In the previous section, we have seen the basic details about electrolysis. In this section, we will see the electrolysis of water.
■ Can pure water conduct electricity?
Let us see the answer in steps:
1. We have seen that for the passage of electricity through a liquid, there must be free ions.
2. Pure water indeed has some free ions. Let us see how they are formed:
• The water ionises to some extent even with out any external influence. The equation is:
H2O ⟶ H+ + (OH)-1
• Each of the H+ ions thus formed will combine with a water molecule to give a hydronium ion (H3O)+
• The equation is: H2O (l) + H2O (l) ⟶ H2O (aq) + H+ (aq) + (OH)-1 (aq)
This is same as: 2H2O (l) ⟶ (H3O)+ (aq) + (OH)-1 (aq)
• But this is a reversible reaction. We have seen details about reversible reactions here.
So we must change the 'type of arrow'. The equation should be written as:
2H2O (l) ⇌ (H3O)+ (aq) + (OH)-1 (aq)
• So there are two reactions taking place simultaneously:
♦ Forward reaction: The water molecules dissociating into ions
♦ Backward reaction: The ions combining together to give back water molecules
• If there is no external influence, these two reactions will be in equilibrium. That means, there will always be some (H3O)+ and (OH)1- ions present in pure water
• Note that 2 water molecules are required to produce one (H3O)+ ion and one (OH)1- ion
• But for our present discussion, we do not need the 'number of water molecules required'.
• What we have to consider is:
The quantity of these ions is very low in pure water. So it does not conduct electricity.
3. But when a little acid is added to the water, the situation changes.
• A large number of (H3O)+ ions will be produced. We have seen the details here.
4. Let us recall the changes that happen when sulphuric acid (H2SO4) is added to water:
• The H2SO4 gives two H+ ions. The equation is:
H2SO4 ⟶ 2H+ + SO42-
• Each of these H+ ions combines with one water molecule to form one hydronium ion (H3O)+. The equation is:
H2O (l) + H+ (aq) ⟶ (H3O)+ (aq)
5. Now there are enough ions to conduct electricity. The ions are:
• Positively charged (H3O)1+ ions
• Negatively charged (SO4)2- ions
• Negatively charged (OH)1- ions
• Note that, the number of (OH)1- ions is still very less. Because, the addition of sulphuric acid does not cause any increase in the number of (OH)1- ions. If we want an increase in the number of (OH)1- ions, we must add a base instead of acid.
• We can make an electrolytic cell with this water (to which a little sulphuric acid is added) as the electrolyte.
6. When the switch is turned on,
• the positive (H3O)1+ ions move towards the negative electrode
• the negative (SO4)2- ions move towards the positive electrode
7. Let us consider the positive (H3O)1+ ions first.
• They reach the negative electrode.
• They will gain electrons. That is., they undergo reduction
• Whatever be the type of cell (galvanic or electrolytic), cathode is where reduction takes place.
• The equation is:
2(H3O)1+ (aq) + 2e- ⟶ H2 (g) + 2H2O (l)
• Thus we get hydrogen in gaseous form at the cathode
8. Now we will consider the negative ions (SO4)2- and (OH)1- .
• As mentioned above, there are only a very small number of (OH)1- ions. So we need not consider them
• The (SO4)2- ions reach the positive electrode.
• They will lose electrons. That is., they undergo oxidation
• Whatever be the type of cell (galvanic or electrolytic), anode is where oxidation takes place.
• So the (SO4)2- ions want to get oxidized. That is., they want to donate their electrons
• But the water molecules (H2O) are more readily oxidized than the (SO4)2- ions
• So the reaction taking place is:
2H2O (l) ⟶ O2 (g) + 4H+ (aq) + 4e-
• Thus we get oxygen in gaseous form at the anode
Next we will discuss about the electrolysis of molten Sodium chloride
1. Sodium chloride in solid state do not conduct electricity. This is because, it's ions have no freedom of movement. But when it is in molten state, the Na+ ions and Cl- ions are free to move. So it will conduct electricity.
2. When the switch is turned on,
• the positive Na+ ions move towards the negative electrode
• the negative Cl- ions move towards the positive electrode
7. Let us consider the positive Na+ ions first.
• They reach the negative electrode.
• They will gain electrons. That is., they undergo reduction
• Whatever be the type of cell (galvanic or electrolytic), cathode is where reduction takes place.
• The equation is:
Na+ + 1e- ⟶ Na0
• Thus we get sodium at the cathode
8. Now we will consider the negative Cl- ions
• The Cl- ions reach the positive electrode.
• They will lose electrons. That is., they undergo oxidation
• Whatever be the type of cell (galvanic or electrolytic), anode is where oxidation takes place.
• So the Cl- ions want to get oxidized. That is., they want to donate their electrons
• So the reaction taking place is:
2Cl- ⟶ Cl20 + 2e-
• Thus we get chlorine at the anode
Next we will discuss about the electrolysis of Sodium chloride solution
1. Sodium chloride in solid state do not conduct electricity. This is because, it's ions have no freedom of movement. But when it is in solution with water, the Na+ ions and Cl- ions are free to move. So the solution will conduct electricity.
2. When the switch is turned on,
• the positive Na+ ions move towards the negative electrode
• the negative Cl- ions move towards the positive electrode
3. Let us consider the positive Na+ ions first.
• They reach the negative electrode.
• They will gain electrons. That is., they undergo reduction
• Whatever be the type of cell (galvanic or electrolytic), cathode is where reduction takes place.
• So the Na+ ions want to get reduced. That is., they want to gain electrons
• But the water molecules (H2O) are more readily reduced than the Na+ ions
• So the reaction taking place is:
2H2O + 2e- ⟶ H2 (g) + 2(OH)-
• Thus we get hydrogen at the cathode
4. Now we will consider the negative Cl- ions
• The Cl- ions reach the positive electrode.
• They will lose electrons. That is., they undergo oxidation
• Whatever be the type of cell (galvanic or electrolytic), anode is where oxidation takes place.
• So the Cl- ions want to get oxidized. That is., they want to donate their electrons
• In this case, the Cl- ions are more readily oxidized than the water molecules
• So the reaction taking place is:
2Cl- ⟶ Cl20 + 2e-
• Thus we get chlorine at the anode
■ Production of metals and non metals
■ Production of chemicals
■ Purification of metals
■ Electroplating
• In this process, metalic objects are coated with other metals
• Some examples are:
♦ Gold plated ornaments
♦ Chromium plated iron handles
♦ Silver plated vessels
Now we will see some solved examples:
Solved example 12.1The solutions of zinc sulphate (ZnSO4), Ferrous sulphate (FeSO4), Copper sulphate (CuSO4) and Silver nitrate (AgNO3) are taken in four different test tubes.
1. Suppose an iron nail is kept immersed in each one. In which test tube the iron nail undergoes a colour change?
2. What is the reaction taking place here?
3. Justify your answer (Refer reactivity series of metals)
Solution:
• Consider any one of the given test tubes. In it, we have a salt solution. The salt is formed from a metal.
• If that metal is above iron in the reactivity series, no reaction will take place.
• If that metal is below iron in the reactivity series, a displacement reaction will take place. That is., the iron will displace the metal from it's salt solution
Now we will consider each test tube:
1. ZnSO4 solution. Iron is below zinc. So there will not be any reaction
2. FeSO4 solution. Both the metals are iron. So there will not be any reaction
3. CuSO4 solution. Iron is above copper. So there will be a displacement reaction
4. AgNO3 solution. Iron is above silver. So there will be a displacement reaction
Now we can write the answers:
Part 1:
The iron undergoes a colour change in the third and the fourth test tubes
Part 2:
The reaction taking place is Displacement reaction
Part 3:
(a) Iron is more reactive than copper. So iron donates electrons and thus get oxidised. These electrons are accepted by the Cu2+ ions present in the solution. The Cu2+ ions thus get reduced to pure copper atoms (Cu). These atoms get deposited on the surface of the iron nail. Thus the iron nail undergoes colour change
(b) Iron is more reactive than silver. So iron donates electrons and thus get oxidised. These electrons are accepted by the Ag+ ions present in the solution. The Ag+ ions thus get reduced to pure silver atoms (Ag). These atoms get deposited on the surface of the iron nail. Thus the iron nail undergoes colour change
Solved example 12.2
Compare the electrolysis of molten potassium chloride and solution of potassium chloride. What are the processes taking place at the cathode and the anode?
Solution:
Potassium chloride (KCl) in solid state do not conduct electricity. This is because, it's ions have no freedom of movement. But when it is in molten state or in solution, the K+ ions and Cl- ions are free to move. So it will conduct electricity. We can write the comparison as follows:
A. Molten Potassium chloride
When the switch is turned on,
• the positive K+ ions move towards the negative electrode
• the negative Cl- ions move towards the positive electrode
Let us consider the positive K+ ions first.
• They reach the negative electrode.
• They will gain electrons. That is., they undergo reduction
• Whatever be the type of cell (galvanic or electrolytic), cathode is where reduction takes place.
• So the ions will get reduced by gaining electrons. The equation is:
K+ + 1e- ⟶ K0
• Thus we get sodium at the cathode
Now we will consider the negative Cl- ions
• The Cl- ions reach the positive electrode (anode).
• They will lose electrons. That is., they undergo oxidation
• Whatever be the type of cell (galvanic or electrolytic), anode is where oxidation takes place.
• So the Cl- ions want to get oxidized. That is., they want to donate their electrons
• So the reaction taking place is:
2Cl- ⟶ Cl20 + 2e-
• Thus we get chlorine at the anode
B. Potassium chloride solution
When the switch is turned on,
• the positive K+ ions move towards the negative electrode
• the negative Cl- ions move towards the positive electrode
Let us consider the positive K+ ions first.
• They reach the negative electrode.
• They will gain electrons. That is., they undergo reduction
• Whatever be the type of cell (galvanic or electrolytic), cathode is where reduction takes place.
• So the K+ ions want to get reduced. That is., they want to gain electrons
• But the water molecules (H2O) are more readily reduced than the K+ ions
• So the reaction taking place is:
2H2O + 2e- ⟶ H2 (g) + 2(OH)-
• Thus we get hydrogen at the cathode
4. Now we will consider the negative Cl- ions
• The Cl- ions reach the positive electrode
• They will lose electrons. That is., they undergo oxidation
• Whatever be the type of cell (galvanic or electrolytic), anode is where oxidation takes place.
• So the Cl- ions want to get oxidized. That is., they want to donate their electrons
• In this case, the Cl- ions are more readily oxidized than the water molecules
• So the reaction taking place is:
2Cl- ⟶ Cl20 + 2e-
• Thus we get chlorine at the anode
Solved example 12.3
You are given a solution of silver nitrate (AgNO3), a solution of magnesium sulphate (MgSO4), a silver rod and a magnesium ribbon. How can you arrange a galvanic cell using these? Write down the reactions taking place at the cathode and the anode
Solution:
1. Take two beakers. Add the magnesium sulphate solution into one beaker.
2. Add the silver nitrate (AgNO3) solution into the other.
3. Keep the magnesium ribbon in the magnesium sulphate solution.
4. Keep the silver rod in the silver nitrate solution.
5. Connect the negative terminal of a voltmeter to the magnesium ribbon. Connect the positive terminal to the silver rod.
6. Connect the solutions in the two beakers by a salt bridge. A long strip of filter paper soaked in KCl solution can be used instead of the salt bridge.
7. Now observe the change in the voltmeter reading.
From the voltmeter, it is clear that electricity is produced in the experiment. This arrangement is the Magnesium - Silver galvanic cell. Now we will write the working principle:
8. Magnesium is more reactive than silver.
So Mg in the ribbon, loses two electrons and become Mg2+. The equation can be written as:
Mg0 (s) ⟶ Mg+2 (aq) + 2e-1
• This is a oxidation reaction. Because the oxidation number of Mg increases from zero to +2.
■ The electrode at which oxidation takes place is called the anode.
9. The newly formed Mg+2 ions gets detached from the ribbon and goes into the solution. But the released electrons stick to the ribbon rod
• Note that many Mg+2 ions are already present in the magnesium sulphate solution because, the aqueous Magnesium sulphate solution exists as a mixture of Mg+2 ions and (SO4)-2 ions.
10. Because of the released electrons, the Mg ribbon becomes negatively charged. These free electrons reach the silver rod through the external circuit
11. These electrons which reach the silver rod flows through the silver rod and reaches the silver nitrate solution.
• In the silver nitrate solution, Ag+1 ions are present. These ions receive the electrons and become Ag atoms. The equation can be written as:
2Ag+1 (aq) + 2e-1 ⟶ 2Ag0 (s)
• This is a reduction reaction. Because the oxidation number of Ag+1 decreases from +1 to zero.
Note the coefficient '2' in front of Ag. This is because, each Mg atom will donate 2 electrons while each Ag atom will donate only one electron. So two Ag+1 ions can benefit from one Mg+2 ion
■ The electrode at which reduction takes place is called the cathode.
12. So both oxidation and reduction takes place in this reaction. Thus it is a redox reaction.
• The transfer of electrons produced by the redox reaction causes the flow of electric current.
• So we have made a galvanic cell using magnesium ribbon and silver rod.
Solved example 12.4
Keep two carbon electrodes immersed in copper sulphate solution. Then pass electricity through the solution.
(i) At which electrode does colour change occur - Anode or cathode?
(ii) Is there any change in the blue colour of the copper sulphate solution?
(iii) Write down the chemical equations for the changes occuring here
Solution:
Part (i):
Colour change occur at cathode.
Part (ii):
The blue colour of the copper sulphate solution fades
Part (iii):
• This is an electrolysis of copper sulphate solution
• One electrode will be connected to the negative of the battery. This will be the negative electrode.
• The positive Cu2+ ions will move to this negative electrode. There they gain electrons and thus undergo reduction
• Whatever be the type of cell (galvanic or electrolytic), cathode is where reduction takes place.
• The equation is:
Cu+2 (aq) + 2e-1 ⟶ Cu0 (s)
• This is a reduction reaction. These Cu atoms get deposited on the cathode. So the colour change occur at the cathode
• Now we will consider the negative ions (SO4)2-
• The (SO4)2- ions reach the positive electrode. There they lose electrons and thus undergo oxidation.
• Whatever be the type of cell (galvanic or electrolytic), anode is where oxidation takes place.
• So the (SO4)2- ions reach the anode.
• They want to donate their electrons
• But the water molecules (H2O) are more readily oxidized than the (SO4)2- ions
• So the reaction taking place is:
2H2O (l) ⟶ O2 (g) + 4H+ (aq) + 4e-
• Thus we get oxygen in gaseous form at the anode
• As the reaction continues, the number of Cu2+ ions decreases. So the blue colour fades
Solved example 12.5
When acidified copper sulphate solution is electrolysed, oxygen is obtained at the anode. What arrangements are to be made for this? Which element is deposited at the cathode?
Solution:
• Take copper sulphate solution in a beaker. Put two carbon electrodes in it. Connect the electrodes to a battery through a switch. See fig.12.14 below:
• When the switch is turned on, a reaction will take place. The working principle is explained in the previous solved example 12.4. We will get copper at the cathode
Solved example 12.6
How many galvanic cells can be made using the metals Ag, Cu, Zn and Mg. When galvanic cells are made using these metals, What will be the nature of reaction in each cell? (Reactivity: Mg > Zn > Cu > Ag)
Solution:
■ There are 6 possible combinations. That is., we can make six different galvanic cells. They are:
(i) Mg-Zn (ii) Mg-Cu (iii) Mg-Ag (iv) Zn-Cu (v) Zn-Ag (iv) Cu-Ag
• Consider the first combination:
♦ Mg is more reactive. So it will be the anode
What ever be the type of cell, oxidation occurs at the anode. The reaction at anode is:
Mg (s) ⟶ Mg2+ (aq) + 2e-
♦ Zn will be the cathode
What ever be the type of cell, reduction occurs at the cathode. The reaction at cathode is:
Zn+2 (aq) + 2e-1 ⟶ Zn (s)
• Consider the second combination:
♦ Mg is more reactive. So it will be the anode. The oxidation reaction at anode is:
Mg (s) ⟶ Mg2+ (aq) + 2e-
♦ Cu will be the cathode. The reduction reaction at cathode is:
Cu+2 (aq) + 2e-1 ⟶ Cu (s)
• Consider the third combination:
♦ Mg is more reactive. So it will be the anode. The oxidation reaction at anode is:
Mg (s) ⟶ Mg2+ (aq) + 2e-
♦ Ag will be the cathode. The reduction reaction at cathode is:
2Ag+1 (aq) + 2e-1 ⟶ 2Ag (s)
• Consider the fourth combination:
♦ Zn is more reactive. So it will be the anode. The oxidation reaction at anode is:
Zn (s) ⟶ Zn2+ (aq) + 2e-
♦ Cu will be the cathode. The reduction reaction at cathode is:
Cu+2 (aq) + 2e-1 ⟶ Cu (s)
• Consider the fifth combination:
♦ Zn is more reactive. So it will be the anode. The oxidation reaction at anode is:
Zn (s) ⟶ Zn2+ (aq) + 2e-
♦ Ag will be the cathode. The reduction reaction at cathode is:
2Ag+1 (aq) + 2e-1 ⟶ 2Ag (s)
• Consider the sixth combination:
♦ Cu is more reactive. So it will be the anode. The oxidation reaction at anode is:
Cu (s) ⟶ Cu2+ (aq) + 2e-
♦ Ag will be the cathode. The reduction reaction at cathode is:
2Ag+1 (aq) + 2e-1 ⟶ 2Ag (s)
We have completed this discussion on Electrochemistry. In the next section, we will see production of metals.
■ Can pure water conduct electricity?
Let us see the answer in steps:
1. We have seen that for the passage of electricity through a liquid, there must be free ions.
2. Pure water indeed has some free ions. Let us see how they are formed:
• The water ionises to some extent even with out any external influence. The equation is:
H2O ⟶ H+ + (OH)-1
• Each of the H+ ions thus formed will combine with a water molecule to give a hydronium ion (H3O)+
• The equation is: H2O (l) + H2O (l) ⟶ H2O (aq) + H+ (aq) + (OH)-1 (aq)
This is same as: 2H2O (l) ⟶ (H3O)+ (aq) + (OH)-1 (aq)
• But this is a reversible reaction. We have seen details about reversible reactions here.
So we must change the 'type of arrow'. The equation should be written as:
2H2O (l) ⇌ (H3O)+ (aq) + (OH)-1 (aq)
• So there are two reactions taking place simultaneously:
♦ Forward reaction: The water molecules dissociating into ions
♦ Backward reaction: The ions combining together to give back water molecules
• If there is no external influence, these two reactions will be in equilibrium. That means, there will always be some (H3O)+ and (OH)1- ions present in pure water
• Note that 2 water molecules are required to produce one (H3O)+ ion and one (OH)1- ion
• But for our present discussion, we do not need the 'number of water molecules required'.
• What we have to consider is:
The quantity of these ions is very low in pure water. So it does not conduct electricity.
3. But when a little acid is added to the water, the situation changes.
• A large number of (H3O)+ ions will be produced. We have seen the details here.
4. Let us recall the changes that happen when sulphuric acid (H2SO4) is added to water:
• The H2SO4 gives two H+ ions. The equation is:
H2SO4 ⟶ 2H+ + SO42-
• Each of these H+ ions combines with one water molecule to form one hydronium ion (H3O)+. The equation is:
H2O (l) + H+ (aq) ⟶ (H3O)+ (aq)
5. Now there are enough ions to conduct electricity. The ions are:
• Positively charged (H3O)1+ ions
• Negatively charged (SO4)2- ions
• Negatively charged (OH)1- ions
• Note that, the number of (OH)1- ions is still very less. Because, the addition of sulphuric acid does not cause any increase in the number of (OH)1- ions. If we want an increase in the number of (OH)1- ions, we must add a base instead of acid.
• We can make an electrolytic cell with this water (to which a little sulphuric acid is added) as the electrolyte.
6. When the switch is turned on,
• the positive (H3O)1+ ions move towards the negative electrode
• the negative (SO4)2- ions move towards the positive electrode
7. Let us consider the positive (H3O)1+ ions first.
• They reach the negative electrode.
• They will gain electrons. That is., they undergo reduction
• Whatever be the type of cell (galvanic or electrolytic), cathode is where reduction takes place.
• The equation is:
2(H3O)1+ (aq) + 2e- ⟶ H2 (g) + 2H2O (l)
• Thus we get hydrogen in gaseous form at the cathode
8. Now we will consider the negative ions (SO4)2- and (OH)1- .
• As mentioned above, there are only a very small number of (OH)1- ions. So we need not consider them
• The (SO4)2- ions reach the positive electrode.
• They will lose electrons. That is., they undergo oxidation
• Whatever be the type of cell (galvanic or electrolytic), anode is where oxidation takes place.
• So the (SO4)2- ions want to get oxidized. That is., they want to donate their electrons
• But the water molecules (H2O) are more readily oxidized than the (SO4)2- ions
• So the reaction taking place is:
2H2O (l) ⟶ O2 (g) + 4H+ (aq) + 4e-
• Thus we get oxygen in gaseous form at the anode
1. Sodium chloride in solid state do not conduct electricity. This is because, it's ions have no freedom of movement. But when it is in molten state, the Na+ ions and Cl- ions are free to move. So it will conduct electricity.
2. When the switch is turned on,
• the positive Na+ ions move towards the negative electrode
• the negative Cl- ions move towards the positive electrode
7. Let us consider the positive Na+ ions first.
• They reach the negative electrode.
• They will gain electrons. That is., they undergo reduction
• Whatever be the type of cell (galvanic or electrolytic), cathode is where reduction takes place.
• The equation is:
Na+ + 1e- ⟶ Na0
• Thus we get sodium at the cathode
8. Now we will consider the negative Cl- ions
• The Cl- ions reach the positive electrode.
• They will lose electrons. That is., they undergo oxidation
• Whatever be the type of cell (galvanic or electrolytic), anode is where oxidation takes place.
• So the Cl- ions want to get oxidized. That is., they want to donate their electrons
• So the reaction taking place is:
2Cl- ⟶ Cl20 + 2e-
• Thus we get chlorine at the anode
1. Sodium chloride in solid state do not conduct electricity. This is because, it's ions have no freedom of movement. But when it is in solution with water, the Na+ ions and Cl- ions are free to move. So the solution will conduct electricity.
2. When the switch is turned on,
• the positive Na+ ions move towards the negative electrode
• the negative Cl- ions move towards the positive electrode
3. Let us consider the positive Na+ ions first.
• They reach the negative electrode.
• They will gain electrons. That is., they undergo reduction
• Whatever be the type of cell (galvanic or electrolytic), cathode is where reduction takes place.
• So the Na+ ions want to get reduced. That is., they want to gain electrons
• But the water molecules (H2O) are more readily reduced than the Na+ ions
• So the reaction taking place is:
2H2O + 2e- ⟶ H2 (g) + 2(OH)-
• Thus we get hydrogen at the cathode
4. Now we will consider the negative Cl- ions
• The Cl- ions reach the positive electrode.
• They will lose electrons. That is., they undergo oxidation
• Whatever be the type of cell (galvanic or electrolytic), anode is where oxidation takes place.
• So the Cl- ions want to get oxidized. That is., they want to donate their electrons
• In this case, the Cl- ions are more readily oxidized than the water molecules
• So the reaction taking place is:
2Cl- ⟶ Cl20 + 2e-
• Thus we get chlorine at the anode
Uses of Electrolysis
The process of electrolysis finds application in the following areas:■ Production of metals and non metals
■ Production of chemicals
■ Purification of metals
■ Electroplating
• In this process, metalic objects are coated with other metals
• Some examples are:
♦ Gold plated ornaments
♦ Chromium plated iron handles
♦ Silver plated vessels
Now we will see some solved examples:
Solved example 12.1The solutions of zinc sulphate (ZnSO4), Ferrous sulphate (FeSO4), Copper sulphate (CuSO4) and Silver nitrate (AgNO3) are taken in four different test tubes.
1. Suppose an iron nail is kept immersed in each one. In which test tube the iron nail undergoes a colour change?
2. What is the reaction taking place here?
3. Justify your answer (Refer reactivity series of metals)
Solution:
• Consider any one of the given test tubes. In it, we have a salt solution. The salt is formed from a metal.
• If that metal is above iron in the reactivity series, no reaction will take place.
• If that metal is below iron in the reactivity series, a displacement reaction will take place. That is., the iron will displace the metal from it's salt solution
Now we will consider each test tube:
1. ZnSO4 solution. Iron is below zinc. So there will not be any reaction
2. FeSO4 solution. Both the metals are iron. So there will not be any reaction
3. CuSO4 solution. Iron is above copper. So there will be a displacement reaction
4. AgNO3 solution. Iron is above silver. So there will be a displacement reaction
Now we can write the answers:
Part 1:
The iron undergoes a colour change in the third and the fourth test tubes
Part 2:
The reaction taking place is Displacement reaction
Part 3:
(a) Iron is more reactive than copper. So iron donates electrons and thus get oxidised. These electrons are accepted by the Cu2+ ions present in the solution. The Cu2+ ions thus get reduced to pure copper atoms (Cu). These atoms get deposited on the surface of the iron nail. Thus the iron nail undergoes colour change
(b) Iron is more reactive than silver. So iron donates electrons and thus get oxidised. These electrons are accepted by the Ag+ ions present in the solution. The Ag+ ions thus get reduced to pure silver atoms (Ag). These atoms get deposited on the surface of the iron nail. Thus the iron nail undergoes colour change
Solved example 12.2
Compare the electrolysis of molten potassium chloride and solution of potassium chloride. What are the processes taking place at the cathode and the anode?
Solution:
Potassium chloride (KCl) in solid state do not conduct electricity. This is because, it's ions have no freedom of movement. But when it is in molten state or in solution, the K+ ions and Cl- ions are free to move. So it will conduct electricity. We can write the comparison as follows:
A. Molten Potassium chloride
When the switch is turned on,
• the positive K+ ions move towards the negative electrode
• the negative Cl- ions move towards the positive electrode
Let us consider the positive K+ ions first.
• They reach the negative electrode.
• They will gain electrons. That is., they undergo reduction
• Whatever be the type of cell (galvanic or electrolytic), cathode is where reduction takes place.
• So the ions will get reduced by gaining electrons. The equation is:
K+ + 1e- ⟶ K0
• Thus we get sodium at the cathode
Now we will consider the negative Cl- ions
• The Cl- ions reach the positive electrode (anode).
• They will lose electrons. That is., they undergo oxidation
• Whatever be the type of cell (galvanic or electrolytic), anode is where oxidation takes place.
• So the Cl- ions want to get oxidized. That is., they want to donate their electrons
• So the reaction taking place is:
2Cl- ⟶ Cl20 + 2e-
• Thus we get chlorine at the anode
When the switch is turned on,
• the positive K+ ions move towards the negative electrode
• the negative Cl- ions move towards the positive electrode
Let us consider the positive K+ ions first.
• They reach the negative electrode.
• They will gain electrons. That is., they undergo reduction
• Whatever be the type of cell (galvanic or electrolytic), cathode is where reduction takes place.
• So the K+ ions want to get reduced. That is., they want to gain electrons
• But the water molecules (H2O) are more readily reduced than the K+ ions
• So the reaction taking place is:
2H2O + 2e- ⟶ H2 (g) + 2(OH)-
• Thus we get hydrogen at the cathode
4. Now we will consider the negative Cl- ions
• The Cl- ions reach the positive electrode
• They will lose electrons. That is., they undergo oxidation
• Whatever be the type of cell (galvanic or electrolytic), anode is where oxidation takes place.
• So the Cl- ions want to get oxidized. That is., they want to donate their electrons
• In this case, the Cl- ions are more readily oxidized than the water molecules
• So the reaction taking place is:
2Cl- ⟶ Cl20 + 2e-
• Thus we get chlorine at the anode
Solved example 12.3
You are given a solution of silver nitrate (AgNO3), a solution of magnesium sulphate (MgSO4), a silver rod and a magnesium ribbon. How can you arrange a galvanic cell using these? Write down the reactions taking place at the cathode and the anode
Solution:
1. Take two beakers. Add the magnesium sulphate solution into one beaker.
2. Add the silver nitrate (AgNO3) solution into the other.
3. Keep the magnesium ribbon in the magnesium sulphate solution.
4. Keep the silver rod in the silver nitrate solution.
5. Connect the negative terminal of a voltmeter to the magnesium ribbon. Connect the positive terminal to the silver rod.
6. Connect the solutions in the two beakers by a salt bridge. A long strip of filter paper soaked in KCl solution can be used instead of the salt bridge.
7. Now observe the change in the voltmeter reading.
From the voltmeter, it is clear that electricity is produced in the experiment. This arrangement is the Magnesium - Silver galvanic cell. Now we will write the working principle:
8. Magnesium is more reactive than silver.
So Mg in the ribbon, loses two electrons and become Mg2+. The equation can be written as:
Mg0 (s) ⟶ Mg+2 (aq) + 2e-1
• This is a oxidation reaction. Because the oxidation number of Mg increases from zero to +2.
■ The electrode at which oxidation takes place is called the anode.
9. The newly formed Mg+2 ions gets detached from the ribbon and goes into the solution. But the released electrons stick to the ribbon rod
• Note that many Mg+2 ions are already present in the magnesium sulphate solution because, the aqueous Magnesium sulphate solution exists as a mixture of Mg+2 ions and (SO4)-2 ions.
10. Because of the released electrons, the Mg ribbon becomes negatively charged. These free electrons reach the silver rod through the external circuit
11. These electrons which reach the silver rod flows through the silver rod and reaches the silver nitrate solution.
• In the silver nitrate solution, Ag+1 ions are present. These ions receive the electrons and become Ag atoms. The equation can be written as:
2Ag+1 (aq) + 2e-1 ⟶ 2Ag0 (s)
• This is a reduction reaction. Because the oxidation number of Ag+1 decreases from +1 to zero.
Note the coefficient '2' in front of Ag. This is because, each Mg atom will donate 2 electrons while each Ag atom will donate only one electron. So two Ag+1 ions can benefit from one Mg+2 ion
■ The electrode at which reduction takes place is called the cathode.
12. So both oxidation and reduction takes place in this reaction. Thus it is a redox reaction.
• The transfer of electrons produced by the redox reaction causes the flow of electric current.
• So we have made a galvanic cell using magnesium ribbon and silver rod.
Solved example 12.4
Keep two carbon electrodes immersed in copper sulphate solution. Then pass electricity through the solution.
(i) At which electrode does colour change occur - Anode or cathode?
(ii) Is there any change in the blue colour of the copper sulphate solution?
(iii) Write down the chemical equations for the changes occuring here
Solution:
Part (i):
Colour change occur at cathode.
Part (ii):
The blue colour of the copper sulphate solution fades
Part (iii):
• This is an electrolysis of copper sulphate solution
• One electrode will be connected to the negative of the battery. This will be the negative electrode.
• The positive Cu2+ ions will move to this negative electrode. There they gain electrons and thus undergo reduction
• Whatever be the type of cell (galvanic or electrolytic), cathode is where reduction takes place.
• The equation is:
Cu+2 (aq) + 2e-1 ⟶ Cu0 (s)
• This is a reduction reaction. These Cu atoms get deposited on the cathode. So the colour change occur at the cathode
• Now we will consider the negative ions (SO4)2-
• The (SO4)2- ions reach the positive electrode. There they lose electrons and thus undergo oxidation.
• Whatever be the type of cell (galvanic or electrolytic), anode is where oxidation takes place.
• So the (SO4)2- ions reach the anode.
• They want to donate their electrons
• But the water molecules (H2O) are more readily oxidized than the (SO4)2- ions
• So the reaction taking place is:
2H2O (l) ⟶ O2 (g) + 4H+ (aq) + 4e-
• Thus we get oxygen in gaseous form at the anode
• As the reaction continues, the number of Cu2+ ions decreases. So the blue colour fades
Solved example 12.5
When acidified copper sulphate solution is electrolysed, oxygen is obtained at the anode. What arrangements are to be made for this? Which element is deposited at the cathode?
Solution:
• Take copper sulphate solution in a beaker. Put two carbon electrodes in it. Connect the electrodes to a battery through a switch. See fig.12.14 below:
• When the switch is turned on, a reaction will take place. The working principle is explained in the previous solved example 12.4. We will get copper at the cathode
Solved example 12.6
How many galvanic cells can be made using the metals Ag, Cu, Zn and Mg. When galvanic cells are made using these metals, What will be the nature of reaction in each cell? (Reactivity: Mg > Zn > Cu > Ag)
Solution:
■ There are 6 possible combinations. That is., we can make six different galvanic cells. They are:
(i) Mg-Zn (ii) Mg-Cu (iii) Mg-Ag (iv) Zn-Cu (v) Zn-Ag (iv) Cu-Ag
• Consider the first combination:
♦ Mg is more reactive. So it will be the anode
What ever be the type of cell, oxidation occurs at the anode. The reaction at anode is:
Mg (s) ⟶ Mg2+ (aq) + 2e-
♦ Zn will be the cathode
What ever be the type of cell, reduction occurs at the cathode. The reaction at cathode is:
Zn+2 (aq) + 2e-1 ⟶ Zn (s)
• Consider the second combination:
♦ Mg is more reactive. So it will be the anode. The oxidation reaction at anode is:
Mg (s) ⟶ Mg2+ (aq) + 2e-
♦ Cu will be the cathode. The reduction reaction at cathode is:
Cu+2 (aq) + 2e-1 ⟶ Cu (s)
• Consider the third combination:
♦ Mg is more reactive. So it will be the anode. The oxidation reaction at anode is:
Mg (s) ⟶ Mg2+ (aq) + 2e-
♦ Ag will be the cathode. The reduction reaction at cathode is:
2Ag+1 (aq) + 2e-1 ⟶ 2Ag (s)
• Consider the fourth combination:
♦ Zn is more reactive. So it will be the anode. The oxidation reaction at anode is:
Zn (s) ⟶ Zn2+ (aq) + 2e-
♦ Cu will be the cathode. The reduction reaction at cathode is:
Cu+2 (aq) + 2e-1 ⟶ Cu (s)
• Consider the fifth combination:
♦ Zn is more reactive. So it will be the anode. The oxidation reaction at anode is:
Zn (s) ⟶ Zn2+ (aq) + 2e-
♦ Ag will be the cathode. The reduction reaction at cathode is:
2Ag+1 (aq) + 2e-1 ⟶ 2Ag (s)
• Consider the sixth combination:
♦ Cu is more reactive. So it will be the anode. The oxidation reaction at anode is:
Cu (s) ⟶ Cu2+ (aq) + 2e-
♦ Ag will be the cathode. The reduction reaction at cathode is:
2Ag+1 (aq) + 2e-1 ⟶ 2Ag (s)
We have completed this discussion on Electrochemistry. In the next section, we will see production of metals.
