Chapter 13.6 - Some Common Chemical Properties Of Alkanes

In the previous section, we completed a discussion on the mechanism of halogenation. We also saw the basics of complete and incomplete combustion. In this section, we will see controlled oxidation. Later in this section we will also see isomerisation, aromatization and reaction with steam.

---

III. Controlled oxidation

This can be written in 4 steps:
1. We have seen two types of reaction of alkanes with oxygen:
    ♦ Reaction when dioxygen supply is sufficient.    
    ♦ Reaction when dioxygen supply is insufficient.
2. Next we will see the reaction when supply of dioxygen is regulated.
• That means, we supply measured quantities of oxygen so that the desired products are obtained.
3. In addition to regulating the quantity of oxygen, we need to take care of some other factors also:
    ♦ Adjust the temperature to the suitable value.    
    ♦ Adjust the pressure to the suitable value.
    ♦ Provide suitable catalysts.
4. Let us see some examples of such reactions:
Example 1:
This can be written in 4 steps:
(i) In the previous section, we have seen complete combustion of methane:
$\rm{CH_4 (g)~+~2O_2 (g)~ \color {green}{\xrightarrow[{}]{combustion}} ~ CO_2 (g)~+~2H_2 O (g)}$
• So for complete combustion, if we take two moles of methane, there must be four moles of dioxygen.
(ii) We also saw incomplete combustion of methane:
$\rm{CH_4 (g)~+~O_2 (g)~ \color {green}{\xrightarrow[{combustion}]{Incomplete}} ~ C (s)~+~2H_2 O (g)}$
• So for incomplete combustion, if we take two moles of methane, there must be only two moles of dioxygen.
(iii) Now consider the following reaction:
$\rm{2 CH_4 ~+~O_2 ~ \color {green}{\xrightarrow[{}]{Cu/523 K/100 atm}} ~ 2CH_3 OH}$
• CH₃OH is methanol
• Here, we are taking two moles of methane but only one mole of dioxygen.
• So we can write:
If we regulate the supply of dioxygen in such a way that, the ratio of methane to dioxygen is 2:1, the oxidation of methane will give methanol.
(iv) Also note that,
    ♦ Temperature needs to be kept at 523 K .
    ♦ Pressure needs to be kept at 100 atm.
    ♦ Cu must be used as catalyst.

Example 2:
This can be written in 2 steps:
(i) In the previous section, we saw incomplete combustion of methane:
$\rm{CH_4 (g)~+~O_2 (g)~ \color {green}{\xrightarrow[{combustion}]{Incomplete}} ~ C (s)~+~2H_2 O (g)}$
(ii) Now consider the following reaction:
$\rm{CH_4 ~+~O_2 ~ \color {green}{\xrightarrow[{Δ}]{Mo_2 O_3}} ~ HCHO}$
• HCHO is methanal
• Here, we are taking the same quantities as in incomplete combustion but Mo₂O₃ is used as catalyst.
• So we can write:
If we regulate the supply of dioxygen in such a way that, the ratio of methane to dioxygen is 1:1, and use Mo₂O₃ as catalyst, the oxidation of methane will give methanal.

Example 3
Controlled oxidation of ethane in the presence of (CH₃COO)₂Mn as catalyst, will give ethanoic acid. The equation is:
$\rm{2CH_3 CH_3 ~+~3 O_2 ~ \color {green}{\xrightarrow[{Δ}]{(CH_3 COO)_2 Mn}} ~ 2CH_3 COOH~+~H_2 O}$ 

Example 4
This can be written in 2 steps:
1. Normally, alkanes resist oxidation. But if there is a tertiary H atom in the alkane, that H atom can be oxidized to the corresponding alcohol. We must use potassium permanganate also in the reaction. The equation is:
$\rm{(CH_3)_3 CH ~+~3 O_2 ~ \color {green}{\xrightarrow[{oxidation}]{KMnO_4}} ~ (CH_3)_3 COH}$
2. Using Lewis structures, the reaction can be explained as shown in fig.13.37 below:

Fig.13.37

---

IV. Isomerisation

This can be written in 3 steps:
1. n-Alkanes can be heated in the presence of anhydrous aluminium chloride and hydrogen chloride gas.
• [The letter “n” in n-Alkanes stands for “normal”. They are straight chain alkanes. They do not have any branches]
• [The word “anhydrous” indicates that, no water molecules are present. If water molecules are present, they will interfere with the reaction and we will not get the desired products]
2. When heated in this way, the n-alkanes isomerise to branched chain alkanes.
• There may be more than one products.
    ♦ But each product will be an isomer of the n-alkane taken initially.
    ♦ Each product will contain branches.
3. Let us see an example:

Fig.13.38

• In fig.13.38(a) above, the n-alkane taken initially is n-Hexane.
• The two products are 2-Methylpentane and 3-Methylpentane.
    ♦ They are both isomers of n-Hexane .
    ♦ These are the major products.
• Some minor products may also form during the reaction. But minor products are usually not reported in organic reactions. In our present case, a possible minor product is shown in fig.b. It is also an isomer of n-Hexane.

---

V. Aromatization

This can be written in 6 steps:
1. Take a n-alkane and subject it to 773 K temperature and 10-20 atm pressure.
• Also use any one of the three catalysts given below:
    ♦ Oxide of vanadium
    ♦ Oxide of molybdenum
    ♦ Oxide of chromium supported over alumina.
• The n-alkane taken, should have at least six C atoms.
2. During this process, the n-alkane will get dehydrogenated. That means, the n-alkane will lose some H atoms.
• When H atoms are lost, some of the single bonds will be converted to double bonds. This is to satisfy the valencies of C atoms.
3. Also during this process, the straight chain of the n-alkane will get cyclised. That means, the straight chain will change into a ring.
4. If the n-alkane taken initially has six C atoms, then the result will be benzene. This is shown in fig.13.39 below:

Fig.13.39

• Though the n-Hexane has a straight chain structure, in the fig above, it is shown in a bent form. While in this bent form, if the two C atoms at the ends of the chain can join together, we will get the cyclic structure benzene.
5. If the n-alkane taken initially has more than six C atoms, then the extra C atoms will form methyl groups around the benzene ring.
• Toleune has one methyl group around a benzene ring. So to obtain toleune, we must subject n-heptane to aromatization.
6. Consider the bent form of n-hexane (C₆H₁₄) in fig.13.39 above. While in this bent form, if only two H atoms from the ends are lost, we will be getting cyclohexane (C₆H₁₂)
• But since the catalysts shown in fig.13.39 are used and also since the specified temperature and pressure are applied, a total of eight H atoms will be lost. Thus we will be getting benzene (C₆H₆) instead of cyclohexane.

---

VI. Reaction with steam

This can be written in 3 steps:
1. Methane will react with steam if a temperature of 1273 K is available.
• Nickel should be also present as a catalyst.
2. The products are carbon monoxide and dihydrogen.
3. This method is used for the industrial preparation of dihydrogen gas.

$\rm{CH_4 ~+~H_2 O ~ \color {green}{\xrightarrow[{Δ}]{Ni}} ~ CO~+~3H_2}$

---

VII. Pyrolysis

This can be written in 5 steps:
1. When higher alkanes are subjected to high temperature, they decompose into lower alkanes, alkenes etc.,
2. Such a decomposition reaction into smaller fragments by the application of heat is called pyrolysis. Another name for this process is cracking.
• An example is shown below:

Fig.13.40

    ♦ In the first case, hexane undergoes pyrolysis to give hexene and dihydrogen.
    ♦ In the second case, hexane undergoes pyrolysis to give butene and ethane.
    ♦ In the third case, hexane undergoes pyrolysis to give propene, ethene and methane.
3. Pyrolysis of alkanes is believed to be a free radical reaction.
(A free radical reaction is a reaction which involves free radicals. For example, in the previous section, we saw that halogenation involves chlorine free radicals and methyl free radicals)
4. Pyrolysis of kerosene will give oil gas.
• In this process, dodecane, a constituent of kerosene, decomposes into heptane and pentene.
• A temperature of 973 K is required for this process.
• Also nickel or palladium must be present as a catalyst.
$\rm{C_{12} H_{26} ~ \color {green}{\xrightarrow[{973 K}]{Pt/Pd/Ni}} ~ C_7 H_{16}~+~C_5 H_{10}~+~\text{other products}}$
    ♦ C₁₂H₂₆ is Dodecane.
    ♦ C₇H₁₆ is Heptane.
    ♦ C₅H₁₀ is Pentene.
5. Pyrolysis of petrol gives petrol gas.

---

We have completed a discussion on the chemical properties of alkanes. In the next section we will see conformations.

Previous

Contents

Next

Copyright©2022 Higher secondary chemistry.blogspot.com

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.

---

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.

---

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


Previous

Contents

Next

Copyright©2021 Higher secondary chemistry.blogspot.com 

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.

---

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.

---

In the next section, we will see electrode potential.


Previous

Contents

Next

Copyright©2021 Higher secondary chemistry.blogspot.com

 

Chapter 8.13 - Half Reaction Method When Medium Is Basic

In the previous section, we saw the steps required for acidic medium. In this section, we will see the steps required for basic medium. 

Example 7
Permanganate ion reacts with bromide ion in basic medium to give manganese
dioxide and bromate ion. Write the balanced ionic equation for the reaction.
Solution:
1. The reactants are: MnO₄^- (aq) and Br^- (aq). (See list of common polyatomic ions)
2. The products are: MnO₂ (s) and BrO₃^- (aq).
3. So the skeletal equation is:
MnO₄^- (aq) + Br^- (aq) → MnO₂ (s) + BrO₃^- (aq)
• We want to balance this equation. It can be written in 12 steps:
Step 1: Write the oxidation numbers of all elements.
• For our present case, we get:
$\mathbf\small{\rm{\overset{+7}{Mn}\,\overset{-2}{O^{-}_4}\,(g)+\overset{-1}{Br^{-}}\,(aq)\rightarrow \overset{+4}{Mn}\,\overset{-2}{O_2}\,(s)+\overset{+5}{Br}\,\overset{-2}{O_3^{-}}\,(aq)}}$
Step 2: Identify the atoms that are oxidized. Also identify the atoms that are reduced.
• For our present case, we see that:
Br is oxidized and Mn is reduced.
Step 3: Write the oxidation half reaction and reduction half reaction.
In our present case,
• The oxidation half reaction is:
Br^- (aq) → BrO₃^- (aq)
• The reduction half reaction is:
MnO₄^- (aq) → MnO₂ (s) (aq)
Step 4: Balance the atoms other than O and H in the oxidation half reaction.
• In our present case, the number of Br atoms are the same on both sides. So it is already balanced.
Step 5: Balance the atoms other than O and H in the reduction half reaction.
• In our present case, the number of Mn atoms are the same on both sides. So it is already balanced.
Step 6: Balance O atoms in the oxidation half reaction by using H₂O molecules. Also balance the H atoms by using H⁺ ions.
• In our present case, there are three excess O atoms on the product side. So we must add three H₂O molecules on the reactant side. We get:
Br^- (aq) + 3H₂O (l) → BrO₃^- (aq)
• Now there are six excess H atoms on the reactant side. So we must put six H⁺ ions on the product side. We get:
Br^- (aq) + 3H₂O (l) → BrO₃^- (aq) + 6H⁺
• To balance the six H⁺ ions, we add six OH^- ions on both sides. We get:
Br^- (aq) + 3H₂O (l) + 6OH^- (aq) → BrO₃^- (aq) + 6H⁺ + 6OH^- (aq)
• The H⁺ and OH^- ions on the product side reacts together to give six H₂O molecules. So the equation becomes:
Br^- (aq) + 3H₂O (l) + 6OH^- (aq) → BrO₃^- (aq) + 6H₂O (l)
• There are H₂O molecules on both sides. So the net equation is:
Br^- (aq) + 6OH^- (aq) → BrO₃^- (aq) + 3H₂O (l)
Step 7: Balance O atoms in the reduction half reaction by using H₂O molecules. Also balance the H atoms by using H⁺ ions.
• In our present case, there are two excess O atoms on the reactant side. So we must add two H₂O molecules on the product side. We get:
MnO₄^- (aq) → MnO₂ (s) (aq) + 2H₂O
• Now there are four extra H atoms on the product side. So we must put four H⁺ ions on the reactant side. We get:
MnO₄^- (aq) + 4H⁺ → MnO₂ (s) (aq) + 2H₂O
• To balance the four H⁺ ions, we add four OH^- ions on both sides. We get:
MnO₄^- (aq) + 4H⁺ (aq) + 4OH^- (aq) → MnO₂ (s) (aq) + 2H₂O + 4OH^- (aq)
• The H⁺ and OH^- ions on the reactant side reacts together to give four H₂O molecules. So the equation becomes:
MnO₄^- (aq) + 4H₂O (l) → MnO₂ (s) (aq) + 2H₂O + 4OH^- (aq)
• There are H₂O molecules on both sides. So the net equation is:
MnO₄^- (aq) + 2H₂O (l) → MnO₂ (s) (aq) + 4OH^- (aq)
Step 8: Balance the charges in the oxidation half reaction, by adding appropriate number of electrons on the appropriate side.
• In our present case, the list of charges are as follows:
Reactant side:
   ♦ 1 No. × 1- (from the Br^- ion) = -1
   ♦ 6 No. × 1- (from the OH^- ion) = -6
   ♦ Total = -7
Product side:
   ♦ 1 No. × 1- (from the BrO₃^- ion) = -1
• So there is an excess charge of -6 on the reactant side. Therefore we must add 6e^- on the product side. We get:
Br^- (aq) + 6OH^- (aq) → BrO₃^- (aq) + 3H₂O (l) + 6e^-
Step 9: Balance the charges in the reduction half reaction, by adding appropriate number of electrons on the appropriate side.
• In our present case, the list of charges are as follows:
Reactant side:
   ♦ 1 No. × 1- (from the MnO₄^- ion) = -1
Product side:
   ♦ 4 No. × 1- (from the OH^- ion) = -4
• So there is an excess charge of -3 on the product side. Therefore we must add 3e^- on the reactant side. We get:
MnO₄^- (aq) + 2H₂O (l) +3e^- → MnO₂ (s) (aq) + 4OH^- (aq)
Step 10: Make the number of electrons the same, in both the half reactions.
• In our present case,
    ♦ The oxidation half reaction has 6e^-.
    ♦ The reduction half reaction has 3e^-.
• So we multiply the reduction half reaction by '3'.
Also we keep the oxidation half reaction as such. We get:
Br^- (aq) + 6OH^- (aq) → BrO₃^- (aq) + 3H₂O (l) + 6e^-
2MnO₄^- (aq) + 4H₂O (l) +6e^- → 2MnO₂ (s) (aq) + 8OH^- (aq)
Step 11: Add the two half reactions together.
• In our present case, we get:
2MnO₄^- (aq) + 4H₂O (l) +6e^- + Br^- (aq) + 6OH^- (aq) → 2MnO₂ (s) (aq) + 8OH^- (aq) + BrO₃^- (aq) + 3H₂O (l) + 6e^-
• So the net equation is:
2MnO₄^- (aq) + H₂O (l) + Br^- (aq) → 2MnO₂ (s) (aq) + 2OH^- (aq) + BrO₃^- (aq)
Step 12: Check the balancing of all atoms. Also check the balancing of charges.
• In our present case, we have the following list for atoms:
Reactant side:
   ♦ Mn - 2 No., O - 9 No., Br - 1 No., H - 2 No. 
Product side:
   ♦ Mn - 2 No., O - 9 No., Br - 1 No., H - 2 No. 
         ✰ So all atoms are balanced.
• We have the following table for charges:
Reactant side:
   ♦ 2 No. × 1- (from the MnO₄^- ion) = -2
   ♦ 1 No. × 1- (from the Br^- ion) = -1
   ♦ Total = -3
Product side:
   ♦ 2 No. × 1- (from the OH^- ion) = -2
   ♦ 1 No. × 1- (from the BrO3- ion) = -1
   ♦ Total = -3
         ✰ All charges are balanced.
◼  So the balanced equation is same as that obtained in step 11:
2MnO₄^- (aq) + H₂O (l) + Br^- (aq) → 2MnO₂ (s) (aq) + 2OH^- (aq) + BrO₃^- (aq)

Example 8
Balance the following equation in basic medium:
CrO₂^- (aq) + ClO^- (aq) → CrO₄^2- (aq) + Cl^- (aq)
Solution:
The given skeletal equation is:
CrO₂^- (aq) + ClO^- (aq) → CrO₄^2- (aq) + Cl^- (aq)
• We want to balance this equation. It can be written in 12 steps:
Step 1: Write the oxidation numbers of all elements.
• For our present case, we get:
$\mathbf\small{\rm{
\overset{+3}{Cr}\,\overset{-2}{O_2^{-}}\,(aq)+\overset{+1}{Cl}\,\overset{-2}{O^{-}}(aq)\rightarrow \overset{+6}{Cr}\,\overset{-2}{O_4^{2-}}\,(aq)
+\overset{-1}{Cl^{-}}(aq)}}$
Step 2: Identify the atoms that are oxidized. Also identify the atoms that are reduced.
• For our present case, we see that:
Cr is oxidized and Cl is reduced.
Step 3: Write the oxidation half reaction and reduction half reaction.
In our present case,
• The oxidation half reaction is:
CrO₂^- (aq) → CrO₄^2- (aq)
• The reduction half reaction is:
ClO^- (aq) → Cl^- (aq)
Step 4: Balance the atoms other than O and H in the oxidation half reaction.
• In our present case, the number of Cr atoms are the same on both sides. So it is already balanced.
Step 5: Balance the atoms other than O and H in the reduction half reaction.
• In our present case, the number of Cl atoms are the same on both sides. So it is already balanced.
Step 6: Balance O atoms in the oxidation half reaction by using H₂O molecules. Also balance the H atoms by using H⁺ ions.
• In our present case, there are two excess O atoms on the product side. So we must add two H₂O molecules on the reactant side. We get:
CrO₂^- (aq) + 2H₂O (l) → CrO₄^2- (aq)
• Now there are four excess H atoms on the reactant side. So we must put four H⁺ ions on the product side. We get:
CrO₂^- (aq) + 2H₂O (l) → CrO₄^2- (aq) + 4H⁺ (aq)
• To balance the four H⁺ ions, we add four OH^- ions on both sides. We get:
CrO₂^- (aq) + 2H₂O (l) + 4OH^- (aq) → CrO₄^2- (aq) + 4H⁺ (aq) + 4OH^- (aq)
• The H⁺ and OH^- ions on the product side reacts together to give four H₂O molecules. So the equation becomes:
CrO₂^- (aq) + 2H₂O (l) + 4OH^- (aq) → CrO₄^2- (aq) + 4H₂O (l)
• There are H₂O molecules on both sides. So the net equation is:
CrO₂^- (aq) + 4OH^- (aq) → CrO₄^2- (aq) + 2H₂O (l)
Step 7: Balance O atoms in the reduction half reaction by using H₂O molecules. Also balance the H atoms by using H⁺ ions.
• In our present case, there is one excess O atom on the reactant side. So we must add one H₂O molecule on the product side. We get:
ClO^- (aq) → Cl^- (aq) + H₂O (l)
• Now there are two extra H atoms on the product side. So we must put two H⁺ ions on the reactant side. We get:
ClO^- (aq) + 2H⁺ → Cl^- (aq) + H₂O (l)
• To balance the two H⁺ ions, we add two OH^- ions on both sides. We get:
ClO^- (aq) + 2H⁺ + 2OH^- (aq) → Cl^- (aq) + H₂O (l) + 2OH^- (aq)
• The H⁺ and OH^- ions on the reactant side reacts together to give two H₂O molecules. So the equation becomes:
ClO^- (aq) + 2H₂O (l) → Cl^- (aq) + H₂O (l) + 2OH^- (aq)
• There are H₂O molecules on both sides. So the net equation is:
ClO^- (aq) + H₂O (l) → Cl^- (aq) + 2OH^- (aq)
Step 8: Balance the charges in the oxidation half reaction, by adding appropriate number of electrons on the appropriate side.
• In our present case, the list of charges are as follows:
Reactant side:
   ♦ 1 No. × 1- (from the CrO₂^- ion) = -1
   ♦ 4 No. × 1- (from the OH^- ion) = -4
   ♦ Total = -5
Product side:
   ♦ 1 No. × 2- (from the CrO₄^2- ion) = -2
• So there is an excess charge of -3 on the reactant side. Therefore we must add 3e^- on the product side. We get:
CrO₂^- (aq) + 4OH^- (aq) → CrO₄^2- (aq) + 2H₂O (l) + 3e^-
Step 9: Balance the charges in the reduction half reaction, by adding appropriate number of electrons on the appropriate side.
• In our present case, the list of charges are as follows:
Reactant side:
   ♦ 1 No. × 1- (from the ClO^- ion) = -1
Product side:
   ♦ 1 No. × 1- (from the Cl^- ion) = -1
   ♦ 2 No. × 1- (from the OH^- ion) = -2
   ♦ Total = -3
• So there is an excess charge of -2 on the product side. Therefore we must add 2e^- on the reactant side. We get:
ClO^- (aq) + H₂O (l) + 2e^- → Cl^- (aq) + 2OH^- (aq)
Step 10: Make the number of electrons the same, in both the half reactions.
• In our present case,
    ♦ The oxidation half reaction has 3e^-.
    ♦ The reduction half reaction has 2e^-.
• So we multiply the oxidation half reaction by '2'.
Also we multiply the reduction half reaction by '3'. We get:
2CrO₂^- (aq) + 8OH^- (aq) → 2CrO₄^2- (aq) + 4H₂O (l) + 6e^-
3ClO^- (aq) + 3H₂O (l) + 6e^- → 3Cl^- (aq) + 6OH^- (aq)
Step 11: Add the two half reactions together.
• In our present case, we get:
2CrO₂^- (aq) + 8OH^- (aq) + 3ClO^- (aq) + 3H₂O (l) + 6e^- → 2CrO₄^2- (aq) + 4H₂O (l) + 6e^- + 3Cl^- (aq) + 6OH^- (aq)
The net equation is:
2CrO₂^- (aq) + 2OH^- (aq) + 3ClO^- (aq) → 2CrO₄^2- (aq) + H₂O (l) + 3Cl^- (aq)
Step 12: Check the balancing of all atoms. Also check the balancing of charges.
• In our present case, we have the following list for atoms:
Reactant side:
   ♦ Cr - 2 No., O - 9 No., Cl - 3 No., H - 2 No. 
Product side:
   ♦ Cr - 2 No., O - 9 No., Cl - 3 No., H - 2 No. 
         ✰ So all atoms are balanced.
• We have the following table for charges:
Reactant side:
   ♦ 2 No. × 1- (from the CrO₂^- ion) = -2
   ♦ 2 No. × 1- (from the OH^- ion) = -2
   ♦ 3 No. × 1- (from the ClO^- ion) = -3
   ♦ Total = -7
Product side:
   ♦ 2 No. × 2- (from the CrO₄^2- ion) = -4
   ♦ 3 No. × 1- (from the 3Cl^- ion) = -3
   ♦ Total = -7
         ✰ All charges are balanced.
◼  So the balanced equation is same as that obtained in step 11:
2CrO₂^- (aq) + 2OH^- (aq) + 3ClO^- (aq) → 2CrO₄^2- (aq) + H₂O (l) + 3Cl^- (aq)

Example 9
Permanganate (VII) ion MnO₄^- in basic solution oxidizes iodide ion I^- to produce molecular iodine (I₂) and manganese (iv) oxide (MnO₂). Write a balanced ionic equation to represent this redox reaction.
Solution:
The skeletal equation is:
MnO₄^- (aq) + I^- (aq) → MnO₂ (s) + I₂ (aq)
• We want to balance this equation. It can be written in 12 steps:
Step 1: Write the oxidation numbers of all elements.
• For our present case, we get:
$\mathbf\small{\rm{\overset{+7}{Mn}\,\overset{-2}{O^{-}_4}\,(g)+\overset{-1}{I^{-}}\,(aq)\rightarrow \overset{+4}{Mn}\,\overset{-2}{O_2}\,(s)+\overset{0}{I_2}\,(aq)}}$
Step 2: Identify the atoms that are oxidized. Also identify the atoms that are reduced.
• For our present case, we see that:
I is oxidized and Mn is reduced.
Step 3: Write the oxidation half reaction and reduction half reaction.
In our present case,
• The oxidation half reaction is:
I^- (aq) → I₂ (aq)
• The reduction half reaction is:
MnO₄^- (aq) → MnO₂ (s) (aq)
Step 4: Balance the atoms other than O and H in the oxidation half reaction.
• In our present case, we put '2' in front of I^- in the reactant side. We get:
2I^- (aq) → I₂ (aq)
Step 5: Balance the atoms other than O and H in the reduction half reaction.
• In our present case, the number of Mn atoms are the same on both sides. So it is already balanced.
Step 6: Balance O atoms in the oxidation half reaction by using H₂O molecules. Also balance the H atoms by using H⁺ ions.
• In our present case, there are no O and H atoms
Step 7: Balance O atoms in the reduction half reaction by using H₂O molecules. Also balance the H atoms by using H⁺ ions.
• In our present case, there are two excess O atoms on the reactant side. So we must add two H₂O molecules on the product side. We get:
MnO₄^- (aq) → MnO₂ (s) (aq) + 2H₂O
• Now there are four extra H atoms on the product side. So we must put four H⁺ ions on the reactant side. We get:
MnO₄^- (aq) + 4H⁺ → MnO₂ (s) (aq) + 2H₂O
• To balance the four H⁺ ions, we add four OH^- ions on both sides. We get:
MnO₄^- (aq) + 4H⁺ (aq) + 4OH^- (aq) → MnO₂ (s) (aq) + 2H₂O + 4OH^- (aq)
• The H⁺ and OH^- ions on the reactant side reacts together to give four H₂O molecules. So the equation becomes:
MnO₄^- (aq) + 4H₂O (l) → MnO₂ (s) (aq) + 2H₂O + 4OH^- (aq)
• There are H₂O molecules on both sides. So the net equation is:
MnO₄^- (aq) + 2H₂O (l) → MnO₂ (s) (aq) + 4OH^- (aq)
Step 8: Balance the charges in the oxidation half reaction, by adding appropriate number of electrons on the appropriate side.
• In our present case, the list of charges are as follows:
Reactant side:
   ♦ 2 No. × 1- (from the I^- ion) = -2
Product side:
   ♦ zero
• So there is an excess charge of -2 on the reactant side. Therefore we must add 2e^- on the product side. We get:
2I^- (aq) → I₂ (aq) + 2e^-
Step 9: Balance the charges in the reduction half reaction, by adding appropriate number of electrons on the appropriate side.
• In our present case, the list of charges are as follows:
Reactant side:
   ♦ 1 No. × 1- (from the MnO₄^- ion) = -1
Product side:
   ♦ 4 No. × 1- (from the OH^- ion) = -4
• So there is an excess charge of -3 on the product side. Therefore we must add 3e^- on the reactant side. We get:
MnO₄^- (aq) + 2H₂O (l) +3e^- → MnO₂ (s) (aq) + 4OH^- (aq)
Step 10: Make the number of electrons the same, in both the half reactions.
• In our present case,
    ♦ The oxidation half reaction has 2e^-.
    ♦ The reduction half reaction has 3e^-.
• So we multiply the oxidation half reaction by '3'.
• Also we multiply the reduction half reaction by '2'.
We get:
6I^- (aq) → 3I₂ (aq) + 6e^-
2MnO₄^- (aq) + 4H₂O (l) +6e^- → 2MnO₂ (s) (aq) + 8OH^- (aq)
Step 11: Add the two half reactions together.
• In our present case, we get:
2MnO₄^- (aq) + 4H₂O (l) +6e^- + 6I^- (aq) → 2MnO₂ (s) (aq) + 8OH^- (aq) + 3I₂ (aq) + 6e^-
So the net equation is:
2MnO₄^- (aq) + 4H₂O (l) + 6I^- (aq) → 2MnO₂ (s) (aq) + 8OH^- (aq) + 3I₂ (aq)
Step 12: Check the balancing of all atoms. Also check the balancing of charges.
• In our present case, we have the following list for atoms:
Reactant side:
   ♦ Mn - 2 No., O - 12 No., I - 6 No., H - 8 No. 
Product side:
   ♦ Mn - 2 No., O - 12 No., I - 6 No., H - 8 No. 
         ✰ So all atoms are balanced.
• We have the following table for charges:
Reactant side:
   ♦ 2 No. × 1- (from the MnO₄^- ion) = -2
   ♦ 6 No. × 1- (from the I^- ion) = -6
   ♦ Total = -8
Product side:
   ♦ 8 No. × 1- (from the OH^- ion) = -8
         ✰ All charges are balanced.
◼  So the balanced equation is same as that obtained in step 11:
2MnO₄^- (aq) + 4H₂O (l) + 6I^- (aq) → 2MnO₂ (s) (aq) + 8OH^- (aq) + 3I₂ (aq)

---

The following link gives three more examples:

Solved example 8.11 - Part (a), Part (b) and Part (c)

---

• We have completed a discussion on half reaction method.
• In the next section, we will see redox reactions as the basis for titrations.

Previous

Contents

Next

Copyright©2021 Higher secondary chemistry.blogspot.com

Chapter 8.12 - Half Reaction Method When Medium Is Acidic

In the previous section, we saw the steps in half reaction method, related to neutral medium. We saw them while analyzing three examples. In this section, we will see the additional steps required for acidic medium.

Example 4:
Write the net ionic equation for the reaction of potassium dichromate(VI),
K₂Cr₂O₇ with sodium sulphite, Na₂SO₃, in an acid solution to give chromium(III) ion and the sulfate ion.
Solution:
1. The reactants are: K₂Cr₂O₇ and Na₂SO₃ 
2. K₂Cr₂O₇ is an ionic compound
   ♦ The ions are: K⁺ and Cr₂O₇^2-
• In the dissolved state, the two ions separate away from each other.
• Cr₂O₇^2- will not dissociate further. It is a polyatomic ion. (A list of such polyatomic ions can be seen here. It is better to remember their formula and names)
3. Na₂SO₃ is an ionic compound
   ♦ The ions are: Na⁺ and SO₃^2-
• In the dissolved state, the two ions separate away from each other.
• SO₃^2- will not dissociate further. It is a polyatomic ion.
4. K⁺ and Na⁺ are spectator ions. They do not take part in the reaction. In the reaction equation, they will appear as such on both sides, and thus will cancel out.
• So the actual reactants are: Cr₂O₇^2- and SO₃^2- 
5. We are given the products: chromium(III) ion and the sulfate ion.
• Chromium(III) ion means, the chromium ion with oxidation state +3. Obviously, it is the Cr³⁺ ion.
• Sulfate ion is known to us from the list. It is the SO₄^2- ion.
6. Steps (1) to (4) give us the reactants. Step (5) gives us the products.
• So now we can write the skeletal equation:
Cr₂O₇^2- + SO₃^2- → Cr³⁺ + SO₄^2-
• We want to balance this equation. It can be written in 12 steps:
Step 1: Write the oxidation numbers of all elements.
• For our present case, we get:
$\mathbf\small{\rm{\overset{+6}{Cr}_2\,\overset{-2}{O_7^{2-}}\,(aq)
+\overset{+4}{S}\,\overset{-2}{O_3^{2-}}\,(aq)\rightarrow
\overset{+3}{Cr^{3+}}\,(aq)+\overset{+6}{S}\,\overset{-2}{O_4^{2-}}\,(aq)}}$
Step 2: Identify the atoms that are oxidized. Also identify the atoms that are reduced.
• For our present case, we see that:
S is oxidized and Cr is reduced.
Step 3: Write the oxidation half reaction and reduction half reaction.
In our present case,
• The oxidation half reaction is:
SO₃^2- (aq) → SO₄^2- (aq)
• The reduction half reaction is:
Cr₂O₇^2- (aq) → Cr³⁺ (aq)
Step 4: Balance the atoms other than O and H in the oxidation half reaction.
• In our present case, the number of S atoms are the same on both sides. So it is already balanced.
Step 5: Balance the atoms other than O and H in the reduction half reaction.
• In our present case, there are two Cr atoms on the reactant side, but only one on the product side. So we must put '2' in front of Cr³⁺ on the product side. We get:
Cr₂O₇^2- (aq) → 2Cr³⁺ (aq)
Step 6: Balance O atoms in the oxidation half reaction by using H₂O molecules. Also balance the H atoms by using H⁺ ions.
• In our present case, SO₄^2- has one O more than SO₃^2-. So we must add one H₂O molecule on the reactant side. We get:
SO₃^2- (aq) + H₂O (l) → SO₄^2- (aq)
• Now there are two extra H atoms on the reactant side. So we must put two H⁺ ions on the product side. We get:
SO₃^2- (aq) + H₂O (l) → SO₄^2- (aq) + 2H⁺ (aq)
Step 7: Balance O atoms in the reduction half reaction by using H₂O molecules. Also balance the H atoms by using H⁺ ions.
• In our present case, Cr₂O₇^2- has seven O more than Cr³⁺. So we must add seven H₂O molecules on the product side. We get:
Cr₂O₇^2- (aq) → 2Cr³⁺ (aq) + 7H₂O
• Now there are fourteen extra H atoms on the product side. So we must put fourteen H⁺ ions on the reactant side. We get:
Cr₂O₇^2- (aq) + 14H⁺ → 2Cr³⁺ (aq) + 7H₂O
Step 8: Balance the charges in the oxidation half reaction, by adding appropriate number of electrons on the appropriate side.
• In our present case, the list of charges are as follows:
Reactant side:
   ♦ 1 No. × 2- (from the SO₃^2- ion) = -2
Product side:
   ♦ 1 No. × 2- (from the SO₄^2- ion) = -2
   ♦ 2 No. × 1+ (from the H⁺ ion) = +2
• So there is an excess charge of -2 on the reactant side. Therefore we must add 2e^- on the product side. We get:
SO₃^2- (aq) + H₂O (l) → SO₄^2- (aq) + 2H⁺ (aq) + 2e^-
Step 9: Balance the charges in the reduction half reaction, by adding appropriate number of electrons on the appropriate side.
• In our present case, the list of charges are as follows:
Reactant side:
   ♦ 1 No. × 2- (from the Cr₂O₇^2- ion) = -2
   ♦ 14 No. × 1+ (from the H⁺ ion) = +14
Product side:
   ♦ 2 No. × 3+ (from the Cr³⁺ ion) = +6
• So there is an excess charge of +6 on the reactant side. Therefore we must add 6e^- on the reactant side. We get:
Cr₂O₇^2- (aq) + 14H⁺ + 6e^- → 2Cr³⁺ (aq) + 7H₂O
Step 10: Make the number of electrons the same, in both the half reactions.
• In our present case,
    ♦ The oxidation half reaction has 2e^-.
    ♦ The reduction half reaction has 6e^-.
• So we multiply the oxidation half reaction by '3'. We get:
3SO₃^2- (aq) + 3H₂O (l) → 3SO₄^2- (aq) + 6H⁺ (aq) + 6e^-
Step 11: Add the two half reactions together.
• In our present case, we get:
Cr₂O₇^2- (aq) + 14H⁺ + 6e^- + 3SO₃^2- (aq) + 3H₂O (l) → 2Cr³⁺ (aq) + 7H₂O  + 3SO₄^2- (aq) + 6H⁺ (aq) + 6e^- 
• The 6e^- on either sides, cancel each other.
• The 6H⁺ on the product side will take away 6 H⁺ from the reaction side. So 8H⁺ will remain on the reaction side.
• The 3H₂O on the reaction side will take away 3H₂O from the product side. So 4H₂O will remain on the product side.
• We get:
Cr₂O₇^2- (aq) + 3SO₃^2- (aq) + 8H⁺  → 2Cr³⁺ (aq) + 3SO₄^2- (aq) + 4H₂O
Step 12: Check the balancing of all atoms. Also check the balancing of charges.
• In our present case, we have the following table for atoms:
Reactant side:
   ♦ Cr - 2 No., O - 16 No., S - 3 No., H - 8 No. 
Product side:
   ♦ Cr - 2 No., O - 16 No., S - 3 No., H - 8 No. 
         ✰ So all atoms are balanced.
• We have the following table for charges:
Reactant side:
   ♦ 1 No. × 2- (from the Cr₂O₇^2- ion) = -2
   ♦ 3 No. × 2- (from the SO₃^2- ions) = -6
   ♦ 8 No. × 1+ (from the H⁺ ions) = +8
   ♦ Total = 0
Product side:
   ♦ 2 No. × 3+ (from the Cr³⁺ ion) = +6
   ♦ 3 No. × 2- (from the SO₄^2- ions) = -6
   ♦ Total = 0
         ✰ All charges are balanced.
◼  So the balanced equation is same as that obtained in step 11:
Cr₂O₇^2- (aq) + 3SO₃^2- (aq) + 8H⁺  → 2Cr³⁺ (aq) + 3SO₄^2- (aq) + 4H₂O

Example 5:
Balance the equation in acidic medium:
MnO₄^- (aq) + Cl^- (aq) → Mn²⁺ (aq) + Cl₂ (g)
Solution:
We want to balance this equation. It can be written in 12 steps:
Step 1: Write the oxidation numbers of all elements.
• For our present case, we get:
$\mathbf\small{\rm{\overset{+7}{Mn}\,\overset{-2}{O_4^{-}}\,(aq)
+\overset{-1}{Cl^{-}}\,(aq)\rightarrow\overset{+2}{Mn^{2+}}\,(aq)
+\overset{0}{Cl_2}\,(g)}}$
Step 2: Identify the atoms that are oxidized. Also identify the atoms that are reduced.
• For our present case, we see that:
Cl is oxidized and Mn is reduced.
Step 3: Write the oxidation half reaction and reduction half reaction.
In our present case,
• The oxidation half reaction is:
Cl^- (aq) → Cl₂ (g)
• The reduction half reaction is:
MnO₄^- (aq) → Mn²⁺ (aq)
Step 4: Balance the atoms other than O and H in the oxidation half reaction.
• In our present case, we put '2' in front of Cl^- on the reactant side. We get:
2Cl^- (aq) → Cl₂ (g)
Step 5: Balance the atoms other than O and H in the reduction half reaction.
• In our present case, the number of Mn atoms are same on both sides. So it is already balanced
Step 6: Balance O atoms in the oxidation half reaction by using H₂O molecules. Also balance the H atoms by using H⁺ ions.
• In our present case, there are no O and H atoms to balance.
Step 7: Balance O atoms in the reduction half reaction by using H₂O molecules. Also balance the H atoms by using H⁺ ions.
• In our present case, there are extra four O atoms on the reactant side. So we must add four H₂O molecules on the product side. We get:
MnO₄^- (aq) → Mn²⁺ (aq) + 4H₂O
• Now there are eight extra H atoms on the product side. So we must put eight H⁺ ions on the reactant side. We get:
MnO₄^- (aq) + 8H⁺ → Mn²⁺ (aq) + 4H₂O
Step 8: Balance the charges in the oxidation half reaction, by adding appropriate number of electrons on the appropriate side.
• In our present case, the list of charges are as follows:
Reactant side:
   ♦ 2 No. × 1- (from the Cl^- ion) = -2
Product side:
   ♦ zero
• So there is an excess charge of -2 on the reactant side. Therefore we must add 2e^- on the product side. We get:
2Cl^- (aq) → Cl₂ (g) + 2e^-
Step 9: Balance the charges in the reduction half reaction, by adding appropriate number of electrons on the appropriate side.
• In our present case, the list of charges are as follows:
Reactant side:
   ♦ 1 No. × 1- (from the MnO₄^- ion) = -1
   ♦ 8 No. × 1+ (from the H⁺ ion) = +8
   ♦ Total = +7
Product side:
   ♦ 1 No. × 2+ (from the Mn²⁺ ion) = +2
• So there is an excess charge of +5 on the reactant side. Therefore we must add 5e^- on the reactant side. We get:
MnO₄^- (aq) + 8H⁺ +5e^- → Mn²⁺ (aq) + 4H₂O
Step 10: Make the number of electrons the same, in both the half reactions.
• In our present case,
    ♦ The oxidation half reaction has 2e^-.
    ♦ The reduction half reaction has 5e^-.
• So we multiply the oxidation half reaction by '5'.
• Also we multiply the reduction half reaction by '2'.
We get:
10Cl^- (aq) → 5Cl₂ (g) + 10e^-
2MnO₄^- (aq) + 16H⁺ + 10e^- → 2Mn²⁺ (aq) + 8H₂O
Step 11: Add the two half reactions together.
• In our present case, we get:
2MnO₄^- (aq) + 16H⁺ + 10e^- + 10Cl^- (aq) → 2Mn²⁺ (aq) + 8H₂O + 5Cl₂ (g) + 10e-
• The 10e^- on either sides, cancel each other. We get:
2MnO₄^- (aq) + 16H⁺ + 10Cl^- (aq) → 2Mn²⁺ (aq) + 8H₂O + 5Cl₂ (g)
Step 12: Check the balancing of all atoms. Also check the balancing of charges.
• In our present case, we have the following table for atoms:
Reactant side:
   ♦ Mn - 2 No., O - 8 No., Cl - 10 No., H - 16 No. 
Product side:
   ♦ Mn - 2 No., O - 8 No., Cl - 10 No., H - 16 No. 
         ✰ So all atoms are balanced.
• We have the following table for charges:
Reactant side:
   ♦ 2 No. × 1- (from the MnO₄^- ion) = -2
   ♦ 10 No. × 1- (from the SO₃^2- ions) = -10
   ♦ 16 No. × 1+ (from the H⁺ ions) = +16
   ♦ Total = +4
Product side:
   ♦ 2 No. × 2+ (from the Mn²⁺ ion) = +4
         ✰ All charges are balanced.
◼  So the balanced equation is same as that obtained in step 11:
2MnO₄^- (aq) + 16H⁺ + 10Cl^- (aq) → 2Mn²⁺ (aq) + 8H₂O + 5Cl₂ (g)

Example 6:
Balance the equation in acidic medium:
Fe²⁺ (aq) + Cr₂O₇^2- → Fe³⁺ (aq) + Cr³⁺ (aq)
Solution:
We want to balance this equation. It can be written in 12 steps:
Step 1: Write the oxidation numbers of all elements.
• For our present case, we get:
$\mathbf\small{\rm{\overset{+2}{Fe^{2+}}\,(aq)+\overset{+6}{Cr}_2\,\overset{-2}{O_7^{2-}}\,(aq)\rightarrow\overset{+3}{Fe^{3+}}\,(aq)
+\overset{+3}{Cr^{3+}}\,(aq)}}$
Step 2: Identify the atoms that are oxidized. Also identify the atoms that are reduced.
• For our present case, we see that:
Fe is oxidized and Cr is reduced.
Step 3: Write the oxidation half reaction and reduction half reaction.
In our present case,
• The oxidation half reaction is:
Fe²⁺ (aq) → Fe³⁺ (aq)
• The reduction half reaction is:
Cr₂O₇^2- → Cr³⁺ (aq)
Step 4: Balance the atoms other than O and H in the oxidation half reaction.
• In our present case, the number of Fe atoms are the same on both sides. So it is already balanced.
Step 5: Balance the atoms other than O and H in the reduction half reaction.
• In our present case, there are two Cr atoms on the reactant side, but only one on the product side. So we must put '2' in front of Cr³⁺ on the product side. We get:
Cr₂O₇^2- (aq) → 2Cr³⁺ (aq)
Step 6: Balance O atoms in the oxidation half reaction by using H₂O molecules. Also balance the H atoms by using H⁺ ions.
• In our present case, there are no O and H atoms to balance.
Step 7: Balance O atoms in the reduction half reaction by using H₂O molecules. Also balance the H atoms by using H⁺ ions.
• In our present case, Cr₂O₇^2- has seven O more than Cr³⁺. So we must add seven H₂O molecules on the product side. We get:
Cr₂O₇^2- (aq) → 2Cr³⁺ (aq) + 7H₂O
• Now there are fourteen extra H atoms on the product side. So we must put fourteen H⁺ ions on the reactant side. We get:
Cr₂O₇^2- (aq) + 14H⁺ → 2Cr³⁺ (aq) + 7H₂O
Step 8: Balance the charges in the oxidation half reaction, by adding appropriate number of electrons on the appropriate side.
• In our present case, the list of charges are as follows:
Reactant side:
   ♦ 1 No. × 2+ (from the Fe²⁺ ion) = +2
Product side:
   ♦ 1 No. × 3+ (from the Fe³⁺ ion) = +3
• So there is an excess charge of +1 on the product side. Therefore we must add 1e^- on the product side. We get:
Fe²⁺ (aq) → Fe³⁺ (aq) + 1e^-
Step 9: Balance the charges in the reduction half reaction, by adding appropriate number of electrons on the appropriate side.
• In our present case, the list of charges are as follows:
Reactant side:
   ♦ 1 No. × 2- (from the Cr₂O₇^2- ion) = -2
   ♦ 14 No. × 1+ (from the H⁺ ion) = +14
Product side:
   ♦ 2 No. × 3+ (from the Cr³⁺ ion) = +6
• So there is an excess charge of +6 on the reactant side. Therefore we must add 6e^- on the reactant side. We get:
Cr₂O₇^2- (aq) + 14H⁺ + 6e^- → 2Cr³⁺ (aq) + 7H₂O
Step 10: Make the number of electrons the same, in both the half reactions.
• In our present case,
    ♦ The oxidation half reaction has 1e^-.
    ♦ The reduction half reaction has 6e^-.
• So we multiply the oxidation half reaction by '6'. We get:
6Fe²⁺ (aq) → 6Fe³⁺ (aq) + 6e^-
Step 11: Add the two half reactions together.
• In our present case, we get:
6Fe²⁺ (aq) + Cr₂O₇^2- (aq) + 14H⁺ + 6e^- → 6Fe³⁺ (aq) + 6e^- + 2Cr³⁺ (aq) + 7H₂O
• The 6e^- on either sides, cancel each other.
• We get:
6Fe²⁺ (aq) + Cr₂O₇^2- (aq) + 14H⁺ → 6Fe³⁺ (aq) + 2Cr³⁺ (aq) + 7H₂O
Step 12: Check the balancing of all atoms. Also check the balancing of charges.
• In our present case, we have the following table for atoms:
Reactant side:
   ♦ Fe - 6 No., Cr - 2 No., O - 7 No., H - 14 No. 
Product side:
   ♦ Fe - 6 No., Cr - 2 No., O - 7 No., H - 14 No. 
         ✰ So all atoms are balanced.
• We have the following table for charges:
Reactant side:
   ♦ 6 No. × 2+ (from the Fe²⁺ ions) = +12
   ♦ 1 No. × 2- (from the Cr₂O₇^2- ion) = -2
   ♦ 14 No. × 1+ (from the H⁺ ions) = +14
   ♦ Total = 24
Product side:
   ♦ 6 No. × 3+ (from the Fe³⁺ ions) = +18
   ♦ 2 No. × 3+ (from the Cr³⁺ ion) = +6
   ♦ Total = 24
         ✰ All charges are balanced.
◼  So the balanced equation is same as that obtained in step 11:
6Fe²⁺ (aq) + Cr₂O₇^2- (aq) + 14H⁺ → 6Fe³⁺ (aq) + 2Cr³⁺ (aq) + 7H₂O

---

The following link gives three more examples:

Solved example 8.10 - Part (a), Part (b) and Part (c)

---

• We have seen the additional steps required in acidic medium when half reaction method is used.
• In the next section, we will see some examples, which take place in basic medium.
   ♦ There we will see the steps required for basic medium.


Previous

Contents

Next

Copyright©2021 Higher secondary chemistry.blogspot.com