Chapter 12.4 - The Copper - Silver Galvanic cell

In the previous section, we saw a galvanic cell made using zinc and copper. In this section, we will see another galvanic cell made using copper and silver.

1. Take two beakers. Add 100 mL of 1 M copper sulphate solution into one beaker. 
2. Add 100 mL of silver nitrate (AgNO₃) solution into the other. See fig.12.7 below:
[We have seen the method to prepare 1 M solution of any given salt. See details here]

Fig.12.7

3. Immerse a copper rod in the copper sulphate solution. 
4. Immerse a silver rod in the silver nitrate solution. 
5. Connect the negative terminal of a voltmeter to the copper rod. Connect the positive terminal to the silver rod. 
6. Connect the solutions in the two beakers by a salt bridge (For details about salt bridge, see the previous section,). 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. Let us see the reason:
8. In the previous experiments we have seen that copper is more reactive than silver. 
So Cu in the copper rod, loses two electrons and become. Cu²⁺. The equation can be written as:
Cu⁰ (s) ⟶ Cu⁺² (aq) + 2e^-1
• This is a oxidation reaction. Because the oxidation number of Cu increases from zero to +2.
• In other words:
Losing electrons is oxidation. So an oxidation reaction is taking place here.
■ The electrode at which oxidation takes place is called the anode.
9. The newly formed Cu⁺² ions gets detached from the copper rod and goes into the solution. But the released electrons stick to the Cu rod
• Note that many Cu⁺² ions are already present in the copper sulphate solution because, the aqueous copper sulphate solution exists as a mixture of Cu⁺² ions and (SO₄)^-2 ions.
10. Because of the released electrons, the Cu rod 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.
• In other words:
Gaining electrons is reduction. So a reduction reaction is taking place here.
• Note the coefficient '2' in front of Ag. This is because, each Cu atom will donate 2 electrons while each Ag atom will donate only one electron. So two Ag⁺¹ ions can benefit from one Cu⁺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.
• If a flow of electric current is obtained from a device, we can call it a cell.
• The cell which converts chemical energy to electrical energy through redox reaction is called Galvanic cell or voltaic cell.
• So we have made a galvanic cell using copper and silver electrodes. In the previous section, we made one with zinc and copper electrodes.

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In this cell also, we can see two interesting facts:

Fact 1:
• Out of the two metals copper and silver, copper is more ready to donate electrons.
• So oxidation takes place at the copper rod.
• We have seen that, the electrode at which oxidation takes place is called the anode.
So we can say this:
■ In the galvanic cell, the more reactive metal will become the anode
The opposite can also be written:
■ In a galvanic cell, the less reactive metal will become the cathode
Fact 2:
• We have seen that the free electrons are first formed when the Cu atoms become Cu⁺² ions. 
• This happens at the anode. 
■ Thus the electrons flow is from the anode to cathode
The above facts are shown in the fig.12.8 below:

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■ We have seen two galvanic cells. At this stage, if we are given two metal rods, we are in a position to determine which will be the anode and which will be the cathode. Let us see some examples:
• We are given 3 metals: A, B and C
• With the 3 metals, 3 combinations are possible: AB, BC and CA. In each of the 3 combinations, the metal which is more reactive will be the anode and the other will be the cathode.
■ Let the 3 metals be Zn, Cu and Ag. Which are the possible combinations? Write the anode and cathode in each?
Solution:
■ There are 3 possible combinations. That is., we can make three different galvanic cells. They are:
(i) Zn-Cu  (ii) Cu-Ag  (iii) Zn-Ag
• Consider the first combination:
    ♦ Zn is more reactive. So it will be the anode
    ♦ Cu will be the cathode            
Consider the second combination:
    ♦ Cu is more reactive. So it will be the anode
    ♦ Ag will be the cathode
Consider the third combination:
    ♦ Zn is more reactive. So it will be the anode
    ♦ Ag will be the cathode    

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We have seen different galvanic cells. Now we will have a short discussion about the 'direction of current'

• In the discussions, we have seen that, the flow of electrons is from the anode to cathode.
    ♦ In a galvanic cell, anode is the negative electrode because the free electrons are first formed there. It is the source of electrons.
    ♦ In a galvanic cell, cathode is the positive electrode cell because the negatively charged electrons flows towards it
■ So we can write:
The flow of electrons is from the negative electrode (anode) to the positive electrode (cathode)
• But the flow of electric current is considered to be from the positive electrode to the negative electrode. 
Why is that so?
• In earlier days when electricity was discovered, the the flow of current was assumed to be from the positive to the negative electrode. On the basis of this assumption, many rules were written down, and many equations were formulated. Later it was discovered that, electrons flow from negative electrode to positive electrode. But then, all rules and equations would have to be rewritten. To avoid such a difficulty, the direction of current is still assumed to be from positive to negative electrodes.
• We will get correct answers if we follow the 'same convention' in all the calculations
• That is., if we consider the convention: 'current flow to be from positive to negative'. We must apply it in all the calculations
• So the current flow which is from positive to negative is considered as 'conventional current'
• The electron flow from negative to positive is considered as 'electron current'
These details are shown in the fig.12.9 below:

Fig.12.9

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• We use Dry cells in our day to day life as a source of electricity. 
    ♦ The name 'Dry cell' indicates that it is dry. So there is no liquid components in it. 
    ♦ Indeed, the electrolyte in it, is in the form of a paste. This will help to avoid spillage. 
    ♦ Some images can be seen here.
• Mercury cells are also of this type. Images can be seen here. 
    ♦ They are used in small appliances like watches, cameras etc., 
■ Dry cells and Mercury cells cannot be recharged. Cells which cannot be recharged are called primary cells
• Another type available in the market of is the nickel-cadmium cell. Images can be seen here. 
    ♦ It can be recharged. 
■ Cells which can be recharged are called secondary cells. 
• Lithium ion cells (images here) used in mobile phones are also of this type

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So we have completed this discussion on Galvanic cells. In the next section, we will see Electrolysis. 

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Chapter 12.3 - The Zinc - Copper Galvanic cell

In the previous section, we saw reason for the displacement of silver from a solution of silver nitrate. In this section, we will see how metals can be used to make electricity.

1. Take two beakers. Add 100 mL of 1 M zinc sulphate solution into one beaker. 
2. Add 100 mL of copper sulphate solution into the other. See fig.12.5 below:
[We have seen the method to prepare 1 M solution of any given salt. See details here]

Fig.12.5

3. Immerse a zinc rod in the zinc sulphate solution. 
4. Immerse a copper rod in the copper sulphate solution. 
5. Connect the negative terminal of a voltmeter to the zinc rod. Connect the positive terminal to the copper 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.

■ Method of making a salt bridge: • Make a paste by mixing two items: (i) gelatin (ii) Potassium chloride (KCl)     ♦ agar agar gel can be used instead of gelatin     ♦ Potassium nitrate (KNO3) or Ammonium chloride (NH4Cl) can be used instead of KCl • Fill this paste into an U-tube • Close the ends of the tube with cotton balls • The salt bridge is ready to use. We will see it's function later in this section

7. Now observe the change in the voltmeter reading.
From the voltmeter, it is clear that electricity is produced in the experiment. Let us see the reason:
8. In the previous experiments we have seen that zinc is more reactive than copper. 
So Zn in the zinc rod, loses two electrons and become. Zn²⁺. The equation can be written as:
Zn⁰ (s) ⟶ Zn⁺² (aq) + 2e^-1
• This is a oxidation reaction. Because the oxidation number of Zn increases from zero to +2.
• In other words:
Losing electrons is oxidation. So an oxidation reaction is taking place here.
■ The electrode at which oxidation takes place is called the anode.
9. The newly formed Zn⁺² ions gets detached from the zinc rod and goes into the solution. But the released electrons stick to the Zn rod
• Note that many Zn⁺² ions are already present in the zinc sulphate solution because, the aqueous zinc sulphate solution exists as a mixture of Zn⁺² ions and (SO₄)^-2 ions.
10. Because of the released electrons, the Zn rod becomes negatively charged. These free electrons reach the copper rod through the external circuit
11. These electrons which reach the copper rod flows through the copper rod and reaches the copper sulphate solution. 
• In the copper solution, Cu⁺² ions are present. These ions receive the electrons and become Cu atoms. The equation can be written as:
Cu⁺² (aq) + 2e^-1 ⟶ Cu⁰ (s)
• This is a reduction reaction. Because the oxidation number of Cu⁺² decreases from +2 to zero.
• In other words:
Gaining electrons is reduction. So a reduction reaction is taking place here.
■ 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.
• If a flow of electric current is obtained from a device, we can call it a cell.
• The cell which converts chemical energy to electrical energy through redox reaction is called Galvanic cell or Voltaic cell.
• They are named after the scientists Luigi Galvani and Alessandro Volta who achieved early developments in such cells.

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In such cells, we can see two interesting facts:

Fact 1:
• Out of the two metals zinc and copper, zinc is more ready to donate electrons. 
• So oxidation takes place at the zinc rod. 
• We have seen that, the electrode at which oxidation takes place is called the anode.
So we can say this:
■ In the galvanic cell, the more reactive metal will become the anode
The opposite can also be written:
■ In a galvanic cell, the less reactive metal will become the cathode
Fact 2:
• We have seen that the free electrons are first formed when the Zn atoms become Zn⁺² ions. 
• This happens at the anode. 
■ Thus in a galvanic cell, the electrons flow is from the anode to cathode
The above facts are shown in the fig.12.6 below: 

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Corrosion of metals

At this stage, we can have a short discussion about the basics of 'corrosion of metals'. 

• We have seen that the zinc atoms become Zn⁺² ions and leave the zinc rod. If the process continues, the zinc rod will soon become useless.

• The zinc atoms left the rod because favorable conditions were available on the surface of the zinc rod.

    ♦ We provided those favorable conditions by connecting it with a copper rod, providing zinc sulphate and copper sulphate solutions etc.,

• If such favorable conditions occur naturally around the surface of any metal, it's atoms would surely leave. 

• For example, the oxygen and water vapour present in the atmospheric air can provide favorable conditions for certain metals. 

• The atoms of those metals will then become ions and will leave the original metal surface. 
• The released electrons will be received by oxygen. The oxygen become negative ions. 
• The 'positive metal ions' and 'negative oxygen ions' together will form a new compound. The rust which we see on the surface of iron is such a compound.           

■ The process of conversion of a metal into it's compounds by continuous interaction with atmospheric air and water vapour is termed corrosion of metals. This is an electrochemical reaction. 
• Metals like potassium, sodium, zinc, copper etc., also undergo corrosion.

Cathodic protection

• In the above discussion, we have seen that the zinc which is more reactive than copper has become the anode.
    ♦ Atoms of the anode loses electrons and become ions. 
• If in the place of copper, we use a metal which is more reactive than zinc, then that metal will become anode. 
    ♦ Zinc will become the cathode and will be saved. 
• So we must make the metal to be protected, the cathode
• This is the basis of cathodic protection.
■ Cathodic protection is a technique used to control the corrosion of a metal surface by making it the cathode of an electrochemical cell. In the simplest form, the metal to be protected is connected to a more easily corroded 'sacrificial metal' to act as the anode
• We will see more technical details of this method in higher classes

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Now we will discuss about the salt bridge
• Consider the cathode compartment in the above galvanic cell. 
    ♦ We have a copper rod in a copper sulphate solution. 
• The copper sulphate solution always exists as a mixture of Cu⁺² and (SO₄)^-2 ions. 
• We have seen that, the electrons reaching the copper rod will be received by the Cu⁺² ions, and they will become Cu atoms. 
• So the positive charges in the solution will continuously decrease. In other words, there will be an excess negative charge (due to the (SO₄)^-2 ions) in the solution. 
• So these ions flow from the copper compartment to the zinc compartment through the salt bridge.
• This will help to maintain the electrical neutrality of the solution

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So we have completed this discussion on Galvanic cell using Zinc and copper. In the next section, we will see another Galvanic cell. 

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Chapter 12.2 - Displacement of Silver by Copper

In the previous section, we saw reason for the displacement of copper from a solution of copper sulphate. In this section, we will see a similar case.

• In a solution of silver nitrate (AgNO₃),we introduce copper. 
    ♦ It is convenient to use copper wire available in the form of a coil. In this way, there will be more copper within a less space. 
• We know that copper occupies a higher position than silver in the reactivity series. Then what will be the result?
Ans: The copper has a greater tendency to give off electrons. So it will become Cu²⁺ ions. 
• The silver, which is present as Ag¹⁺ ions will receive these electrons and will become Ag atoms. Thus we will get pure silver.
    ♦ Note that silver ions (Ag¹⁺) have only one positive charge. 
    ♦ So two silver ions can benefit from one copper atom. 
    ♦ Because each copper atom will give donate two elecrons and become Cu²⁺.   
• This is a displacement reaction. Because, silver has been displaced from it's solution. 

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Now we will see an experiment to demonstrate the above findings. 

• Prepare some AgNO₃ solution in a beaker. Dip a copper wire in it. See fig.12.4 below:

Fig.12.4

Note down the observations:
1. Initially the copper wire has it's own reddish brown appearance. 
• But after the experiment, the wire is covered with deposits of silver
2. Initially, the solution is colourless.
• But after the experiment, it's colour changes to blue
■ Let us see the explanation for the above two observations:
• When the copper wire is dipped in the solution, a reaction takes place. The equation for this reaction is:
Cu (s) + 2AgNO₃ (aq) ⟶ Cu(NO₃)₂ (aq) + 2Ag (s)
• We can see that pure silver (Ag) is formed as one of the products. 
• This is because, the Cu atoms, which is more reactive than Ag, donated electrons and became Cu²⁺ ions. These electrons were received by the Ag¹⁺ ions, and they became Ag atoms.
• The newly formed Ag atoms stick to the surface of copper wire. This is shown in the enlarged view inside the yellow square in fig.12.4 above. The enlarged view shows what is actually happening in a small square portion on the surface of the copper wire
• The formation of the blue colour of the solution after the experiment is due to the presence of Cu²⁺ ions.
• A video of the experiment can be seen here. 
• It is interesting to note that, the deposited silver can be separated from the copper wire, just by shaking the wire 

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• We must analyse the above reaction in terms of Oxidation and Reduction

• We have learned about oxidation and reduction in a previous lesson. Details here.

• Consider the equation again:
Cu (s) + 2AgNO₃ (aq) ⟶ Cu(NO₃)₂ (aq) + 2Ag (s)

We will write the oxidation state of each component on the reactants side as well as the products side.

Reactants:

1. Cu is an atom of the element. So it's oxidation number will be obviously zero. We can write: Cu⁰

2. AgNO₃ is an ionic compound. Let us write it as a combination: (Ag)(NO₃)

• The nitrate ion (NO₃)^-1 has a known oxidation number of -1

• So the oxidation number of Ag in AgNO₃ must be +1

• Thus we can write: Ag⁺¹NO₃^-1 

3. Cu(NO₃)₂ is similar to AgNO₃

• Cu(NO₃)₂ is an ionic compound. Let us write it as a combination: (Cu)[(NO₃)₂]

• The nitrate ion (NO₃)^-1 has a known oxidation number of -1

• So there must be two (NO₃)^-1 groups to compensate for the +2 oxidation number of Cu

• Thus we can write: Cu⁺²(NO₃)₂^-1 

4. Ag is an atom of the element. So it's oxidation number will be obviously zero. We can write: Ag⁰

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• So the equation can be written as:

Cu⁰ (s) + 2(Ag⁺¹NO₃^-1) ⟶ Cu⁺²(NO₃)₂^-1 + Ag⁰

• In the above equation, we can see that:
    ♦ oxidation number of Cu increased from 0 to +2
    ♦ oxidation number of Ag decreased from +1 to 0
• So we can write:
    ♦ Cu is oxidized
        ★ In other words, Cu which was in it's pure form, is oxidized to a positive ion
    ♦ Ag is reduced
        ★ In other words, positive Ag ion is reduced to pure silver
■ In this reaction, oxidation and reduction take place simultaneously. It is a redox reaction.
■ Note the following points:
• Copper is more reactive than silver. Copper is situated above silver in the reactivity series
• As a result of the reaction:
    ♦ Cu donated electrons and got oxidized
    ♦ Ag received electrons and got reduced

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Now we are in a position to tell whether a displacement reaction will take place or not. See the example below:

■ We are given the following solutions:
• Magnesium sulphate • Copper sulphate • Zinc sulphate • Ferrous sulphate • Silver nitrate
■ Also we are given the following metals in their pure forms:
• Magnesium • Copper • Zinc • Iron • Silver
■ Let us now try all the possible combinations:
A. Magnesium sulphate solution
1. Magnesium: 
If we dip pure magnesium in magnesium sulphate solution, no reaction will take place. Since the displacement reaction is not possible, we put a '✘' mark in the table 12.1 given below. This mark is put at the intersection of:
• Row corresponding to magnesium sulphate
• Column corresponding to magnesium
2. Copper: 
If we dip pure copper in magnesium sulphate solution, displacement reaction will not take place. This is because, copper is less reactive than magnesium. Since the displacement reaction is not possible, we put a '✘' mark in the table 12.1 given below. This mark is put at the intersection of:
• Row corresponding to magnesium sulphate
• Column corresponding to copper
3. Zinc: 
If we dip pure zinc in magnesium sulphate solution, displacement reaction will not take place. This is because, zinc is less reactive than magnesium. Since the displacement reaction is not possible, we put a '✘' mark in the table 12.1 given below. This mark is put at the intersection of:
• Row corresponding to magnesium sulphate
• Column corresponding to zinc
4. Iron: 
If we dip pure iron in magnesium sulphate solution, displacement reaction will not take place. This is because, iron is is less reactive than magnesium. Since the displacement reaction is not possible, we put a '✘' mark in the table 12.1 given below. This mark is put at the intersection of:
• Row corresponding to magnesium sulphate
• Column corresponding to iron
5. Silver: If we dip pure silver in magnesium sulphate solution, displacement reaction will not take place. This is because, silver is less reactive than magnesium. Since the displacement reaction is not possible, we put a '✘' mark in the table 12.1 given below. This mark is put at the intersection of:
• Row corresponding to magnesium sulphate
• Column corresponding to silver
B. Copper sulphate solution
1. Magnesium:
If we dip pure magnesium in copper sulphate solution, displacement reaction will take place. Magnesium will displace copper from the solution. This displaced copper will get deposited. This is because, magnesium is more reactive than copper. Since the displacement reaction is possible, we put a '✔' mark in the table 12.1 given below. This mark is put at the intersection of:
• Row corresponding to copper sulphate
• Column corresponding to magnesium   
2. Copper: 
If we dip pure copper in copper sulphate solution, no reaction will take place. Since the displacement reaction is not possible, we put a '✘' mark in the table 12.1 given below. This mark is put at the intersection of:
• Row corresponding to copper sulphate
• Column corresponding to copper
3. Zinc:
If we dip pure zinc in copper sulphate solution, displacement reaction will take place. Zinc will displace copper from the solution. This displaced copper will get deposited. This is because, zinc is more reactive than copper. Since the displacement reaction is possible, we put a '✔' mark in the table 12.1 given below. This mark is put at the intersection of:
• Row corresponding to copper sulphate
• Column corresponding to zinc

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In this way appropriate marks can be put filled up in the table as shown below:

Table 12.1

	Mg	Cu	Zn	Fe	Ag
Magnesium sulphate	✘	✘	✘	✘	✘
Copper sulphate	✔	✘	✔	✔	✘
Zinc sulphate	✔	✘	✘	✘	✘
Ferrous sulphate	✔	✘	✔	✘	✘
Silver nitrate	✔	✔	✔	✔	✘

In the above table, we find that:
• Mg has 4 '✔' marks
• Zn has 3 '✔' marks
• Iron has 2 '✔' marks
• Copper has 1 '✔' mark
• Silver has 0 '✔' mark
■ So we can arrange them in the decreasing order of reactivity:
Mg > Zn > Fe > Cu > Ag
■ This agrees with the order in the actual reactivity series that we saw in the previous section

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In the next section, we will see how metals can be used to produce electricity. 

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