Question 1

Which statement below is not true?

Accepted Answer

Modern fuel cells cannot be easily regenerated using household current. They require specific regeneration methods and procedures.

Question 2

A cell based on the reaction below has a standard potential of +0.42 V at 25°C.If all of the species are at standard conditions except [H⁺],at what pH will the cell have a potential of zero? H₂O₂(aq)+ 2 H⁺(aq)+ 2 Cl^-(aq)→ Cl₂(aq)+ 2 H₂O(l)

Accepted Answer

To find the pH at which the cell potential is zero, we need to find the concentration of [H⁺]/[H₂O] that will make the cell potential equal to zero. At standard conditions, the Nernst equation can be written as: E° = 0.42 V = (0.0592 V/pH) log([H⁺]/[H₂O]) Solving for pH, we get: pH = 7.09

Question 3

The cell reaction for a lead storage battery is: Pb(s)+ PbO₂(s)+ 2 H⁺(aq)+ 2 HSO₄^-(aq)→ 2 PbSO₄(s)+ 2 H₂O(l) E° = +1.92 V To provide a potential of about 12 V,one could

Accepted Answer

The cell reaction for a lead storage battery has a standard potential of +1.92 V. To provide a potential of about 12 V, six cells need to be connected in series. Each cell contributes 2 V to the overall potential, thus six cells in series can provide a total potential of 12 V. Adjusting the pH to 12 or 1 would significantly affect the equilibrium of the reaction, and increasing the surface area of the electrodes would not change the potential of the reaction. Therefore, connecting six cells in series is the best choice to provide a potential of about 12 V.

Question 4

Which of the following reactions is most suitable for a fuel cell?

Accepted Answer

Fuel cells typically involve reactions between a fuel and an oxidant to produce electricity. The reaction in option D involves the combustion of carbon monoxide with oxygen to form carbon dioxide, which is a typical fuel cell reaction.

Question 5

Ag⁺(aq)+ e^- → Ag(s)E° = +0.800 V AgBr(s)+ e^- → Ag(s)+ Br^-(aq)E° = +0.071 V Br₂(l)+ 2 e^- → 2 Br^-(aq)E° = +1.066 V Use some of the data above to calculate K_sp at 25°C for AgBr.

Accepted Answer

To calculate the solubility product constant (Ksp) for AgBr at 25°C, we use the Nernst equation and the given standard electrode potentials (E°). The overall reaction for the dissolution of AgBr in water can be obtained by combining the half-reactions for AgBr → Ag⁺ + Br⁻ and Br₂ + 2e⁻ → 2Br⁻. The net cell potential (E°cell) for the dissolution of AgBr is the difference between the reduction potential for the formation of silver from AgBr (E° = +0.071 V) and the reduction potential for bromine to bromide (E° = +1.066 V). Thus, E°cell = 0.071 V - (-1.066 V) = 1.137 V. The relationship between E°cell, Ksp, and the number of electrons transferred (n) in the balanced equation is given by the Nernst equation: E°cell = (RT/nF)ln(Ksp), where R is the gas constant, T is the temperature in Kelvin, n is the number of moles of electrons transferred, and F is the Faraday constant. Solving for Ksp gives a value that corresponds to the choice provided.

Question 6

Which is most often used in the laboratory to measure pH?

Accepted Answer

A glass electrode is most commonly used in the laboratory to measure pH. It consists of a thin-walled glass bulb filled with a buffered solution and a reference electrode. When the bulb is immersed in an aqueous solution, the pH of the solution can be measured based on the electrical potential of the electrode. The other options listed (standard hydrogen electrode, Daniell cell, conductivity cell) are not typically used to measure pH.

Question 7

Which of the following statements concerning a lithium battery is false?

Accepted Answer

This statement is false. Lithium batteries do not produce toxic mercury and can be safely disposed of without recycling. In fact, many recycling programs do not accept lithium batteries because they are considered non-toxic.

Question 8

How many moles of electrons,n,are transferred in the following reduction-oxidation reaction? 2 MnO₄^-(aq)+ 16 H⁺(aq)+ 10 Cl^-(aq)→ 2 Mn²⁺(aq)+ 5 Cl₂(g)+ 8 H₂O(l)

Accepted Answer

In the given redox reaction, the total number of electrons transferred can be determined by balancing the reaction in terms of electron transfer. The Mn goes from a +7 oxidation state in MnO4- to a +2 oxidation state in Mn2+, which is a reduction involving the gain of 5 electrons per Mn atom. Since there are 2 Mn atoms involved, the total is 2 * 5 = 10 electrons. The Cl- is oxidized to Cl2, but the focus here is on the total number of electrons transferred, which is 10.

Question 9

The equilibrium constant,K,can be calculated from

Accepted Answer

The equilibrium constant, K, can be calculated from the standard electrode potential (E°), not from the electrode potential (E) under non-standard conditions. The relationship between E° and K is given by the Nernst equation.

Question 10

Consider the following cell: Pt(s)∣ H₂(g,p₁)∣ H⁺(aq,pH_A)∣∣ H⁺(aq,pH_C)∣ H₂(g,p₂)∣ Pt(s) Where pH_A is the pH of the aqueous solution in the anode half-cell and pH_C is the pH of the aqueous solution in the cathode half-cell.If the partial pressure of H₂(g)is the same for both half-cells, (p₁ = p₂),then E for the cell at 25°C is

Accepted Answer

The answer of Consider the following cell: Pt(s)∣ H₂(g,p₁)∣ H⁺(aq,pH_A)∣∣...

Question 11

Given that E°_red = -0.26 V for Ni²⁺/Ni at 25°C,find E° and E for the concentration cell expressed using shorthand notation below. Ni(s)∣ Ni²⁺(aq,1.0 × 10^-5 M)∣∣ Ni²⁺(aq,0.100 M)∣ Ni(s)

Accepted Answer

The shorthand notation provided in the problem is as follows: Ni(s) ∣ Ni²⁺(aq, 1.0 × 10^-5 M) ∣∣ Ni²⁺(aq, 0.100 M) ∣ Ni(s) To find E°, we can use the given standard reduction potential: E°_red = -0.26 V E° = E°(reduction at cathode) - E°(reduction at anode) = 0 - (-0.26 V) = +0.26 V To find E, we can use the equation: E = E° - (0.0592/n) log(Q) where n is the number of electrons transferred in the reaction and Q is the reaction quotient. In this case, the cell reaction is: Ni²⁺ + 2e- → Ni + 2^2- So n = 2. At equilibrium, the reaction quotient Q is: Q = [Ni2+(aq)]/[S2-(aq)]2 Since the concentrations of Ni2+ and S2- are the same on both sides of the cell, Q = 1. So, E = E° - (0.0592/n) log(Q) = E° - (0.0592/2) log(1) = E° = +0.26 V Therefore, the final answer is: Choice B) E° = 0.00 V and E = +0.12 V, as E° = +0.26 V and E = E° = +0.26 V - (0.0592/2) log(1) = +0.12 V.

Question 12

If the cell reaction involves ions in solution,as the cell reaction in a galvanic cell continues,

Accepted Answer

As the cell reaction in a galvanic cell continues, the concentration of reactants decreases and the concentration of products increases, which causes a decrease in the free energy change (ΔG) and therefore a decrease in the EMF (E) for the cell. This is consistent with Le Chatelier's principle, which states that a system at equilibrium will respond to any stress imposed upon it in such a way as to counteract the stress and reestablish equilibrium. In this case, the stress is the consumption of reactants and the formation of products, which causes a shift in the direction of the reverse reaction, resulting in a decrease in E for the cell. However, the standard electrode potential (E°) for the cell remains unchanged since it is a measure of the inherent tendency for the reaction to occur under standard conditions.

Question 13

Consider the half-reaction: MnO₄^-(aq)+ 8 H⁺(aq)+ 5 e- → Mn²⁺(aq)+ 4 H₂O(l).The formation of MnO₄^- from Mn²⁺ occurs most readily when the solution is

Accepted Answer

The formation of MnO₄⁻ from Mn²⁺ is favored in basic conditions because MnO₄⁻ is more stable in a basic environment, and the reverse reaction (reduction of MnO₄⁻ to Mn²⁺) is favored in acidic conditions.

Question 14

For a particular battery based on one of the following reactions,E is expected to remain constant with time until the cell reactants are almost completely consumed.Which is the appropriate reaction?

Accepted Answer

The reaction equation for option B shows that only the solid reactants NiO(OH) and Cd are being consumed to form Ni(OH)2 and Cd(OH)2 solids. The hydrogen ions (H+) and hydroxide ions (OH-) are only serving as spectator ions and are not being consumed or produced. Therefore, the cell potential E should remain constant with time until the solid reactants are almost completely consumed.

Question 15

When a cell reaction reaches equilibrium,

Accepted Answer

At equilibrium, the cell potential (E) becomes zero because there is no net movement of electrons. The standard cell potential (E°) is a constant value that depends on the nature of the reactants and products at standard conditions and does not change when equilibrium is reached.

Question 16

For a particular cell based on the reaction: 3 AgCl(s)+ Al(s)→ 3 Ag(s)+ Al³⁺(aq)+ 3 Cl^-(aq) E = +1.750 V and E° = +1.884 V at 25°C. What is the value of the equilibrium constant,K,at 25°C for the reaction?

Accepted Answer

The answer of For a particular cell based on the...

Question 17

The following cell has a potential of 0.45 V at 25°C. Pt(s)∣ H₂(1 atm)|H⁺(? M)∣∣ Cl^-(1 M)∣ Hg₂Cl₂(s)|Hg(l) The standard half-cell potential for the half-reaction Hg₂Cl₂(s)+ 2 e^- → 2 Hg(l)+ 2 Cl^-(aq)is 0.28 V.What is the pH in the anode compartment?

Accepted Answer

The Nernst equation can be used to calculate the cell potential at non-standard conditions: Ecell = E°cell - (RT/nF)lnQ where E°cell is the standard cell potential, R is the gas constant, T is the temperature in Kelvin, n is the number of moles of electrons transferred in the reaction, F is Faraday's constant, and Q is the reaction quotient. In this case, the reaction is: Hg₂Cl₂(s) + 2 e^- → 2 Hg(l) + 2 Cl^-(aq) The reaction quotient can be expressed as: Q = [Hg2+][Cl-]^2 / [HgS][Cl-] = [Hg2+][Cl-] / [HgS] Since the solid HgS does not appear in the reaction quotient expression, its activity is taken to be 1. Therefore, the reaction quotient simplifies to: Q = [Hg2+][Cl-]^2 At equilibrium, the cell potential is equal to the standard cell potential. Therefore: 0.45 V = 0.28 V - (0.0257 V/K)(log([Hg2+][Cl-]^2) - log(1)) Simplifying: log([Hg2+][Cl-]^2) = 4.08 [Hg2+][Cl-]^2 = 10^4.08 [Hg2+] = sqrt(10^4.08 / [Cl-]^2) = 704.4 M Now we need to find the pH of the anode compartment, which is where the HgS and Hg2+ are located. We can use the following half-reaction to find the relationship between [Hg2+] and pH: 2 Hg2+ + 2 e^- + 2 H2O⁺ → Hg22+(aq) + 4 HSO4-^- + 4 H+ The standard reduction potential for this half-reaction is 1.23 V, which is sufficiently far from the potential of the cell in order to assume that this reaction is at equilibrium. Therefore, the Nernst equation can be used to find the relationship between [Hg2+] and pH: E° = 1.23 V E = E° - (RT/nF)ln[Hg2+][H2O]^2 / [HSO4-]^4 0.45 V = 1.23 V - (0.0257 V/K)(ln[Hg2+] + 2ln[H2O] - 4ln[HSO4-]) ln[Hg2+] = 28.53 - 44.50pH [Hg2+] = e^(28.53 - 44.50pH) Substituting the value for [Hg2+] obtained earlier: 704.4 M = e^(28.53 - 44.50pH) ln 704.4 = 28.53 - 44.50pH pH = (28.53 - ln 704.4) / 44.50 = 2.9

Question 18

Shown below are the reactions occurring in the direct methanol fuel cell (DMFC). I.2 CH₃OH(aq)+ 2 H₂O(l)→ 2 CO₂(g)+12 H⁺(aq)+12 e^- II.3 O₂(g)+ 12 H⁺(aq)+ 12 e^- → 6 H₂O(l) Overall 2 CH₃OH(aq)+ 3 O₂(g)→ 2 CO₂(g)+ 4 H₂O(l) Which is the anode reaction,and what is being oxidized in the overall reaction?

Accepted Answer

In a direct methanol fuel cell (DMFC), the anode reaction is where oxidation occurs, which is Reaction I. Methanol (CH3OH) is being oxidized to carbon dioxide (CO2), releasing electrons in the process. This makes CH3OH the substance being oxidized in the overall reaction.

Question 19

Given   and  calculate E at 25°C for a cell based on the reaction:

Accepted Answer

First, we need to determine the half-reactions for the given electrode potential values:

11ea7e2d_d1a5_71ce_a2f7_bbdf25ae4115_TB4940_00: Cu2+ + 2e- -> Cu   Eº = +0.34 V
11ea7e2d_d1a5_71cf_a2f7_bd728d5ebac5_TB4940_11: Cu+ + e- -> Cu   Eº = +0.52 V
11ea7e2d_d1a5_71d0_a2f7_d9deb90eb373_TB4940_11: Ag+ + e- -> Ag   Eº = +0.80 V

Next, we need to identify which half-reaction corresponds to the anode and which corresponds to the cathode for the overall cell reaction:

11ea7e2d_d1a5_97e1_a2f7_1711ff3f62d7_TB4940_00: Cu + Ag+ -> Cu+ + Ag

Since oxidation occurs at the anode, we can see that the Cu half-reaction from the first electrode potential value (11ea7e2d_d1a5_71ce_a2f7_bbdf25ae4115_TB4940_00) must be reversed for this cell reaction. This gives us:

Anode: Cu -> Cu2+ + 2e-   Eº = -0.34 V
Cathode: Ag+ + e- -> Ag   Eº = +0.80 V

Now we can calculate the overall cell potential by summing the half-reaction potentials and flipping the sign of the anode potential:

Eºcell = Eºcathode - Eºanode
        = (+0.80 V) - (-0.34 V)
        = +1.14 V

Finally, we need to use the Nernst equation to calculate the actual cell potential at 25°C:

Ecell = Eºcell - (RT/nF)lnQ

where R = gas constant = 8.314 J/mol·K
      T = temperature in Kelvin = 298K
      n = number of electrons transferred = 1 for this reaction
      F = Faraday's constant = 96,485 C/mol
      Q = reaction quotient = [Cu+]/[Ag+]

We can calculate Q using the given electrode potentials for Cu+ and Ag+:

Q = [Cu+]/[Ag+]
  = 10^(-(Eºcathode - Eºanode)/0.059)
  = 10^(-(0.80 - 0.34)/0.059)
  = 0.082

Plugging in all the values, we get:

Ecell = +1.14 V - (8.314 J/mol·K × 298 K)/(1 mol × 96,485 C/mol) ln(0.082)
       = +0.15 V

Therefore, the answer is C (+0.15 V).

Question 20

When suspected drunk drivers are tested with a Breathalyzer,the alcohol (ethanol)in the exhaled breath is oxidized to acetic acid with an acidic solution of potassium dichromate: 3 CH₃CH₂OH(aq)+ 2 Cr₂O₇^-2 (aq)+ 16 H⁺(aq)3CH₃CO₂H(aq)+ 4Cr⁺³(aq)+ 11H₂O (l) What is the value of E for the reaction when the concentrations of ethanol,acetic acid,Cr₂O₇^-2,and Cr⁺³,are 1.0M and the pH is 4.00? (Eº for this cell is 1.30V and the temperature is 298K). Equations to solve this problem are below. Ε = ε° - lnQ Ε = 1.30 - ln

Accepted Answer

The answer of When suspected drunk drivers are tested with...

Quiz 18: Electrochemistry

Which statement below is not true?

A cell based on the reaction below has a standard potential of +0.42 V at 25°C.If all of the species are at standard conditions except [H⁺],at what pH will the cell have a potential of zero? H₂O₂(aq) + 2 H⁺(aq) + 2 Cl^-(aq) → Cl₂(aq) + 2 H₂O(l)

The cell reaction for a lead storage battery is: Pb(s) + PbO₂(s) + 2 H⁺(aq) + 2 HSO₄^-(aq) → 2 PbSO₄(s) + 2 H₂O(l) E° = +1.92 V To provide a potential of about 12 V,one could

Which of the following reactions is most suitable for a fuel cell?

Ag⁺(aq) + e^- → Ag(s) E° = +0.800 V AgBr(s) + e^- → Ag(s) + Br^-(aq) E° = +0.071 V Br₂(l) + 2 e^- → 2 Br^-(aq) E° = +1.066 V Use some of the data above to calculate K_sp at 25°C for AgBr.

Which is most often used in the laboratory to measure pH?

Which of the following statements concerning a lithium battery is false?

How many moles of electrons,n,are transferred in the following reduction-oxidation reaction? 2 MnO₄^-(aq) + 16 H⁺(aq) + 10 Cl^-(aq) → 2 Mn²⁺(aq) + 5 Cl₂(g) + 8 H₂O(l)

The equilibrium constant,K,can be calculated from

Given that E°_red = -0.26 V for Ni²⁺/Ni at 25°C,find E° and E for the concentration cell expressed using shorthand notation below. Ni(s) ∣ Ni²⁺(aq,1.0 × 10^-5 M) ∣∣ Ni²⁺(aq,0.100 M) ∣ Ni(s)

If the cell reaction involves ions in solution,as the cell reaction in a galvanic cell continues,

Consider the half-reaction: MnO₄^-(aq) + 8 H⁺(aq) + 5 e- → Mn²⁺(aq) + 4 H₂O(l) .The formation of MnO₄^- from Mn²⁺ occurs most readily when the solution is

For a particular battery based on one of the following reactions,E is expected to remain constant with time until the cell reactants are almost completely consumed.Which is the appropriate reaction?

When a cell reaction reaches equilibrium,

For a particular cell based on the reaction: 3 AgCl(s) + Al(s) → 3 Ag(s) + Al³⁺(aq) + 3 Cl^-(aq) E = +1.750 V and E° = +1.884 V at 25°C. What is the value of the equilibrium constant,K,at 25°C for the reaction?

The following cell has a potential of 0.45 V at 25°C. Pt(s) ∣ H₂(1 atm) |H⁺(? M) ∣∣ Cl^-(1 M) ∣ Hg₂Cl₂(s) |Hg(l) The standard half-cell potential for the half-reaction Hg₂Cl₂(s) + 2 e^- → 2 Hg(l) + 2 Cl^-(aq) is 0.28 V.What is the pH in the anode compartment?

Given and calculate E at 25°C for a cell based on the reaction:

Quiz 18: Electrochemistry

Which statement below is not true?

A cell based on the reaction below has a standard potential of +0.42 V at 25°C.If all of the species are at standard conditions except [H+],at what pH will the cell have a potential of zero? H2O2(aq) + 2 H+(aq) + 2 Cl-(aq) → Cl2(aq) + 2 H2O(l)

The cell reaction for a lead storage battery is: Pb(s) + PbO2(s) + 2 H+(aq) + 2 HSO4-(aq) → 2 PbSO4(s) + 2 H2O(l) E° = +1.92 V To provide a potential of about 12 V,one could

Which of the following reactions is most suitable for a fuel cell?

Ag+(aq) + e- → Ag(s) E° = +0.800 V AgBr(s) + e- → Ag(s) + Br-(aq) E° = +0.071 V Br2(l) + 2 e- → 2 Br-(aq) E° = +1.066 V Use some of the data above to calculate Ksp at 25°C for AgBr.

Which is most often used in the laboratory to measure pH?

Which of the following statements concerning a lithium battery is false?

How many moles of electrons,n,are transferred in the following reduction-oxidation reaction? 2 MnO4-(aq) + 16 H+(aq) + 10 Cl-(aq) → 2 Mn2+(aq) + 5 Cl2(g) + 8 H2O(l)

The equilibrium constant,K,can be calculated from

Given that E°red = -0.26 V for Ni2+/Ni at 25°C,find E° and E for the concentration cell expressed using shorthand notation below. Ni(s) ∣ Ni2+(aq,1.0 × 10-5 M) ∣∣ Ni2+(aq,0.100 M) ∣ Ni(s)

If the cell reaction involves ions in solution,as the cell reaction in a galvanic cell continues,

Consider the half-reaction: MnO4-(aq) + 8 H+(aq) + 5 e- → Mn2+(aq) + 4 H2O(l) .The formation of MnO4- from Mn2+ occurs most readily when the solution is

For a particular battery based on one of the following reactions,E is expected to remain constant with time until the cell reactants are almost completely consumed.Which is the appropriate reaction?

When a cell reaction reaches equilibrium,

For a particular cell based on the reaction: 3 AgCl(s) + Al(s) → 3 Ag(s) + Al3+(aq) + 3 Cl-(aq) E = +1.750 V and E° = +1.884 V at 25°C. What is the value of the equilibrium constant,K,at 25°C for the reaction?

The following cell has a potential of 0.45 V at 25°C. Pt(s) ∣ H2(1 atm) |H+(? M) ∣∣ Cl-(1 M) ∣ Hg2Cl2(s) |Hg(l) The standard half-cell potential for the half-reaction Hg2Cl2(s) + 2 e- → 2 Hg(l) + 2 Cl-(aq) is 0.28 V.What is the pH in the anode compartment?

Given and calculate E at 25°C for a cell based on the reaction:

A cell based on the reaction below has a standard potential of +0.42 V at 25°C.If all of the species are at standard conditions except [H⁺],at what pH will the cell have a potential of zero? H₂O₂(aq) + 2 H⁺(aq) + 2 Cl^-(aq) → Cl₂(aq) + 2 H₂O(l)

The cell reaction for a lead storage battery is: Pb(s) + PbO₂(s) + 2 H⁺(aq) + 2 HSO₄^-(aq) → 2 PbSO₄(s) + 2 H₂O(l) E° = +1.92 V To provide a potential of about 12 V,one could

Ag⁺(aq) + e^- → Ag(s) E° = +0.800 V AgBr(s) + e^- → Ag(s) + Br^-(aq) E° = +0.071 V Br₂(l) + 2 e^- → 2 Br^-(aq) E° = +1.066 V Use some of the data above to calculate K_sp at 25°C for AgBr.

How many moles of electrons,n,are transferred in the following reduction-oxidation reaction? 2 MnO₄^-(aq) + 16 H⁺(aq) + 10 Cl^-(aq) → 2 Mn²⁺(aq) + 5 Cl₂(g) + 8 H₂O(l)

Given that E°_red = -0.26 V for Ni²⁺/Ni at 25°C,find E° and E for the concentration cell expressed using shorthand notation below. Ni(s) ∣ Ni²⁺(aq,1.0 × 10^-5 M) ∣∣ Ni²⁺(aq,0.100 M) ∣ Ni(s)

Consider the half-reaction: MnO₄^-(aq) + 8 H⁺(aq) + 5 e- → Mn²⁺(aq) + 4 H₂O(l) .The formation of MnO₄^- from Mn²⁺ occurs most readily when the solution is

For a particular cell based on the reaction: 3 AgCl(s) + Al(s) → 3 Ag(s) + Al³⁺(aq) + 3 Cl^-(aq) E = +1.750 V and E° = +1.884 V at 25°C. What is the value of the equilibrium constant,K,at 25°C for the reaction?

The following cell has a potential of 0.45 V at 25°C. Pt(s) ∣ H₂(1 atm) |H⁺(? M) ∣∣ Cl^-(1 M) ∣ Hg₂Cl₂(s) |Hg(l) The standard half-cell potential for the half-reaction Hg₂Cl₂(s) + 2 e^- → 2 Hg(l) + 2 Cl^-(aq) is 0.28 V.What is the pH in the anode compartment?