Concept explainers
A constant
(a)
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.
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.
Half-cell reaction for cell-2 was given below.
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.
From the cell reaction it was known that two mole of electron will be needed to produce one mole of copper,
So
Since time was given the amperes of current flowing through the circuit can be calculated by the equation given below
To find: The amount of copper deposited and current consumed in a CuCl2 cell connected in series with the cell containing AgNO3.
Answer to Problem 18.54QP
Since cells containing AgNO3 and CuCl2 were connected in series, the same amount of electricity will pass through both of them. So first let us calculate the coulombs of charges passed through the AgNO3 cell, from the quantity of silver deposited.
Since one mole of electron is needed for the reduction of one mole of Ag+
Then
As state above the coulombs of current passing through the CuCl2 cell also will be
Then
Explanation of Solution
Since cells containing AgNO3 and CuCl2 were connected in series, the same amount of electricity will pass through both of them. So first let us calculate the coulombs of charges passed through the AgNO3 cell, which resulted in the deposition of 2.0 g of silver.
Since one mole of electron is needed for the reduction of one mole of Ag+
Then
As state above the coulombs of charges passing through the CuCl2 cell also will be
In the present case the quantity of silver deposited upon electrolysis of AgNO3 was given, from that the quantity of charges consumed was calculated and utilized for the determination of mass of Cu deposited in the second cell.
(b)
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.
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.
Half-cell reaction for cell-2 was given below.
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.
From the cell reaction it was known that two mole of electron will be needed to produce one mole of copper,
So
Since time was given the amperes of current flowing through the circuit can be calculated by the equation given below
To find: The amount of copper deposited and current consumed in a CuCl2 cell connected in series with the cell containing AgNO3.
Answer to Problem 18.54QP
The amount of current passing through the CuCl2 cell for 3.75 h can calculate as follows
Time = 3.75 h or 13500 s
Explanation of Solution
The amount of current passing through the CuCl2 cell for 3.75 h can calculate as follows
Time = 3.75h or 13500s.
On dividing the coulombs of charges produced by time in seconds the amount of current passing through the second cell was calculated as 0.1325A.
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Chapter 18 Solutions
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