Need to write the cell reaction for the electrolysis of aqueous AgNO 3 solution and calculate the amount of charges used to deposit 0.67g of silver. Concept introduction: Since the electrolysis of AgNO 3 takes in aqueous solution, reduction of Ag + was done by the electrons obtained by the oxidation of water. Mass of the metal produced was given, from that number of moles of Ag produced can be calculated. For the reduction of one moles of Ag + ion, one mole of electron was need. So the number of mole of silver produced is equal to the number of moles of electrons needed. The coulomb of charges can be attained by the multiplication of moles of electron with Faraday constant. Number of moles of silver (n) = Weight in g Atomic Mass in g mole -1 Number of moles of silver produced = Number of moles of electrons needed Charges = Number of moles of electrons×Faraday constant To find: Cell reaction of the electrolysis of AgNO 3 in aqueous solution and number of charges need to deposit 0.67g of silver.

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Answer 1

Question

Chapter 18, Problem 18.52QP

(a)

Interpretation Introduction

Interpretation:

Need to write the cell reaction for the electrolysis of aqueous AgNO₃ solution and calculate the amount of charges used to deposit 0.67g of silver.

Concept introduction:

Since the electrolysis of AgNO₃ takes in aqueous solution, reduction of Ag⁺ was done by the electrons obtained by the oxidation of water. Mass of the metal produced was given, from that number of moles of Ag produced can be calculated. For the reduction of one moles of Ag⁺ ion, one mole of electron was need. So the number of mole of silver produced is equal to the number of moles of electrons needed. The coulomb of charges can be attained by the multiplication of moles of electron with Faraday constant.

Number of moles of silver (n) = Weight in gAtomic Mass in g mole-1Number of moles of silver produced = Number of moles of electrons neededCharges = Number of moles of electrons×Faraday constant

To find: Cell reaction of the electrolysis of AgNO₃ in aqueous solution and number of charges need to deposit 0.67g of silver.

(b)

Interpretation Introduction

Interpretation:

Need to write the cell reaction for the electrolysis of aqueous AgNO₃ solution and calculate the amount of charges used to deposit 0.67g of silver.

Concept introduction:

Since the electrolysis of AgNO₃ takes in aqueous solution, reduction of Ag⁺ was done by the electrons obtained by the oxidation of water. Mass of the metal produced was given, from that number of moles of Ag produced can be calculated. For the reduction of one moles of Ag⁺ ion, one mole of electron was need. So the number of mole of silver produced is equal to the number of moles of electrons needed. The coulomb of charges can be attained by the multiplication of moles of electron with Faraday constant.

Number of moles of silver (n) = Weight in gAtomic Mass in g mole-1Number of moles of silver produced = Number of moles of electrons neededCharges = Number of moles of electrons×Faraday constant

To find: Cell reaction of the electrolysis of AgNO₃ in aqueous solution and number of charges need to deposit 0.67g of silver.

(c)

Interpretation Introduction

Interpretation:

Need to write the cell reaction for the electrolysis of aqueous AgNO₃ solution and calculate the amount of charges used to deposit 0.67g of silver.

Concept introduction:

Since the electrolysis of AgNO₃ takes in aqueous solution, reduction of Ag⁺ was done by the electrons obtained by the oxidation of water. Mass of the metal produced was given, from that number of moles of Ag produced can be calculated. For the reduction of one moles of Ag⁺ ion, one mole of electron was need. So the number of mole of silver produced is equal to the number of moles of electrons needed. The coulomb of charges can be attained by the multiplication of moles of electron with Faraday constant.

Number of moles of silver (n) = Weight in gAtomic Mass in g mole-1Number of moles of silver produced = Number of moles of electrons neededCharges = Number of moles of electrons×Faraday constant

To find: Cell reaction of the electrolysis of AgNO₃ in aqueous solution and number of charges need to deposit 0.67g of silver.

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6. Consider the following exothermic reaction below. 2Cu2+(aq) +41 (aq)2Cul(s) + 12(aq) a. If Cul is added, there will be a shift left/shift right/no shift (circle one). b. If Cu2+ is added, there will be a shift left/shift right/no shift (circle one). c. If a solution of AgNO3 is added, there will be a shift left/shift right/no shift (circle one). d. If the solvent hexane (C6H14) is added, there will be a shift left/shift right/no shift (circle one). Hint: one of the reaction species is more soluble in hexane than in water. e. If the reaction is cooled, there will be a shift left/shift right/no shift (circle one). f. Which of the changes above will change the equilibrium constant, K?

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Answer 2

Question

Chapter 18, Problem 18.52QP

(a)

Interpretation Introduction

Interpretation:

Need to write the cell reaction for the electrolysis of aqueous AgNO₃ solution and calculate the amount of charges used to deposit 0.67g of silver.

Concept introduction:

Since the electrolysis of AgNO₃ takes in aqueous solution, reduction of Ag⁺ was done by the electrons obtained by the oxidation of water. Mass of the metal produced was given, from that number of moles of Ag produced can be calculated. For the reduction of one moles of Ag⁺ ion, one mole of electron was need. So the number of mole of silver produced is equal to the number of moles of electrons needed. The coulomb of charges can be attained by the multiplication of moles of electron with Faraday constant.

Number of moles of silver (n) = Weight in gAtomic Mass in g mole-1Number of moles of silver produced = Number of moles of electrons neededCharges = Number of moles of electrons×Faraday constant

To find: Cell reaction of the electrolysis of AgNO₃ in aqueous solution and number of charges need to deposit 0.67g of silver.

(b)

Interpretation Introduction

Interpretation:

Need to write the cell reaction for the electrolysis of aqueous AgNO₃ solution and calculate the amount of charges used to deposit 0.67g of silver.

Concept introduction:

Since the electrolysis of AgNO₃ takes in aqueous solution, reduction of Ag⁺ was done by the electrons obtained by the oxidation of water. Mass of the metal produced was given, from that number of moles of Ag produced can be calculated. For the reduction of one moles of Ag⁺ ion, one mole of electron was need. So the number of mole of silver produced is equal to the number of moles of electrons needed. The coulomb of charges can be attained by the multiplication of moles of electron with Faraday constant.

Number of moles of silver (n) = Weight in gAtomic Mass in g mole-1Number of moles of silver produced = Number of moles of electrons neededCharges = Number of moles of electrons×Faraday constant

To find: Cell reaction of the electrolysis of AgNO₃ in aqueous solution and number of charges need to deposit 0.67g of silver.

(c)

Interpretation Introduction

Interpretation:

Need to write the cell reaction for the electrolysis of aqueous AgNO₃ solution and calculate the amount of charges used to deposit 0.67g of silver.

Concept introduction:

Since the electrolysis of AgNO₃ takes in aqueous solution, reduction of Ag⁺ was done by the electrons obtained by the oxidation of water. Mass of the metal produced was given, from that number of moles of Ag produced can be calculated. For the reduction of one moles of Ag⁺ ion, one mole of electron was need. So the number of mole of silver produced is equal to the number of moles of electrons needed. The coulomb of charges can be attained by the multiplication of moles of electron with Faraday constant.

Number of moles of silver (n) = Weight in gAtomic Mass in g mole-1Number of moles of silver produced = Number of moles of electrons neededCharges = Number of moles of electrons×Faraday constant

To find: Cell reaction of the electrolysis of AgNO₃ in aqueous solution and number of charges need to deposit 0.67g of silver.

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Answer 3

Question

Chapter 18, Problem 18.52QP

(a)

Interpretation Introduction

Interpretation:

Need to write the cell reaction for the electrolysis of aqueous AgNO₃ solution and calculate the amount of charges used to deposit 0.67g of silver.

Concept introduction:

Since the electrolysis of AgNO₃ takes in aqueous solution, reduction of Ag⁺ was done by the electrons obtained by the oxidation of water. Mass of the metal produced was given, from that number of moles of Ag produced can be calculated. For the reduction of one moles of Ag⁺ ion, one mole of electron was need. So the number of mole of silver produced is equal to the number of moles of electrons needed. The coulomb of charges can be attained by the multiplication of moles of electron with Faraday constant.

Number of moles of silver (n) = Weight in gAtomic Mass in g mole-1Number of moles of silver produced = Number of moles of electrons neededCharges = Number of moles of electrons×Faraday constant

To find: Cell reaction of the electrolysis of AgNO₃ in aqueous solution and number of charges need to deposit 0.67g of silver.

(b)

Interpretation Introduction

Interpretation:

Need to write the cell reaction for the electrolysis of aqueous AgNO₃ solution and calculate the amount of charges used to deposit 0.67g of silver.

Concept introduction:

Since the electrolysis of AgNO₃ takes in aqueous solution, reduction of Ag⁺ was done by the electrons obtained by the oxidation of water. Mass of the metal produced was given, from that number of moles of Ag produced can be calculated. For the reduction of one moles of Ag⁺ ion, one mole of electron was need. So the number of mole of silver produced is equal to the number of moles of electrons needed. The coulomb of charges can be attained by the multiplication of moles of electron with Faraday constant.

Number of moles of silver (n) = Weight in gAtomic Mass in g mole-1Number of moles of silver produced = Number of moles of electrons neededCharges = Number of moles of electrons×Faraday constant

To find: Cell reaction of the electrolysis of AgNO₃ in aqueous solution and number of charges need to deposit 0.67g of silver.

(c)

Interpretation Introduction

Interpretation:

Need to write the cell reaction for the electrolysis of aqueous AgNO₃ solution and calculate the amount of charges used to deposit 0.67g of silver.

Concept introduction:

Since the electrolysis of AgNO₃ takes in aqueous solution, reduction of Ag⁺ was done by the electrons obtained by the oxidation of water. Mass of the metal produced was given, from that number of moles of Ag produced can be calculated. For the reduction of one moles of Ag⁺ ion, one mole of electron was need. So the number of mole of silver produced is equal to the number of moles of electrons needed. The coulomb of charges can be attained by the multiplication of moles of electron with Faraday constant.

Number of moles of silver (n) = Weight in gAtomic Mass in g mole-1Number of moles of silver produced = Number of moles of electrons neededCharges = Number of moles of electrons×Faraday constant

To find: Cell reaction of the electrolysis of AgNO₃ in aqueous solution and number of charges need to deposit 0.67g of silver.

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Answer 4

Question

Chapter 18, Problem 18.52QP

(a)

Interpretation Introduction

Interpretation:

Need to write the cell reaction for the electrolysis of aqueous AgNO₃ solution and calculate the amount of charges used to deposit 0.67g of silver.

Concept introduction:

Since the electrolysis of AgNO₃ takes in aqueous solution, reduction of Ag⁺ was done by the electrons obtained by the oxidation of water. Mass of the metal produced was given, from that number of moles of Ag produced can be calculated. For the reduction of one moles of Ag⁺ ion, one mole of electron was need. So the number of mole of silver produced is equal to the number of moles of electrons needed. The coulomb of charges can be attained by the multiplication of moles of electron with Faraday constant.

Number of moles of silver (n) = Weight in gAtomic Mass in g mole-1Number of moles of silver produced = Number of moles of electrons neededCharges = Number of moles of electrons×Faraday constant

To find: Cell reaction of the electrolysis of AgNO₃ in aqueous solution and number of charges need to deposit 0.67g of silver.

(b)

Interpretation Introduction

Interpretation:

Need to write the cell reaction for the electrolysis of aqueous AgNO₃ solution and calculate the amount of charges used to deposit 0.67g of silver.

Concept introduction:

Since the electrolysis of AgNO₃ takes in aqueous solution, reduction of Ag⁺ was done by the electrons obtained by the oxidation of water. Mass of the metal produced was given, from that number of moles of Ag produced can be calculated. For the reduction of one moles of Ag⁺ ion, one mole of electron was need. So the number of mole of silver produced is equal to the number of moles of electrons needed. The coulomb of charges can be attained by the multiplication of moles of electron with Faraday constant.

Number of moles of silver (n) = Weight in gAtomic Mass in g mole-1Number of moles of silver produced = Number of moles of electrons neededCharges = Number of moles of electrons×Faraday constant

To find: Cell reaction of the electrolysis of AgNO₃ in aqueous solution and number of charges need to deposit 0.67g of silver.

(c)

Interpretation Introduction

Interpretation:

Need to write the cell reaction for the electrolysis of aqueous AgNO₃ solution and calculate the amount of charges used to deposit 0.67g of silver.

Concept introduction:

Since the electrolysis of AgNO₃ takes in aqueous solution, reduction of Ag⁺ was done by the electrons obtained by the oxidation of water. Mass of the metal produced was given, from that number of moles of Ag produced can be calculated. For the reduction of one moles of Ag⁺ ion, one mole of electron was need. So the number of mole of silver produced is equal to the number of moles of electrons needed. The coulomb of charges can be attained by the multiplication of moles of electron with Faraday constant.

Number of moles of silver (n) = Weight in gAtomic Mass in g mole-1Number of moles of silver produced = Number of moles of electrons neededCharges = Number of moles of electrons×Faraday constant

To find: Cell reaction of the electrolysis of AgNO₃ in aqueous solution and number of charges need to deposit 0.67g of silver.

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Answer 5

Chapter 18, Problem 18.52QP

(a)

Interpretation Introduction

Interpretation:

Need to write the cell reaction for the electrolysis of aqueous AgNO₃ solution and calculate the amount of charges used to deposit 0.67g of silver.

Concept introduction:

Since the electrolysis of AgNO₃ takes in aqueous solution, reduction of Ag⁺ was done by the electrons obtained by the oxidation of water. Mass of the metal produced was given, from that number of moles of Ag produced can be calculated. For the reduction of one moles of Ag⁺ ion, one mole of electron was need. So the number of mole of silver produced is equal to the number of moles of electrons needed. The coulomb of charges can be attained by the multiplication of moles of electron with Faraday constant.

Number of moles of silver (n) = Weight in gAtomic Mass in g mole-1Number of moles of silver produced = Number of moles of electrons neededCharges = Number of moles of electrons×Faraday constant

To find: Cell reaction of the electrolysis of AgNO₃ in aqueous solution and number of charges need to deposit 0.67g of silver.

(b)

Interpretation Introduction

Interpretation:

Need to write the cell reaction for the electrolysis of aqueous AgNO₃ solution and calculate the amount of charges used to deposit 0.67g of silver.

Concept introduction:

Since the electrolysis of AgNO₃ takes in aqueous solution, reduction of Ag⁺ was done by the electrons obtained by the oxidation of water. Mass of the metal produced was given, from that number of moles of Ag produced can be calculated. For the reduction of one moles of Ag⁺ ion, one mole of electron was need. So the number of mole of silver produced is equal to the number of moles of electrons needed. The coulomb of charges can be attained by the multiplication of moles of electron with Faraday constant.

Number of moles of silver (n) = Weight in gAtomic Mass in g mole-1Number of moles of silver produced = Number of moles of electrons neededCharges = Number of moles of electrons×Faraday constant

To find: Cell reaction of the electrolysis of AgNO₃ in aqueous solution and number of charges need to deposit 0.67g of silver.

(c)

Interpretation Introduction

Interpretation:

Need to write the cell reaction for the electrolysis of aqueous AgNO₃ solution and calculate the amount of charges used to deposit 0.67g of silver.

Concept introduction:

Since the electrolysis of AgNO₃ takes in aqueous solution, reduction of Ag⁺ was done by the electrons obtained by the oxidation of water. Mass of the metal produced was given, from that number of moles of Ag produced can be calculated. For the reduction of one moles of Ag⁺ ion, one mole of electron was need. So the number of mole of silver produced is equal to the number of moles of electrons needed. The coulomb of charges can be attained by the multiplication of moles of electron with Faraday constant.

Number of moles of silver (n) = Weight in gAtomic Mass in g mole-1Number of moles of silver produced = Number of moles of electrons neededCharges = Number of moles of electrons×Faraday constant

To find: Cell reaction of the electrolysis of AgNO₃ in aqueous solution and number of charges need to deposit 0.67g of silver.

Answer 6

Chapter 18, Problem 18.52QP

(a)

Interpretation Introduction

Interpretation:

Need to write the cell reaction for the electrolysis of aqueous AgNO₃ solution and calculate the amount of charges used to deposit 0.67g of silver.

Concept introduction:

Since the electrolysis of AgNO₃ takes in aqueous solution, reduction of Ag⁺ was done by the electrons obtained by the oxidation of water. Mass of the metal produced was given, from that number of moles of Ag produced can be calculated. For the reduction of one moles of Ag⁺ ion, one mole of electron was need. So the number of mole of silver produced is equal to the number of moles of electrons needed. The coulomb of charges can be attained by the multiplication of moles of electron with Faraday constant.

Number of moles of silver (n) = Weight in gAtomic Mass in g mole-1Number of moles of silver produced = Number of moles of electrons neededCharges = Number of moles of electrons×Faraday constant

To find: Cell reaction of the electrolysis of AgNO₃ in aqueous solution and number of charges need to deposit 0.67g of silver.

Answer 7

Chapter 18, Problem 18.52QP

(a)

Interpretation Introduction

Interpretation:

Need to write the cell reaction for the electrolysis of aqueous AgNO₃ solution and calculate the amount of charges used to deposit 0.67g of silver.

Concept introduction:

Since the electrolysis of AgNO₃ takes in aqueous solution, reduction of Ag⁺ was done by the electrons obtained by the oxidation of water. Mass of the metal produced was given, from that number of moles of Ag produced can be calculated. For the reduction of one moles of Ag⁺ ion, one mole of electron was need. So the number of mole of silver produced is equal to the number of moles of electrons needed. The coulomb of charges can be attained by the multiplication of moles of electron with Faraday constant.

Number of moles of silver (n) = Weight in gAtomic Mass in g mole-1Number of moles of silver produced = Number of moles of electrons neededCharges = Number of moles of electrons×Faraday constant

To find: Cell reaction of the electrolysis of AgNO₃ in aqueous solution and number of charges need to deposit 0.67g of silver.

Answer 8

(b)

Interpretation Introduction

Interpretation:

Need to write the cell reaction for the electrolysis of aqueous AgNO₃ solution and calculate the amount of charges used to deposit 0.67g of silver.

Concept introduction:

Since the electrolysis of AgNO₃ takes in aqueous solution, reduction of Ag⁺ was done by the electrons obtained by the oxidation of water. Mass of the metal produced was given, from that number of moles of Ag produced can be calculated. For the reduction of one moles of Ag⁺ ion, one mole of electron was need. So the number of mole of silver produced is equal to the number of moles of electrons needed. The coulomb of charges can be attained by the multiplication of moles of electron with Faraday constant.

Number of moles of silver (n) = Weight in gAtomic Mass in g mole-1Number of moles of silver produced = Number of moles of electrons neededCharges = Number of moles of electrons×Faraday constant

To find: Cell reaction of the electrolysis of AgNO₃ in aqueous solution and number of charges need to deposit 0.67g of silver.

Answer 9

(b)

Interpretation Introduction

Interpretation:

Need to write the cell reaction for the electrolysis of aqueous AgNO₃ solution and calculate the amount of charges used to deposit 0.67g of silver.

Concept introduction:

Since the electrolysis of AgNO₃ takes in aqueous solution, reduction of Ag⁺ was done by the electrons obtained by the oxidation of water. Mass of the metal produced was given, from that number of moles of Ag produced can be calculated. For the reduction of one moles of Ag⁺ ion, one mole of electron was need. So the number of mole of silver produced is equal to the number of moles of electrons needed. The coulomb of charges can be attained by the multiplication of moles of electron with Faraday constant.

Number of moles of silver (n) = Weight in gAtomic Mass in g mole-1Number of moles of silver produced = Number of moles of electrons neededCharges = Number of moles of electrons×Faraday constant

To find: Cell reaction of the electrolysis of AgNO₃ in aqueous solution and number of charges need to deposit 0.67g of silver.

Answer 10

(c)

Interpretation Introduction

Interpretation:

Need to write the cell reaction for the electrolysis of aqueous AgNO₃ solution and calculate the amount of charges used to deposit 0.67g of silver.

Concept introduction:

Since the electrolysis of AgNO₃ takes in aqueous solution, reduction of Ag⁺ was done by the electrons obtained by the oxidation of water. Mass of the metal produced was given, from that number of moles of Ag produced can be calculated. For the reduction of one moles of Ag⁺ ion, one mole of electron was need. So the number of mole of silver produced is equal to the number of moles of electrons needed. The coulomb of charges can be attained by the multiplication of moles of electron with Faraday constant.

Number of moles of silver (n) = Weight in gAtomic Mass in g mole-1Number of moles of silver produced = Number of moles of electrons neededCharges = Number of moles of electrons×Faraday constant

To find: Cell reaction of the electrolysis of AgNO₃ in aqueous solution and number of charges need to deposit 0.67g of silver.

Answer 11

(c)

Interpretation Introduction

Interpretation:

Need to write the cell reaction for the electrolysis of aqueous AgNO₃ solution and calculate the amount of charges used to deposit 0.67g of silver.

Concept introduction:

Since the electrolysis of AgNO₃ takes in aqueous solution, reduction of Ag⁺ was done by the electrons obtained by the oxidation of water. Mass of the metal produced was given, from that number of moles of Ag produced can be calculated. For the reduction of one moles of Ag⁺ ion, one mole of electron was need. So the number of mole of silver produced is equal to the number of moles of electrons needed. The coulomb of charges can be attained by the multiplication of moles of electron with Faraday constant.

Number of moles of silver (n) = Weight in gAtomic Mass in g mole-1Number of moles of silver produced = Number of moles of electrons neededCharges = Number of moles of electrons×Faraday constant

To find: Cell reaction of the electrolysis of AgNO₃ in aqueous solution and number of charges need to deposit 0.67g of silver.

Answer 12

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Answer 13

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Answer 14

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Answer 15

Ch. 18.1 - Permanganate ion and iodide ion react in basic...Ch. 18.1 - Use the half-reaction method to balance the...Ch. 18.1 - Use the half-reaction method to balance the...Ch. 18.1 - Prob. 1PPC Ch. 18.1 - Prob. 18.1.1SR Ch. 18.1 - Prob. 18.1.2SR Ch. 18.3 - A galvanic cell consists of an Mg electrode in a...Ch. 18.3 - Determine the overall cell reaction and Ecell (at...Ch. 18.3 - A galvanic cell with Ecell = 0.30 V can be...Ch. 18.3 - Prob. 2PPC

Solution Summary: The author explains the cell reaction of the electrolysis of AgNO 3 in aqueous solution and calculate the amount of charges used to deposit 0.67g of silver.

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