Question 1

Choose the INCORRECT statement.

Accepted Answer

Primary batteries cannot be recharged by passing electricity through the battery.

Question 2

Determine E°cell for the reaction: Pb²⁺ + Zn → Pb + Zn²⁺. The half reactions are: Pb²⁺(aq) + 2 e- → Pb(s) E° = -0.125 V Zn²⁺(aq) + 2 e- → Zn(s) E° = -0.763 V

Accepted Answer

To determine E°cell we need to use the formula: E°cell = E°cathode - E°anode Where E°cathode is the standard reduction potential of the cathode (reduction half reaction) and E°anode is the standard reduction potential of the anode (oxidation half reaction). In this case, the reduction half reaction is: Pb²⁺(aq) + 2 e- → Pb(s) E° = -0.125 V And the oxidation half reaction is: Zn(s) → Zn²⁺(aq) + 2 e- E° = -0.763 V Notice that we need to reverse the oxidation half reaction to obtain a reduction half reaction: Zn²⁺(aq) + 2 e- → Zn(s) E° = +0.763 V Now we can calculate E°cell: E°cell = E°cathode - E°anode E°cell = -0.125 V - (+0.763 V) E°cell = -0.888 V However, this value is negative because the reaction is not spontaneous under standard conditions. Therefore, we need to reverse the sign of E°cell: E°cell = +0.888 V Therefore, the best choice is A, with an explanation that the value of E°cell is calculated by subtracting the E° of the anode from the E° of the cathode, and in this case, the answer is +0.888 V.

Question 3

Calculate the Ksp of lead iodide from the following standard electrode potentials, at 25°C: 2 e- + PbI₂(S) → Pb(s) + 2 I-(aq) E° = -0.365 V 2 e- + Pb²⁺(aq) → Pb(s) E° = -0.126 V

Accepted Answer

The Ksp can be calculated using the Nernst equation and the given standard electrode potentials. The reaction for PbI2 dissolving is PbI2(s) ⇌ Pb2+(aq) + 2I-(aq). The cell potential for this reaction is E°cell = E°(Pb2+/Pb) - E°(PbI2/Pb). Substituting the given values, E°cell = -0.126 V - (-0.365 V) = 0.239 V. Using the relation ΔG° = -nFE°cell and ΔG° = -RTlnK, we find Ksp = 8 × 10^-9.

Question 4

If the voltage of an electrochemical cell is negative then the cell reaction is:

Accepted Answer

A negative voltage indicates that the cell reaction is not spontaneous, meaning it requires energy input to occur.

Question 5

Choose the INCORRECT statement.

Accepted Answer

One Faraday of charge will plate out one mole of Cu from a solution of CuSO4(aq), not two moles.

Question 6

For the reaction: Mg(s) + AgNO₃(aq) → Ag(s) + Mg(NO₃)₂(aq) Ag+(aq) + e- → Ag(s) E° = 0.800 V Mg₂+(aq) + 2 e- → Mg(s) E° = -2.356 V Is the reaction spontaneous and why?

Accepted Answer

The reaction is spontaneous because the overall cell potential is positive (0.800 V - (-2.356 V) = 3.156 V), indicating that the reaction will proceed spontaneously in the direction written.

Question 7

What is the concentration of Al³⁺ in the following cell at 25°C if the cell voltage is 1.486 V? Al ∣ Al³⁺(?) || Sn²⁺(3.21 × 10^-4 M) ∣ Sn Al³⁺ + 3 e- → Al(s) E° = -1.676 V Sn²⁺ + 2 e- → Sn(s) E° = -0.137 V

Accepted Answer

The Nernst equation is:E = E° - (RT/nF)lnQwhere:E is the cell voltageE° is the standard cell potentialR is the gas constant (8.314 J/mol·K)T is the temperature in Kelvinn is the number of moles of electrons transferredF is the Faraday constant (96,485 C/mol)Q is the reaction quotientIn this case, the cell voltage is 1.486 V, the standard cell potential is -1.676 V - (-0.137 V) = -1.539 V, the temperature is 25°C (298 K), and the number of moles of electrons transferred is 3.Substituting these values into the Nernst equation, we get:1.486 V = -1.539 V - (8.314 J/mol·K * 298 K / (3 mol * 96,485 C/mol))lnQSolving for Q, we get:Q = 1.6 × 10^-5The reaction quotient is equal to the concentration of Al³⁺ divided by the concentration of Sn²⁺.Therefore, the concentration of Al³⁺ is:[Al³⁺] = Q * [Sn²⁺] = 1.6 × 10^-5 M * 3.21 × 10^-4 M = 5.14 × 10^-9 M

Question 8

Glass electrodes are used for direct measurement of:

Accepted Answer

Glass electrodes are commonly used for direct measurement of pH, as they can detect the voltage change produced by a change in the concentration of H+ ions in solution.

Question 9

Consider the cell: Ni ∣ Ni²⁺ (?M) || Cu²⁺(0.136 M) ∣ Cu Ni²⁺/Ni E° = -0.257 V Cu²⁺/Cu E° = 0.340 V The measured potential of the cell is 0.621 V. What is [Ni²⁺] at 25°C?

Accepted Answer

The cell potential is calculated using the Nernst equation. The standard cell potential (E°cell) is 0.340 V - (-0.257 V) = 0.597 V. The measured potential is 0.621 V, which is higher than the standard potential, indicating a lower concentration of Ni²⁺. Using the Nernst equation, solve for [Ni²⁺]: 0.621 V = 0.597 V - (0.0592/2)log([Ni²⁺]/0.136). Solving gives [Ni²⁺] = 2 × 10⁻² M.

Question 10

The lithium-ion battery is based on:

Accepted Answer

The answer of The lithium-ion battery is based on:
...

Question 11

Choose the INCORRECT statement.

Accepted Answer

The incorrect statement is E because the active metal used in cathodic protection is actually called a sacrificial anode, not a sacrificial cathode.

Question 12

What is the cell diagram for the spontaneous cell involving the Fe³⁺ ∣ Fe²⁺ (0.771V) half cell and the Zn²⁺ ∣ Zn (-0.763 V) half cell?

Accepted Answer

The answer of What is the cell diagram for the...

Question 13

In the cell Fe²⁺ ∣ Fe³⁺ || Cu²⁺ ∣ Cu which will increase the cell voltage the most?

Accepted Answer

The answer of In the cell Fe²⁺ ∣ Fe³⁺ ||...

Question 14

Accepted Answer

The standard calomel electrode (IV) and the silver/silver chloride electrode (I) are commonly used as reference electrodes.

Question 15

Determine E°cell for the reaction: 2 Al + 3 Zn²⁺ → 2 Al³⁺ + 3 Zn. The half reactions are: Al³⁺(aq) + 3 e- → Al(s) E° = -1.676 V Zn²⁺(aq) + 2 e- → Zn(s) E° = -0.763 V

Accepted Answer

The correct answer is A because the E°cell is calculated by subtracting the E° of the anode from the E° of the cathode. In this case, the anode is the Al³⁺/Al half-reaction and the cathode is the Zn²⁺/Zn half-reaction. Therefore, E°cell = E°cathode - E°anode = (-0.763 V) - (-1.676 V) = 0.913 V.

Question 16

What is the concentration of Cu²⁺ in the following cell at 25°C if the cell voltage is 1.253 V? Zn ∣ Zn²⁺ (0.125 M) || Cu²⁺(?) ∣ Cu Cu²⁺ + 2 e- → Cu(s) E° = 0.340 V Zn²⁺ + 2 e- → Zn(s) E° = -0.763 V

Accepted Answer

The answer of What is the concentration of Cu²⁺ in...

Question 17

Determine E°cell for the reaction: 2 Ag+ + Mg → 2 Ag + Mg2+. The half reactions are: Mg2+(aq) + 2 e- → Mg(s) E° = -2.356 V
Ag+(aq) + e- → Ag(s) E° = 0.800 V

Accepted Answer

The answer of Determine E°cell for the reaction: 2 Ag+...

Question 18

Choose the INCORRECT statement.

Accepted Answer

When Ecell < 0, the reaction is non-spontaneous.

Question 19

Determine the Ksp at 25°C of AgCl(s). The half-reactions are: Ag⁺(aq) + 1 e- → Ag(s) E° = +0.771 V AgCl(s) + 1 e- → Ag(s) + Cl- E° = +0.222 V

Accepted Answer

The answer of Determine the Ksp at 25°C of AgCl(s)....

Question 20

For the reaction: Mg(s) + AgNO₃(aq) → Ag(s) + Mg(NO₃)₂(aq) Ag⁺(aq) + e- → Ag(s) E° = 0.800 V Mg₂+(aq) + 2 e- → Mg(s) E° = -2.356 V Estimate Keq at 25°C.

Accepted Answer

The standard cell potential (E°cell) is calculated as E°cell = E°cathode - E°anode = 0.800 V - (-2.356 V) = 3.156 V. Using the Nernst equation, ΔG° = -nFE°cell and ΔG° = -RTlnKeq, we find Keq = e^(nFE°cell/RT). With n = 2, F = 96485 C/mol, R = 8.314 J/mol·K, and T = 298 K, Keq ≈ 5.7 × 10^106.

Quiz 20: Electrochemistry

Choose the INCORRECT statement.

Determine E°cell for the reaction: Pb²⁺ + Zn → Pb + Zn²⁺. The half reactions are: Pb²⁺(aq) + 2 e- → Pb(s) E° = -0.125 V Zn²⁺(aq) + 2 e- → Zn(s) E° = -0.763 V

Calculate the Ksp of lead iodide from the following standard electrode potentials, at 25°C: 2 e- + PbI₂(S) → Pb(s) + 2 I-(aq) E° = -0.365 V 2 e- + Pb²⁺(aq) → Pb(s) E° = -0.126 V

If the voltage of an electrochemical cell is negative then the cell reaction is:

Choose the INCORRECT statement.

For the reaction: Mg(s) + AgNO₃(aq) → Ag(s) + Mg(NO₃) ₂(aq) Ag+(aq) + e- → Ag(s) E° = 0.800 V Mg₂+(aq) + 2 e- → Mg(s) E° = -2.356 V Is the reaction spontaneous and why?

What is the concentration of Al³⁺ in the following cell at 25°C if the cell voltage is 1.486 V? Al ∣ Al³⁺(?) || Sn²⁺(3.21 × 10^-4 M) ∣ Sn Al³⁺ + 3 e- → Al(s) E° = -1.676 V Sn²⁺ + 2 e- → Sn(s) E° = -0.137 V

Glass electrodes are used for direct measurement of:

Consider the cell: Ni ∣ Ni²⁺ (?M) || Cu²⁺(0.136 M) ∣ Cu Ni²⁺/Ni E° = -0.257 V Cu²⁺/Cu E° = 0.340 V The measured potential of the cell is 0.621 V. What is [Ni²⁺] at 25°C?

The lithium-ion battery is based on:

Choose the INCORRECT statement.

What is the cell diagram for the spontaneous cell involving the Fe³⁺ ∣ Fe²⁺ (0.771V) half cell and the Zn²⁺ ∣ Zn (-0.763 V) half cell?

In the cell Fe²⁺ ∣ Fe³⁺ || Cu²⁺ ∣ Cu which will increase the cell voltage the most?

Which of the following are commonly used as alternate standard electrodes: I. Ag(s) | AgCl(s) | Cl- (1.0 M) II. Hg(l) | HgCl₂(s) | Cl- (1.0 M) III. Ag(s) | AgCl(s) | Cl- (0.1 M) IV. Hg(l) | Hg₂Cl₂(s) | Cl- (1.0 M) V. Hg(s) | Hg₂Cl₂(s) | Cl- (1.0 M)

Determine E°cell for the reaction: 2 Al + 3 Zn²⁺ → 2 Al³⁺ + 3 Zn. The half reactions are: Al³⁺(aq) + 3 e- → Al(s) E° = -1.676 V Zn²⁺(aq) + 2 e- → Zn(s) E° = -0.763 V

What is the concentration of Cu²⁺ in the following cell at 25°C if the cell voltage is 1.253 V? Zn ∣ Zn²⁺ (0.125 M) || Cu²⁺(?) ∣ Cu Cu²⁺ + 2 e- → Cu(s) E° = 0.340 V Zn²⁺ + 2 e- → Zn(s) E° = -0.763 V

Determine E°cell for the reaction: 2 Ag+ + Mg → 2 Ag + Mg2+. The half reactions are: Mg2+(aq) + 2 e- → Mg(s) E° = -2.356 V Ag+(aq) + e- → Ag(s) E° = 0.800 V

Choose the INCORRECT statement.

Determine the Ksp at 25°C of AgCl(s) . The half-reactions are: Ag⁺(aq) + 1 e- → Ag(s) E° = +0.771 V AgCl(s) + 1 e- → Ag(s) + Cl- E° = +0.222 V

For the reaction: Mg(s) + AgNO₃(aq) → Ag(s) + Mg(NO₃) ₂(aq) Ag⁺(aq) + e- → Ag(s) E° = 0.800 V Mg₂+(aq) + 2 e- → Mg(s) E° = -2.356 V Estimate Keq at 25°C.

Quiz 20: Electrochemistry

Choose the INCORRECT statement.

Determine E°cell for the reaction: Pb2+ + Zn → Pb + Zn2+. The half reactions are: Pb2+(aq) + 2 e- → Pb(s) E° = -0.125 V Zn2+(aq) + 2 e- → Zn(s) E° = -0.763 V

Calculate the Ksp of lead iodide from the following standard electrode potentials, at 25°C: 2 e- + PbI2(S) → Pb(s) + 2 I-(aq) E° = -0.365 V 2 e- + Pb2+(aq) → Pb(s) E° = -0.126 V

If the voltage of an electrochemical cell is negative then the cell reaction is:

Choose the INCORRECT statement.

For the reaction: Mg(s) + AgNO3(aq) → Ag(s) + Mg(NO3) 2(aq) Ag+(aq) + e- → Ag(s) E° = 0.800 V Mg2+(aq) + 2 e- → Mg(s) E° = -2.356 V Is the reaction spontaneous and why?

What is the concentration of Al3+ in the following cell at 25°C if the cell voltage is 1.486 V? Al ∣ Al3+(?) || Sn2+(3.21 × 10-4 M) ∣ Sn Al3+ + 3 e- → Al(s) E° = -1.676 V Sn2+ + 2 e- → Sn(s) E° = -0.137 V

Glass electrodes are used for direct measurement of:

Consider the cell: Ni ∣ Ni2+ (?M) || Cu2+ (0.136 M) ∣ Cu Ni2+/Ni E° = -0.257 V Cu2+/Cu E° = 0.340 V The measured potential of the cell is 0.621 V. What is [Ni2+] at 25°C?

The lithium-ion battery is based on:

Choose the INCORRECT statement.

What is the cell diagram for the spontaneous cell involving the Fe3+ ∣ Fe2+ (0.771V) half cell and the Zn2+ ∣ Zn (-0.763 V) half cell?

In the cell Fe2+ ∣ Fe3+ || Cu2+ ∣ Cu which will increase the cell voltage the most?

Which of the following are commonly used as alternate standard electrodes: I. Ag(s) | AgCl(s) | Cl- (1.0 M) II. Hg(l) | HgCl2(s) | Cl- (1.0 M) III. Ag(s) | AgCl(s) | Cl- (0.1 M) IV. Hg(l) | Hg2Cl2(s) | Cl- (1.0 M) V. Hg(s) | Hg2Cl2(s) | Cl- (1.0 M)

Determine E°cell for the reaction: 2 Al + 3 Zn2+ → 2 Al3+ + 3 Zn. The half reactions are: Al3+(aq) + 3 e- → Al(s) E° = -1.676 V Zn2+(aq) + 2 e- → Zn(s) E° = -0.763 V

What is the concentration of Cu2+ in the following cell at 25°C if the cell voltage is 1.253 V? Zn ∣ Zn2+ (0.125 M) || Cu2+(?) ∣ Cu Cu2+ + 2 e- → Cu(s) E° = 0.340 V Zn2+ + 2 e- → Zn(s) E° = -0.763 V

Determine E°cell for the reaction: 2 Ag+ + Mg → 2 Ag + Mg2+. The half reactions are: Mg2+(aq) + 2 e- → Mg(s) E° = -2.356 V Ag+(aq) + e- → Ag(s) E° = 0.800 V

Choose the INCORRECT statement.

Determine the Ksp at 25°C of AgCl(s) . The half-reactions are: Ag+(aq) + 1 e- → Ag(s) E° = +0.771 V AgCl(s) + 1 e- → Ag(s) + Cl- E° = +0.222 V

For the reaction: Mg(s) + AgNO3(aq) → Ag(s) + Mg(NO3) 2(aq) Ag+(aq) + e- → Ag(s) E° = 0.800 V Mg2+(aq) + 2 e- → Mg(s) E° = -2.356 V Estimate Keq at 25°C.

Determine E°cell for the reaction: Pb²⁺ + Zn → Pb + Zn²⁺. The half reactions are: Pb²⁺(aq) + 2 e- → Pb(s) E° = -0.125 V Zn²⁺(aq) + 2 e- → Zn(s) E° = -0.763 V

Calculate the Ksp of lead iodide from the following standard electrode potentials, at 25°C: 2 e- + PbI₂(S) → Pb(s) + 2 I-(aq) E° = -0.365 V 2 e- + Pb²⁺(aq) → Pb(s) E° = -0.126 V

For the reaction: Mg(s) + AgNO₃(aq) → Ag(s) + Mg(NO₃) ₂(aq) Ag+(aq) + e- → Ag(s) E° = 0.800 V Mg₂+(aq) + 2 e- → Mg(s) E° = -2.356 V Is the reaction spontaneous and why?

What is the concentration of Al³⁺ in the following cell at 25°C if the cell voltage is 1.486 V? Al ∣ Al³⁺(?) || Sn²⁺(3.21 × 10^-4 M) ∣ Sn Al³⁺ + 3 e- → Al(s) E° = -1.676 V Sn²⁺ + 2 e- → Sn(s) E° = -0.137 V

Consider the cell: Ni ∣ Ni²⁺ (?M) || Cu²⁺(0.136 M) ∣ Cu Ni²⁺/Ni E° = -0.257 V Cu²⁺/Cu E° = 0.340 V The measured potential of the cell is 0.621 V. What is [Ni²⁺] at 25°C?

What is the cell diagram for the spontaneous cell involving the Fe³⁺ ∣ Fe²⁺ (0.771V) half cell and the Zn²⁺ ∣ Zn (-0.763 V) half cell?

In the cell Fe²⁺ ∣ Fe³⁺ || Cu²⁺ ∣ Cu which will increase the cell voltage the most?

Which of the following are commonly used as alternate standard electrodes: I. Ag(s) | AgCl(s) | Cl- (1.0 M) II. Hg(l) | HgCl₂(s) | Cl- (1.0 M) III. Ag(s) | AgCl(s) | Cl- (0.1 M) IV. Hg(l) | Hg₂Cl₂(s) | Cl- (1.0 M) V. Hg(s) | Hg₂Cl₂(s) | Cl- (1.0 M)

Determine E°cell for the reaction: 2 Al + 3 Zn²⁺ → 2 Al³⁺ + 3 Zn. The half reactions are: Al³⁺(aq) + 3 e- → Al(s) E° = -1.676 V Zn²⁺(aq) + 2 e- → Zn(s) E° = -0.763 V

What is the concentration of Cu²⁺ in the following cell at 25°C if the cell voltage is 1.253 V? Zn ∣ Zn²⁺ (0.125 M) || Cu²⁺(?) ∣ Cu Cu²⁺ + 2 e- → Cu(s) E° = 0.340 V Zn²⁺ + 2 e- → Zn(s) E° = -0.763 V

Determine the Ksp at 25°C of AgCl(s) . The half-reactions are: Ag⁺(aq) + 1 e- → Ag(s) E° = +0.771 V AgCl(s) + 1 e- → Ag(s) + Cl- E° = +0.222 V

For the reaction: Mg(s) + AgNO₃(aq) → Ag(s) + Mg(NO₃) ₂(aq) Ag⁺(aq) + e- → Ag(s) E° = 0.800 V Mg₂+(aq) + 2 e- → Mg(s) E° = -2.356 V Estimate Keq at 25°C.