Chapter 19, Problem 19.52QP

(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.

  $\begin{array}{l}\text{Number}\text{of}\text{moles of silver (n)}\text{=}\frac{\text{Weight in g}}{\text{Atomic}{\text{Mass in g mole}}^{\text{-1}}}\\ \text{Number}\text{of}\text{moles of silver produced =}\text{Number}\text{of}\text{moles of electrons needed}\\ \text{Charges}\text{=}\text{Number}\text{of}\text{moles of electrons×Faraday}\text{constant}\end{array}$

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.

  $\begin{array}{l}\text{Number}\text{of}\text{moles of silver (n)}\text{=}\frac{\text{Weight in g}}{\text{Atomic}{\text{Mass in g mole}}^{\text{-1}}}\\ \text{Number}\text{of}\text{moles of silver produced =}\text{Number}\text{of}\text{moles of electrons needed}\\ \text{Charges}\text{=}\text{Number}\text{of}\text{moles of electrons×Faraday}\text{constant}\end{array}$

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.

  $\begin{array}{l}\text{Number}\text{of}\text{moles of silver (n)}\text{=}\frac{\text{Weight in g}}{\text{Atomic}{\text{Mass in g mole}}^{\text{-1}}}\\ \text{Number}\text{of}\text{moles of silver produced =}\text{Number}\text{of}\text{moles of electrons needed}\\ \text{Charges}\text{=}\text{Number}\text{of}\text{moles of electrons×Faraday}\text{constant}\end{array}$

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