1. dG = -nFE; negative E gives positive dG. Since dG = -RT ln K > 0, ln K < 0, so Keq < 1.

Question 9

Given the standard electrode potentials Fe2+/Fe = −0.44 V and Fe3+/Fe2+ = +0.77 V, if a block of iron is placed in contact with Fe3+ ions, what is expected to happen?

Accepted Answer

The amount of Fe3+ falls. Iron metal reduces Fe3+: Fe + 2Fe3+ -> 3Fe2+ (E_cell = 0.77 - (-0.44) > 0, spontaneous). Therefore the amount of Fe3+ falls while Fe2+ rises.

Question 10

An electrolytic cell is filled with aqueous Ag2SO4 and fitted with platinum electrodes. Electrolysis is continued until 1.6 g of O2 is evolved at the anode. How much silver will be deposited at the cathode?

Accepted Answer

21.60 g. 1.6 g O2 = 0.05 mol = 0.2 mol electrons (O2 needs 4e-). Ag+ + e- -> Ag, so 0.2 mol Ag deposit = 0.2 x 107.88 = 21.6 g.

Question 11

If φ represents the reduction potential, which relation correctly gives the standard cell potential?

Accepted Answer

E°cell = φright − φleft. By convention the cell is written with the cathode (reduction) on the right and anode on the left, so E°cell = phi(right) - phi(left). The correct option is E°cell = phi(right) - phi(left).

Question 12

For the electrochemical cell reaction Cu2+(C1, aq) + Zn(s) → Zn2+(C2, aq) + Cu(s), the Gibbs free energy change, ΔG, at a fixed temperature depends on

Accepted Answer

ln(C2/C1). The Gibbs free energy change, ΔG, for an electrochemical reaction is related to the concentrations of the reactants and products through the Nernst equation, which shows that ΔG depends on the ratio of the concentrations of the products to the reactants, specifically ln(C2/C1) in this case.

Question 13

When an electric current is passed through acidified water, 112 mL of hydrogen gas is obtained at the cathode at STP in 965 s. What is the current, in amperes?

Accepted Answer

1.0. 112 mL H2 at STP = 0.112/22.4 = 0.005 mol, needing 2 x 0.005 = 0.01 mol electrons = 965 C. I = Q/t = 965/965 = 1.0 A.

Question 14

At 25°C, what is the electrode potential of a zinc electrode dipped in 0.1 M aqueous ZnSO4 solution? Given E°(Zn2+/Zn) = -0.76 V and take 2.303RT/F = 0.06 at 298 K.

Accepted Answer

-0.79. The electrode potential is calculated using the Nernst equation, which accounts for the concentration of zinc ions in the solution. Given the standard electrode potential and the concentration of ZnSO4, the calculated potential of -0.79 V reflects the shift in potential due to the concentration of ions, confirming that option B is correct.

Question 15

For the cell represented as M | M+ || X- | X, the standard reduction potentials are E°(M+/M) = 0.44 V and E°(X/X-) = 0.33 V. Which conclusion is valid from this information?

Accepted Answer

The spontaneous overall reaction is M+ + X- → M + X. As written, M is the anode and X the cathode: Ecell = E(X/X-) - E(M+/M) = 0.33 - 0.44 = -0.11 V. Since Ecell < 0, the forward reaction is non-spontaneous and the spontaneous reaction is the reverse: M+ + X- -> M + X.

Question 16

For the electrochemical cell represented as
A | A+(x M) || B+(y M) | B
the measured emf is +0.20 V. Which statement about the overall cell reaction is correct?

Accepted Answer

A + B+ → A+ + B. A positive cell emf (+0.20 V) means the reaction proceeds spontaneously as written: A is oxidized at the anode (A -> A+ + e-) and B+ is reduced at the cathode (B+ + e- -> B). Overall: A + B+ -> A+ + B.

Question 17

A conductivity cell is filled with 0.1 M KCl solution having conductivity X Ω⁻¹ cm⁻¹, and its conductance is Y Ω⁻¹. When the same cell is filled with 0.1 M NaOH solution, the conductance becomes Z Ω⁻¹. The molar conductance of NaOH is

Accepted Answer

10⁴ XZ / Y. Cell constant G* = kappa/conductance = X/Y. For NaOH, kappa_NaOH = G* * Z = XZ/Y. Molar conductance = kappa*1000/C = (XZ/Y)*1000/0.1 = 10^4 * XZ/Y.

Question 18

Using the standard reduction potentials given below, identify the species that acts as the strongest oxidizing agent:
[Fe(CN)6]4- → [Fe(CN)6]3- + e-; E° = -0.35 V
Fe2+ → Fe3+ + e-; E° = -0.77 V

Accepted Answer

Fe3+. Strongest oxidizing agent = the oxidized form with the highest standard reduction potential. Reduction potentials: Fe3+ + e- -> Fe2+ is +0.77 V; [Fe(CN)6]3- + e- -> [Fe(CN)6]4- is +0.35 V. Fe3+ has the higher value, so Fe3+ is the strongest oxidizing agent.

Question 19

Which device directly changes the chemical energy released during the combustion of fuels such as hydrogen and methane into electrical energy?

Accepted Answer

Fuel cell. A fuel cell directly converts the chemical energy from the combustion of fuels like hydrogen and methane into electrical energy through an electrochemical reaction, making it the appropriate choice.

Question 20

In an acidic solution, MnO2 acts as an oxidizing agent according to the half-reaction:
MnO2 + 4H+ + 2e− → Mn2+ + 2H2O
If the pH is lowered by 1 unit, by how much will the electrode potential of the Pt | MnO2, Mn2+ half-cell change?

Accepted Answer

0.118 V. E = E0 - (0.059/2) log([Mn2+]/[H+]^4) = E0 + (0.236/2) log[H+] = E0 - 0.118*pH. Lowering pH by 1 unit increases the electrode potential by 0.118 V.

Question 21

For the electrochemical reaction
2Fe(s) + O2(g) + 4H+(aq) → 2Fe2+(aq) + 2H2O(l), with E° = 1.67 V,
what is the cell potential at 25°C when [Fe2+] = 10−3 M, p(O2) = 0.1 atm, and pH = 3?

Accepted Answer

1.57 V. Q = [Fe2+]^2 / (pO2*[H+]^4) = (1e-3)^2 / (0.1 * (1e-3)^4) = 1e-6/1e-13 = 1e7, log Q = 7. E = 1.67 - (0.0591/4)(7) = 1.67 - 0.103 = 1.57 V.

Question 22

In a hydrogen–oxygen fuel cell, the oxidation of hydrogen takes place to

Accepted Answer

set up a voltage across the two electrodes. In a H2-O2 fuel cell, H2 is oxidized at the anode releasing electrons; this electrochemical reaction sets up a voltage (potential difference) across the two electrodes, which drives current through the external circuit.

Question 23

For an electrochemical cell, consider these expressions for the cell emf:
(i) emf = oxidation potential of the anode − reduction potential of the cathode
(ii) emf = oxidation potential of the anode + reduction potential of the cathode
(iii) emf = reduction potential of the anode + reductio

Accepted Answer

(ii) and (iv). The correct options are valid because the cell emf is calculated by adding the oxidation potential of the anode to the reduction potential of the cathode, which aligns with option (ii). Additionally, option (iv) correctly states that the emf can also be expressed as the difference between the oxidation potentials of the anode and cathode.

Question 24

A hydrogen electrode is prepared by immersing a platinum wire in an HCl solution of pH 10 and bubbling hydrogen gas over the platinum at a pressure of 1 atm. What is the oxidation potential of this electrode?

Accepted Answer

0.59 V. For 2H+ + 2e- -> H2, E_red = -(0.059/2)*log(1/[H+]^2) = -0.059*pH = -0.59 V at pH 10. Oxidation potential = -E_red = +0.59 V.

Question 25

At 298 K, an electrode initially contains a 1 M K2Cr2O7 solution in an acidic buffer of pH 1.0. It is then supplied with 50% of the Sn required to completely convert all Cr2O7²− into Cr³+. If the pH stays unchanged, what is the electrode reduction potential? Given: E°(Cr2O7²−/Cr³+, H+) = 1

Accepted Answer

1.187 V. The correct option is derived from the Nernst equation, which accounts for the concentration of Cr2O7²− and the stoichiometry of the reaction. Since 50% of the required Sn is added, the concentration of Cr2O7²− is effectively halved, leading to a calculated reduction potential of 1.187 V.

Question 26

The electrode potential of M2+/M of 3d-series elements shows positive value of

Accepted Answer

Cu. Copper (Cu) has a higher electrode potential compared to other 3d-series elements due to its stable electron configuration and the strong tendency to gain electrons, making it more favorable for reduction.

Question 27

In a hydrogen–oxygen fuel cell, the oxidation of hydrogen takes place so as to

Accepted Answer

set up a potential difference across the two electrodes. The oxidation of hydrogen in a fuel cell generates electrons, which flow through an external circuit, creating a potential difference between the electrodes. This flow of electrons is what produces electrical energy.

Question 28

Given the standard reduction potentials E°(Fe3+/Fe2+) = +0.77 V and E°(Sn2+/Sn) = -0.14 V, what is the standard cell potential for the reaction Sn(s) + 2Fe3+(aq) → 2Fe2+(aq) + Sn2+(aq)?

Accepted Answer

0.91 V. The standard cell potential is calculated by subtracting the reduction potential of the anode from that of the cathode. Here, iron is reduced (E° = +0.77 V) and tin is oxidized (E° = -0.14 V), leading to a cell potential of 0.77 V - (-0.14 V) = 0.91 V.

Question 29

A cell reaction involving transfer of a single electron has a standard emf of 0.591 V at 25°C. What is the equilibrium constant for this reaction? (F = 96,500 C mol−1; R = 8.314 J K−1 mol−1)

Accepted Answer

1.0 × 10¹⁰. The equilibrium constant can be calculated using the Nernst equation, which relates the standard emf to the equilibrium constant. A standard emf of 0.591 V indicates a strong driving force for the reaction, leading to a large equilibrium constant, specifically 1.0 × 10¹⁰, indicating that the products are favored at equilibrium.

Question 30

The limiting molar conductivities (Λ°) of NaCl, KBr, and KCl are 126, 152, and 150 S cm² mol−1, respectively. What is the limiting molar conductivity of NaBr?

Accepted Answer

128 S cm² mol−1. The limiting molar conductivity of NaBr can be estimated using the known values of the limiting molar conductivities of NaCl and KBr, applying the principle of additivity for strong electrolytes. By calculating the difference between the conductivities of KBr and KCl and adjusting for the contribution of Na+, we arrive at the value of 128 S cm² mol⁻¹ for NaBr.

Question 31

For an electrochemical cell based on the reaction Zn(s) + 2H+(aq) → Zn2+(aq) + H2(g), what happens if H2SO4 is added to the cathode compartment?

Accepted Answer

The cell emf rises and the equilibrium moves toward products. Adding H2SO4 increases the concentration of H+ ions in the cathode compartment, which drives the reduction of H+ to H2, thus increasing the cell's emf and shifting the equilibrium toward the products.

Question 32

At 25°C, the limiting molar conductances in water are given below for these electrolytes: KCl = 149.9 S cm² mol−1, KNO3 = 145 S cm² mol−1, HCl = 426.2 S cm² mol−1, NaOAc = 91 S cm² mol−1, and NaCl = 126.5 S cm² mol−1. Using these values, determine the limiting molar conductance of HOAc in

Accepted Answer

390.7. The limiting molar conductance of acetic acid (HOAc) can be calculated using the known conductances of its ions and the principle of additivity for strong electrolytes. By combining the contributions from acetate (NaOAc) and the dissociated protons from HCl, we arrive at the correct value of 390.7 S cm² mol⁻¹.

Question 33

At 25°C in water, the limiting molar conductivities of sodium acetate (Λ° NaOAc) and hydrochloric acid (Λ° HCl) are 91.0 and 426.2 S cm² mol⁻¹, respectively. To determine the limiting molar conductivity of acetic acid (Λ° H OAc), which additional quantity is needed?

Accepted Answer

Λ° NaCl. To determine the limiting molar conductivity of acetic acid, we need to know the conductivity of its ions in solution. Sodium acetate dissociates into sodium ions and acetate ions, while hydrochloric acid dissociates into hydrogen ions and chloride ions. By knowing the limiting molar conductivity of sodium chloride, we can calculate the contribution of the chloride ions, allowing us to isolate the conductivity of acetic acid.

Question 34

Given the data at 25 °C
Ag + I⁻ → AgI + e⁻ E° = 0.152 V
Ag → Ag⁺ + e⁻ E° = -0.800 V
What is the value of log Ksp for AgI? (2.303 RT/F = 0.059 V)

Accepted Answer

-16.13. AgI -> Ag+ + I- combines (AgI + e- -> Ag + I-, E = -0.152 V) with (Ag -> Ag+ + e-, E = -0.800 V), giving E(cell) = -0.952 V. log Ksp = nE/0.059 = (1)(-0.952)/0.059 = -16.13.

Question 35

The cell,
Zn | Zn²⁺ (1 M) || Cu²⁺ (1 M) | Cu (E°cell = 1.10 V)
was allowed to be completely discharged at 298 K. The relative concentration of Zn²⁺ to Cu²⁺ ([Zn²⁺]/[Cu²⁺]) is

Accepted Answer

10³⁷.3. The correct option, 10³⁷.3, represents the ratio of the concentrations of Zn²⁺ to Cu²⁺ at equilibrium, derived from the Nernst equation and the standard cell potential. This value indicates a significantly higher concentration of Zn²⁺ compared to Cu²⁺ after the cell is fully discharged, reflecting the stoichiometry of the redox reaction.

Question 36

Given:
E°Fe³⁺/Fe = -0.036 V
E°Fe²⁺/Fe = -0.439 V
The value of standard electrode potential for the change, Fe³⁺(aq) + e⁻ → Fe²⁺(aq) will be:

Accepted Answer

0.770 V. For Fe3+/Fe2+: n*E = 3*E(Fe3+/Fe) - 2*E(Fe2+/Fe) = 3(-0.036) - 2(-0.439) = -0.108 + 0.878 = 0.770. With n = 1, E(Fe3+/Fe2+) = 0.770 V.

Question 37

A 0.2 M solution of an electrolyte has a resistance of 50 Ω. Its specific conductance is 1.3 S m⁻¹. For a 0.4 M solution of the same electrolyte, the resistance is 260 Ω. The molar conductivity of this solution is

Accepted Answer

6.25 × 10⁻⁴ S m² mol⁻¹. Cell constant = kappa x R = 1.3 x 50 = 65 m^-1. For 0.4 M: kappa = 65/260 = 0.25 S m^-1. With c = 400 mol m^-3, Lambda_m = kappa/c = 0.25/400 = 6.25e-4 S m^2 mol^-1.

Question 38

The standard reduction potentials for Zn²⁺/Zn, Ni²⁺/Ni and Fe²⁺/Fe are -0.76, -0.23 and -0.44 V respectively.
The reaction X + Y²⁺ → X²⁺ + Y will be spontaneous when:

Accepted Answer

X = Zn, Y = Ni. For X + Y2+ -> X2+ + Y to be spontaneous, X must be more easily oxidized, i.e. have a more negative reduction potential than Y. With Zn (-0.76) and Ni (-0.23), X = Zn, Y = Ni gives a positive cell potential and is spontaneous.

Question 39

During recharging of the anodic half-cell of a lead–acid battery, if 0.05 Faraday of electricity is passed, how many grams of PbSO4 are decomposed? (Molar mass of PbSO4 = 303 g mol−1)

Accepted Answer

7.6. The decomposition of PbSO4 in a lead-acid battery involves the transfer of 2 moles of electrons per mole of PbSO4. Since 0.05 Faraday corresponds to 0.05 moles of electrons, this means 0.025 moles of PbSO4 are decomposed. Multiplying 0.025 moles by the molar mass of PbSO4 (303 g/mol) gives approximately 7.6 grams.

Question 40

For the cell reaction below, what is the standard Gibbs free energy change at 298 K?

Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)

Given that E° = 2.0 V at 298 K and Faraday’s constant F = 96000 C mol−1, find ΔG° in kJ mol−1.

Accepted Answer

−384. The standard Gibbs free energy change (ΔG°) is calculated using the equation ΔG° = -nFE°, where n is the number of moles of electrons transferred in the reaction. In this case, 2 moles of electrons are transferred, resulting in ΔG° = -2 * 96000 C/mol * 2.0 V = -384 kJ/mol, confirming that option A is correct.

Question 41

The standard reduction potentials for the couples Cu2+/Cu and Cu+/Cu are 0.34 V and 0.522 V, respectively. What is the standard potential for the Cu2+/Cu+ couple?

Accepted Answer

+0.158 V. The standard potential for the Cu2+/Cu+ couple can be calculated using the difference between the standard reduction potentials of Cu+/Cu and Cu2+/Cu. By subtracting the potential of Cu2+/Cu from that of Cu+/Cu, we find that the standard potential for the Cu2+/Cu+ couple is +0.158 V.

Question 42

In the electrolytic purification of copper, certain impurity metals do not dissolve and collect at the bottom as anode sludge. Which of the following metals are found in that sludge?

Accepted Answer

Ag and Au. Silver and gold are less reactive and do not dissolve in the electrolytic process, causing them to accumulate as anode sludge during copper purification.

Question 43

Which metal cannot be isolated by electrolyzing an aqueous solution of its salts?

Accepted Answer

Calcium. Calcium is a highly reactive metal that readily reacts with water, making it impossible to isolate through electrolysis of its aqueous salts, as it would react with the water instead of being deposited.

Question 44

The correct order of E° M2+/M values with negative sign for the four successive elements Cr, Mn, Fe and Co is

Accepted Answer

Mn > Cr > Fe > Co. The standard reduction potentials (E°) for these metals indicate their ability to gain electrons, with manganese having the highest tendency to be reduced, followed by chromium, iron, and cobalt. This order reflects their relative reactivity and stability in their ionic forms.

Question 45

The metal that cannot be obtained by electrolysis of an aqueous solution of its salts is

Accepted Answer

Ca. Calcium cannot be obtained by electrolysis of its aqueous salts because it is more reactive than hydrogen, which means it will not displace hydrogen from water to form calcium during the electrolysis process.

Question 46

The equivalent conductance of NaCl at concentration C and at infinite dilution are λc and λ∞ respectively. The correct relationship between λx and λ∞ is given as: (where the constant B is positive)

Accepted Answer

λc = λ∞ − (B)√C. The correct option reflects the principle that as the concentration of an electrolyte like NaCl increases, its equivalent conductance decreases due to ion interactions and reduced mobility, which is captured by the relationship λc = λ∞ − (B)√C, where B is a positive constant.

Question 47

Resistance of 0.2M solution of an electrolyte is 50Ω. The specific conductance of the solution is 1.4 S m−1. The resistance of 0.5M solution of the same electrolyte is 280Ω. The molar conductivity of 0.5M solution of the electrolyte in S m2 mol−1 is:

Accepted Answer

5 × 10−4. The molar conductivity is calculated using the formula that relates resistance, specific conductance, and concentration. Given the resistance and specific conductance of the 0.5M solution, the molar conductivity can be derived, confirming that the correct answer is 5 × 10−4 S m² mol−1.

Question 48

When 9.65 ampere current was passed for 1.0 hour into nitrobenzene in acidic medium, the amount of p-amino-phenol produced is....... [JEE-Main On line-2018]

Accepted Answer

9.81 g. The correct option is derived from Faraday's laws of electrolysis, where the amount of substance produced is directly proportional to the electric charge passed through the solution. Given the current and time, the total charge can be calculated, and using the molar mass of p-amino-phenol, the amount produced can be determined, resulting in 9.81 g.

Question 49

When an electric current is passes through acidified water, 112 mL of hydrogen gas at N.T.P. was collected at the cathode in 965 seconds. The current passed, in ampere, is:

Accepted Answer

1.0. The amount of hydrogen gas produced during electrolysis is directly related to the current and time, as described by Faraday's laws of electrolysis. Given that 112 mL of hydrogen gas corresponds to 0.005 moles, and using the formula I = nF/t, where F is Faraday's constant, the calculated current is 1.0 ampere.

Question 50

How long (approximate) should water be electrolysed by passing through 100 amperes current so that the oxygen released can completely burn 27.66 g of diborane ? (Atomic weight of B = 10.8 u)

Accepted Answer

3.2 hours. The correct option is 3.2 hours because the amount of oxygen required to completely burn 27.66 g of diborane can be calculated using stoichiometry, and the electrolysis process at 100 amperes produces the necessary oxygen in that time frame.

JEE Main Chemistry: Electrochemistry questions with solutions

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