Chapter 8.16 - Electrode Potential

In the previous section, we saw the Galvanic cell. We also saw anode, cathode and redox couple. In this section, we will see electrode potential.

Electrode potential can be explained in 13 steps:
1. In the Galvanic cell, we saw that, Zn is readily oxidized.
• The readiness to get oxidized will have an impact on the effectiveness of the cell.
2. To ‘get oxidized’ means ‘to donate electrons’.
• So readiness to get oxidized means: readiness to donate electrons.
• If it has more readiness to donate electrons, there will be more number of electrons in the circuit. Then the efficiency of the cell will increase.
3. So it is clear that, at the anode, we must have an atom which has high readiness to get oxidized. This will ensure a good supply of electrons.
4. Recall that, the atom which gets oxidized is the reducing agent.
• So we can write:
The atom at the anode must be a good reducing agent.
5. In a redox couple, we will be having an oxidizing agent and a corresponding reducing agent.
• We saw that, if the reducing agent is strong, the oxidizing agent will be weak and vice versa.
• We must select a redox couple in such a way that:
    ♦ The reducing agent in that couple is strong.
    ♦ The oxidizing agent in that couple is weak.
• We must use that couple at the anode.
6. Having a strong reducing agent at the anode is not good enough. The electrons supplied by that reducing agent must be readily pulled by the atoms at the cathode.
• An atom which is ‘ready to pull electrons’ means that, it is ‘ready to get reduced’  • In the Galvanic cell, we saw that, Cu²⁺ is readily reduced.
• The readiness to get reduced will have an impact on the effectiveness of the cell.
7. If the atoms at the cathode have greater readiness to get reduced, more electrons can reach the cathode.
• The efficiency of the cell will increase.
8. So it is clear that, at the cathode, we must have an atom which has high readiness to get reduced. This will ensure a good flow of electrons.
9. Recall that, the atom which gets reduced is the oxidizing agent.
• So we can write:
The atom at the cathode must be a good oxidizing agent.
10. In a redox couple, we will be having an oxidizing agent and a corresponding reducing agent.
• We saw that, if the reducing agent is strong, the oxidizing agent will be weak and vice versa.
• We must select a redox couple in such a way that:
    ♦ The reducing agent in that couple is weak.
    ♦ The oxidizing agent in that couple is strong.
• We must use that couple at the cathode.
11. Let us write a summary:
• At the anode, we must use a redox couple in such a way that:
    ♦ The reducing agent in that couple is strong.
    ♦ The oxidizing agent in that couple is weak.
• At the cathode, we must use a redox couple in such a way that:
    ♦ The reducing agent in that couple is weak.
    ♦ The oxidizing agent in that couple is strong.
12. Here arises a problem. It can be written in steps:
(i) We have a few redox couples available to be used.
• The reducing agents in all of those couples are known to be strong.
   ♦ We want one of those couples to be used at the anode.
• We would want the couple with the strongest reducing agent.
   ♦ But all of them are know to be strong. Which one is the strongest?
(ii) We have a few redox couples available to be used
• The oxidizing agents in all of those couples are known to be strong.
   ♦ We want one of those couples to be used at the cathode.
• We would want the couple with the strongest oxidizing agent.
   ♦ But all of them are know to be strong. Which one is the strongest?
13. To answer such questions, scientists have prepared a table. This table shows the relative strengths of various redox couples. This table is known as The Standard Electrode Potential table. We will learn what ‘standard electrode potential’ is in higher classes. At present, we only need those values to make comparisons.
• Let us see how this table can be used. It can be written in 6 steps:
(i) As we move along the table from bottom to top, the potential value increases.
• So the bottom most value is the largest negative value.
• As we move up, the values become less and less negative. That means, they are increasing.
• At the hydrogen couple, the value becomes zero.
• After that, the values are positive.
• The top most couples has the largest positive values.
(ii) We know that, in each couple, there is an oxidizing agent and the corresponding reducing agent.
• As we move along the table from bottom to top, the oxidizing agent in the couples become stronger and stronger.   
• As we move along the table from top to bottom, the reducing agent in the couples become stronger and stronger.
(iii) Imagine that, we have a few redox couples available to be used at the anode.
• We know that, to be used at the anode, reducing agent in the couple should be strong.
• So, the couples that we intend to try for anode, must be located towards the bottom of the table.
• Also, the one among them, which is at the bottom most position, must be selected.     
(iv) Imagine that, we have a few redox couples available to be used at the cathode.
• We know that, to be used at the cathode, oxidizing agent in the couple should be strong.
• So, the couples that we intend to try for cathode, must be located towards the top of the table.
• Also, the one among them, which is at the top most position, must be selected.
(v) We said that, as we move from bottom to top, the oxidizing agent in the couple becomes stronger and stronger.
• Also we said that, as we move from top to bottom, the reducing agent in the couple becomes stronger and stronger.
• If this is true, we can assume that,
   ♦ The oxidizing agent in the top most couple is the strongest oxidizing agent.
         ✰ The reducing agent in the top most couple is the weakest reducing agent.
   ♦ The oxidizing agent in the bottom most couple is the weakest oxidizing agent.
         ✰ The reducing agent in the bottom most couple is the strongest reducing agent.
(vi) Let us check whether our assumptions are correct:
◼ Consider the top most couple:
F₂ (g) + 2e^- → 2F^-
• Fluorine (F₂) can pull electrons and become F^- ions. Two F^- ions can donate electrons and become F₂
   ♦ So F₂ is an oxidizing agent and F^- is a reducing agent.
• F is the most electronegative element in the periodic table. It can pull electrons very strongly.
   ♦ So naturally F₂ will top the list of oxidizing agents.
• Theoretically, F^- can supply it’s extra electron. So theoretically, it is a reducing agent. But since it is the strongest electronegative element, F^- will not donate any electrons.
   ♦ That means, it is the weakest reducing agent.
• Thus this redox couple occupies the top most position in the table.
◼ Consider the bottom most couple:
Li⁺ + e^- → Li (s)
• This has the opposite characteristics when compared to F.
• Lithium (Li) can donate electrons and become Li⁺ ions. This Li⁺ ions can accept electrons and become Li.
   ♦ So Li is a reducing agent and Li⁺ is an oxidizing agent.
• Li is a highly electropositive element in the periodic table. It can donate electrons very easily.
   ♦ So naturally Li will be towards the top of the list of reducing agents.
• Theoretically, Li⁺ can accept one electron. So theoretically, it is an oxidizing agent. But since it is a strong electropositive element, Li⁺ will not accept any electrons.
   ♦ That means, it is the weakest oxidizing agent.
• Thus this redox couple occupies the bottom position in the table.

Let us see a solved example:
Solved example 8.12
Given the standard electrode potentials,
K⁺ |K = –2.93V, Ag⁺|Ag = 0.80V, Hg²⁺|Hg = 0.79V, Mg²⁺|Mg = –2.37V, Cr³⁺|Cr = –0.74V
arrange these metals in their increasing order of reducing power.
Solution:
• We know that, the reducing power increases as we move from top to bottom along the table
• From top to bottom, the standard electrode potential decreases
• So to arrange in the increasing order of reducing power, they should be arranged in the decreasing order of standard electrode potential. Thus we get:
Ag⁺|Ag (0.80 V)  <  Hg²⁺|Hg (0.79 V)  <  Cr³⁺|Cr (–0.74 V)  <  Mg²⁺|Mg (–2.37 V)  <  K⁺|K (–2.93 V)

Solved example 8.13
Arrange the following metals in the order in which they displace each other from the solution of their salts.
Al, Cu, Fe, Mg and Zn.
Solution:
1. We are given five metals. Take any two of them.
   ♦ Take a strip of the left side metal.
   ♦ Take a solution made using the salt of the right side metal.
• Put the strip in the solution.
   ♦ The right side metal should get deposited on the strip.
2. So the left side metal has the ability to displace the right side metal from the ‘solution of the salt of the right side metal’
3. For example, consider Cu, Ag
• Put a strip of Cu in silver nitrate solution. Silver will get deposited on the Cu strip.
• Here, Ag has a greater power to pull electrons than Cu.
• So Ag has greater oxidizing power than Cu
• This is same as:
Cu has a greater reducing power than Ag
4. So in the table of standard electrode potential, Cu will be below Ag.
• So the given metals should be arranged in such a way that, any metal should be below all other metals on the right side.
5. We see that, among the given five metals,
Mg occupies the bottom most position.
   ♦ Above Mg, comes Al
   ♦ Above Al, comes Zn
   ♦ Above Zn comes, Fe
   ♦ Above Fe comes Cu
• So the order is: Mg, Al, Zn, Fe, Cu.
6. We can write:
    ♦ Mg will displace Al, Zn, Fe, Cu.
    ♦ Al will displace Zn, Fe, Cu. But not Mg.
    ♦ Zn will displace Fe, Cu. But not Mg and Al.
    ♦ Fe will displace Cu. But not Mg, Al, Zn, Fe.

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Representation of Galvanic cell

This can be written in 7 steps:
1. We know that, the Galvanic cell has two parts.
   ♦ Oxidation takes place at the part on the left side.
   ♦ Reduction takes place at the part on the right side.
2. We call each part as a half cell.
   ♦ The part on the left side is called oxidation half cell.
   ♦ The part on the right side is called reduction half cell.
3. The half cells are represented by the reactions taking place in them.
• Usually, we write a reaction in the form of an equation. We write all the reactants and products involved in that reaction.
• But here,we do not write all the reactants and products. We write only the oxidized form and reduced form.
4. For example, in our present Galvanic cell, the reaction taking place at the oxidation half cell is: Zn → Zn²⁺ + 2e^-
• But for representing this half cell, we write only Zn (the reduced form) and Zn²⁺ (the oxidized form).
• Zn is oxidized to Zn²⁺. So we write Zn first and after that, Zn²⁺
• The two are separated by a vertical line.
• Thus the oxidation half cell is represented as: Zn|Zn²⁺   
5. Similarly, in our present Galvanic cell, the reaction taking place at the reduction half cell is: Cu²⁺ + 2e^- → Cu
• But for representing this half cell, we write only Cu (the reduced form) and Cu²⁺ (the oxidized form).
• Cu²⁺ is reduced to Cu. So we write Cu²⁺ first and after that, Cu.
• The two are separated by a vertical line.
• Thus the reduction half cell is represented as: Cu|Cu²⁺
6. Now we can represent the Galvanic cell as a whole.
• For that,
   ♦ We write the oxidation half cell first.
   ♦ Then we write the reduction half cell.
   ♦ The two half cells are separated by two vertical lines.
         ✰ These two vertical lines represent the salt bridge.
7. Thus our present Galvanic cell can be represented as:
Zn|Zn²⁺|| Cu|Cu²⁺

Solved example 8.14
Depict the galvanic cell in which the reaction Zn(s) + 2Ag⁺ (aq) → Zn²⁺ (aq) +2Ag(s) takes place, Further show:
(i) which of the electrode is negatively charged,
(ii) the carriers of the current in the cell, and
(iii) individual reaction at each electrode.
Solution:
• In the given equation, we see that, Zn is being oxidized and Ag⁺ is being reduced.
• So the two half cells can be represented as:
   ♦ Oxidation half cell: Zn|Zn²⁺
   ♦ Reduction half cell: Ag⁺|Ag
• For representing the cell as a whole, we write the oxidation half cell first and then the reduction half cell. The two are separated by two vertical lines. Thus we get:  Zn|Zn²⁺ || Ag⁺|Ag
Part (i):
1. We know that, anode is the electrode at which oxidation takes place. We also know that anode is the negatively charged electrode.
2. In our present case, Zn|Zn²⁺ is written on the left side. So it is the anode. So Zn|Zn²⁺ is the negatively charged electrode.
3. We can think in this way also:
Each Zn atom in the Zn electrode, donates two electrons and dissolve into the solution as Zn²⁺ (aq) ions. These donated electrons accumulate on the Zn strip, thus making it the negatively charged electrode.
Part (ii):
1. Electrons flow from the Zn strip to the Ag strip (-ve to +ve). So current flows from Ag strip to Zn strip (+ve to -ve).
2. In the solution, charges are transferred through ions.
Part (iii):
1. The reaction at the anode is:
Zn → Zn²⁺ + 2e^-      
2. The reaction at the cathode is:
Ag⁺ + e^- → Ag

Solved example 8.15
Using the standard electrode potentials given in the Table 8.1, predict if the
reaction between the following is feasible:
(a) Fe³⁺ (aq) and I^- (aq)
(b) Ag⁺ (aq) and Cu(s)
(c) Fe³⁺ (aq) and Br^- (aq)
(d) Ag (s) and Fe³⁺ (aq)
(e) Br₂ (aq) and Fe²⁺ (aq).
Solution:
Part (a): Fe³⁺ (aq) and I^- (aq)
1. If the reaction is to occur, Fe³⁺ must pull electron from I^-.
• We can think of a competition between Fe³⁺ and I^-.
    ♦ Fe³⁺ wants to pull electron from I^-.
    ♦ I^- wants to keep the electron. It is also pulling the electron.
2. A species which pulls electrons is an oxidizing agent.
• So if Fe³⁺ is a stronger oxidizing agent, it will succeed in pulling the electron.
• In the table, we see that:
    ♦ Fe³⁺ + e^- → Fe²⁺ is above
    ♦ I₂ + 2e^- → 2I^- is below
3. In the given system, there are Fe³⁺ and I^-
• The equation for I involves I₂ on the left side. But what we have in the system is I^-. So imagine that, I^- has lost electron and I₂ molecules are formed.
• So now in the system, there are Fe³⁺, I₂ and e^-
    ♦ Fe³⁺ will grab the newly formed e^-
    ♦ I₂ cannot grab the newly formed e^-
• This is because, Fe³⁺ is above I₂. The one which is above, is a stronger oxidizing agent and so, can pull electrons from the one which is below.
• Thus the Fe³⁺ will get electrons. I^- will be oxidized to I₂. So, this reaction is feasible.

Part (b): Ag⁺ (aq) and Cu(s)
1. If the reaction is to occur, Ag⁺ must pull electron from Cu.
• We can think of a competition between Ag⁺ and Cu.
    ♦ Ag⁺ wants to pull electron from Cu.
    ♦ Cu wants to keep the electron. It is also pulling the electron.
2. A species which pulls electrons is an oxidizing agent.
• So if Ag⁺ is a stronger oxidizing agent, it will succeed in pulling the electron.
• In the table, we see that:
    ♦ Ag⁺ + e^- → Ag is above.
    ♦ Cu²⁺ + 2e^- → Cu is below.
3. In the given system, there are Ag⁺ and Cu.
• The equation for Cu involves Cu²⁺ on the left side. But what we have in the system is Cu. So imagine that, Cu has lost electrons and Cu²⁺ ions are formed.
• So now in the system, there are Ag⁺, Cu²⁺ and e^-
    ♦ Ag⁺ will grab the newly formed e^-
    ♦ Cu²⁺ cannot grab the newly formed e^-
• This is because, Ag⁺ is above Cu²⁺. The one which is above, is a stronger oxidizing agent and so, can pull electrons from the one which is below.
• Thus the Ag⁺ will get electrons. Cu will be oxidized to Cu²⁺. So this reaction is feasible.

Part (c): Fe³⁺ (aq) and Br^- (aq)
1. If the reaction is to occur, Fe³⁺ must pull electron from Br^-.
• We can think of a competition between Fe³⁺ and Br^-.
    ♦ Fe³⁺ wants to pull electron from Br^-.
    ♦ Br^- wants to keep the electron. It is also pulling the electron.
2. A species which pulls electrons is an oxidizing agent.
• So if Fe³⁺ is a stronger oxidizing agent, it will succeed in pulling the electron.
• In the table, we see that:
    ♦ Fe³⁺ + e^- → Fe²⁺ is below
    ♦ Br₂ + 2e^- → 2Br^- is above
3. In the given system, there are Fe³⁺ and Br^-
• The equation for Br^- involves Br₂ on the left side. But what we have in the system is Br^-. So imagine that, Br^- has lost electron and Br₂ molecules are formed.
• So now in the system, there are Fe³⁺, Br₂ and e^-
    ♦ Fe³⁺ cannot grab the newly formed e^-
    ♦ Br₂ will grab the newly formed e^-
• This is because, Fe³⁺ is below Br₂. The one which is below, is a weaker oxidizing agent and so, cannot pull electrons from the one which is above.
• So what we imagined is wrong. The Fe³⁺ will not get electrons. This reaction is not feasible.

Part (d): Ag (s) and Fe³⁺ (aq)
1. If the reaction is to occur, Fe³⁺ must pull electron from Ag.
• We can think of a competition between Fe³⁺ and Ag.
    ♦ Fe³⁺ wants to pull electron from Ag.
    ♦ Ag wants to keep the electron. It is also pulling the electron.
2. A species which pulls electrons is an oxidizing agent.
• So if Fe³⁺ is a stronger oxidizing agent, it will succeed in pulling the electron.
• In the table, we see that:
    ♦ Fe³⁺ + e^- → Fe²⁺ is below
    ♦ Ag⁺ + e^- → Ag is above
3. In the given system, there are Fe³⁺ and Ag.
• The equation for Ag involves Ag⁺ on the left side. But what we have in the system is Ag. So imagine that, Ag has lost electron and Ag⁺ is formed.
So now in the system, there are Fe³⁺, Ag⁺ and e^-
    ♦ Fe³⁺ cannot grab the newly formed e^-
    ♦ Ag⁺ will grab the newly formed e^-
• This is because, Fe³⁺ is below Ag⁺. The one which is below, is a weaker oxidizing agent and so, cannot pull electrons from the one which is above.
• So what we imagined is wrong. The Fe³⁺ will not get electrons. This reaction is not feasible.

Part (e): Br₂ (s) and Fe²⁺ (aq)
1. If the reaction is to occur, Fe²⁺ must lose one electron and become Fe³⁺. This electron must be accepted by Br₂.
• We can think of a competition between Fe²⁺ and Br₂.
    ♦ Fe²⁺ wants to lose electron to Br₂.
    ♦ Br₂ does not want to accept the electron.
2. A species which loses electrons is a reducing agent.
• So if Fe²⁺ is a stronger reducing agent, it will succeed in donating the electron.
• In the table, we see that:
    ♦ Fe³⁺ + e^- → Fe²⁺ is below
    ♦ Br₂ + 2e^- → 2Br^- is above
3. In the given system, there are Fe²⁺ and Br₂.
• The equation for Fe²⁺ involves Fe³⁺ on the left side. But what we have in the system is Fe²⁺. So imagine that, Fe²⁺ has lost electron and Fe³⁺ is formed.
• So now in the system, there are Fe³⁺, Br₂ and e^-
    ♦ Fe³⁺ cannot grab the newly formed e^-
    ♦ Br₂ will grab the newly formed e^-
• This is because, Br₂ is above Fe³⁺. The one which is above, is a stronger oxidizing agent and so, can pull electrons from the one which is below.
• Thus the Fe²⁺ will lose electrons and so, this reaction is feasible.

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We have completed the discussions in this chapter. In the next section, we will see some solved examples related to this chapter in general.


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