Question:

How much electricity in terms of Faraday is required to produce: (i) 20 g of Ca from molten CaCl² (ii) 40 g of Al from molten Al2​O₃ [Given : Molar mass of calcium & Aluminium are 40 g mol^-1 & 27 g mol^-1 respectively]

Answer:

(i) 20 g of Ca from molten CaCl₂

Step 1: Calculate the moles of Ca:

Moles of Ca = 20 g / 40 g mol^-1

Moles of Ca = 0.5 mol

Step 2: Calculate the Faraday required:

Faraday required = 0.5 mol × 96,485 C mol^-1

Faraday required = 48,243 C

(ii) 40 g of Al from molten Al2O3

Step 1: Calculate the moles of Al:

Moles of Al = 40 g / 27 g mol^-1

Moles of Al = 1.48 mol

Step 2: Calculate the Faraday required:

Faraday required = 1.48 mol × 96,485 C mol^-1

Faraday required = 143,854 C

Question:

The resistance of a conductivity cell containing 0.001M KCl solution at 298 K is 1500Ω. What is the cell constant if conductivity of 0.001M KCl solution at 298 K is 0.146×10^-3S cm^-1?

Answer:

Step 1: Calculate the specific conductance of 0.001M KCl solution at 298 K. Specific conductance = Conductivity/Concentration = 0.146×10^-3S cm^-1/0.001M = 146 S cm²/m

Step 2: Calculate the cell constant. Cell constant = Resistance/Specific conductance = 1500Ω/146 S cm²/m = 10.27 Ω-1cm²/m

Question:

Three electrolytic cells A,B,C containing solutions of ZnSO₄,AgNO₃ and CuSO₃, respectively are connected in series. A steady current of 1.5 amperes was passed through them until 1.45 g of silver deposited at the cathode of cell B. How long did the current flow? What mass of copper and zinc were deposited?

Answer:

1. Calculate the moles of silver deposited at the cathode of cell B: 1.45 g of AgNO3 = 1.45 g/169.87 g/mol = 0.0085 mol

2. Calculate the total charge passed through the cells: 1.5 amperes x 1 hour = 1.5 Coulombs

3. Calculate the time the current flowed: 1.5 Coulombs/0.0085 mol = 176.47 hours

4. Calculate the mass of zinc deposited: 0.0085 mol of ZnSO4 x 65.38 g/mol = 0.55 g

5. Calculate the mass of copper deposited: 0.0085 mol of CuSO3 x 159.61 g/mol = 1.35 g

Question:

Depict the galvanic cell in which the reaction Zn(s)+2Ag⁺(aq)→Zn²⁺(aq)+2Ag(s) takes place. Further show: (i) Which of the electrode is negatively charged? (ii) The carriers of the current in the cell. (iii) Individual reaction at each electrode.

Answer:

Answer:

Depiction of Galvanic Cell:

(i) The electrode at which the reaction Zn(s)→Zn²⁺(aq) takes place is negatively charged.

(ii) The carriers of current in the cell are electrons.

(iii) Individual reactions at each electrode:

At the Anode (Negatively Charged Electrode): Zn(s)→Zn²⁺(aq) + 2^e-

At the Cathode (Positively Charged Electrode): 2Ag⁺(aq) + 2^e- → 2Ag(s)

Question:

The conductivity of sodium chloride at 298 K has been determined at different concentrations and the results are given below: Concentration/M 0.001 0.010 0.020 0.050 0.100 10²k/Sm^-1 1.237 11.85 23.15 55.53 106.74 Calculate ∧m​ for all concentrations and draw a plot between ∧m​ and c^1/2. Find the value of ∧m⁰.

Answer:

Step 1: Calculate ∧m​ for all concentrations using the formula ∧m​ = √(c × 10²k/Sm^-1).

Step 2: Plot a graph between ∧m​ and c^1/2.

Step 3: Calculate the value of ∧m⁰ by drawing a line of best fit through the graph and extrapolating to c^1/2=0.

The value of ∧m⁰ can be calculated from the graph.

Question:

The conductivity of 0.20 M solution of KCl at 298 K is 0.0248 S cm^-1. Calculate its molar conductivity.

Answer:

1. Convert the given temperature from Kelvin to Celsius: 298 K = 25°C

2. Calculate the molar concentration of KCl: 0.20 M

3. Calculate the molar conductivity: 0.0248 S cm^-1 x (1000 cm/m) x (1 mol/1000 mmol) = 0.248 mS cm² mol^-1

4. Calculate the molar conductivity: 0.248 mS cm² mol^-1 x (1 mol/1000 mmol) = 0.000248 mS cm² mmol^-1

Question:

Write the Nernst equation and emf of the following cells at 298 K: (i) Mg(s)∣Mg⁺2(0.001 M)∥Cu⁺2(0.0001 M)∣Cu(s). (ii) Fe(s)∣Fe⁺2(0.001 M)∥H+(1M)∣H₂​(g)(1bar)∣Pt(s) (iii) Sn(s)∣Sn²⁺(0.050 M)∥H+(0.020 M)∣H₂​(g)(1bar)∣Pt(s) (iv) Pt(s)∣Br2​(l)∣Br^-(0.010 M)∥H+(0.030 M)∣H₂​(g)(1bar)∣Pt(s).

Answer:

(i) Nernst equation: E = E° - (2.303RT/nF)lnQ emf = 0.0591 V

(ii) Nernst equation: E = E° - (2.303RT/nF)lnQ emf = 0.0248 V

(iii) Nernst equation: E = E° - (2.303RT/nF)lnQ emf = -0.0554 V

(iv) Nernst equation: E = E° - (2.303RT/nF)lnQ emf = 0.0417 V

Question:

Calculate the standard cell potentials of galvanic cell in which the following reactions take place: (i) 2Cr(s)+3Cd²⁺(aq)→2Cr³⁺(aq)+3Cd (ii) Fe²⁺(aq)+Ag⁺(aq)→Fe³⁺(aq)+Ag(s) Calculate the Δr​G^⊖ and equilibrium constant of the reactions

Answer:

Step 1: Calculate the standard cell potentials of the galvanic cell by using the equation E°cell = E°cathode - E°anode.

Step 2: Determine the standard reduction potentials (E°) of each reaction by using the standard reduction potential table.

Step 3: Calculate the Δr​G^⊖ of each reaction by using the equation Δr​G^⊖ = -nFE°, where n is the number of moles of electrons transferred in the reaction and F is the Faraday constant (F = 96,485 C/mol).

Step 4: Calculate the equilibrium constant of each reaction by using the equation K = ^e-Δr​G^⊖/RT, where R is the gas constant (R = 8.314 J/mol K) and T is the temperature in Kelvin.

Question:

How much charge is required for the following reductions? (i) 1 mol of Al³⁺ to Al (ii) 1 mol of Cu²⁺ to Cu (iii) 1 mol of MnO4​^- to Mn²⁺

Answer:

(i) 1 mol of Al³⁺ to Al requires 3 moles of electrons, or 3 Faradays of charge.

(ii) 1 mol of Cu²⁺ to Cu requires 2 moles of electrons, or 2 Faradays of charge.

(iii) 1 mol of MnO₄^- to Mn²⁺ requires 5 moles of electrons, or 5 Faradays of charge.

Question:

Conductivity of 0.00241 M acetic acid is 7.896×10^-5Scm^-1. Calculate its molar conductivity. If ∧m0​ for acetic acid is 390.5 S cm²mol^-1, what is its dissociation constant?

Answer:

1. Molar conductivity (Λm) = Conductivity (Λ) / Concentration (c)

Λm = 7.896×10^-5Scm^-1 / 0.00241 M Λm = 3.28×10^-3 S cm² mol^-1

2. Dissociation constant (K) = Λm / Λm0

K = 3.28×10^-3 S cm² mol^-1 / 390.5 S cm² mol^-1 K = 8.38×10^-5

Question:

In the button cells widely used in watches and other devices the following reaction takes place: Zn(s)+Ag₂O(s)+H₂​O(l)→Zn²⁺(aq)+2Ag(s)+2OH^-(aq) Determine Δr​G^θ and Eθ for the reaction.

Answer:

1. Calculate the standard enthalpy change for the reaction, ΔH^θ: ΔH^θ = [2(-109.3 kJ/mol)] + (-285.8 kJ/mol) - [0 kJ/mol] ΔH^θ = -494.1 kJ/mol

2. Calculate the standard entropy change for the reaction, ΔS^θ: ΔS^θ = [2(87.2 J/K mol)] + (70.1 J/K mol) - [0 J/K mol] ΔS^θ = 174.3 J/K mol

3. Calculate the standard Gibbs free energy change for the reaction, ΔG^θ: ΔG^θ = ΔH^θ - TΔS^θ ΔG^θ = -494.1 kJ/mol - (298 K)(174.3 J/K mol) ΔG^θ = -836.9 kJ/mol

4. Calculate the standard electrode potential for the reaction, Eθ: Eθ = -[ΔG^θ/(2F)] Eθ = -[-836.9 kJ/mol/(2)(96485 C/mol)] Eθ = +1.07 V

Question:

Define conductivity and molar conductivity for the solution of an electrolyte. Discuss their variation with concentration.

Answer:

Conductivity: Conductivity is a measure of the ability of a material to conduct electricity. It is usually expressed as the reciprocal of electrical resistivity. Conductivity is measured in Siemens per meter (S/m).

Molar Conductivity: Molar conductivity is the conductivity of a solution divided by its molar concentration. It is usually expressed in Siemens per molar (S/M).

Variation with Concentration: The conductivity and molar conductivity of an electrolyte solution both increase with increasing concentration. This is because the higher concentration of ions in the solution increases the number of ions available to carry current, resulting in a higher conductivity. The molar conductivity increases more rapidly than the conductivity because the increase in the number of ions is greater than the increase in concentration.

Question:

Arrange the following metals in the order in which they displace each other from the solution of their salts. Al,Cu,Fe,Mg and Zn

Answer:

Answer:

1. Mg
2. Fe
3. Al
4. Zn
5. Cu

Question:

How much electricity is required in coulomb for the oxidation of: (i) 1 mol of H₂​O to O₂​? (ii) 1 mol of FeO to Fe₂​O₃?

Answer:

(i) 1 mol of H₂O to O₂

Step 1: Determine the number of moles of electrons involved in the reaction.

The reaction is 2H₂O (l) → 2H₂ (g) + O₂ (g). This reaction involves 4 moles of electrons.

Step 2: Calculate the number of coulombs of electricity required.

1 mole of electrons is equal to 96,500 coulombs of electricity. Therefore, 4 moles of electrons is equal to 386,000 coulombs of electricity.

Therefore, the amount of electricity required for the oxidation of 1 mol of H₂O to O₂ is 386,000 coulombs.

(ii) 1 mol of FeO to Fe₂O3

Step 1: Determine the number of moles of electrons involved in the reaction.

The reaction is FeO (s) → Fe₂O3 (s). This reaction involves 8 moles of electrons.

Step 2: Calculate the number of coulombs of electricity required.

1 mole of electrons is equal to 96,500 coulombs of electricity. Therefore, 8 moles of electrons is equal to 772,000 coulombs of electricity.

Therefore, the amount of electricity required for the oxidation of 1 mol of FeO to Fe₂O3 is 772,000 coulombs.

Question:

Using the standard electrode potentials given in Table 3.1, predict if the reaction between the following is feasible: (i) Fe³⁺(aq) and I^-(aq) (ii) Ag⁺(aq) and Cu(s) (iii) Fe³⁺(aq) and Br^-(aq) (iv) Ag(s) and Fe³⁺(aq) (v) Br2​(aq) and Fe²⁺(aq)

Answer:

(i) Fe³⁺(aq) + I^-(aq) → Fe²⁺(aq) + I₂(aq) Fe³⁺/Fe₂+ E° = -0.77V I−/I₂ E° = +0.54V The reaction is feasible because the standard electrode potential for the reaction is positive (+0.77V).

(ii) Ag⁺(aq) + Cu(s) → Ag(s) + Cu²⁺(aq) Ag+/Ag E° = +0.80V Cu²⁺/Cu E° = +0.34V The reaction is feasible because the standard electrode potential for the reaction is positive (+1.14V).

(iii) Fe³⁺(aq) + Br^-(aq) → Fe²⁺(aq) + Br2(aq) Fe³⁺/Fe₂+ E° = -0.77V Br−/Br2 E° = +1.09V The reaction is feasible because the standard electrode potential for the reaction is positive (+0.32V).

(iv) Ag(s) + Fe³⁺(aq) → Ag⁺(aq) + Fe(s) Ag+/Ag E° = +0.80V Fe³⁺/Fe E° = +0.77V The reaction is feasible because the standard electrode potential for the reaction is positive (+1.57V).

(v) Br2​(aq) + Fe²⁺(aq) → Br−(aq) + Fe³⁺(aq) Br2/Br− E° = +1.09V Fe₂+/Fe³⁺ E° = -0.77V The reaction is feasible because the standard electrode potential for the reaction is positive (+1.86V).

Question:

Predict the products of electrolysis in each of the following: (i) An aqueous solution of AgNO₃ with silver electrodes. (ii) An aqueous solution of AgNO₃ with platinum electrodes. (iii) A dilute solution of H₂​SO₄ with platinum electrodes. (iv) An aqueous solution of CuCl² with platinum electrodes

Answer:

(i) An aqueous solution of AgNO₃ with silver electrodes: At the silver electrode (cathode): Reduction of silver ions to silver metal, Ag+ (aq) + ^e- → Ag (s) At the silver electrode (anode): Oxidation of water to oxygen gas and hydrogen ions, 2H₂O (l) → O₂ (g) + 4H⁺ (aq) + 4^e-

(ii) An aqueous solution of AgNO₃ with platinum electrodes: At the platinum electrode (cathode): Reduction of silver ions to silver metal, Ag+ (aq) + ^e- → Ag (s) At the platinum electrode (anode): Oxidation of water to oxygen gas and hydrogen ions, 2H₂O (l) → O₂ (g) + 4H⁺ (aq) + 4^e-

(iii) A dilute solution of H₂​SO₄ with platinum electrodes: At the platinum electrode (cathode): Reduction of hydrogen ions to hydrogen gas, 2H+ (aq) + 2^e- → H₂ (g) At the platinum electrode (anode): Oxidation of water to oxygen gas and hydrogen ions, 2H₂O (l) → O₂ (g) + 4H⁺ (aq) + 4^e-

(iv) An aqueous solution of CuCl² with platinum electrodes: At the platinum electrode (cathode): Reduction of copper ions to copper metal, Cu²⁺ (aq) + 2^e- → Cu (s) At the platinum electrode (anode): Oxidation of chloride ions to chlorine gas, 2Cl- (aq) → Cl₂ (g) + 2^e-

Question:

Given the standard electrode potentials, K⁺/K=−2.93 V,Ag⁺/Ag=0.80 V, Hg²⁺/Hg=0.79 V, Mg²⁺/Mg=−2.37V,Cr³⁺/Cr=−0.74 V Arrange these metals in their increasing order of reducing power.

Answer:

1. Arrange the standard electrode potentials in descending order: K⁺/K = -2.93 V, Mg²⁺/Mg = -2.37 V, Cr³⁺/Cr = -0.74 V, Ag⁺/Ag = 0.80 V, Hg²⁺/Hg = 0.79 V

2. Arrange the metals in increasing order of reducing power: Mg, Cr, Ag, Hg, K