Chapter
stringclasses 18
values | sentence_range
stringlengths 3
9
| Text
stringlengths 7
7.34k
|
|---|---|---|
1 | 2005-2008 | 13)
Rationalised 2023-24
39
Electrochemistry
If the circuit in Daniell cell (Fig 2 1) is closed then we note that the reaction
Zn(s) + Cu2+(aq) ® Zn2+(aq) + Cu(s)
(2 1)
takes place and as time passes, the concentration of Zn2+ keeps
on increasing while the concentration of Cu2+ keeps on decreasing |
1 | 2006-2009 | 2 1) is closed then we note that the reaction
Zn(s) + Cu2+(aq) ® Zn2+(aq) + Cu(s)
(2 1)
takes place and as time passes, the concentration of Zn2+ keeps
on increasing while the concentration of Cu2+ keeps on decreasing At the same time voltage of the cell as read on the voltmeter keeps
on decreasing |
1 | 2007-2010 | 1) is closed then we note that the reaction
Zn(s) + Cu2+(aq) ® Zn2+(aq) + Cu(s)
(2 1)
takes place and as time passes, the concentration of Zn2+ keeps
on increasing while the concentration of Cu2+ keeps on decreasing At the same time voltage of the cell as read on the voltmeter keeps
on decreasing After some time, we shall note that there is no change
in the concentration of Cu2+ and Zn2+ ions and at the same time,
voltmeter gives zero reading |
1 | 2008-2011 | 1)
takes place and as time passes, the concentration of Zn2+ keeps
on increasing while the concentration of Cu2+ keeps on decreasing At the same time voltage of the cell as read on the voltmeter keeps
on decreasing After some time, we shall note that there is no change
in the concentration of Cu2+ and Zn2+ ions and at the same time,
voltmeter gives zero reading This indicates that equilibrium has been
attained |
1 | 2009-2012 | At the same time voltage of the cell as read on the voltmeter keeps
on decreasing After some time, we shall note that there is no change
in the concentration of Cu2+ and Zn2+ ions and at the same time,
voltmeter gives zero reading This indicates that equilibrium has been
attained In this situation the Nernst equation may be written as:
E(cell) = 0 =
(
)
o
Ecell
–
2 |
1 | 2010-2013 | After some time, we shall note that there is no change
in the concentration of Cu2+ and Zn2+ ions and at the same time,
voltmeter gives zero reading This indicates that equilibrium has been
attained In this situation the Nernst equation may be written as:
E(cell) = 0 =
(
)
o
Ecell
–
2 303
2
log [Zn
]
[Cu
]
2
2
RT
F
+
+
or
(
)
o
Ecell
=
2
2
2 |
1 | 2011-2014 | This indicates that equilibrium has been
attained In this situation the Nernst equation may be written as:
E(cell) = 0 =
(
)
o
Ecell
–
2 303
2
log [Zn
]
[Cu
]
2
2
RT
F
+
+
or
(
)
o
Ecell
=
2
2
2 303
[Zn
]
log
2
[Cu
]
RT
F
But at equilibrium,
[
]
[
]
Zn
Cu
2
2
+
+ = Kc for the reaction 2 |
1 | 2012-2015 | In this situation the Nernst equation may be written as:
E(cell) = 0 =
(
)
o
Ecell
–
2 303
2
log [Zn
]
[Cu
]
2
2
RT
F
+
+
or
(
)
o
Ecell
=
2
2
2 303
[Zn
]
log
2
[Cu
]
RT
F
But at equilibrium,
[
]
[
]
Zn
Cu
2
2
+
+ = Kc for the reaction 2 1
and at T = 298K the above equation can be written as
(
)
o
Ecell
= 0 059
2 |
1 | 2013-2016 | 303
2
log [Zn
]
[Cu
]
2
2
RT
F
+
+
or
(
)
o
Ecell
=
2
2
2 303
[Zn
]
log
2
[Cu
]
RT
F
But at equilibrium,
[
]
[
]
Zn
Cu
2
2
+
+ = Kc for the reaction 2 1
and at T = 298K the above equation can be written as
(
)
o
Ecell
= 0 059
2 V log KC = 1 |
1 | 2014-2017 | 303
[Zn
]
log
2
[Cu
]
RT
F
But at equilibrium,
[
]
[
]
Zn
Cu
2
2
+
+ = Kc for the reaction 2 1
and at T = 298K the above equation can be written as
(
)
o
Ecell
= 0 059
2 V log KC = 1 1 V (
(
)
o
Ecell
= 1 |
1 | 2015-2018 | 1
and at T = 298K the above equation can be written as
(
)
o
Ecell
= 0 059
2 V log KC = 1 1 V (
(
)
o
Ecell
= 1 1V)
log KC = (1 |
1 | 2016-2019 | V log KC = 1 1 V (
(
)
o
Ecell
= 1 1V)
log KC = (1 1V × 2)
37 |
1 | 2017-2020 | 1 V (
(
)
o
Ecell
= 1 1V)
log KC = (1 1V × 2)
37 288
0 |
1 | 2018-2021 | 1V)
log KC = (1 1V × 2)
37 288
0 059 V
KC = 2 × 1037 at 298K |
1 | 2019-2022 | 1V × 2)
37 288
0 059 V
KC = 2 × 1037 at 298K In general,
(
)
o
Ecell
= 2 |
1 | 2020-2023 | 288
0 059 V
KC = 2 × 1037 at 298K In general,
(
)
o
Ecell
= 2 303RT
nF
log KC
(2 |
1 | 2021-2024 | 059 V
KC = 2 × 1037 at 298K In general,
(
)
o
Ecell
= 2 303RT
nF
log KC
(2 14)
Thus, Eq |
1 | 2022-2025 | In general,
(
)
o
Ecell
= 2 303RT
nF
log KC
(2 14)
Thus, Eq (2 |
1 | 2023-2026 | 303RT
nF
log KC
(2 14)
Thus, Eq (2 14) gives a relationship between equilibrium constant
of the reaction and standard potential of the cell in which that reaction
takes place |
1 | 2024-2027 | 14)
Thus, Eq (2 14) gives a relationship between equilibrium constant
of the reaction and standard potential of the cell in which that reaction
takes place Thus, equilibrium constants of the reaction, difficult to
measure otherwise, can be calculated from the corresponding Eo value
of the cell |
1 | 2025-2028 | (2 14) gives a relationship between equilibrium constant
of the reaction and standard potential of the cell in which that reaction
takes place Thus, equilibrium constants of the reaction, difficult to
measure otherwise, can be calculated from the corresponding Eo value
of the cell 2 |
1 | 2026-2029 | 14) gives a relationship between equilibrium constant
of the reaction and standard potential of the cell in which that reaction
takes place Thus, equilibrium constants of the reaction, difficult to
measure otherwise, can be calculated from the corresponding Eo value
of the cell 2 3 |
1 | 2027-2030 | Thus, equilibrium constants of the reaction, difficult to
measure otherwise, can be calculated from the corresponding Eo value
of the cell 2 3 1 Equilibrium
Constant
from Nernst
Equation
Example 2 |
1 | 2028-2031 | 2 3 1 Equilibrium
Constant
from Nernst
Equation
Example 2 1
Example 2 |
1 | 2029-2032 | 3 1 Equilibrium
Constant
from Nernst
Equation
Example 2 1
Example 2 1
Example 2 |
1 | 2030-2033 | 1 Equilibrium
Constant
from Nernst
Equation
Example 2 1
Example 2 1
Example 2 1
Example 2 |
1 | 2031-2034 | 1
Example 2 1
Example 2 1
Example 2 1
Example 2 |
1 | 2032-2035 | 1
Example 2 1
Example 2 1
Example 2 1
Represent the cell in which the following reaction takes place
Mg(s) + 2Ag+(0 |
1 | 2033-2036 | 1
Example 2 1
Example 2 1
Represent the cell in which the following reaction takes place
Mg(s) + 2Ag+(0 0001M) ® Mg2+(0 |
1 | 2034-2037 | 1
Example 2 1
Represent the cell in which the following reaction takes place
Mg(s) + 2Ag+(0 0001M) ® Mg2+(0 130M) + 2Ag(s)
Calculate its E(cell) if
(
)
o
Ecell
= 3 |
1 | 2035-2038 | 1
Represent the cell in which the following reaction takes place
Mg(s) + 2Ag+(0 0001M) ® Mg2+(0 130M) + 2Ag(s)
Calculate its E(cell) if
(
)
o
Ecell
= 3 17 V |
1 | 2036-2039 | 0001M) ® Mg2+(0 130M) + 2Ag(s)
Calculate its E(cell) if
(
)
o
Ecell
= 3 17 V The cell can be written as Mgú Mg2+(0 |
1 | 2037-2040 | 130M) + 2Ag(s)
Calculate its E(cell) if
(
)
o
Ecell
= 3 17 V The cell can be written as Mgú Mg2+(0 130M)úú Ag+(0 |
1 | 2038-2041 | 17 V The cell can be written as Mgú Mg2+(0 130M)úú Ag+(0 0001M)ú Ag
Ecell
=
(
)
+
+
2
o
cell
2
Mg
–RT
2Fln
Ag
E
= 3 |
1 | 2039-2042 | The cell can be written as Mgú Mg2+(0 130M)úú Ag+(0 0001M)ú Ag
Ecell
=
(
)
+
+
2
o
cell
2
Mg
–RT
2Fln
Ag
E
= 3 17 V –
0 059
2
0 0001 2 |
1 | 2040-2043 | 130M)úú Ag+(0 0001M)ú Ag
Ecell
=
(
)
+
+
2
o
cell
2
Mg
–RT
2Fln
Ag
E
= 3 17 V –
0 059
2
0 0001 2 log
( |
1 | 2041-2044 | 0001M)ú Ag
Ecell
=
(
)
+
+
2
o
cell
2
Mg
–RT
2Fln
Ag
E
= 3 17 V –
0 059
2
0 0001 2 log
( )
V
0 |
1 | 2042-2045 | 17 V –
0 059
2
0 0001 2 log
( )
V
0 130
= 3 |
1 | 2043-2046 | log
( )
V
0 130
= 3 17 V – 0 |
1 | 2044-2047 | )
V
0 130
= 3 17 V – 0 21V = 2 |
1 | 2045-2048 | 130
= 3 17 V – 0 21V = 2 96 V |
1 | 2046-2049 | 17 V – 0 21V = 2 96 V Solution
Solution
Solution
Solution
Solution
Rationalised 2023-24
40
Chemistry
The standard electrode potential for Daniell cell is 1 |
1 | 2047-2050 | 21V = 2 96 V Solution
Solution
Solution
Solution
Solution
Rationalised 2023-24
40
Chemistry
The standard electrode potential for Daniell cell is 1 1V |
1 | 2048-2051 | 96 V Solution
Solution
Solution
Solution
Solution
Rationalised 2023-24
40
Chemistry
The standard electrode potential for Daniell cell is 1 1V Calculate
the standard Gibbs energy for the reaction:
Zn(s) + Cu2+(aq) ¾® Zn2+(aq) + Cu(s)
DrGo = – nF
o
(cell)
E
n in the above equation is 2, F = 96487 C mol–1 and
(
)
o
Ecell
= 1 |
1 | 2049-2052 | Solution
Solution
Solution
Solution
Solution
Rationalised 2023-24
40
Chemistry
The standard electrode potential for Daniell cell is 1 1V Calculate
the standard Gibbs energy for the reaction:
Zn(s) + Cu2+(aq) ¾® Zn2+(aq) + Cu(s)
DrGo = – nF
o
(cell)
E
n in the above equation is 2, F = 96487 C mol–1 and
(
)
o
Ecell
= 1 1 V
Therefore, DrGo = – 2 × 1 |
1 | 2050-2053 | 1V Calculate
the standard Gibbs energy for the reaction:
Zn(s) + Cu2+(aq) ¾® Zn2+(aq) + Cu(s)
DrGo = – nF
o
(cell)
E
n in the above equation is 2, F = 96487 C mol–1 and
(
)
o
Ecell
= 1 1 V
Therefore, DrGo = – 2 × 1 1V × 96487 C mol–1
= – 21227 J mol–1
= – 212 |
1 | 2051-2054 | Calculate
the standard Gibbs energy for the reaction:
Zn(s) + Cu2+(aq) ¾® Zn2+(aq) + Cu(s)
DrGo = – nF
o
(cell)
E
n in the above equation is 2, F = 96487 C mol–1 and
(
)
o
Ecell
= 1 1 V
Therefore, DrGo = – 2 × 1 1V × 96487 C mol–1
= – 21227 J mol–1
= – 212 27 kJ mol–1
Example 2 |
1 | 2052-2055 | 1 V
Therefore, DrGo = – 2 × 1 1V × 96487 C mol–1
= – 21227 J mol–1
= – 212 27 kJ mol–1
Example 2 3
Example 2 |
1 | 2053-2056 | 1V × 96487 C mol–1
= – 21227 J mol–1
= – 212 27 kJ mol–1
Example 2 3
Example 2 3
Example 2 |
1 | 2054-2057 | 27 kJ mol–1
Example 2 3
Example 2 3
Example 2 3
Example 2 |
1 | 2055-2058 | 3
Example 2 3
Example 2 3
Example 2 3
Example 2 |
1 | 2056-2059 | 3
Example 2 3
Example 2 3
Example 2 3
Solution
Solution
Solution
Solution
Solution
Electrical work done in one second is equal to electrical potential
multiplied by total charge passed |
1 | 2057-2060 | 3
Example 2 3
Example 2 3
Solution
Solution
Solution
Solution
Solution
Electrical work done in one second is equal to electrical potential
multiplied by total charge passed If we want to obtain maximum work
from a galvanic cell then charge has to be passed reversibly |
1 | 2058-2061 | 3
Example 2 3
Solution
Solution
Solution
Solution
Solution
Electrical work done in one second is equal to electrical potential
multiplied by total charge passed If we want to obtain maximum work
from a galvanic cell then charge has to be passed reversibly The
reversible work done by a galvanic cell is equal to decrease in its Gibbs
energy and therefore, if the emf of the cell is E and nF is the amount
of charge passed and DrG is the Gibbs energy of the reaction, then
DrG = – nFE(cell)
(2 |
1 | 2059-2062 | 3
Solution
Solution
Solution
Solution
Solution
Electrical work done in one second is equal to electrical potential
multiplied by total charge passed If we want to obtain maximum work
from a galvanic cell then charge has to be passed reversibly The
reversible work done by a galvanic cell is equal to decrease in its Gibbs
energy and therefore, if the emf of the cell is E and nF is the amount
of charge passed and DrG is the Gibbs energy of the reaction, then
DrG = – nFE(cell)
(2 15)
It may be remembered that E(cell) is an intensive parameter but DrG
is an extensive thermodynamic property and the value depends on n |
1 | 2060-2063 | If we want to obtain maximum work
from a galvanic cell then charge has to be passed reversibly The
reversible work done by a galvanic cell is equal to decrease in its Gibbs
energy and therefore, if the emf of the cell is E and nF is the amount
of charge passed and DrG is the Gibbs energy of the reaction, then
DrG = – nFE(cell)
(2 15)
It may be remembered that E(cell) is an intensive parameter but DrG
is an extensive thermodynamic property and the value depends on n Thus, if we write the reaction
Zn(s) + Cu2+(aq) ¾® Zn2+(aq) + Cu(s)
(2 |
1 | 2061-2064 | The
reversible work done by a galvanic cell is equal to decrease in its Gibbs
energy and therefore, if the emf of the cell is E and nF is the amount
of charge passed and DrG is the Gibbs energy of the reaction, then
DrG = – nFE(cell)
(2 15)
It may be remembered that E(cell) is an intensive parameter but DrG
is an extensive thermodynamic property and the value depends on n Thus, if we write the reaction
Zn(s) + Cu2+(aq) ¾® Zn2+(aq) + Cu(s)
(2 1)
DrG = – 2FE(cell)
but when we write the reaction
2 Zn (s) + 2 Cu2+(aq) ¾®2 Zn2+(aq) + 2Cu(s)
DrG = – 4FE(cell)
If the concentration of all the reacting species is unity, then
E(cell) =
(
)
o
Ecell
and we have
DrGo = – nF
o
E(cell)
(2 |
1 | 2062-2065 | 15)
It may be remembered that E(cell) is an intensive parameter but DrG
is an extensive thermodynamic property and the value depends on n Thus, if we write the reaction
Zn(s) + Cu2+(aq) ¾® Zn2+(aq) + Cu(s)
(2 1)
DrG = – 2FE(cell)
but when we write the reaction
2 Zn (s) + 2 Cu2+(aq) ¾®2 Zn2+(aq) + 2Cu(s)
DrG = – 4FE(cell)
If the concentration of all the reacting species is unity, then
E(cell) =
(
)
o
Ecell
and we have
DrGo = – nF
o
E(cell)
(2 16)
Thus, from the measurement of
(
)
o
Ecell
we can obtain an important
thermodynamic quantity, DrGo, standard Gibbs energy of the reaction |
1 | 2063-2066 | Thus, if we write the reaction
Zn(s) + Cu2+(aq) ¾® Zn2+(aq) + Cu(s)
(2 1)
DrG = – 2FE(cell)
but when we write the reaction
2 Zn (s) + 2 Cu2+(aq) ¾®2 Zn2+(aq) + 2Cu(s)
DrG = – 4FE(cell)
If the concentration of all the reacting species is unity, then
E(cell) =
(
)
o
Ecell
and we have
DrGo = – nF
o
E(cell)
(2 16)
Thus, from the measurement of
(
)
o
Ecell
we can obtain an important
thermodynamic quantity, DrGo, standard Gibbs energy of the reaction From the latter we can calculate equilibrium constant by the equation:
DrGo = –RT ln K |
1 | 2064-2067 | 1)
DrG = – 2FE(cell)
but when we write the reaction
2 Zn (s) + 2 Cu2+(aq) ¾®2 Zn2+(aq) + 2Cu(s)
DrG = – 4FE(cell)
If the concentration of all the reacting species is unity, then
E(cell) =
(
)
o
Ecell
and we have
DrGo = – nF
o
E(cell)
(2 16)
Thus, from the measurement of
(
)
o
Ecell
we can obtain an important
thermodynamic quantity, DrGo, standard Gibbs energy of the reaction From the latter we can calculate equilibrium constant by the equation:
DrGo = –RT ln K 2 |
1 | 2065-2068 | 16)
Thus, from the measurement of
(
)
o
Ecell
we can obtain an important
thermodynamic quantity, DrGo, standard Gibbs energy of the reaction From the latter we can calculate equilibrium constant by the equation:
DrGo = –RT ln K 2 3 |
1 | 2066-2069 | From the latter we can calculate equilibrium constant by the equation:
DrGo = –RT ln K 2 3 2 Electro-
chemical
Cell and
Gibbs
Energy of
the Reaction
Calculate the equilibrium constant of the reaction:
Cu(s) + 2Ag+(aq) ® Cu2+(aq) + 2Ag(s)
(
)
o
Ecell
= 0 |
1 | 2067-2070 | 2 3 2 Electro-
chemical
Cell and
Gibbs
Energy of
the Reaction
Calculate the equilibrium constant of the reaction:
Cu(s) + 2Ag+(aq) ® Cu2+(aq) + 2Ag(s)
(
)
o
Ecell
= 0 46 V
(
)
o
Ecell
= 0 059
2 |
1 | 2068-2071 | 3 2 Electro-
chemical
Cell and
Gibbs
Energy of
the Reaction
Calculate the equilibrium constant of the reaction:
Cu(s) + 2Ag+(aq) ® Cu2+(aq) + 2Ag(s)
(
)
o
Ecell
= 0 46 V
(
)
o
Ecell
= 0 059
2 V log KC = 0 |
1 | 2069-2072 | 2 Electro-
chemical
Cell and
Gibbs
Energy of
the Reaction
Calculate the equilibrium constant of the reaction:
Cu(s) + 2Ag+(aq) ® Cu2+(aq) + 2Ag(s)
(
)
o
Ecell
= 0 46 V
(
)
o
Ecell
= 0 059
2 V log KC = 0 46 V or
log
KC =
0 46 |
1 | 2070-2073 | 46 V
(
)
o
Ecell
= 0 059
2 V log KC = 0 46 V or
log
KC =
0 46 0 0592
V
V
×
= 15 |
1 | 2071-2074 | V log KC = 0 46 V or
log
KC =
0 46 0 0592
V
V
×
= 15 6
KC = 3 |
1 | 2072-2075 | 46 V or
log
KC =
0 46 0 0592
V
V
×
= 15 6
KC = 3 92 × 1015
Example 2 |
1 | 2073-2076 | 0 0592
V
V
×
= 15 6
KC = 3 92 × 1015
Example 2 2
Example 2 |
1 | 2074-2077 | 6
KC = 3 92 × 1015
Example 2 2
Example 2 2
Example 2 |
1 | 2075-2078 | 92 × 1015
Example 2 2
Example 2 2
Example 2 2
Example 2 |
1 | 2076-2079 | 2
Example 2 2
Example 2 2
Example 2 2
Example 2 |
1 | 2077-2080 | 2
Example 2 2
Example 2 2
Example 2 2
Solution
Solution
Solution
Solution
Solution
Rationalised 2023-24
41
Electrochemistry
It is necessary to define a few terms before we consider the subject of
conductance of electricity through electrolytic solutions |
1 | 2078-2081 | 2
Example 2 2
Example 2 2
Solution
Solution
Solution
Solution
Solution
Rationalised 2023-24
41
Electrochemistry
It is necessary to define a few terms before we consider the subject of
conductance of electricity through electrolytic solutions The electrical
resistance is represented by the symbol ‘R’ and it is measured in ohm (W)
which in terms of SI base units is equal to (kg m2)/(S3 A2) |
1 | 2079-2082 | 2
Example 2 2
Solution
Solution
Solution
Solution
Solution
Rationalised 2023-24
41
Electrochemistry
It is necessary to define a few terms before we consider the subject of
conductance of electricity through electrolytic solutions The electrical
resistance is represented by the symbol ‘R’ and it is measured in ohm (W)
which in terms of SI base units is equal to (kg m2)/(S3 A2) It can be
measured with the help of a Wheatstone bridge with which you are
familiar from your study of physics |
1 | 2080-2083 | 2
Solution
Solution
Solution
Solution
Solution
Rationalised 2023-24
41
Electrochemistry
It is necessary to define a few terms before we consider the subject of
conductance of electricity through electrolytic solutions The electrical
resistance is represented by the symbol ‘R’ and it is measured in ohm (W)
which in terms of SI base units is equal to (kg m2)/(S3 A2) It can be
measured with the help of a Wheatstone bridge with which you are
familiar from your study of physics The electrical resistance of any object
is directly proportional to its length, l, and inversely proportional to its
area of cross section, A |
1 | 2081-2084 | The electrical
resistance is represented by the symbol ‘R’ and it is measured in ohm (W)
which in terms of SI base units is equal to (kg m2)/(S3 A2) It can be
measured with the help of a Wheatstone bridge with which you are
familiar from your study of physics The electrical resistance of any object
is directly proportional to its length, l, and inversely proportional to its
area of cross section, A That is,
R µ l
A or R = r l
A
(2 |
1 | 2082-2085 | It can be
measured with the help of a Wheatstone bridge with which you are
familiar from your study of physics The electrical resistance of any object
is directly proportional to its length, l, and inversely proportional to its
area of cross section, A That is,
R µ l
A or R = r l
A
(2 17)
The constant of proportionality, r (Greek, rho), is called resistivity
(specific resistance) |
1 | 2083-2086 | The electrical resistance of any object
is directly proportional to its length, l, and inversely proportional to its
area of cross section, A That is,
R µ l
A or R = r l
A
(2 17)
The constant of proportionality, r (Greek, rho), is called resistivity
(specific resistance) Its SI units are ohm metre (W m) and quite often
its submultiple, ohm centimetre (W cm) is also used |
1 | 2084-2087 | That is,
R µ l
A or R = r l
A
(2 17)
The constant of proportionality, r (Greek, rho), is called resistivity
(specific resistance) Its SI units are ohm metre (W m) and quite often
its submultiple, ohm centimetre (W cm) is also used IUPAC recommends
the use of the term resistivity over specific resistance and hence in the
rest of the book we shall use the term resistivity |
1 | 2085-2088 | 17)
The constant of proportionality, r (Greek, rho), is called resistivity
(specific resistance) Its SI units are ohm metre (W m) and quite often
its submultiple, ohm centimetre (W cm) is also used IUPAC recommends
the use of the term resistivity over specific resistance and hence in the
rest of the book we shall use the term resistivity Physically, the resistivity
for a substance is its resistance when it is one metre long and its area
of cross section is one m2 |
1 | 2086-2089 | Its SI units are ohm metre (W m) and quite often
its submultiple, ohm centimetre (W cm) is also used IUPAC recommends
the use of the term resistivity over specific resistance and hence in the
rest of the book we shall use the term resistivity Physically, the resistivity
for a substance is its resistance when it is one metre long and its area
of cross section is one m2 It can be seen that:
1 W m = 100 W cm or 1 W cm = 0 |
1 | 2087-2090 | IUPAC recommends
the use of the term resistivity over specific resistance and hence in the
rest of the book we shall use the term resistivity Physically, the resistivity
for a substance is its resistance when it is one metre long and its area
of cross section is one m2 It can be seen that:
1 W m = 100 W cm or 1 W cm = 0 01 W m
The inverse of resistance, R, is called conductance, G, and we have
the relation:
G = 1
R = ρ
A=κ
A
l
l
(2 |
1 | 2088-2091 | Physically, the resistivity
for a substance is its resistance when it is one metre long and its area
of cross section is one m2 It can be seen that:
1 W m = 100 W cm or 1 W cm = 0 01 W m
The inverse of resistance, R, is called conductance, G, and we have
the relation:
G = 1
R = ρ
A=κ
A
l
l
(2 18)
The SI unit of conductance is siemens, represented by the symbol
‘S’ and is equal to ohm–1 (also known as mho) or W–1 |
1 | 2089-2092 | It can be seen that:
1 W m = 100 W cm or 1 W cm = 0 01 W m
The inverse of resistance, R, is called conductance, G, and we have
the relation:
G = 1
R = ρ
A=κ
A
l
l
(2 18)
The SI unit of conductance is siemens, represented by the symbol
‘S’ and is equal to ohm–1 (also known as mho) or W–1 The inverse of
resistivity, called conductivity (specific conductance) is represented by
the symbol, k (Greek, kappa) |
1 | 2090-2093 | 01 W m
The inverse of resistance, R, is called conductance, G, and we have
the relation:
G = 1
R = ρ
A=κ
A
l
l
(2 18)
The SI unit of conductance is siemens, represented by the symbol
‘S’ and is equal to ohm–1 (also known as mho) or W–1 The inverse of
resistivity, called conductivity (specific conductance) is represented by
the symbol, k (Greek, kappa) IUPAC has recommended the use of term
conductivity over specific conductance and hence we shall use the term
conductivity in the rest of the book |
1 | 2091-2094 | 18)
The SI unit of conductance is siemens, represented by the symbol
‘S’ and is equal to ohm–1 (also known as mho) or W–1 The inverse of
resistivity, called conductivity (specific conductance) is represented by
the symbol, k (Greek, kappa) IUPAC has recommended the use of term
conductivity over specific conductance and hence we shall use the term
conductivity in the rest of the book The SI units of conductivity are
S m–1 but quite often, k is expressed in S cm–1 |
1 | 2092-2095 | The inverse of
resistivity, called conductivity (specific conductance) is represented by
the symbol, k (Greek, kappa) IUPAC has recommended the use of term
conductivity over specific conductance and hence we shall use the term
conductivity in the rest of the book The SI units of conductivity are
S m–1 but quite often, k is expressed in S cm–1 Conductivity of a
material in S m–1 is its conductance when it is 1 m long and its area
of cross section is 1 m2 |
1 | 2093-2096 | IUPAC has recommended the use of term
conductivity over specific conductance and hence we shall use the term
conductivity in the rest of the book The SI units of conductivity are
S m–1 but quite often, k is expressed in S cm–1 Conductivity of a
material in S m–1 is its conductance when it is 1 m long and its area
of cross section is 1 m2 It may be noted that 1 S cm–1 = 100 S m–1 |
1 | 2094-2097 | The SI units of conductivity are
S m–1 but quite often, k is expressed in S cm–1 Conductivity of a
material in S m–1 is its conductance when it is 1 m long and its area
of cross section is 1 m2 It may be noted that 1 S cm–1 = 100 S m–1 2 |
1 | 2095-2098 | Conductivity of a
material in S m–1 is its conductance when it is 1 m long and its area
of cross section is 1 m2 It may be noted that 1 S cm–1 = 100 S m–1 2 4
2 |
1 | 2096-2099 | It may be noted that 1 S cm–1 = 100 S m–1 2 4
2 4
2 |
1 | 2097-2100 | 2 4
2 4
2 4
2 |
1 | 2098-2101 | 4
2 4
2 4
2 4
2 |
1 | 2099-2102 | 4
2 4
2 4
2 4 Conductance
Conductance
Conductance
Conductance
Conductance
of Electrolytic
of Electrolytic
of Electrolytic
of Electrolytic
of Electrolytic
Solutions
Solutions
Solutions
Solutions
Solutions
Intext Questions
Intext Questions
Intext Questions
Intext Questions
Intext Questions
2 |
1 | 2100-2103 | 4
2 4
2 4 Conductance
Conductance
Conductance
Conductance
Conductance
of Electrolytic
of Electrolytic
of Electrolytic
of Electrolytic
of Electrolytic
Solutions
Solutions
Solutions
Solutions
Solutions
Intext Questions
Intext Questions
Intext Questions
Intext Questions
Intext Questions
2 4 Calculate the potential of hydrogen electrode in contact with a solution
whose pH is 10 |
1 | 2101-2104 | 4
2 4 Conductance
Conductance
Conductance
Conductance
Conductance
of Electrolytic
of Electrolytic
of Electrolytic
of Electrolytic
of Electrolytic
Solutions
Solutions
Solutions
Solutions
Solutions
Intext Questions
Intext Questions
Intext Questions
Intext Questions
Intext Questions
2 4 Calculate the potential of hydrogen electrode in contact with a solution
whose pH is 10 2 |
1 | 2102-2105 | 4 Conductance
Conductance
Conductance
Conductance
Conductance
of Electrolytic
of Electrolytic
of Electrolytic
of Electrolytic
of Electrolytic
Solutions
Solutions
Solutions
Solutions
Solutions
Intext Questions
Intext Questions
Intext Questions
Intext Questions
Intext Questions
2 4 Calculate the potential of hydrogen electrode in contact with a solution
whose pH is 10 2 5 Calculate the emf of the cell in which the following reaction takes place:
Ni(s) + 2Ag+ (0 |
1 | 2103-2106 | 4 Calculate the potential of hydrogen electrode in contact with a solution
whose pH is 10 2 5 Calculate the emf of the cell in which the following reaction takes place:
Ni(s) + 2Ag+ (0 002 M) ® Ni2+ (0 |
1 | 2104-2107 | 2 5 Calculate the emf of the cell in which the following reaction takes place:
Ni(s) + 2Ag+ (0 002 M) ® Ni2+ (0 160 M) + 2Ag(s)
Given that
o
Ecell
= 1 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.