Question 1

What is the best description of the following reaction? PbO₂ (s) + H₂O (l) + 2 e^- → PbO (s) + 2 OH^- (aq)

Accepted Answer

The reaction involves the reduction of PbO₂ to PbO. The presence of electrons on the product side indicates a reduction reaction.

Question 2

The position of equilibrium for the cell Cr₂O₇^2- (aq) + 6 Fe²⁺(aq) + 14 H⁺(aq) → 2 Cr³⁺ (aq) + 6 Fe³⁺ (aq) + 7 H₂O (l) Favours the formation of products. Identify the half-cell processes that occur at the anode and cathode.

Accepted Answer

The position of equilibrium favors the formation of products, suggesting that reduction reactions occur at the cathode and oxidation reactions occur at the anode. The reduction half-reaction involves the reduction of Cr, and the oxidation half-reaction involves the oxidation of Fe. Therefore, the cathode is Cr₂O₇^2-/Cr³⁺ and the anode is Fe³⁺/Fe²⁺.

Question 3

Determine the limiting molar conductivity of an aqueous solution of magnesium chloride, MgCl₂ (aq).

Accepted Answer

The limiting molar conductivity of an electrolyte is the molar conductivity when the concentration of the electrolyte approaches zero. The molar conductivity of MgCl2 at infinite dilution is given as 25.8 mS cm^2 mol^-1. Since the question asks for the value in units of mS m^-1 mol^-1, we need to convert the units by multiplying by 10 to get 258 mS m^-1 mol^-1. Then, we need to divide by 10 to the power of 3 to get 25.8 mS m^-1 mol^-1. Therefore, the correct answer is C: 25.8 mS m^-1 mol^-1.

Question 4

The conductivity of an aqueous solution of ammonium sulphate, (NH₄)₂SO₄ (aq) is 0.316 S m^-1. Calculate the concentration of the solution.

Accepted Answer

The molar conductivity of (NH4)2SO4 is 129.8 S cm2 mol-1. The concentration of the solution is therefore:Concentration = Conductivity / Molar conductivity= 0.316 S m-1 / 129.8 S cm2 mol-1= 0.00244 mol dm-3= 0.00244 M= 2.44 mM

Question 5

The conductivity of a solution of ethanoic acid, CH₃COOH, is 5)05 × 10^-4 S m^-1. Calculate the pH of the solution.

Accepted Answer

Ethanoic acid is a weak acid, and its conductivity is related to the concentration of ions in solution. Given the conductivity, the concentration of H+ ions can be estimated, leading to a pH of approximately 4.9.

Question 6

The conductivity of a solution of methanoic acid, HCOOH, which was made by dissolving 0.921 g of acid in water and making up the volume to 1.00 dm³, was measured to be 5.05 × 10^-4 S m^-1 at a temperature of 298 K. Determine the acid dissociation constant of methanoic acid at this temperature.

Accepted Answer

The correct answer is A because the acid dissociation constant of methanoic acid at 298 K is 0.234.

Question 7

Use the data for the standard reduction potentials given in Table 17.2 to predict which of the following is the strongest oxidising agent in acidic solution: OCl^- (aq), Cr₂O₇^- (aq), H₂O₂ (aq) or MnO₄^- (aq)

Accepted Answer

The answer of Use the data for the standard reduction...

Question 8

By considering the standard reduction potentials given in Table 7.2, decide which of the following metals could be used as a sacrificial coating to protect nickel metal, Ni (s).

Accepted Answer

The metal that can be used as a sacrificial coating to protect nickel metal should have a lower standard reduction potential than nickel (-0.25V). Among the given metals, magnesium has the lowest standard reduction potential (-2.37V) compared to nickel, making it the best choice for sacrificial coating. Copper (0.34V), silver (0.80V), and tin (-0.14V) all have higher or almost equal reduction potentials to nickel and therefore would not act as good sacrificial coatings.

Question 9

Calculate the standard cell potential for the reaction 2 MnO₄^-(aq) + 5 Sn²⁺ (aq) + 16 H⁺ (aq) → 2 Mn²⁺ (aq) + 5 Sn⁴⁺ (aq) + 8 H₂O At a temperature of 298 K.

Accepted Answer

The standard cell potential is calculated using the standard reduction potentials of the half-reactions. The reduction potential for MnO₄⁻ to Mn²⁺ is +1.51 V, and for Sn⁴⁺ to Sn²⁺ is +0.15 V. The cell potential is the difference: 1.51 V - 0.15 V = +1.36 V, which rounds to +1.37 V.

Question 10

Determine the standard cell potential at 298 K for the reaction 2 OH^- (aq) + Cl₂ (g) → Cl^- (aq) + OCl^- (aq) + H₂O (l) Using the data for the standard reduction potentials in Table 17.2.

Accepted Answer

1. Identify the half-reactions: 
    a. Reduction: Cl2 (g) + 2e- → 2Cl- (aq)        E°red = +1.36 V (from Table 17.2)
    b. Oxidation: 2OH- (aq) → 1/2 O2 (g) + H2O (l) + 2e-    E°red = -0.40 V (from Table 17.2)
    
2. Add the two half-reactions to get the overall reaction. The Cl2 (g) reduces the OH- (aq) to produce Cl- (aq) and water.

2OH- (aq) + Cl2 (g) → 2Cl- (aq) + 1/2 O2 (g) + H2O (l)
  
3. Flip the oxidation half-reaction to get:

O2 (g) + 2H2O (l) + 4e- → 4OH- (aq)   E°red = +0.40 V (flip the sign)

4. Multiply the reduction by 2 and add to the oxidation:
    
    2Cl2 (g) + 4OH- (aq) → 4Cl- (aq) + O2 (g) + 2H2O (l)     ∆E°cell = +1.36 V - 2(-0.40 V) = +2.16 V 
    
5. Check if the reaction is spontaneous by comparing ∆E°cell to zero. If it's positive, it's spontaneous. If it's negative, it's not spontaneous.

∆G° = - nF∆E°cell  (where n = number of moles of electrons transferred, F = Faraday's constant = 96,485 C/mol) 
    ∆G° = - 4 mol x 96,485 C/mol x (+2.16 V) = - 8.31 x 10^5 J/mol 
    
    Because ∆G° is negative, the reaction is spontaneous.

Therefore, the standard cell potential at 298K for the reaction is +2.16V.

However, the given answer choices are all slightly different. Looking closer, we can see that the reactions in the answer choices are not balanced or written in the same way as our answer. Nonetheless, answer choice A is closest to our result (+0.47V). It's possible that there was an error in the question or answer choices.

Question 11

Determine the amount of electrical work that can be done by the system in the reaction Cu (s) + Zn²⁺ (aq) → Cu²⁺ (aq) + Zn (s) For which

Accepted Answer

The correct answer is A because the reaction is spontaneous and the standard cell potential is positive.

Question 12

Determine the standard Gibbs energy change at 298 K for the reaction Hg₂²⁺ (aq) + 2 Cl^- (aq) → Hg₂Cl₂ (s) Using the data for the standard reduction potentials in Table 17.2.

Accepted Answer

The answer of Determine the standard Gibbs energy change at...

Question 13

The standard Gibbs energy change for the reaction H⁺ (aq) + OH^- (aq) → H₂O (l) Is -80.1 kJ mol^-1 at 298 K. Determine the standard reaction potential for the cell Pt (s) | O₂ (g) | H⁺ (aq) ║ OH^- (aq) | O₂ (g) | Pt (s) At this temperature.

Accepted Answer

The standard reaction potential (E°) can be calculated using the relation ΔG° = -nFE°, where ΔG° is the standard Gibbs energy change, n is the number of moles of electrons transferred, and F is the Faraday constant (96485 C/mol). For this reaction, n = 2. Solving for E° gives E° = -ΔG°/(nF) = 80.1 kJ/mol / (2 * 96485 C/mol) = 0.83 V.

Question 14

Using the standard reduction potentials in Table 17.2 to calculate the equilibrium constant for the reaction Fe³⁺(aq) + V²⁺ (aq) → Fe²⁺(aq) + V³⁺ (aq) At a temperature of 298 K.

Accepted Answer

We can use the Nernst equation to calculate the standard cell potential and then use it to calculate the equilibrium constant.

The half-reactions are: 
FeS1U1P13 + 2e- → FeS1U1P12 (E° = -0.535 V) 
VS1U1P13+S1S1P0 + e- → VS1U1P12+S1S1P0 (E° = 0.34 V)

Adding the two gives: 
FeS1U1P13+ S1S1P0(aq) + VS1U1P12+S1S1P0 (aq) → FeS1U1P12+ S1S1P0(aq) + VS1U1P13+S1S1P0 (aq)
E°cell = -0.535 + 0.34 = -0.195 V

At 298 K, the Nernst equation is: 
Ecell = E°cell - (RT/nF)ln(Q)
where Q is the reaction quotient and n is the number of electrons transferred (in this case, 3).

The equilibrium constant is related to the reaction quotient by: 
K = exp(-nF∆G°/RT)
where ∆G° is the standard free energy change, which is related to E°cell by:
∆G° = -nFE°cell

Substituting these equations and simplifying, we get: 
K = exp((nF/RT)(E°cell))
K = exp((3 * 96485 / (8.314 * 298))(-0.195))
K = 2.6 × 10^17

Therefore, the best choice is C.

Question 15

The standard cell potential for the reaction 6 [Fe(CN)₆]^3- (aq) + 2 Cr³⁺ (aq) + 7 H₂O (aq) → Cr₂O₇^2- (aq) + 14 H⁺ + 6 [Fe(CN)₆]^4- (aq) Has a value of -0.97 V at 298 K. Predict whether the reaction, as written, will be spontaneous in the forward direction at this temperature.

Accepted Answer

A negative standard cell potential indicates that the reaction is non-spontaneous in the forward direction and spontaneous in the backward direction.

Question 16

Use the Nernst equation to calculate the cell potential for the cell Al (s) | Al³⁺ (aq, c = 0.20 mol dm^-3) ║ Cu²⁺ (aq, c = 0.10 mol dm^-3) | Cu (s) At a temperature of 298 K.

Accepted Answer

The answer of Use the Nernst equation to calculate the...

Question 17

Determine the change in standard Gibbs energy at 298 K for the reaction PbO₂ (s) + Pb (s) + 2 H₂SO₄ (aq) → 2 PbSO₄ (aq) + 2 H₂O (l) Which this basis of the lead-acid battery.

Accepted Answer

The correct answer is A because the change in standard Gibbs energy for the reaction is -394 kJ mol^-1. This can be calculated using the standard Gibbs energy of formation of the reactants and products.

Question 18

A concentration cell is formed by dipping two copper, Cu (s) electrodes into copper sulfate, CuSO₄ (aq), solutions of different concentrations, and joining the half cells by a salt bridge. Determine the ratio of the concentrations of the copper sulfate solutions necessary to create a potential difference between the electrodes of 0.10 V.

Accepted Answer

The Nernst equation relates the concentration ratio to the potential difference. For a concentration cell with a potential difference of 0.10 V, the concentration ratio is approximately 2400.

Question 19

Use the standard reduction potentials for the Ag(s)/AgCl(s) and Ag⁺(aq)/Ag(s) half cells in Table 16.2 to determine the solubility product of silver chloride, AgCl, at 298 K.

Accepted Answer

The solubility product (Ksp) of AgCl can be calculated using the Nernst equation and the standard reduction potentials for the Ag/AgCl and Ag+/Ag half-cells. The correct Ksp value is 1.6 x 10^-10.

Question 20

Determine the limiting molar conductivity of an aqueous solution of ammonium chloride, NH₄Cl (aq).

Accepted Answer

The answer of Determine the limiting molar conductivity of an...

Quiz 16: Electrochemistry

What is the best description of the following reaction? PbO₂ (s) + H₂O (l) + 2 e^- → PbO (s) + 2 OH^- (aq)

The position of equilibrium for the cell Cr₂O₇^2- (aq) + 6 Fe²⁺(aq) + 14 H⁺(aq) → 2 Cr³⁺ (aq) + 6 Fe³⁺ (aq) + 7 H₂O (l) Favours the formation of products. Identify the half-cell processes that occur at the anode and cathode.

Determine the limiting molar conductivity of an aqueous solution of magnesium chloride, MgCl₂ (aq) .

The conductivity of an aqueous solution of ammonium sulphate, (NH₄) ₂SO₄ (aq) is 0.316 S m^-1. Calculate the concentration of the solution.

The conductivity of a solution of ethanoic acid, CH₃COOH, is 5) 05 × 10^-4 S m^-1. Calculate the pH of the solution.

The conductivity of a solution of methanoic acid, HCOOH, which was made by dissolving 0.921 g of acid in water and making up the volume to 1.00 dm³, was measured to be 5.05 × 10^-4 S m^-1 at a temperature of 298 K. Determine the acid dissociation constant of methanoic acid at this temperature.

Use the data for the standard reduction potentials given in Table 17.2 to predict which of the following is the strongest oxidising agent in acidic solution: OCl^- (aq) , Cr₂O₇^- (aq) , H₂O₂ (aq) or MnO₄^- (aq)

By considering the standard reduction potentials given in Table 7.2, decide which of the following metals could be used as a sacrificial coating to protect nickel metal, Ni (s) .

Calculate the standard cell potential for the reaction 2 MnO₄^-(aq) + 5 Sn²⁺ (aq) + 16 H⁺ (aq) → 2 Mn²⁺ (aq) + 5 Sn⁴⁺ (aq) + 8 H₂O At a temperature of 298 K.

Determine the standard cell potential at 298 K for the reaction 2 OH^- (aq) + Cl₂ (g) → Cl^- (aq) + OCl^- (aq) + H₂O (l) Using the data for the standard reduction potentials in Table 17.2.

Determine the amount of electrical work that can be done by the system in the reaction Cu (s) + Zn²⁺ (aq) → Cu²⁺ (aq) + Zn (s) For which

Determine the standard Gibbs energy change at 298 K for the reaction Hg₂²⁺ (aq) + 2 Cl^- (aq) → Hg₂Cl₂ (s) Using the data for the standard reduction potentials in Table 17.2.

The standard Gibbs energy change for the reaction H⁺ (aq) + OH^- (aq) → H₂O (l) Is -80.1 kJ mol^-1 at 298 K. Determine the standard reaction potential for the cell Pt (s) | O₂ (g) | H⁺ (aq) ║ OH^- (aq) | O₂ (g) | Pt (s) At this temperature.

Using the standard reduction potentials in Table 17.2 to calculate the equilibrium constant for the reaction Fe³⁺(aq) + V²⁺ (aq) → Fe²⁺(aq) + V³⁺ (aq) At a temperature of 298 K.

The standard cell potential for the reaction 6 [Fe(CN) ₆]^3- (aq) + 2 Cr³⁺ (aq) + 7 H₂O (aq) → Cr₂O₇^2- (aq) + 14 H⁺ + 6 [Fe(CN) ₆]^4- (aq) Has a value of -0.97 V at 298 K. Predict whether the reaction, as written, will be spontaneous in the forward direction at this temperature.

Use the Nernst equation to calculate the cell potential for the cell Al (s) | Al³⁺ (aq, c = 0.20 mol dm^-3) ║ Cu²⁺ (aq, c = 0.10 mol dm^-3) | Cu (s) At a temperature of 298 K.

Determine the change in standard Gibbs energy at 298 K for the reaction PbO₂ (s) + Pb (s) + 2 H₂SO₄ (aq) → 2 PbSO₄ (aq) + 2 H₂O (l) Which this basis of the lead-acid battery.

Use the standard reduction potentials for the Ag(s) /AgCl(s) and Ag⁺(aq) /Ag(s) half cells in Table 16.2 to determine the solubility product of silver chloride, AgCl, at 298 K.

Determine the limiting molar conductivity of an aqueous solution of ammonium chloride, NH₄Cl (aq) .

Quiz 16: Electrochemistry

What is the best description of the following reaction? PbO2 (s) + H2O (l) + 2 e- → PbO (s) + 2 OH- (aq)

The position of equilibrium for the cell Cr2O72- (aq) + 6 Fe2+(aq) + 14 H+(aq) → 2 Cr3+ (aq) + 6 Fe3+ (aq) + 7 H2O (l) Favours the formation of products. Identify the half-cell processes that occur at the anode and cathode.

Determine the limiting molar conductivity of an aqueous solution of magnesium chloride, MgCl2 (aq) .

The conductivity of an aqueous solution of ammonium sulphate, (NH4) 2SO4 (aq) is 0.316 S m-1. Calculate the concentration of the solution.

The conductivity of a solution of ethanoic acid, CH3COOH, is 5) 05 × 10-4 S m-1. Calculate the pH of the solution.

The conductivity of a solution of methanoic acid, HCOOH, which was made by dissolving 0.921 g of acid in water and making up the volume to 1.00 dm3, was measured to be 5.05 × 10-4 S m-1 at a temperature of 298 K. Determine the acid dissociation constant of methanoic acid at this temperature.

Use the data for the standard reduction potentials given in Table 17.2 to predict which of the following is the strongest oxidising agent in acidic solution: OCl- (aq) , Cr2O7- (aq) , H2O2 (aq) or MnO4- (aq)

By considering the standard reduction potentials given in Table 7.2, decide which of the following metals could be used as a sacrificial coating to protect nickel metal, Ni (s) .

Calculate the standard cell potential for the reaction 2 MnO4-(aq) + 5 Sn2+ (aq) + 16 H+ (aq) → 2 Mn2+ (aq) + 5 Sn4+ (aq) + 8 H2O At a temperature of 298 K.

Determine the standard cell potential at 298 K for the reaction 2 OH- (aq) + Cl2 (g) → Cl- (aq) + OCl- (aq) + H2O (l) Using the data for the standard reduction potentials in Table 17.2.

Determine the amount of electrical work that can be done by the system in the reaction Cu (s) + Zn2+ (aq) → Cu2+ (aq) + Zn (s) For which

Determine the standard Gibbs energy change at 298 K for the reaction Hg22+ (aq) + 2 Cl- (aq) → Hg2Cl2 (s) Using the data for the standard reduction potentials in Table 17.2.

The standard Gibbs energy change for the reaction H+ (aq) + OH- (aq) → H2O (l) Is -80.1 kJ mol-1 at 298 K. Determine the standard reaction potential for the cell Pt (s) | O2 (g) | H+ (aq) ║ OH- (aq) | O2 (g) | Pt (s) At this temperature.

Using the standard reduction potentials in Table 17.2 to calculate the equilibrium constant for the reaction Fe3+ (aq) + V2+ (aq) → Fe2+ (aq) + V3+ (aq) At a temperature of 298 K.

The standard cell potential for the reaction 6 [Fe(CN) 6]3- (aq) + 2 Cr3+ (aq) + 7 H2O (aq) → Cr2O72- (aq) + 14 H+ + 6 [Fe(CN) 6]4- (aq) Has a value of -0.97 V at 298 K. Predict whether the reaction, as written, will be spontaneous in the forward direction at this temperature.

Use the Nernst equation to calculate the cell potential for the cell Al (s) | Al3+ (aq, c = 0.20 mol dm-3) ║ Cu2+ (aq, c = 0.10 mol dm-3) | Cu (s) At a temperature of 298 K.

Determine the change in standard Gibbs energy at 298 K for the reaction PbO2 (s) + Pb (s) + 2 H2SO4 (aq) → 2 PbSO4 (aq) + 2 H2O (l) Which this basis of the lead-acid battery.

Use the standard reduction potentials for the Ag(s) /AgCl(s) and Ag+(aq) /Ag(s) half cells in Table 16.2 to determine the solubility product of silver chloride, AgCl, at 298 K.

Determine the limiting molar conductivity of an aqueous solution of ammonium chloride, NH4Cl (aq) .

What is the best description of the following reaction? PbO₂ (s) + H₂O (l) + 2 e^- → PbO (s) + 2 OH^- (aq)

The position of equilibrium for the cell Cr₂O₇^2- (aq) + 6 Fe²⁺(aq) + 14 H⁺(aq) → 2 Cr³⁺ (aq) + 6 Fe³⁺ (aq) + 7 H₂O (l) Favours the formation of products. Identify the half-cell processes that occur at the anode and cathode.

Determine the limiting molar conductivity of an aqueous solution of magnesium chloride, MgCl₂ (aq) .

The conductivity of an aqueous solution of ammonium sulphate, (NH₄) ₂SO₄ (aq) is 0.316 S m^-1. Calculate the concentration of the solution.

The conductivity of a solution of ethanoic acid, CH₃COOH, is 5) 05 × 10^-4 S m^-1. Calculate the pH of the solution.

The conductivity of a solution of methanoic acid, HCOOH, which was made by dissolving 0.921 g of acid in water and making up the volume to 1.00 dm³, was measured to be 5.05 × 10^-4 S m^-1 at a temperature of 298 K. Determine the acid dissociation constant of methanoic acid at this temperature.

Use the data for the standard reduction potentials given in Table 17.2 to predict which of the following is the strongest oxidising agent in acidic solution: OCl^- (aq) , Cr₂O₇^- (aq) , H₂O₂ (aq) or MnO₄^- (aq)

Calculate the standard cell potential for the reaction 2 MnO₄^-(aq) + 5 Sn²⁺ (aq) + 16 H⁺ (aq) → 2 Mn²⁺ (aq) + 5 Sn⁴⁺ (aq) + 8 H₂O At a temperature of 298 K.

Determine the standard cell potential at 298 K for the reaction 2 OH^- (aq) + Cl₂ (g) → Cl^- (aq) + OCl^- (aq) + H₂O (l) Using the data for the standard reduction potentials in Table 17.2.

Determine the amount of electrical work that can be done by the system in the reaction Cu (s) + Zn²⁺ (aq) → Cu²⁺ (aq) + Zn (s) For which

Determine the standard Gibbs energy change at 298 K for the reaction Hg₂²⁺ (aq) + 2 Cl^- (aq) → Hg₂Cl₂ (s) Using the data for the standard reduction potentials in Table 17.2.

The standard Gibbs energy change for the reaction H⁺ (aq) + OH^- (aq) → H₂O (l) Is -80.1 kJ mol^-1 at 298 K. Determine the standard reaction potential for the cell Pt (s) | O₂ (g) | H⁺ (aq) ║ OH^- (aq) | O₂ (g) | Pt (s) At this temperature.

Using the standard reduction potentials in Table 17.2 to calculate the equilibrium constant for the reaction Fe³⁺(aq) + V²⁺ (aq) → Fe²⁺(aq) + V³⁺ (aq) At a temperature of 298 K.

The standard cell potential for the reaction 6 [Fe(CN) ₆]^3- (aq) + 2 Cr³⁺ (aq) + 7 H₂O (aq) → Cr₂O₇^2- (aq) + 14 H⁺ + 6 [Fe(CN) ₆]^4- (aq) Has a value of -0.97 V at 298 K. Predict whether the reaction, as written, will be spontaneous in the forward direction at this temperature.

Use the Nernst equation to calculate the cell potential for the cell Al (s) | Al³⁺ (aq, c = 0.20 mol dm^-3) ║ Cu²⁺ (aq, c = 0.10 mol dm^-3) | Cu (s) At a temperature of 298 K.

Determine the change in standard Gibbs energy at 298 K for the reaction PbO₂ (s) + Pb (s) + 2 H₂SO₄ (aq) → 2 PbSO₄ (aq) + 2 H₂O (l) Which this basis of the lead-acid battery.

Use the standard reduction potentials for the Ag(s) /AgCl(s) and Ag⁺(aq) /Ag(s) half cells in Table 16.2 to determine the solubility product of silver chloride, AgCl, at 298 K.

Determine the limiting molar conductivity of an aqueous solution of ammonium chloride, NH₄Cl (aq) .