Question 1

Based on the following information, Cl₂(g)+ 2 e^- → 2 Cl^-(aq)E° = +1.36 V Mg²⁺(aq)+ 2 e^- → 2 Mg(s)E° = -2.37 V Which of the following chemical species is the strongest reducing agent?

Accepted Answer

The strongest reducing agent is the species with the most negative reduction potential, as it is more likely to donate electrons and be oxidized. Mg has a reduction potential of -2.37 V, making it the strongest reducing agent of the given species.

Question 2

The gas OF₂ can be produced from the electrolysis of an aqueous solution of KF,as shown in the equation below. OF₂(g)+ 2 H⁺(aq)+ 4 e^- → H₂O(l)+ 2 F^-(aq)E° = +2.15 V Using the given standard reduction potential,calculate the amount of OF₂ that is produced,and the electrode at which the OF₂ is produced,upon the passage of 0.480 faradays through an aqueous KF solution.

Accepted Answer

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Question 3

For a galvanic cell that uses the following two half-reactions, Cr₂O₇^2-(aq)+ 14 H⁺(aq)+ 6 e^- → 2 Cr³⁺(aq)+ 7 H₂O(l) Pb(s)→ Pb²⁺(aq)+ 2 e^- How many moles of Pb(s)are oxidized by three moles of Cr₂O₇^2-?

Accepted Answer

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Question 4

A galvanic cell consists of one half-cell that contains Ag(s)and Ag⁺(aq),and one half-cell that contains Cu(s)and Cu²⁺(aq).What species are produced at the electrodes under standard conditions? Ag⁺(aq)+ e^- → Ag(s)E° = +0.80 V Cu²⁺(aq)+ 2 e^- → Cu(s)E° = +0.34 V

Accepted Answer

The cathode is where reduction occurs, and the anode is where oxidation occurs. The given reactions show Ag+ being reduced to Ag(s) and Cu2+ being reduced to Cu(s), with their respective standard electrode potentials. Since Ag+ has a higher potential (+0.80 V) than Cu2+ (+0.34 V), Ag+ will be reduced at the cathode, forming Ag(s). Consequently, Cu(s) will oxidize to Cu2+ at the anode, completing the cell.

Question 5

The electrolysis of molten AlCl₃ for 3.25 hr with an electrical current of 15.0 A produces ________ g of aluminum metal.

Accepted Answer

The mass of aluminum produced can be calculated using Faraday's law of electrolysis. The molar mass of aluminum (Al) is 26.98 g/mol, and it has a valency of 3, meaning 3 moles of electrons are required to produce 1 mole of aluminum. The total charge (Q) passed through the electrolyte is current (I) times time (t), Q = I × t = 15.0 A × 3.25 × 3600 s = 175500 C. Since 1 mole of electrons corresponds to 96500 C (Faraday constant), the moles of electrons used are 175500 C / 96500 C/mol = 1.818 moles. Since it takes 3 moles of electrons to produce 1 mole of Al, the moles of Al produced are 1.818 moles of electrons / 3 = 0.606 moles of Al. Finally, the mass of Al produced is 0.606 moles × 26.98 g/mol = 16.4 g.

Question 6

Consider the galvanic cell,Zn(s)∣ Zn²⁺(aq)∣∣ Pb²⁺(aq)∣ Pb(s).Which one of the following changes to the cell would cause the cell potential to increase (i.e. ,become more positive)?

Accepted Answer

An increase in the concentration of Pb²⁺ would shift the equilibrium towards the side with a lower concentration of Pb²⁺, which in this case is the right side. This would increase the amount of Pb2+ ions available for reduction at the cathode, increasing the cell potential. Option A would have the opposite effect, shifting the equilibrium towards the left side and decreasing the cell potential. Changes in the mass of Zn(s) would not affect the concentrations of the species in solution, and therefore would not affect the cell potential.

Question 7

A galvanic cell consists of a Mg²⁺/Mg half-cell and a standard hydrogen electrode.If the Mg²⁺/Mg half-cell standard cell functions as the anode,and the standard cell potential is 2.37 V,what is the standard reduction potential for the Mg²⁺/Mg half-cell?

Accepted Answer

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Question 8

A voltaic cell is constructed with two Zn²⁺-Zn electrodes,where the half-reaction is Zn²⁺(aq)+ 2e^- → Zn(s)E° = -0.763 V The concentrations of zinc ion in the two compartments are 5.50 M and 1.11 × 10^-² M,respectively.The cell emf is ________ V.

Accepted Answer

The cell emf can be calculated using the Nernst equation:

Ecell = E°cell - (RT/nF)ln(Q)

Where E°cell is the standard cell potential, R is the gas constant, T is the temperature in Kelvin, n is the number of electrons transferred in the balanced equation, F is Faraday's constant, and Q is the reaction quotient.

For this cell, n = 2 because two electrons are transferred in the half-reaction. 
The reaction quotient can be expressed as:

Q = [Zn2+]1/[Zn2+]2

where [Zn2+]1 and [Zn2+]2 are the concentrations of zinc ions in compartments 1 and 2, respectively.

Plugging in the values:

Ecell = -0.763 V - (0.0257 V/K)(298 K)/2(96,485 C/mol)(ln(5.50/1.11×10^12)) 
Ecell = 0.0798 V

Therefore, the answer is C, 0.0798 V.

Question 9

The most useful ore of aluminum is bauxite,in which Al is present as hydrated oxides,Al₂O₃ · xH₂O The number of kilowatt-hours of electricity required to produce 4.00 kg of aluminum from electrolysis of compounds from bauxite is ________ when the applied emf is 5.00 V.

Accepted Answer

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Question 10

What is the balanced chemical equation for the galvanic cell reaction expressed using shorthand notation below? Al(s)∣ Al³⁺(aq)∣∣ Fe²⁺(aq)∣ Fe(s)

Accepted Answer

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Question 11

Consider the following standard reduction potentials, Ni²⁺(aq)+ 2 e^- → Ni(s)E° = -0.26 V I₂(s)+ 2 e^- → 2 I^-(aq)E° = +0.54 V Under standard conditions,

Accepted Answer

The greater the reduction potential, the stronger the oxidizing agent; and the smaller the reduction potential, the stronger the reducing agent. From the given reduction potentials, we can see that I₂(s) has a higher reduction potential (+0.54 V) than Ni²⁺(aq) (-0.26 V), which means that I₂(s) is a stronger oxidizing agent than Ni²⁺(aq). Similarly, I^-(aq) has a lower reduction potential (-0.54 V) than Ni(s) (-0.26 V), which means that Ni(s) is a stronger reducing agent than I^-(aq).

Question 12

A voltaic cell is constructed with two silver-silver chloride electrodes,where the half-reaction is AgCl(s)+ e^- → Ag(s)+ Cl^-(aq)E° = +0.222 V The concentrations of chloride ion in the two compartments are 0.0222 M and 2.22 M,respectively.The cell emf is ________ V.

Accepted Answer

The Nernst equation for the given half-reaction is Ecell = E° - (0.0592/n) log Q, where n is the number of electrons transferred and Q is the reaction quotient. In this case, n is 1 since only 1 electron is transferred. The reaction quotient can be calculated as Q = [Ag+]/[Cl-]. At standard conditions, both compartment have [Ag+] = [Cl-] = 1 M, so Q = 1. At non-standard conditions, the anode compartment has [Cl-] = 0.0222 M and the cathode compartment has [Cl-] = 2.22 M, so Q = 0.01. Thus, Ecell = 0.222 - (0.0592/1) log 0.01 = 0.118 V. The correct answer is B.

Question 13

The standard cell potential (E°)of a voltaic cell constructed using the cell reaction below is 0.76 V: Zn(s)+ 2H⁺(aq)→ Zn²⁺(aq)+ H₂(g) With P_H₂ = 1.0 atm and [Zn²⁺] = 1.0 M,the cell potential is 0.66 V.The concentration of H⁺ in the cathode compartment is ________ M.

Accepted Answer

The Nernst equation is:E = E° - (RT/nF)lnQwhere:E is the cell potentialE° 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 transferred in the cell reactionF is the Faraday constant (96,485 C/mol)Q is the reaction quotientIn this case, the cell reaction is:Zn(s) + 2H+(aq) → Zn2+(aq) + H2(g)and the reaction quotient is:Q = [Zn2+][H2]/[H+]2Substituting the given values into the Nernst equation, we get:0.66 V = 0.76 V - (0.0257 V/K)(298 K)/2(96,485 C/mol)ln([Zn2+][H2]/[H+]2)Solving for [H+], we get:[H+] = 4.2 × 10-4 M

Question 14

Given that E°_red = -1.66 V for Al³⁺/Al at 25°C,find E° and E for the concentration cell expressed using shorthand notation below. Al(s)∣ Al³⁺(1.0 × 10^-5 M)∣∣ Al³⁺(0.100 M)∣ Al(s)

Accepted Answer

The shorthand notation for the concentration cell can be written as: Al(s)∣ Al³⁺(0.100 M)∣∣ Al³⁺(1.0 × 10^-5 M)∣ Al(s) The standard potential, E°, for the cell can be calculated using the Nernst equation: Ecell = E° - (0.0592 V/n) log(Q) where Q is the reaction quotient and n is the number of electrons transferred in the reaction. For this cell, the reaction is: Al³⁺ + 3e- → Al + 13H+ + P Using the concentrations given in the shorthand notation, the reaction quotient is: Q = [Al3+][H+]13/[Al³⁺] = 1.0 × 1013 Plugging in the values: E° = Ecell + (0.0592 V/n) log(Q) E° = (-1.66 V) + (0.0592 V/3) log(1.0 × 1013) E° = 0.00 V Therefore, the answer is either A or B. The voltage, E, for the cell can be calculated using the Nernst equation: Ecell = E° - (0.0592 V/n) log(Q) + (0.0592 V/n) log([H+]left/[H+]right) where [H+]left and [H+]right are the hydrogen ion concentrations on the left and right sides of the cell, respectively. For this cell, the hydrogen ion concentrations are: [H+]left = 1.0 × 10-5 M [H+]right = 0.100 M Plugging in the values: E = Ecell + (0.0592 V/n) log(Q) - (0.0592 V/n) log([H+]left/[H+]right) E = (0.00 V) + (0.0592 V/3) log(1.0 × 1013) - (0.0592 V/3) log(1.0 × 10-5/0.100) E = 0.12 V Therefore, the answer is B: E° = 0.00 V and E = +0.12 V.

Question 15

The standard emf for the cell using the overall cell reaction below is +0.48 V: Zn(s)+ Ni²⁺(aq)→ Zn²⁺(aq)+ Ni(s) The emf generated by the cell when [Ni²⁺] = 2.50 M and [Zn²⁺] = 0.100 M is ________ V.

Accepted Answer

To calculate the emf generated by the cell, we can use the Nernst equation:
E = E° - (RT/nF) * ln(Q)
where:
E° = standard emf (given as +0.48 V)
R = gas constant (8.31 J/mol·K)
T = temperature in Kelvin (assumed to be 298 K)
n = number of moles of electrons transferred in the balanced cell reaction (2 in this case)
F = Faraday's constant (96,485 C/mol)
Q = reaction quotient

The reaction quotient can be calculated using the concentrations of the species involved in the cell reaction:
Q = [Ni]/[Zn] * ([S1U1P12+]^2 * [S1P0]^2) / 1
(Note that the concentration of solid Zn and Ni are not included in the Q expression)

Plugging in the values:
Q = (2.50 M)/(0.100 M) * (1.0*10^-24)^2 / 1
(Note that we can assume the concentration of water is constant at 1 M)

Simplifying:
Q = 6.25*10^-21

Now we can plug this value and the other given values into the Nernst equation to find the emf:
E = 0.48 V - (8.31 J/mol·K / (2*96,485 C/mol)) * ln(6.25*10^-21)
E = 0.52 V

Therefore, the answer is choice C (0.52 V).

Question 16

The town of Natrium,West Virginia,derives its name from the sodium produced in the electrolysis of molten sodium chloride (NaCl)mined from ancient salt deposits.The number of kilowatt-hours of electricity required to produce 4.60 kg of metallic sodium from the electrolysis of molten NaCl(s)is ________ when the applied emf is 4.50 V.

Accepted Answer

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Question 17

How many kilowatt-hours of electricity are used to produce 3.00 kg of magnesium in the electrolysis of molten MgCl₂ with an applied emf of 4.50 V?

Accepted Answer

The amount of electricity used in electrolysis can be calculated by multiplying the applied emf by the amount of time the process takes and the current used. The current can be determined by Faraday's Law:

Current (I) = (mass of substance produced) / (molar mass of substance × number of electrons transferred during reaction × Faraday's constant)

For magnesium, the number of electrons transferred during the reaction is 2 and the molar mass is 24.31 g/mol. The Faraday constant is 96,485 C/mol.

I = (3.00 kg ÷ 24.31 g/mol) × (2 electrons ÷ mol) × (96,485 C/mol) = 23,421 C

To find the time it takes to produce the magnesium, we can use the equation:

Charge (Q) = I × t

We know that Q = 23,421 C and the applied emf is 4.50 V. So,

Q = emf × t × I

23,421 C = 4.50 V × t × I

t = (23,421 C) ÷ (4.50 V × I)

t = (23,421 C) ÷ (4.50 V × 1.19 A)

t = 4,000 s

Now that we know the time, we can calculate the amount of electricity used:

Electricity (E) = emf × I × t

E = 4.50 V × 1.19 A × 4,000 s

E = 21,420 J = 5.95 Wh = 0.00595 kWh

Therefore, the answer is D: 29.8 kWh (rounded to three significant figures).

Question 18

Using the following standard reduction potentials, Fe³⁺(aq)+ e^- → Fe²⁺(aq)E° = +0.77 V Ni²⁺(aq)+ 2 e^- → Ni(s)E° = -0.26 V Calculate the standard cell potential for the galvanic cell reaction given below,and determine whether or not this reaction is spontaneous under standard conditions. Ni²⁺(aq)+ 2 Fe²⁺(aq)→ 2 Fe³⁺(aq)+ Ni(s)

Accepted Answer

To calculate the standard cell potential (E°), we need to add the reduction potential of the reduction half-reaction (Ni²⁺(aq)+ 2 e^- → Ni(s), E° = -0.26 V) to the oxidation potential of the oxidation half-reaction (Fe³⁺(aq)+ e^- → Fe²⁺(aq), E° = +0.77 V). E°cell = E°reduction + E°oxidation E°cell = (-0.26 V) + (+0.77 V) E°cell = -0.98 V Since E°cell is negative, the reaction is nonspontaneous under standard conditions.

Question 19

What is the balanced equation for the galvanic cell reaction expressed using shorthand notation below? Ni(s)∣ Ni²⁺(aq)∣∣ Cl₂(g)∣ Cl^-(aq)∣ C(s)

Accepted Answer

In the shorthand notation, the anode half-cell is Ni(s)∣ Ni²⁺(aq)∣∣ Cl₂(g) and the cathode half-cell is C(s)∣∣ Cl^-(aq)∣. The oxidation half-reaction occurs at the anode and involves the oxidation of Ni(s) to Ni²⁺(aq), which releases two electrons. The reduction half-reaction occurs at the cathode and involves the reduction of Cl^-(aq) to Cl₂(g), which accepts two electrons. Therefore, the balanced equation for the galvanic cell reaction is Ni(s)+ 2 Cl^-(aq)→ Ni²⁺(aq)+ 2 Cl₂(g), which corresponds to choice B.

Question 20

The standard emf for the cell using the overall cell reaction below is +2.20 V: 2Al(s)+ 3I₂(s)→ 2Al³⁺(aq)+ 6I^-(aq) The emf generated by the cell when [Al³⁺] = 4.5 × 10^-³ M and [I^-] = 0.15 M is

Accepted Answer

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Quiz 18: Electrochemistry

Based on the following information, Cl₂(g) + 2 e^- → 2 Cl^-(aq) E° = +1.36 V Mg²⁺(aq) + 2 e^- → 2 Mg(s) E° = -2.37 V Which of the following chemical species is the strongest reducing agent?

For a galvanic cell that uses the following two half-reactions, Cr₂O₇^2-(aq) + 14 H⁺(aq) + 6 e^- → 2 Cr³⁺(aq) + 7 H₂O(l) Pb(s) → Pb²⁺(aq) + 2 e^- How many moles of Pb(s) are oxidized by three moles of Cr₂O₇^2-?

A galvanic cell consists of one half-cell that contains Ag(s) and Ag⁺(aq) ,and one half-cell that contains Cu(s) and Cu²⁺(aq) .What species are produced at the electrodes under standard conditions? Ag⁺(aq) + e^- → Ag(s) E° = +0.80 V Cu²⁺(aq) + 2 e^- → Cu(s) E° = +0.34 V

The electrolysis of molten AlCl₃ for 3.25 hr with an electrical current of 15.0 A produces ________ g of aluminum metal.

Consider the galvanic cell,Zn(s) ∣ Zn²⁺(aq) ∣∣ Pb²⁺(aq) ∣ Pb(s) .Which one of the following changes to the cell would cause the cell potential to increase (i.e. ,become more positive) ?

A galvanic cell consists of a Mg²⁺/Mg half-cell and a standard hydrogen electrode.If the Mg²⁺/Mg half-cell standard cell functions as the anode,and the standard cell potential is 2.37 V,what is the standard reduction potential for the Mg²⁺/Mg half-cell?

A voltaic cell is constructed with two Zn²⁺-Zn electrodes,where the half-reaction is Zn²⁺(aq) + 2e^- → Zn(s) E° = -0.763 V The concentrations of zinc ion in the two compartments are 5.50 M and 1.11 × 10^-² M,respectively.The cell emf is ________ V.

The most useful ore of aluminum is bauxite,in which Al is present as hydrated oxides,Al₂O₃ · xH₂O The number of kilowatt-hours of electricity required to produce 4.00 kg of aluminum from electrolysis of compounds from bauxite is ________ when the applied emf is 5.00 V.

What is the balanced chemical equation for the galvanic cell reaction expressed using shorthand notation below? Al(s) ∣ Al³⁺(aq) ∣∣ Fe²⁺(aq) ∣ Fe(s)

Consider the following standard reduction potentials, Ni²⁺(aq) + 2 e^- → Ni(s) E° = -0.26 V I₂(s) + 2 e^- → 2 I^-(aq) E° = +0.54 V Under standard conditions,

A voltaic cell is constructed with two silver-silver chloride electrodes,where the half-reaction is AgCl(s) + e^- → Ag(s) + Cl^-(aq) E° = +0.222 V The concentrations of chloride ion in the two compartments are 0.0222 M and 2.22 M,respectively.The cell emf is ________ V.

The standard cell potential (E°) of a voltaic cell constructed using the cell reaction below is 0.76 V: Zn(s) + 2H⁺(aq) → Zn²⁺(aq) + H₂(g) With P_H₂ = 1.0 atm and [Zn²⁺] = 1.0 M,the cell potential is 0.66 V.The concentration of H⁺ in the cathode compartment is ________ M.

Given that E°_red = -1.66 V for Al³⁺/Al at 25°C,find E° and E for the concentration cell expressed using shorthand notation below. Al(s) ∣ Al³⁺(1.0 × 10^-5 M) ∣∣ Al³⁺(0.100 M) ∣ Al(s)

The standard emf for the cell using the overall cell reaction below is +0.48 V: Zn(s) + Ni²⁺(aq) → Zn²⁺(aq) + Ni(s) The emf generated by the cell when [Ni²⁺] = 2.50 M and [Zn²⁺] = 0.100 M is ________ V.

How many kilowatt-hours of electricity are used to produce 3.00 kg of magnesium in the electrolysis of molten MgCl₂ with an applied emf of 4.50 V?

What is the balanced equation for the galvanic cell reaction expressed using shorthand notation below? Ni(s) ∣ Ni²⁺(aq) ∣∣ Cl₂(g) ∣ Cl^-(aq) ∣ C(s)

The standard emf for the cell using the overall cell reaction below is +2.20 V: 2Al(s) + 3I₂(s) → 2Al³⁺(aq) + 6I^-(aq) The emf generated by the cell when [Al³⁺] = 4.5 × 10^-³ M and [I^-] = 0.15 M is

Quiz 18: Electrochemistry

Based on the following information, Cl2(g) + 2 e- → 2 Cl-(aq) E° = +1.36 V Mg2+(aq) + 2 e- → 2 Mg(s) E° = -2.37 V Which of the following chemical species is the strongest reducing agent?

For a galvanic cell that uses the following two half-reactions, Cr2O72-(aq) + 14 H+(aq) + 6 e- → 2 Cr3+(aq) + 7 H2O(l) Pb(s) → Pb2+(aq) + 2 e- How many moles of Pb(s) are oxidized by three moles of Cr2O72-?

A galvanic cell consists of one half-cell that contains Ag(s) and Ag+(aq) ,and one half-cell that contains Cu(s) and Cu2+(aq) .What species are produced at the electrodes under standard conditions? Ag+(aq) + e- → Ag(s) E° = +0.80 V Cu2+(aq) + 2 e- → Cu(s) E° = +0.34 V

The electrolysis of molten AlCl3 for 3.25 hr with an electrical current of 15.0 A produces ________ g of aluminum metal.

Consider the galvanic cell,Zn(s) ∣ Zn2+(aq) ∣∣ Pb2+(aq) ∣ Pb(s) .Which one of the following changes to the cell would cause the cell potential to increase (i.e. ,become more positive) ?

A galvanic cell consists of a Mg2+/Mg half-cell and a standard hydrogen electrode.If the Mg2+/Mg half-cell standard cell functions as the anode,and the standard cell potential is 2.37 V,what is the standard reduction potential for the Mg2+/Mg half-cell?

A voltaic cell is constructed with two Zn2+-Zn electrodes,where the half-reaction is Zn2+(aq) + 2e- → Zn(s) E° = -0.763 V The concentrations of zinc ion in the two compartments are 5.50 M and 1.11 × 10-2 M,respectively.The cell emf is ________ V.

The most useful ore of aluminum is bauxite,in which Al is present as hydrated oxides,Al2O3 · xH2O The number of kilowatt-hours of electricity required to produce 4.00 kg of aluminum from electrolysis of compounds from bauxite is ________ when the applied emf is 5.00 V.

What is the balanced chemical equation for the galvanic cell reaction expressed using shorthand notation below? Al(s) ∣ Al3+(aq) ∣∣ Fe2+(aq) ∣ Fe(s)

Consider the following standard reduction potentials, Ni2+(aq) + 2 e- → Ni(s) E° = -0.26 V I2(s) + 2 e- → 2 I-(aq) E° = +0.54 V Under standard conditions,

A voltaic cell is constructed with two silver-silver chloride electrodes,where the half-reaction is AgCl(s) + e- → Ag(s) + Cl-(aq) E° = +0.222 V The concentrations of chloride ion in the two compartments are 0.0222 M and 2.22 M,respectively.The cell emf is ________ V.

The standard cell potential (E°) of a voltaic cell constructed using the cell reaction below is 0.76 V: Zn(s) + 2H+(aq) → Zn2+(aq) + H2(g) With PH2 = 1.0 atm and [Zn2+] = 1.0 M,the cell potential is 0.66 V.The concentration of H+ in the cathode compartment is ________ M.

Given that E°red = -1.66 V for Al3+/Al at 25°C,find E° and E for the concentration cell expressed using shorthand notation below. Al(s) ∣ Al3+(1.0 × 10-5 M) ∣∣ Al3+(0.100 M) ∣ Al(s)

The standard emf for the cell using the overall cell reaction below is +0.48 V: Zn(s) + Ni2+(aq) → Zn2+(aq) + Ni(s) The emf generated by the cell when [Ni2+] = 2.50 M and [Zn2+] = 0.100 M is ________ V.

How many kilowatt-hours of electricity are used to produce 3.00 kg of magnesium in the electrolysis of molten MgCl2 with an applied emf of 4.50 V?

What is the balanced equation for the galvanic cell reaction expressed using shorthand notation below? Ni(s) ∣ Ni2+(aq) ∣∣ Cl2(g) ∣ Cl-(aq) ∣ C(s)

The standard emf for the cell using the overall cell reaction below is +2.20 V: 2Al(s) + 3I2(s) → 2Al3+(aq) + 6I-(aq) The emf generated by the cell when [Al3+] = 4.5 × 10-3 M and [I-] = 0.15 M is

Based on the following information, Cl₂(g) + 2 e^- → 2 Cl^-(aq) E° = +1.36 V Mg²⁺(aq) + 2 e^- → 2 Mg(s) E° = -2.37 V Which of the following chemical species is the strongest reducing agent?

For a galvanic cell that uses the following two half-reactions, Cr₂O₇^2-(aq) + 14 H⁺(aq) + 6 e^- → 2 Cr³⁺(aq) + 7 H₂O(l) Pb(s) → Pb²⁺(aq) + 2 e^- How many moles of Pb(s) are oxidized by three moles of Cr₂O₇^2-?

A galvanic cell consists of one half-cell that contains Ag(s) and Ag⁺(aq) ,and one half-cell that contains Cu(s) and Cu²⁺(aq) .What species are produced at the electrodes under standard conditions? Ag⁺(aq) + e^- → Ag(s) E° = +0.80 V Cu²⁺(aq) + 2 e^- → Cu(s) E° = +0.34 V

The electrolysis of molten AlCl₃ for 3.25 hr with an electrical current of 15.0 A produces ________ g of aluminum metal.

Consider the galvanic cell,Zn(s) ∣ Zn²⁺(aq) ∣∣ Pb²⁺(aq) ∣ Pb(s) .Which one of the following changes to the cell would cause the cell potential to increase (i.e. ,become more positive) ?

A galvanic cell consists of a Mg²⁺/Mg half-cell and a standard hydrogen electrode.If the Mg²⁺/Mg half-cell standard cell functions as the anode,and the standard cell potential is 2.37 V,what is the standard reduction potential for the Mg²⁺/Mg half-cell?

A voltaic cell is constructed with two Zn²⁺-Zn electrodes,where the half-reaction is Zn²⁺(aq) + 2e^- → Zn(s) E° = -0.763 V The concentrations of zinc ion in the two compartments are 5.50 M and 1.11 × 10^-² M,respectively.The cell emf is ________ V.

The most useful ore of aluminum is bauxite,in which Al is present as hydrated oxides,Al₂O₃ · xH₂O The number of kilowatt-hours of electricity required to produce 4.00 kg of aluminum from electrolysis of compounds from bauxite is ________ when the applied emf is 5.00 V.

What is the balanced chemical equation for the galvanic cell reaction expressed using shorthand notation below? Al(s) ∣ Al³⁺(aq) ∣∣ Fe²⁺(aq) ∣ Fe(s)

Consider the following standard reduction potentials, Ni²⁺(aq) + 2 e^- → Ni(s) E° = -0.26 V I₂(s) + 2 e^- → 2 I^-(aq) E° = +0.54 V Under standard conditions,

A voltaic cell is constructed with two silver-silver chloride electrodes,where the half-reaction is AgCl(s) + e^- → Ag(s) + Cl^-(aq) E° = +0.222 V The concentrations of chloride ion in the two compartments are 0.0222 M and 2.22 M,respectively.The cell emf is ________ V.

The standard cell potential (E°) of a voltaic cell constructed using the cell reaction below is 0.76 V: Zn(s) + 2H⁺(aq) → Zn²⁺(aq) + H₂(g) With P_H₂ = 1.0 atm and [Zn²⁺] = 1.0 M,the cell potential is 0.66 V.The concentration of H⁺ in the cathode compartment is ________ M.

Given that E°_red = -1.66 V for Al³⁺/Al at 25°C,find E° and E for the concentration cell expressed using shorthand notation below. Al(s) ∣ Al³⁺(1.0 × 10^-5 M) ∣∣ Al³⁺(0.100 M) ∣ Al(s)

The standard emf for the cell using the overall cell reaction below is +0.48 V: Zn(s) + Ni²⁺(aq) → Zn²⁺(aq) + Ni(s) The emf generated by the cell when [Ni²⁺] = 2.50 M and [Zn²⁺] = 0.100 M is ________ V.

How many kilowatt-hours of electricity are used to produce 3.00 kg of magnesium in the electrolysis of molten MgCl₂ with an applied emf of 4.50 V?

What is the balanced equation for the galvanic cell reaction expressed using shorthand notation below? Ni(s) ∣ Ni²⁺(aq) ∣∣ Cl₂(g) ∣ Cl^-(aq) ∣ C(s)

The standard emf for the cell using the overall cell reaction below is +2.20 V: 2Al(s) + 3I₂(s) → 2Al³⁺(aq) + 6I^-(aq) The emf generated by the cell when [Al³⁺] = 4.5 × 10^-³ M and [I^-] = 0.15 M is