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1 | 2405-2408 | m
= 5
1
3
1
4 95 10
Scm
1000cm
0 001028 mol L
L
c
= 48 15 S cm3 mol–1
a
=
2
1
2
1
48 15 Scm mol
390 |
1 | 2406-2409 | 5
1
3
1
4 95 10
Scm
1000cm
0 001028 mol L
L
c
= 48 15 S cm3 mol–1
a
=
2
1
2
1
48 15 Scm mol
390 5 Scm mol
m
m
= 0 |
1 | 2407-2410 | 15 S cm3 mol–1
a
=
2
1
2
1
48 15 Scm mol
390 5 Scm mol
m
m
= 0 1233
k
=
|
1 | 2408-2411 | 15 Scm mol
390 5 Scm mol
m
m
= 0 1233
k
=
( |
1 | 2409-2412 | 5 Scm mol
m
m
= 0 1233
k
=
( )
1c |
1 | 2410-2413 | 1233
k
=
( )
1c –1
2
2
0 001028molL
0 1233
1 0 1233
= 1 |
1 | 2411-2414 | ( )
1c –1
2
2
0 001028molL
0 1233
1 0 1233
= 1 78 × 10–5 mol L–1
Example 2 |
1 | 2412-2415 | )
1c –1
2
2
0 001028molL
0 1233
1 0 1233
= 1 78 × 10–5 mol L–1
Example 2 8
Example 2 |
1 | 2413-2416 | –1
2
2
0 001028molL
0 1233
1 0 1233
= 1 78 × 10–5 mol L–1
Example 2 8
Example 2 8
Example 2 |
1 | 2414-2417 | 78 × 10–5 mol L–1
Example 2 8
Example 2 8
Example 2 8
Example 2 |
1 | 2415-2418 | 8
Example 2 8
Example 2 8
Example 2 8
Example 2 |
1 | 2416-2419 | 8
Example 2 8
Example 2 8
Example 2 8
Solution
Solution
Solution
Solution
Solution
Example 2 |
1 | 2417-2420 | 8
Example 2 8
Example 2 8
Solution
Solution
Solution
Solution
Solution
Example 2 9
Example 2 |
1 | 2418-2421 | 8
Example 2 8
Solution
Solution
Solution
Solution
Solution
Example 2 9
Example 2 9
Example 2 |
1 | 2419-2422 | 8
Solution
Solution
Solution
Solution
Solution
Example 2 9
Example 2 9
Example 2 9
Example 2 |
1 | 2420-2423 | 9
Example 2 9
Example 2 9
Example 2 9
Example 2 |
1 | 2421-2424 | 9
Example 2 9
Example 2 9
Example 2 9
Solution
Solution
Solution
Solution
Solution
Example 2 |
1 | 2422-2425 | 9
Example 2 9
Example 2 9
Solution
Solution
Solution
Solution
Solution
Example 2 7
Example 2 |
1 | 2423-2426 | 9
Example 2 9
Solution
Solution
Solution
Solution
Solution
Example 2 7
Example 2 7
Example 2 |
1 | 2424-2427 | 9
Solution
Solution
Solution
Solution
Solution
Example 2 7
Example 2 7
Example 2 7
Example 2 |
1 | 2425-2428 | 7
Example 2 7
Example 2 7
Example 2 7
Example 2 |
1 | 2426-2429 | 7
Example 2 7
Example 2 7
Example 2 7
Solution
Solution
Solution
Solution
Solution
Rationalised 2023-24
51
Electrochemistry
In an electrolytic cell external source of voltage is used to bring about
a chemical reaction |
1 | 2427-2430 | 7
Example 2 7
Example 2 7
Solution
Solution
Solution
Solution
Solution
Rationalised 2023-24
51
Electrochemistry
In an electrolytic cell external source of voltage is used to bring about
a chemical reaction The electrochemical processes are of great importance
in the laboratory and the chemical industry |
1 | 2428-2431 | 7
Example 2 7
Solution
Solution
Solution
Solution
Solution
Rationalised 2023-24
51
Electrochemistry
In an electrolytic cell external source of voltage is used to bring about
a chemical reaction The electrochemical processes are of great importance
in the laboratory and the chemical industry One of the simplest electrolytic
cell consists of two copper strips dipping in an aqueous solution of
copper sulphate |
1 | 2429-2432 | 7
Solution
Solution
Solution
Solution
Solution
Rationalised 2023-24
51
Electrochemistry
In an electrolytic cell external source of voltage is used to bring about
a chemical reaction The electrochemical processes are of great importance
in the laboratory and the chemical industry One of the simplest electrolytic
cell consists of two copper strips dipping in an aqueous solution of
copper sulphate If a DC voltage is applied to the two electrodes, then
Cu 2+ ions discharge at the cathode (negatively charged) and the following
reaction takes place:
Cu2+(aq) + 2e– ® Cu (s)
(2 |
1 | 2430-2433 | The electrochemical processes are of great importance
in the laboratory and the chemical industry One of the simplest electrolytic
cell consists of two copper strips dipping in an aqueous solution of
copper sulphate If a DC voltage is applied to the two electrodes, then
Cu 2+ ions discharge at the cathode (negatively charged) and the following
reaction takes place:
Cu2+(aq) + 2e– ® Cu (s)
(2 28)
Copper metal is deposited on the cathode |
1 | 2431-2434 | One of the simplest electrolytic
cell consists of two copper strips dipping in an aqueous solution of
copper sulphate If a DC voltage is applied to the two electrodes, then
Cu 2+ ions discharge at the cathode (negatively charged) and the following
reaction takes place:
Cu2+(aq) + 2e– ® Cu (s)
(2 28)
Copper metal is deposited on the cathode At the anode, copper is
converted into Cu2+ ions by the reaction:
Cu(s) ® Cu2+(s) + 2e–
(2 |
1 | 2432-2435 | If a DC voltage is applied to the two electrodes, then
Cu 2+ ions discharge at the cathode (negatively charged) and the following
reaction takes place:
Cu2+(aq) + 2e– ® Cu (s)
(2 28)
Copper metal is deposited on the cathode At the anode, copper is
converted into Cu2+ ions by the reaction:
Cu(s) ® Cu2+(s) + 2e–
(2 29)
Thus copper is dissolved (oxidised) at anode and deposited
(reduced) at cathode |
1 | 2433-2436 | 28)
Copper metal is deposited on the cathode At the anode, copper is
converted into Cu2+ ions by the reaction:
Cu(s) ® Cu2+(s) + 2e–
(2 29)
Thus copper is dissolved (oxidised) at anode and deposited
(reduced) at cathode This is the basis for an industrial process in
which impure copper is converted into copper of high purity |
1 | 2434-2437 | At the anode, copper is
converted into Cu2+ ions by the reaction:
Cu(s) ® Cu2+(s) + 2e–
(2 29)
Thus copper is dissolved (oxidised) at anode and deposited
(reduced) at cathode This is the basis for an industrial process in
which impure copper is converted into copper of high purity The
impure copper is made an anode that dissolves on passing current
and pure copper is deposited at the cathode |
1 | 2435-2438 | 29)
Thus copper is dissolved (oxidised) at anode and deposited
(reduced) at cathode This is the basis for an industrial process in
which impure copper is converted into copper of high purity The
impure copper is made an anode that dissolves on passing current
and pure copper is deposited at the cathode Many metals like Na, Mg,
Al, etc |
1 | 2436-2439 | This is the basis for an industrial process in
which impure copper is converted into copper of high purity The
impure copper is made an anode that dissolves on passing current
and pure copper is deposited at the cathode Many metals like Na, Mg,
Al, etc are produced on large scale by electrochemical reduction of
their respective cations where no suitable chemical reducing agents
are available for this purpose |
1 | 2437-2440 | The
impure copper is made an anode that dissolves on passing current
and pure copper is deposited at the cathode Many metals like Na, Mg,
Al, etc are produced on large scale by electrochemical reduction of
their respective cations where no suitable chemical reducing agents
are available for this purpose Sodium and magnesium metals are produced by the electrolysis of
their fused chlorides and aluminium is produced by electrolysis of
aluminium oxide in presence of cryolite |
1 | 2438-2441 | Many metals like Na, Mg,
Al, etc are produced on large scale by electrochemical reduction of
their respective cations where no suitable chemical reducing agents
are available for this purpose Sodium and magnesium metals are produced by the electrolysis of
their fused chlorides and aluminium is produced by electrolysis of
aluminium oxide in presence of cryolite Quantitative Aspects of Electrolysis
Michael Faraday was the first scientist who described the quantitative
aspects of electrolysis |
1 | 2439-2442 | are produced on large scale by electrochemical reduction of
their respective cations where no suitable chemical reducing agents
are available for this purpose Sodium and magnesium metals are produced by the electrolysis of
their fused chlorides and aluminium is produced by electrolysis of
aluminium oxide in presence of cryolite Quantitative Aspects of Electrolysis
Michael Faraday was the first scientist who described the quantitative
aspects of electrolysis Now Faraday’s laws also flow from what has
been discussed earlier |
1 | 2440-2443 | Sodium and magnesium metals are produced by the electrolysis of
their fused chlorides and aluminium is produced by electrolysis of
aluminium oxide in presence of cryolite Quantitative Aspects of Electrolysis
Michael Faraday was the first scientist who described the quantitative
aspects of electrolysis Now Faraday’s laws also flow from what has
been discussed earlier Faraday’s Laws of Electrolysis
After his extensive investigations on electrolysis of solutions and melts
of electrolytes, Faraday published his results during 1833-34 in the
form of the following well known Faraday’s two laws of electrolysis:
(i) First Law: The amount of chemical reaction which occurs at any
electrode during electrolysis by a current is proportional to the
quantity of electricity passed through the electrolyte (solution or
melt) |
1 | 2441-2444 | Quantitative Aspects of Electrolysis
Michael Faraday was the first scientist who described the quantitative
aspects of electrolysis Now Faraday’s laws also flow from what has
been discussed earlier Faraday’s Laws of Electrolysis
After his extensive investigations on electrolysis of solutions and melts
of electrolytes, Faraday published his results during 1833-34 in the
form of the following well known Faraday’s two laws of electrolysis:
(i) First Law: The amount of chemical reaction which occurs at any
electrode during electrolysis by a current is proportional to the
quantity of electricity passed through the electrolyte (solution or
melt) (ii) Second Law: The amounts of different substances liberated by the
same quantity of electricity passing through the electrolytic solution
are proportional to their chemical equivalent weights (Atomic Mass
of Metal ÷ Number of electrons required to reduce the cation) |
1 | 2442-2445 | Now Faraday’s laws also flow from what has
been discussed earlier Faraday’s Laws of Electrolysis
After his extensive investigations on electrolysis of solutions and melts
of electrolytes, Faraday published his results during 1833-34 in the
form of the following well known Faraday’s two laws of electrolysis:
(i) First Law: The amount of chemical reaction which occurs at any
electrode during electrolysis by a current is proportional to the
quantity of electricity passed through the electrolyte (solution or
melt) (ii) Second Law: The amounts of different substances liberated by the
same quantity of electricity passing through the electrolytic solution
are proportional to their chemical equivalent weights (Atomic Mass
of Metal ÷ Number of electrons required to reduce the cation) 2 |
1 | 2443-2446 | Faraday’s Laws of Electrolysis
After his extensive investigations on electrolysis of solutions and melts
of electrolytes, Faraday published his results during 1833-34 in the
form of the following well known Faraday’s two laws of electrolysis:
(i) First Law: The amount of chemical reaction which occurs at any
electrode during electrolysis by a current is proportional to the
quantity of electricity passed through the electrolyte (solution or
melt) (ii) Second Law: The amounts of different substances liberated by the
same quantity of electricity passing through the electrolytic solution
are proportional to their chemical equivalent weights (Atomic Mass
of Metal ÷ Number of electrons required to reduce the cation) 2 5
2 |
1 | 2444-2447 | (ii) Second Law: The amounts of different substances liberated by the
same quantity of electricity passing through the electrolytic solution
are proportional to their chemical equivalent weights (Atomic Mass
of Metal ÷ Number of electrons required to reduce the cation) 2 5
2 5
2 |
1 | 2445-2448 | 2 5
2 5
2 5
2 |
1 | 2446-2449 | 5
2 5
2 5
2 5
2 |
1 | 2447-2450 | 5
2 5
2 5
2 5 Electrolytic
Electrolytic
Electrolytic
Electrolytic
Electrolytic
Cells and
Cells and
Cells and
Cells and
Cells and
Electrolysis
Electrolysis
Electrolysis
Electrolysis
Electrolysis
Intext Questions
Intext Questions
Intext Questions
Intext Questions
Intext Questions
2 |
1 | 2448-2451 | 5
2 5
2 5 Electrolytic
Electrolytic
Electrolytic
Electrolytic
Electrolytic
Cells and
Cells and
Cells and
Cells and
Cells and
Electrolysis
Electrolysis
Electrolysis
Electrolysis
Electrolysis
Intext Questions
Intext Questions
Intext Questions
Intext Questions
Intext Questions
2 7 Why does the conductivity of a solution decrease with dilution |
1 | 2449-2452 | 5
2 5 Electrolytic
Electrolytic
Electrolytic
Electrolytic
Electrolytic
Cells and
Cells and
Cells and
Cells and
Cells and
Electrolysis
Electrolysis
Electrolysis
Electrolysis
Electrolysis
Intext Questions
Intext Questions
Intext Questions
Intext Questions
Intext Questions
2 7 Why does the conductivity of a solution decrease with dilution 2 |
1 | 2450-2453 | 5 Electrolytic
Electrolytic
Electrolytic
Electrolytic
Electrolytic
Cells and
Cells and
Cells and
Cells and
Cells and
Electrolysis
Electrolysis
Electrolysis
Electrolysis
Electrolysis
Intext Questions
Intext Questions
Intext Questions
Intext Questions
Intext Questions
2 7 Why does the conductivity of a solution decrease with dilution 2 8 Suggest a way to determine the L°m value of water |
1 | 2451-2454 | 7 Why does the conductivity of a solution decrease with dilution 2 8 Suggest a way to determine the L°m value of water 2 |
1 | 2452-2455 | 2 8 Suggest a way to determine the L°m value of water 2 9 The molar conductivity of 0 |
1 | 2453-2456 | 8 Suggest a way to determine the L°m value of water 2 9 The molar conductivity of 0 025 mol L–1 methanoic acid is 46 |
1 | 2454-2457 | 2 9 The molar conductivity of 0 025 mol L–1 methanoic acid is 46 1 S cm2 mol–1 |
1 | 2455-2458 | 9 The molar conductivity of 0 025 mol L–1 methanoic acid is 46 1 S cm2 mol–1 Calculate its degree of dissociation and dissociation constant |
1 | 2456-2459 | 025 mol L–1 methanoic acid is 46 1 S cm2 mol–1 Calculate its degree of dissociation and dissociation constant Given l0(H+)
= 349 |
1 | 2457-2460 | 1 S cm2 mol–1 Calculate its degree of dissociation and dissociation constant Given l0(H+)
= 349 6 S cm2 mol–1 and l0 (HCOO–) = 54 |
1 | 2458-2461 | Calculate its degree of dissociation and dissociation constant Given l0(H+)
= 349 6 S cm2 mol–1 and l0 (HCOO–) = 54 6 S cm2 mol–1 |
1 | 2459-2462 | Given l0(H+)
= 349 6 S cm2 mol–1 and l0 (HCOO–) = 54 6 S cm2 mol–1 Rationalised 2023-24
52
Chemistry
There were no constant current sources available during Faraday’s
times |
1 | 2460-2463 | 6 S cm2 mol–1 and l0 (HCOO–) = 54 6 S cm2 mol–1 Rationalised 2023-24
52
Chemistry
There were no constant current sources available during Faraday’s
times The general practice was to put a coulometer (a standard electrolytic
cell) for determining the quantity of electricity passed from the amount
of metal (generally silver or copper) deposited or consumed |
1 | 2461-2464 | 6 S cm2 mol–1 Rationalised 2023-24
52
Chemistry
There were no constant current sources available during Faraday’s
times The general practice was to put a coulometer (a standard electrolytic
cell) for determining the quantity of electricity passed from the amount
of metal (generally silver or copper) deposited or consumed However,
coulometers are now obsolete and we now have constant current (I)
sources available and the quantity of electricity Q, passed is given by
Q = It
Q is in coloumbs when I is in ampere and t is in second |
1 | 2462-2465 | Rationalised 2023-24
52
Chemistry
There were no constant current sources available during Faraday’s
times The general practice was to put a coulometer (a standard electrolytic
cell) for determining the quantity of electricity passed from the amount
of metal (generally silver or copper) deposited or consumed However,
coulometers are now obsolete and we now have constant current (I)
sources available and the quantity of electricity Q, passed is given by
Q = It
Q is in coloumbs when I is in ampere and t is in second The amount of electricity (or charge) required for oxidation or
reduction depends on the stoichiometry of the electrode reaction |
1 | 2463-2466 | The general practice was to put a coulometer (a standard electrolytic
cell) for determining the quantity of electricity passed from the amount
of metal (generally silver or copper) deposited or consumed However,
coulometers are now obsolete and we now have constant current (I)
sources available and the quantity of electricity Q, passed is given by
Q = It
Q is in coloumbs when I is in ampere and t is in second The amount of electricity (or charge) required for oxidation or
reduction depends on the stoichiometry of the electrode reaction For
example, in the reaction:
Ag +(aq) + e– ® Ag(s)
(2 |
1 | 2464-2467 | However,
coulometers are now obsolete and we now have constant current (I)
sources available and the quantity of electricity Q, passed is given by
Q = It
Q is in coloumbs when I is in ampere and t is in second The amount of electricity (or charge) required for oxidation or
reduction depends on the stoichiometry of the electrode reaction For
example, in the reaction:
Ag +(aq) + e– ® Ag(s)
(2 30)
One mole of the electron is required for the reduction of one mole
of silver ions |
1 | 2465-2468 | The amount of electricity (or charge) required for oxidation or
reduction depends on the stoichiometry of the electrode reaction For
example, in the reaction:
Ag +(aq) + e– ® Ag(s)
(2 30)
One mole of the electron is required for the reduction of one mole
of silver ions We know that charge on one electron is equal to 1 |
1 | 2466-2469 | For
example, in the reaction:
Ag +(aq) + e– ® Ag(s)
(2 30)
One mole of the electron is required for the reduction of one mole
of silver ions We know that charge on one electron is equal to 1 6021 × 10–19C |
1 | 2467-2470 | 30)
One mole of the electron is required for the reduction of one mole
of silver ions We know that charge on one electron is equal to 1 6021 × 10–19C Therefore, the charge on one mole of electrons is equal to:
NA × 1 |
1 | 2468-2471 | We know that charge on one electron is equal to 1 6021 × 10–19C Therefore, the charge on one mole of electrons is equal to:
NA × 1 6021 × 10–19 C = 6 |
1 | 2469-2472 | 6021 × 10–19C Therefore, the charge on one mole of electrons is equal to:
NA × 1 6021 × 10–19 C = 6 02 × 1023 mol–1 × 1 |
1 | 2470-2473 | Therefore, the charge on one mole of electrons is equal to:
NA × 1 6021 × 10–19 C = 6 02 × 1023 mol–1 × 1 6021 × 10–19
C = 96487 C mol–1
This quantity of electricity is called Faraday and is represented by
the symbol F |
1 | 2471-2474 | 6021 × 10–19 C = 6 02 × 1023 mol–1 × 1 6021 × 10–19
C = 96487 C mol–1
This quantity of electricity is called Faraday and is represented by
the symbol F For approximate calculations we use 1F ≃ 96500 C mol–1 |
1 | 2472-2475 | 02 × 1023 mol–1 × 1 6021 × 10–19
C = 96487 C mol–1
This quantity of electricity is called Faraday and is represented by
the symbol F For approximate calculations we use 1F ≃ 96500 C mol–1 For the electrode reactions:
Mg2+(l) + 2e– ¾® Mg(s)
(2 |
1 | 2473-2476 | 6021 × 10–19
C = 96487 C mol–1
This quantity of electricity is called Faraday and is represented by
the symbol F For approximate calculations we use 1F ≃ 96500 C mol–1 For the electrode reactions:
Mg2+(l) + 2e– ¾® Mg(s)
(2 31)
Al3+(l) + 3e– ¾® Al(s)
(2 |
1 | 2474-2477 | For approximate calculations we use 1F ≃ 96500 C mol–1 For the electrode reactions:
Mg2+(l) + 2e– ¾® Mg(s)
(2 31)
Al3+(l) + 3e– ¾® Al(s)
(2 32)
It is obvious that one mole of Mg2+ and Al3+ require 2 mol of electrons
(2F) and 3 mol of electrons (3F) respectively |
1 | 2475-2478 | For the electrode reactions:
Mg2+(l) + 2e– ¾® Mg(s)
(2 31)
Al3+(l) + 3e– ¾® Al(s)
(2 32)
It is obvious that one mole of Mg2+ and Al3+ require 2 mol of electrons
(2F) and 3 mol of electrons (3F) respectively The charge passed through
the electrolytic cell during electrolysis is equal to the product of current
in amperes and time in seconds |
1 | 2476-2479 | 31)
Al3+(l) + 3e– ¾® Al(s)
(2 32)
It is obvious that one mole of Mg2+ and Al3+ require 2 mol of electrons
(2F) and 3 mol of electrons (3F) respectively The charge passed through
the electrolytic cell during electrolysis is equal to the product of current
in amperes and time in seconds In commercial production of metals,
current as high as 50,000 amperes are used that amounts to about
0 |
1 | 2477-2480 | 32)
It is obvious that one mole of Mg2+ and Al3+ require 2 mol of electrons
(2F) and 3 mol of electrons (3F) respectively The charge passed through
the electrolytic cell during electrolysis is equal to the product of current
in amperes and time in seconds In commercial production of metals,
current as high as 50,000 amperes are used that amounts to about
0 518 F per second |
1 | 2478-2481 | The charge passed through
the electrolytic cell during electrolysis is equal to the product of current
in amperes and time in seconds In commercial production of metals,
current as high as 50,000 amperes are used that amounts to about
0 518 F per second A solution of CuSO4 is electrolysed for 10 minutes with a current of
1 |
1 | 2479-2482 | In commercial production of metals,
current as high as 50,000 amperes are used that amounts to about
0 518 F per second A solution of CuSO4 is electrolysed for 10 minutes with a current of
1 5 amperes |
1 | 2480-2483 | 518 F per second A solution of CuSO4 is electrolysed for 10 minutes with a current of
1 5 amperes What is the mass of copper deposited at the cathode |
1 | 2481-2484 | A solution of CuSO4 is electrolysed for 10 minutes with a current of
1 5 amperes What is the mass of copper deposited at the cathode t = 600 s charge = current × time = 1 |
1 | 2482-2485 | 5 amperes What is the mass of copper deposited at the cathode t = 600 s charge = current × time = 1 5 A × 600 s = 900 C
According to the reaction:
Cu2+(aq) + 2e– = Cu(s)
We require 2F or 2 × 96487 C to deposit 1 mol or 63 g of Cu |
1 | 2483-2486 | What is the mass of copper deposited at the cathode t = 600 s charge = current × time = 1 5 A × 600 s = 900 C
According to the reaction:
Cu2+(aq) + 2e– = Cu(s)
We require 2F or 2 × 96487 C to deposit 1 mol or 63 g of Cu For 900 C, the mass of Cu deposited
= (63 g mol–1 × 900 C)/(2 × 96487 C mol–1) = 0 |
1 | 2484-2487 | t = 600 s charge = current × time = 1 5 A × 600 s = 900 C
According to the reaction:
Cu2+(aq) + 2e– = Cu(s)
We require 2F or 2 × 96487 C to deposit 1 mol or 63 g of Cu For 900 C, the mass of Cu deposited
= (63 g mol–1 × 900 C)/(2 × 96487 C mol–1) = 0 2938 g |
1 | 2485-2488 | 5 A × 600 s = 900 C
According to the reaction:
Cu2+(aq) + 2e– = Cu(s)
We require 2F or 2 × 96487 C to deposit 1 mol or 63 g of Cu For 900 C, the mass of Cu deposited
= (63 g mol–1 × 900 C)/(2 × 96487 C mol–1) = 0 2938 g Example 2 |
1 | 2486-2489 | For 900 C, the mass of Cu deposited
= (63 g mol–1 × 900 C)/(2 × 96487 C mol–1) = 0 2938 g Example 2 10
Example 2 |
1 | 2487-2490 | 2938 g Example 2 10
Example 2 10
Example 2 |
1 | 2488-2491 | Example 2 10
Example 2 10
Example 2 10
Example 2 |
1 | 2489-2492 | 10
Example 2 10
Example 2 10
Example 2 10
Example 2 |
1 | 2490-2493 | 10
Example 2 10
Example 2 10
Example 2 10
Solution
Solution
Solution
Solution
Solution
Products of electrolysis depend on the nature of material being
electrolysed and the type of electrodes being used |
1 | 2491-2494 | 10
Example 2 10
Example 2 10
Solution
Solution
Solution
Solution
Solution
Products of electrolysis depend on the nature of material being
electrolysed and the type of electrodes being used If the electrode is
inert (e |
1 | 2492-2495 | 10
Example 2 10
Solution
Solution
Solution
Solution
Solution
Products of electrolysis depend on the nature of material being
electrolysed and the type of electrodes being used If the electrode is
inert (e g |
1 | 2493-2496 | 10
Solution
Solution
Solution
Solution
Solution
Products of electrolysis depend on the nature of material being
electrolysed and the type of electrodes being used If the electrode is
inert (e g , platinum or gold), it does not participate in the chemical
reaction and acts only as source or sink for electrons |
1 | 2494-2497 | If the electrode is
inert (e g , platinum or gold), it does not participate in the chemical
reaction and acts only as source or sink for electrons On the other
hand, if the electrode is reactive, it participates in the electrode reaction |
1 | 2495-2498 | g , platinum or gold), it does not participate in the chemical
reaction and acts only as source or sink for electrons On the other
hand, if the electrode is reactive, it participates in the electrode reaction Thus, the products of electrolysis may be different for reactive and inert
2 |
1 | 2496-2499 | , platinum or gold), it does not participate in the chemical
reaction and acts only as source or sink for electrons On the other
hand, if the electrode is reactive, it participates in the electrode reaction Thus, the products of electrolysis may be different for reactive and inert
2 5 |
1 | 2497-2500 | On the other
hand, if the electrode is reactive, it participates in the electrode reaction Thus, the products of electrolysis may be different for reactive and inert
2 5 1 Products of
Electrolysis
Rationalised 2023-24
53
Electrochemistry
electrodes |
1 | 2498-2501 | Thus, the products of electrolysis may be different for reactive and inert
2 5 1 Products of
Electrolysis
Rationalised 2023-24
53
Electrochemistry
electrodes The products of electrolysis depend on the different oxidising
and reducing species present in the electrolytic cell and their standard
electrode potentials |
1 | 2499-2502 | 5 1 Products of
Electrolysis
Rationalised 2023-24
53
Electrochemistry
electrodes The products of electrolysis depend on the different oxidising
and reducing species present in the electrolytic cell and their standard
electrode potentials Moreover, some of the electrochemical processes
although feasible, are so slow kinetically that at lower voltages these do
not seem to take place and extra potential (called overpotential) has to
be applied, which makes such process more difficult to occur |
1 | 2500-2503 | 1 Products of
Electrolysis
Rationalised 2023-24
53
Electrochemistry
electrodes The products of electrolysis depend on the different oxidising
and reducing species present in the electrolytic cell and their standard
electrode potentials Moreover, some of the electrochemical processes
although feasible, are so slow kinetically that at lower voltages these do
not seem to take place and extra potential (called overpotential) has to
be applied, which makes such process more difficult to occur For example, if we use molten NaCl, the products of electrolysis are
sodium metal and Cl2 gas |
1 | 2501-2504 | The products of electrolysis depend on the different oxidising
and reducing species present in the electrolytic cell and their standard
electrode potentials Moreover, some of the electrochemical processes
although feasible, are so slow kinetically that at lower voltages these do
not seem to take place and extra potential (called overpotential) has to
be applied, which makes such process more difficult to occur For example, if we use molten NaCl, the products of electrolysis are
sodium metal and Cl2 gas Here we have only one cation (Na+) which is
reduced at the cathode (Na+ + e– ® Na) and one anion (Cl–) which is
oxidised at the anode (Cl– ® ½Cl2 + e– ) |
1 | 2502-2505 | Moreover, some of the electrochemical processes
although feasible, are so slow kinetically that at lower voltages these do
not seem to take place and extra potential (called overpotential) has to
be applied, which makes such process more difficult to occur For example, if we use molten NaCl, the products of electrolysis are
sodium metal and Cl2 gas Here we have only one cation (Na+) which is
reduced at the cathode (Na+ + e– ® Na) and one anion (Cl–) which is
oxidised at the anode (Cl– ® ½Cl2 + e– ) During the electrolysis of aqueous
sodium chloride solution, the products are NaOH, Cl2 and H2 |
1 | 2503-2506 | For example, if we use molten NaCl, the products of electrolysis are
sodium metal and Cl2 gas Here we have only one cation (Na+) which is
reduced at the cathode (Na+ + e– ® Na) and one anion (Cl–) which is
oxidised at the anode (Cl– ® ½Cl2 + e– ) During the electrolysis of aqueous
sodium chloride solution, the products are NaOH, Cl2 and H2 In this
case besides Na+ and Cl– ions we also have H+ and OH– ions along with
the solvent molecules, H2O |
1 | 2504-2507 | Here we have only one cation (Na+) which is
reduced at the cathode (Na+ + e– ® Na) and one anion (Cl–) which is
oxidised at the anode (Cl– ® ½Cl2 + e– ) During the electrolysis of aqueous
sodium chloride solution, the products are NaOH, Cl2 and H2 In this
case besides Na+ and Cl– ions we also have H+ and OH– ions along with
the solvent molecules, H2O At the cathode there is competition between the following reduction
reactions:
Na+ (aq) + e– ® Na (s)
(
)
o
Ecell
= – 2 |
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