Chapter 8.15 - The Galvanic Cell

In the previous section, we completed a discussion on redox titration. In this section, we will see redox reactions and electrode processes.

• Recall the experiment that we saw in fig.8.1 in section 8.1. We saw that, copper gets deposited on the zinc strip.
• We placed that zinc strip in copper nitrate solution. Instead of copper nitrate solution, we can use copper sulphate solution. Then also, we will get a deposit of copper.
• We saw that, there is a transfer of electrons from Zn atoms to Cu²⁺ ions. So the electrons start the journey from Zn atoms in the zinc strip. Their journey ends at the Cu²⁺ ions.
• If we can make this ‘travel of electrons’ to take place through a wire, there will be an electric current in that wire. Such an electric current can be used to light bulbs, run motors etc.,
• So our next task is to develop an arrangement so that, the ‘travel of electrons’ takes place through a wire. It can be explained in 11 steps:

1. In fig.8.9(a) below, we have,
   ♦ Zinc sulphate solution in the left side beaker.
         ✰ In this solution, a zinc strip is placed.
   ♦ Copper sulphate solution in the right side beaker.
         ✰ In this solution, a copper strip is placed.

Fig.8.9

2. We connect the two strips through a wire.
• A switch and a volt meter are provided in the wire
3. The two solutions are connected by a salt bridge.
• A salt bridge is a glass tube filled with a specially prepared material.
• This material is prepared by boiling a solution of KCl (potassium chloride) or NH₄NO₃ (ammonium nitrate).
• Before boiling, agar agar is added. So when cooled, it becomes a jelly like substance.
• This substance is filled into the glass tube and the ends of the tube are plugged with soaked cotton.
   ♦ Electrons cannot pass through a salt bridge.
   ♦ But ions can easily pass through it.
4. Now the system is ready for trial. When we turn on the switch, we get a reading on the voltmeter. This indicates that, an electric current is flowing through the wire.
• If instead of the voltmeter, we use a small bulb, it will glow.
• The following steps from (5) to (10) will explain the cause of the electric current.
5. Consider the following two equations:
Zn²⁺ (aq) + 2e^- → Zn (s)
Cu²⁺ (aq) + 2e^- → Cu (s)
• We see that:
   ♦ Zn²⁺ is trying to attract two electrons and become solid Zn atom.
   ♦ Cu²⁺ is trying to attract two electrons and become solid Cu atom.
• But the attraction by Cu²⁺ is greater. So Cu²⁺ will win over Zn²⁺.
6. The Zn²⁺ ions in the zinc sulphate solution has already lost two electrons. They have no electrons to spare.
• But the Zn atoms in the zinc strip has electrons to spare. The Cu²⁺ ions in the right side beaker will pull the electrons from the zinc strip.
• The zinc atoms cannot resist the pull because, Cu²⁺ is stronger. So the Zn atoms give away two electrons each and become Zn²⁺ ions.
   ♦ These ions cannot exist in solid form.
   ♦ So they go into the solution as Zn²⁺ (aq)
7. The electrons given away by Zn atoms, pass through the wire and reach the copper strip.
• Remember that metals are good conductors. So the electrons flow towards the bottom of the copper strip.
• The Cu²⁺ ions attract these electrons and become Cu atoms. Cu atoms are solid atoms. The newly formed Cu atoms stick to the copper strip.
8. Thus we have two results:
• On the left beaker, more and more solid Zn atoms dissolve into the solution as Zn²⁺ ions.
• On the right beaker, more and more solid Cu atoms are deposited on the strip.
9. So sizes of the strips change:
   ♦ The zinc strip becomes smaller and smaller.
   ♦ The copper strip becomes larger and larger.
10. At the same time, we get a continuous flow of electrons through the wire.
   ♦ The flow of electrons is from left to right.
   ♦ So by convention, electric current is from right to left
11. Now we will see the function of the salt bridge. It can be written in 8 steps:
(i) Consider fig.b. In the left side beaker, we have Zn²⁺ ions (shown as green dots) and SO₄^2- ions (shown as red dots).
• In (6) we saw that, in the left side beaker, more and more Zn (s) goes into the solution as Zn²⁺ (aq). So there is a continuous increase of green dots. That means, there is a continuous increase of positive charge in the solution.
(ii) In the right side beaker, we have Cu²⁺ ions (shown as yellow dots) and SO₄^2- ions (shown as red dots).
• In (7) we saw that, in the right side beaker, more and more Cu²⁺ ions gets converted into Cu atoms and stick to the copper strip. So there is a continuous decrease of yellow dots. That means, there is a continuous decrease of positive charge in the solution.
• Decrease in positive charge is same as increase in negative charge.
(iii) Based on (i) and (ii), we can write:
   ♦ The solution on the left side becomes positively charged.
   ♦ The solution on the right side becomes negatively charged.
(iv) We obtained a flow of electrons in the wire because of the build up of electrons (negative charges) on the left strip.
• This build up of excess negative charge on the left side created an electric     potential. This potential is the cause of the electric current from left to right in the wire.
(v) But in (iii), we see that, there is a build up of excess negative charge on the right side. This will create an electric potential which will create a flow of current from right to left
(vi) So the two potentials have opposite  directions. They will cancel each other.
Thus the flow of current in the wire will come to a stop.
(vii) The salt bridge will help us to prevent this situation. Remember that, the salt bridge contains KCl. Since it is in jelly form, it will contain K⁺ ions (brown dots) and Cl^- ions (purple dots).
• So, when the number of positive ions in the left side increase, the Cl^- ions flow from the salt bridge into the left side solution. This is indicated by the black arrow in the left side solution. This flow of Cl^- ions will help to prevent the build up of excess positive charge on the left side solution.
• At the same time, excess SO₄^2- ions on the right side solution, will flow into the salt bridge. This is indicated by the black arrow in the right side solution. This flow of SO₄^2- ions will help to prevent the build up of excess negative charge on the right side solution.
(viii) Thus the salt bridge helps to maintain the electrical neutrality of the solutions. Since no excess charges are produced in either solutions, the opposite potential mentioned in (v) will not form. So the current can flow freely from left to right along the wire.

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We have seen the working of a Galvanic cell. Now we will see some theoretical details about this cell.

Anode and Cathode

• First we will see anode and cathode. It can be written in 2 steps:
1. In the Galvanic cell, the metal strips act as electrodes. In our present Galvanic cell, the zinc strip and the copper strip are the electrodes.
    ♦ The electrode at which oxidation takes place, is called anode.
    ♦ The electrode at which reduction takes place, is called cathode.
2. In our present case,
• The oxidation reaction is: Zn → Zn²⁺ + 2e^-
    ♦ This takes place at the Zn strip.
    ♦ So the Zn strip is the anode.
• The reduction reaction is: Cu²⁺ + 2e^- → Cu
    ♦ This takes place at the Cu strip.
    ♦ So the Cu strip is the cathode.

Redox couple

Now we will see redox couple. It can be written in 10 steps:
1. Consider a reducing agent like Zn. It can donate electrons and cause the  reduction of ions like Cu²⁺. The donation of electrons by Zn can be written as a half reaction equation:
Zn → Zn²⁺ + 2e^-
2. Consider Zn²⁺ on the right side of the above equation.
• This Zn²⁺ is ready to accept two electrons and become Zn.
3. We are aware of the fact that, if Cu and Zn are the only atoms present in the system, Zn²⁺ cannot get any electrons because Cu²⁺ is a stronger electron puller than Zn²⁺.
• But instead of Cu, if some other element which is a weaker electron puller, is present along with Zn²⁺, then this Zn²⁺ will accept two electrons and become Zn.
That means, Zn²⁺ can act as an oxidizing agent.
4. So we can write:
• In the half reaction equation in (1),
   ♦ Zn on the left side is a reducing agent
   ♦ Zn²⁺ on the right side is an oxidizing agent.
5. Let us see another example
Consider an oxidizing agent like Ag⁺. It can accept electrons and cause the oxidation of atoms like Cu. The acceptance of electrons by Ag can be written as a half reaction equation:
Ag⁺ + 1e^- → Ag
6. Consider Ag on the right side of the above equation.
• This Ag is ready to donate one electron and become Ag⁺.
7. We are aware of the fact that, if Cu and Ag are the only atoms present in the system, Ag cannot donate any electrons because Cu²⁺ is a weaker electron puller than Ag⁺.
• But instead of Cu, if some other element which is a stronger electron puller, is present along with Ag, then this Ag will donate one electron and become Ag⁺.
That means, Ag can act as a reducing agent.
8. So we can write:
• In the half reaction equation in (5),
   ♦ Ag⁺ on the left side is an oxidizing agent
   ♦ Ag on the right side is a reducing agent
9. Let us write a summary:
• In (4), we wrote:
   ♦ Zn on the left side is a reducing agent
   ♦ Zn²⁺ on the right side is an oxidizing agent.
• In (8) we wrote:
   ♦ Ag⁺ on the left side is an oxidizing agent
   ♦ Ag on the right side is a reducing agent
10. Now we can write the definition of redox couple:
An oxidizing agent and a reducing agent which appear on opposite sides of a half reaction equation, together form a redox couple.
Example:
   ♦ Zn and Zn²⁺ together form a redox couple.
   ♦ Ag⁺ and Ag together form a redox couple.
• So in a redox couple, there will be both oxidizing agent and reducing agent.
11. In a redox couple,
    ♦ If the oxidizing agent is strong,
    ♦ The corresponding reducing agent will be weak.
• This can be explained by using chlorine as an example. It can be written in 5 steps:
(i) We have: Cl₂ (g) + 2e^- → 2Cl^-
(ii) Cl₂ is the oxidizing agent because it pulls electrons from other atoms.
• Cl₂ is a strong oxidizing agent because, it can exert a strong pulling force on electrons.
(iii) Once it has acquired the electrons, it becomes the reducing agent Cl^-
• Cl^- is a reducing agent because, it can supply electrons.
(iv) But the Cl^- does not part with it’s electron very easily.
• This is obvious. An atom which is a good electron puller, cannot be a good electron giver. It will try to keep the electrons to itself.
(v) So it is clear that, If the oxidizing agent is strong, the corresponding reducing agent will be weak.
12. In a redox couple,
    ♦ If the reducing agent is strong,
    ♦ The corresponding oxidizing agent will be weak.
• This can be explained by using sodium as an example. It can be written in 5 steps:
(i) We have: Na⁺ + e^- → Na (s)
(ii) Na is the reducing agent because it supplies electrons to other atoms.
• Na is a strong reducing agent because, it can readily supply electrons.
(iii) Once it has lost the electrons, it becomes the oxidizing agent Na⁺
• Na⁺ is a oxidizing agent because, it can pull electrons from other atoms and become Na.
(iv) But the Na⁺ does not want any electrons. This is obvious. An atom which is a good electron supplier, cannot be a good electron acceptor. So it is clear that, If the reducing agent is strong, the corresponding oxidizing agent will be weak.

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In the next section, we will see electrode potential.


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