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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Consider a solution of 0.00304 moles of 4-nitrobenzoic acid (pKa = 3.442) dissolved in 25 mL water and titrated with 0.0991 M NaOH. Calculate the pH at the equivalence point

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 15

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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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