Chapter 19, Problem 19.54QP

(a)

Interpretation Introduction

Interpretation:

Need to calculate the amount of copper deposited and current needed to electrolyze the cell containing CuCl₂ connected in series with the cell containing AgNO₃.

Concept introduction:

In the given case two electrolytic cell containing AgNO₃ and CuCl₂ were connected in series, so the quantity of electricity flowing through both of the cell will be same

Half-cell reaction for cell-1 was given below.

${\text{Cathode (reduction) Ag}}^{\text{+}}{}_{\text{(aq)}}{\text{+ e}}^{\text{-}}\stackrel{}{\to }{\text{Ag}}_{\text{(s)}}$

The amount of the silver deposited was given, from which the amount of electricity flows though both of the cell can be calculated in steps. Number of moles of silver deposited was calculated first.  Since one mole of electron is needed to reduce one mole of Ag⁺_. Therefore number of moles of electron is equal to number of moles of sliver. The coulombs of electron passing can be calculated by the equation given below.

$\text{Number}\text{of}\text{moles}\text{of}\text{silver}\text{=}\frac{\text{weight}\text{(g)}}{\text{Atomic}\text{mass}\text{(g}{\text{mole}}^{\text{-1}}\text{)}}$

$\begin{array}{l}\text{Number}\text{of}\text{moles of electrons = Number of moles of Silver}\\ \text{then}\\ \text{Charges}\text{=}\text{Number}\text{of}\text{moles of electrons×Faraday}{\text{constant (96500C/mole e}}^{-}\text{)}\end{array}$

Half-cell reaction for cell-2 was given below.

${\text{Cathode (reduction) Cu}}^{\text{2+}}{}_{\text{(aq)}}{\text{+2e}}^{\text{-}}\stackrel{}{\to }{\text{Cu}}_{\text{(s)}}$

Since they are connected in series, same amount of charges will be passing through both the cell. From the calculated charges the number of moles of electrons utilized can be calculated.

$\text{Number}\text{of}\text{moles of electrons (}n\text{)}\text{=}\text{Charges}\text{×Faraday}{\text{constant (96500C/mole e}}^{-}\text{)}$

From the cell reaction it was known that two mole of electron will be needed to produce one mole of copper,

So 

$\begin{array}{l}numberofmolesof\text{copper}=nmoleof{e}^{-}\times \frac{\text{1mole}\text{Cu}}{2mole{e}^{-}}\\ \text{Weight}\text{of}\text{Copper}=numberofmolesof\text{copper }\times \text{ atomic mass}\end{array}$

Since time was given the amperes of current flowing through the circuit can be calculated by the equation given below

$\text{Current (A) =}\frac{\text{charges (C)}}{\text{time(s)}}$

To find: The amount of copper deposited and current consumed in a CuCl₂ cell connected in series with the cell containing AgNO_3.

(b)

Interpretation Introduction

Interpretation:

Need to calculate the amount of copper deposited and current needed to electrolyze the cell containing CuCl₂ connected in series with the cell containing AgNO₃. 

Concept introduction:

In the given case two electrolytic cell containing AgNO₃ and CuCl₂ were connected in series, so the quantity of electricity flowing through both of the cell will be same

Half-cell reaction for cell-1 was given below.

${\text{Cathode (reduction) Ag}}^{\text{+}}{}_{\text{(aq)}}{\text{+ e}}^{\text{-}}\stackrel{}{\to }{\text{Ag}}_{\text{(s)}}$

The amount of the silver deposited was given, from which the amount of electricity flows though both of the cell can be calculated in steps. Number of moles of silver deposited was calculated first.  Since one mole of electron is needed to reduce one mole of Ag⁺_. Therefore number of moles of electron is equal to number of moles of sliver. The coulombs of electron passing can be calculated by the equation given below.

$\text{Number}\text{of}\text{moles}\text{of}\text{silver}\text{=}\frac{\text{weight}\text{(g)}}{\text{Atomic}\text{mass}\text{(g}{\text{mole}}^{\text{-1}}\text{)}}$

$\begin{array}{l}\text{Number}\text{of}\text{moles of electrons = Number of moles of Silver}\\ \text{then}\\ \text{Charges}\text{=}\text{Number}\text{of}\text{moles of electrons×Faraday}{\text{constant (96500C/mole e}}^{-}\text{)}\end{array}$

Half-cell reaction for cell-2 was given below.

${\text{Cathode (reduction) Cu}}^{\text{2+}}{}_{\text{(aq)}}{\text{+2e}}^{\text{-}}\stackrel{}{\to }{\text{Cu}}_{\text{(s)}}$

Since they are connected in series, same amount of charges will be passing through both the cell. From the calculated charges the number of moles of electrons utilized can be calculated.

$\text{Number}\text{of}\text{moles of electrons (}n\text{)}\text{=}\text{Charges}\text{×Faraday}{\text{constant (96500C/mole e}}^{-}\text{)}$

From the cell reaction it was known that two mole of electron will be needed to produce one mole of copper,

So

$\begin{array}{l}numberofmolesof\text{copper}=nmoleof{e}^{-}\times \frac{\text{1mole}\text{Cu}}{2mole{e}^{-}}\\ \text{Weight}\text{of}\text{Copper}=numberofmolesof\text{copper }\times \text{ atomic mass}\end{array}$

Since time was given the amperes of current flowing through the circuit can be calculated by the equation given below

$\text{Current (A) =}\frac{\text{charges (C)}}{\text{time(s)}}$

To find: The amount of copper deposited and current consumed in a CuCl₂ cell connected in series with the cell containing AgNO_3.