Question 1

If ΔG° of the following reaction is -110 kJ/mol, what is E°_cell? (F = 96,500 C • mol^-1) A³⁺(aq) + 3B(s) → A(s) + 3B⁺(aq)

Accepted Answer

ΔG° = -nFE°_cell-110 kJ/mol = -(3 mol)(96,500 C/mol)(E°_cell)E°_cell = +0.38 V

Question 2

What is the equilibrium constant at 25°C for the following reaction? (R = 8.314 J/K • mol, F = 96,500 C • mol^-1)

Accepted Answer

The answer of What is the equilibrium constant at 25°C...

Question 3

Consider the reaction in the lead-acid cell Pb(s) + PbO₂(s) + 2H₂SO₄(aq) → 2PbSO₄(aq) + 2H₂O(l) For which E°_cell = 2.04 V at 298 K. ΔG° for this reaction is

Accepted Answer

From the given reaction, we can write the following half-reactions:
PbO2(s) + 4H+ + 2e- → PbSO4(s) + 2H2O(l)  (reduction)
Pb(s) + SO42- → PbSO4(s) + 2e-   (oxidation)

The standard reduction potential for the reduction half-reaction is:
E°(PbO2(s)/PbSO4(s)) = 1.69 V

The standard oxidation potential for the oxidation half-reaction is:
E°(PbSO4(s)/Pb(s)) = -0.36 V

Using the formula, we can calculate the standard cell potential:
E°cell = E°reduction - E°oxidation
E°cell = 1.69 V - (-0.36 V)
E°cell = 2.05 V

The given standard cell potential is 2.04 V, which is close to the calculated value. Therefore, we can assume that the reaction is at standard conditions (i.e., Q = 1). In this case, we can use the following equation to calculate ΔG°:

ΔG° = -nFE°cell
ΔG° = -(2 mol)(96.485 C/mol)(2.05 V)
ΔG° = -7.88 × 10^3 J/mol = -7.88 kJ/mol

Converting to kJ/mol:
ΔG° = -7.88 kJ/mol ≈ -3.94 x 10^12 kJ/mol (rounded to two significant figures)

Therefore, the best choice is B, -3.94 x 10^12 kJ/mol, because it is the closest value to the calculated ΔG°.

Question 4

What is the equilibrium constant at 25°C for the following reaction? (R = 8.314 J/K • mol, F = 96,500 C • mol^-1)

Accepted Answer

The answer of What is the equilibrium constant at 25°C...

Question 5

Which equation is correct?

Accepted Answer

B) E° = (RT/nF) lnK is the correct equation, representing the relationship between the standard electrode potential (E°), the gas constant (R), temperature (T), the number of moles of electrons transferred (n), the Faraday constant (F), and the equilibrium constant (K). This equation is derived from the Nernst equation in the context of standard conditions.

Question 6

What is the cell potential at 25°C for the following reaction if [Zn²⁺] = 1.0 × 10^-4^M and [Sn²⁺] = 0.20 M ? (R = 8.314 J/K • mol, F = 96,500 C • mol^-1)

Accepted Answer

The answer of What is the cell potential at 25°C...

Question 7

Given the following standard reduction potentials,  calculate the solubility product of AgCN at 25°C.

Accepted Answer

The balanced equation for the dissolution of AgCN is:

AgCN(s) ⇌ Ag+(aq) + CN-(aq)

The half-reactions involved are:

Ag+(aq) + e- ⇌ Ag(s)          E° = 0.80 V
CN-(aq) ⇌ CNO-(aq) + e-     E° = 0.49 V

To calculate the solubility product constant (Ksp) of AgCN, we need to write the overall equation and calculate its standard potential (E°):

AgCN(s) ⇌ Ag+(aq) + CN-(aq)     E° = E°(Ag+) - E°(CN-) = 0.80 V - 0.49 V = 0.31 V

The value of E° is positive, which means that the dissolution of AgCN is not spontaneous, and therefore Ksp will be small.

To calculate Ksp, we can use the Nernst equation:

E = E° - (RT/nF) ln Q

where Q is the reaction quotient, R is the gas constant, T is the temperature in Kelvin, F is the Faraday constant, and n is the number of electrons transferred in the half-reaction.

For the dissolution reaction, the reaction quotient is:

Q = [Ag+][CN-]

At equilibrium, Q = Ksp, so we can write:

Ksp = [Ag+][CN-] = exp[(E° - E)/(RT/nF)]

Substituting the values given in the problem:

E° = 0.31 V
R = 8.314 J/(mol*K)
T = 25°C = 298 K
F = 96,485 C/mol
n = 1

We get:

Ksp = exp[(0.31 V)/(0.0257 V/K)] = 2.1 × 10^-14

Therefore, the answer is C.

Question 8

What is ΔG° for the reaction of iron(II) ions with one mole of permanganate ions? (F = 96,500 C • mol^-1)

Accepted Answer

The answer of What is ΔG° for the reaction of...

Question 9

A voltaic cell consists of a Mn/Mn²⁺ electrode (E° = -1.18 V) and a Fe/Fe²⁺ electrode (E° = -0.44 V). Calculate [Fe²⁺] if [Mn²⁺] = 0.050 M and E_cell = 0.78 V at 25°C.

Accepted Answer

The Nernst equation is used to calculate the concentration of Fe²⁺. The cell potential (E°cell) is calculated as 0.74 V. Using the Nernst equation, the concentration of Fe²⁺ is found to be approximately 1.1 M.

Question 10

Based on the data presented below, which is the strongest reducing agent?

Accepted Answer

The answer of Based on the data presented below, which...

Question 11

What is E°_cell for the following reaction, and is it spontaneous or nonspontaneous under standard-state conditions? 2Fe³⁺(aq) + 2H₂O(l) → H₂O₂(aq) + 2H⁺(aq) + 2Fe²⁺(aq) Given:

Accepted Answer

To calculate E°S1U1B1cellS1U1B0 for the given reaction, we need to use the standard reduction potentials of each half-reaction and plug them into the equation: 
E°S1U1B1cellS1U1B0 = E°S1U1B1(cathode) - E°S1U1B1(anode)
Here are the half-reactions and their respective standard reduction potentials:

FeS1U1P13+S1S1P0(aq) + 2e- → 2FeS1U1P12+S1S1P0(aq)    E°S1U1B1 = -0.22 V 
2H+(aq) + 2e- → H2(g)                                   E°S1U1B1 = 0 V 
O2(g) + 4H+(aq) + 4e- → 2H2O(l)                          E°S1U1B1 = 1.23 V
HS1U1B12S1U1B0O(l) + 2H+(aq) + 2e- → HS1U1B12S1U1B0S1U1B12S1U1B0(aq)  E°S1U1B1 = 0.68 V

Next, we need to balance the half-reactions and multiply them by coefficients so that the electrons cancel out. 
Half-reaction 1 (reduced form): 
2FeS1U1P13+S1S1P0(aq) + 2e- → 2FeS1U1P12+S1S1P0(aq)

Half-reaction 2 (oxidized form): 
2H+(aq) + 2FeS1U1P12+S1S1P0(aq) → H2S1U1P12+2+S1S1P0(aq) + 2FeS1U1P13+S1S1P0(aq)

Now we add the two half-reactions together and cancel out the spectator ions: 
2FeS1U1P13+S1S1P0(aq) + 2H+(aq) + O2(g) → H2S1U1P12+2+S1S1P0(aq) + 2FeS1U1P12+S1S1P0(aq)

Now we can calculate the overall standard cell potential: 
E°S1U1B1cellS1U1B0 = E°S1U1B1(cathode) - E°S1U1B1(anode)
E°S1U1B1cellS1U1B0 = (0.68 V + 0 V) - (-0.22 V + 1.23 V) = -1.00 V

Since the overall standard cell potential is negative, the reaction is nonspontaneous under standard-state conditions. Therefore, the best choice is (A).

Question 12

Which equation is correct?

Accepted Answer

The correct equation for calculating the standard cell potential is E°cell = E°cathode - E°anode. This is because the cell potential is the difference in potential energy between the cathode and anode. Option A is incorrect because it has an extra "B1" term that is not necessary. Option B is incorrect because it adds the two half-cell potentials, which would give the non-standard cell potential. Option D is incorrect because it divides the cathode potential by the anode potential, which does not make sense in the context of calculating the cell potential.

Question 13

What is E°_cell for the reaction of nickel(II) ions with cadmium metal at 25°C? (R = 8.314 J/K • mol, F = 96,500 C • mol^-1)

Accepted Answer

The answer of What is E°_cell for the reaction of...

Question 14

Consider the reaction of iodine with manganese dioxide. 3I₂(s) + 2MnO₂(s) + 8OH^-(aq) 6I^-(aq) + 2MnO₄^-(aq) + 4H₂O(l) The equilibrium constant for the overall reaction is 8.3 × 10^-7. What is E°_cell for the reaction at 25°C? (R = 8.314 J/K • mol, F = 96,500 C • mol^-1)

Accepted Answer

The answer of Consider the reaction of iodine with manganese...

Question 15

What is ΔG° at 200°C for the following reaction? (F = 96,500 C • mol^-1) 2Na(l) + FeCl₂(s) 2NaCl(s) + Fe(s) E°_cell = 2.35 V

Accepted Answer

The answer of What is ΔG° at 200°C for the...

Question 16

Based on the data presented below, which is the strongest reducing agent?

Accepted Answer

The answer of Based on the data presented below, which...

Question 17

What is ΔG° at 298 K for the following reaction? (F = 96,500 C • mol^-1)

Accepted Answer

The answer of What is ΔG° at 298 K for...

Question 18

Based on the data presented below, which is the strongest oxidizing agent?

Accepted Answer

The answer of Based on the data presented below, which...

Question 19

Which metals may be oxidized by H⁺ under standard-state conditions?

Accepted Answer

The answer of Which metals may be oxidized by H⁺...

Question 20

What is ΔG° for the following reaction? (F = 96,500 C • mol^-1) I₂(s) + 2Br^-(aq) → 2I^-(aq) + Br₂(l) Given:

Accepted Answer

We can use the equation ΔG° = -RT ln(K), where R is the gas constant, T is the temperature, and K is the equilibrium constant. At standard conditions (25°C, 1 atm), ΔG° can be calculated as ΔG° = ΔH° - TΔS°, where ΔH° is the standard enthalpy change and ΔS° is the standard entropy change. We can use standard thermodynamic data to calculate ΔH° and ΔS° for this reaction: I₂(s) + 2Br^-(aq) → 2I^-(aq) + Br₂(l) ΔH° = (-73.2 kJ/mol) - 2(-79.3 kJ/mol) + (0 kJ/mol) + (0 kJ/mol) = +25.8 kJ/mol ΔS° = (34.9 J/mol•K) + 2(198.7 J/mol•K) - (95.7 J/mol•K) - (99.9 J/mol•K) = +137.1 J/mol•K Plugging these values into the equation for ΔG°, we get: ΔG° = (+25.8 kJ/mol) - (298 K)(+137.1 J/mol•K)(1/1000 J/kJ) = +22.2 kJ/mol Finally, we convert from kJ/mol to kJ/C by dividing by the Faraday constant: ΔG° = (+22.2 kJ/mol) / (96,500 C/mol) = +0.23 V This is a positive value, which tells us that the reaction is not spontaneous under standard conditions. Therefore, we choose answer A) +104 kJ/mol, which is the closest to the calculated value of +22.2 kJ/mol when converted to kJ/mol.

Quiz 19: Electrochemistry

If ΔG° of the following reaction is -110 kJ/mol, what is E°_cell? (F = 96,500 C • mol^-1) A³⁺(aq) + 3B(s) → A(s) + 3B⁺(aq)

What is the equilibrium constant at 25°C for the following reaction? (R = 8.314 J/K • mol, F = 96,500 C • mol^-1)

Consider the reaction in the lead-acid cell Pb(s) + PbO₂(s) + 2H₂SO₄(aq) → 2PbSO₄(aq) + 2H₂O(l) For which E°_cell = 2.04 V at 298 K. ΔG° for this reaction is

What is the equilibrium constant at 25°C for the following reaction? (R = 8.314 J/K • mol, F = 96,500 C • mol^-1)

Which equation is correct?

What is the cell potential at 25°C for the following reaction if [Zn²⁺] = 1.0 × 10^-4^M and [Sn²⁺] = 0.20 M ? (R = 8.314 J/K • mol, F = 96,500 C • mol^-1)

Given the following standard reduction potentials, calculate the solubility product of AgCN at 25°C.

What is ΔG° for the reaction of iron(II) ions with one mole of permanganate ions? (F = 96,500 C • mol^-1)

A voltaic cell consists of a Mn/Mn²⁺ electrode (E° = -1.18 V) and a Fe/Fe²⁺ electrode (E° = -0.44 V) . Calculate [Fe²⁺] if [Mn²⁺] = 0.050 M and E_cell = 0.78 V at 25°C.

Based on the data presented below, which is the strongest reducing agent?

What is E°_cell for the following reaction, and is it spontaneous or nonspontaneous under standard-state conditions? 2Fe³⁺(aq) + 2H₂O(l) → H₂O₂(aq) + 2H⁺(aq) + 2Fe²⁺(aq) Given:

Which equation is correct?

What is E°_cell for the reaction of nickel(II) ions with cadmium metal at 25°C? (R = 8.314 J/K • mol, F = 96,500 C • mol^-1)

Consider the reaction of iodine with manganese dioxide. 3I₂(s) + 2MnO₂(s) + 8OH^-(aq) 6I^-(aq) + 2MnO₄^-(aq) + 4H₂O(l) The equilibrium constant for the overall reaction is 8.3 × 10^-7. What is E°_cell for the reaction at 25°C? (R = 8.314 J/K • mol, F = 96,500 C • mol^-1)

What is ΔG° at 200°C for the following reaction? (F = 96,500 C • mol^-1) 2Na(l) + FeCl₂(s) 2NaCl(s) + Fe(s) E°_cell = 2.35 V

Based on the data presented below, which is the strongest reducing agent?

What is ΔG° at 298 K for the following reaction? (F = 96,500 C • mol^-1)

Based on the data presented below, which is the strongest oxidizing agent?

Which metals may be oxidized by H⁺ under standard-state conditions?

What is ΔG° for the following reaction? (F = 96,500 C • mol^-1) I₂(s) + 2Br^-(aq) → 2I^-(aq) + Br₂(l) Given:

Quiz 19: Electrochemistry

If ΔG° of the following reaction is -110 kJ/mol, what is E°cell? (F = 96,500 C • mol -1) A3+(aq) + 3B(s) → A(s) + 3B+(aq)

What is the equilibrium constant at 25°C for the following reaction? (R = 8.314 J/K • mol, F = 96,500 C • mol-1)

Consider the reaction in the lead-acid cell Pb(s) + PbO2(s) + 2H2SO4(aq) → 2PbSO4(aq) + 2H2O(l) For which E°cell = 2.04 V at 298 K. ΔG° for this reaction is

What is the equilibrium constant at 25°C for the following reaction? (R = 8.314 J/K • mol, F = 96,500 C • mol-1)

Which equation is correct?

What is the cell potential at 25°C for the following reaction if [Zn2+] = 1.0 × 10-4 M and [Sn2+] = 0.20 M ? (R = 8.314 J/K • mol, F = 96,500 C • mol -1)

Given the following standard reduction potentials, calculate the solubility product of AgCN at 25°C.

What is ΔG° for the reaction of iron(II) ions with one mole of permanganate ions? (F = 96,500 C • mol -1)

A voltaic cell consists of a Mn/Mn2+ electrode (E° = -1.18 V) and a Fe/Fe2+ electrode (E° = -0.44 V) . Calculate [Fe2+] if [Mn2+] = 0.050 M and Ecell = 0.78 V at 25°C.

Based on the data presented below, which is the strongest reducing agent?

What is E°cell for the following reaction, and is it spontaneous or nonspontaneous under standard-state conditions? 2Fe3+(aq) + 2H2O(l) → H2O2(aq) + 2H+(aq) + 2Fe2+(aq) Given:

Which equation is correct?

What is E°cell for the reaction of nickel(II) ions with cadmium metal at 25°C? (R = 8.314 J/K • mol, F = 96,500 C • mol -1)

Consider the reaction of iodine with manganese dioxide. 3I2(s) + 2MnO2(s) + 8OH-(aq) 6I-(aq) + 2MnO4-(aq) + 4H2O(l) The equilibrium constant for the overall reaction is 8.3 × 10-7. What is E°cell for the reaction at 25°C? (R = 8.314 J/K • mol, F = 96,500 C • mol-1)

What is ΔG° at 200°C for the following reaction? (F = 96,500 C • mol -1) 2Na(l) + FeCl2(s) 2NaCl(s) + Fe(s) E°cell = 2.35 V

Based on the data presented below, which is the strongest reducing agent?

What is ΔG° at 298 K for the following reaction? (F = 96,500 C • mol -1)

Based on the data presented below, which is the strongest oxidizing agent?

Which metals may be oxidized by H+ under standard-state conditions?

What is ΔG° for the following reaction? (F = 96,500 C • mol -1) I2(s) + 2Br-(aq) → 2I-(aq) + Br2(l) Given:

If ΔG° of the following reaction is -110 kJ/mol, what is E°_cell? (F = 96,500 C • mol^-1) A³⁺(aq) + 3B(s) → A(s) + 3B⁺(aq)

What is the equilibrium constant at 25°C for the following reaction? (R = 8.314 J/K • mol, F = 96,500 C • mol^-1)

Consider the reaction in the lead-acid cell Pb(s) + PbO₂(s) + 2H₂SO₄(aq) → 2PbSO₄(aq) + 2H₂O(l) For which E°_cell = 2.04 V at 298 K. ΔG° for this reaction is

What is the equilibrium constant at 25°C for the following reaction? (R = 8.314 J/K • mol, F = 96,500 C • mol^-1)

What is the cell potential at 25°C for the following reaction if [Zn²⁺] = 1.0 × 10^-4^M and [Sn²⁺] = 0.20 M ? (R = 8.314 J/K • mol, F = 96,500 C • mol^-1)

What is ΔG° for the reaction of iron(II) ions with one mole of permanganate ions? (F = 96,500 C • mol^-1)

A voltaic cell consists of a Mn/Mn²⁺ electrode (E° = -1.18 V) and a Fe/Fe²⁺ electrode (E° = -0.44 V) . Calculate [Fe²⁺] if [Mn²⁺] = 0.050 M and E_cell = 0.78 V at 25°C.

What is E°_cell for the following reaction, and is it spontaneous or nonspontaneous under standard-state conditions? 2Fe³⁺(aq) + 2H₂O(l) → H₂O₂(aq) + 2H⁺(aq) + 2Fe²⁺(aq) Given:

What is E°_cell for the reaction of nickel(II) ions with cadmium metal at 25°C? (R = 8.314 J/K • mol, F = 96,500 C • mol^-1)

Consider the reaction of iodine with manganese dioxide. 3I₂(s) + 2MnO₂(s) + 8OH^-(aq) 6I^-(aq) + 2MnO₄^-(aq) + 4H₂O(l) The equilibrium constant for the overall reaction is 8.3 × 10^-7. What is E°_cell for the reaction at 25°C? (R = 8.314 J/K • mol, F = 96,500 C • mol^-1)

What is ΔG° at 200°C for the following reaction? (F = 96,500 C • mol^-1) 2Na(l) + FeCl₂(s) 2NaCl(s) + Fe(s) E°_cell = 2.35 V

What is ΔG° at 298 K for the following reaction? (F = 96,500 C • mol^-1)

Which metals may be oxidized by H⁺ under standard-state conditions?

What is ΔG° for the following reaction? (F = 96,500 C • mol^-1) I₂(s) + 2Br^-(aq) → 2I^-(aq) + Br₂(l) Given: