RBSE Solutions Class 11 Chemistry Chapter 8 Oxidation-Reduction Reactions

RBSE Class 11 Chemistry Chapter 8 Text Book Questions

RBSE Class 11 Chemistry Chapter 8 Multiple Choice Questions

 

Question 1. Oxidation is a process :
(a) In which electrons are accepted
(b) In which electronegative element is added
(c) In which electro positive element is added
(d) In which oxygen is released
Answer: (b) In which electronegative element is added
In simple words: Oxidation is a chemical process where an atom, molecule, or ion loses electrons, or its oxidation state increases. It can also involve adding an electronegative element or removing an electropositive element.

🎯 Exam Tip: Remember the common definitions of oxidation and reduction, often summarized by "OIL RIG" (Oxidation Is Loss, Reduction Is Gain of electrons).

 

Question 3. The oxidation number of carbon in diamond is :
(a) Zero
(b) +4
(c) -4
(d) +2
Answer: (a) Zero
In simple words: In its elemental form, like diamond or graphite, carbon does not share electrons unevenly with other atoms, so its oxidation number is zero. This is a fundamental rule for all elements.

🎯 Exam Tip: The oxidation state of an element in its pure, uncombined form (like diamond, \( \mathrm{O_2} \), \( \mathrm{N_2} \), \( \mathrm{Fe} \)) is always zero, as there is no electronegativity difference to cause electron sharing imbalance.

 

Question 4. In the following reaction \( 2\mathrm{FeCl_3} +\mathrm{SnCl_2} \rightarrow 2\mathrm{FeCl_2} +\mathrm{SnCl_4} \) The substance which get reduced is :
(a) \( \mathrm{Sn^{2+}} \)
(b) \( \mathrm{Fe^{2+}} \)
(c) \( \mathrm{Sn^{4+}} \)
(d) \( \mathrm{Fe^{3+}} \)
Answer: (d) \( \mathrm{Fe^{3+}} \)
In simple words: In this reaction, iron changes its state from \( \mathrm{Fe^{3+}} \) to \( \mathrm{Fe^{2+}} \). This means it gains an electron, which is called reduction. Tin changes from \( \mathrm{Sn^{2+}} \) to \( \mathrm{Sn^{4+}} \), losing electrons, so it gets oxidized.

🎯 Exam Tip: To identify reduction, look for a decrease in the oxidation number or a gain of electrons. For oxidation, look for an increase in oxidation number or a loss of electrons.

 

Question 5. Which of the following is a redox reaction?
(a) \( 2\mathrm{FeCl_3} + \mathrm{SnCl_2} \rightarrow 4\ 2\mathrm{FeCl_2} + \mathrm{SnCl_4} \)
(b) \( \mathrm{ASNO_3} + \mathrm{HCl} \rightarrow \mathrm{AsCl} + \mathrm{HNO_3} \)
(c) \( 2\mathrm{KI} + \mathrm{Pb(NO)_3} \rightarrow 2\mathrm{KNO_3} +\mathrm{PbI_2} \)
(d) \( \mathrm{BaCl_2} + \mathrm{H_2SO_4} \rightarrow \mathrm{BaSO_4} + 2\mathrm{HCl} \)
Answer: (a) \( 2\mathrm{FeCl_3} + \mathrm{SnCl_2} \rightarrow 4\ 2\mathrm{FeCl_2} + \mathrm{SnCl_4} \)
In simple words: A redox reaction is one where both oxidation and reduction happen at the same time. In option (a), iron's oxidation state changes from +3 to +2 (reduction), and tin's changes from +2 to +4 (oxidation). This makes it a redox reaction.

🎯 Exam Tip: To spot a redox reaction, check the oxidation states of all elements involved. If any element's oxidation state changes, it's a redox reaction.

 

Question 7. Which of two F or I shows both positive and negative oxidation states ?
Answer: Iodine (I) shows both positive and negative oxidation states. Fluorine (F), being the most electronegative element, almost always shows an oxidation state of -1 in its compounds, except in its elemental form where it's 0. Iodine, however, can form compounds with more electronegative elements like oxygen and fluorine, leading to positive oxidation states such as +1, +3, +5, and +7, in addition to its common negative state of -1.
In simple words: Iodine can have both positive and negative oxidation numbers, but fluorine usually only has a negative one. Fluorine is so strong at pulling electrons that it almost always ends up with a -1 charge when it's in a compound.

🎯 Exam Tip: Remember that highly electronegative elements like fluorine tend to have only negative oxidation states, while less electronegative halogens (Cl, Br, I) can exhibit positive oxidation states when bonded to more electronegative atoms like oxygen or fluorine itself.

 

Question 8. Calculate the value of x in the following equation : \( \mathrm{ClO^-} + \mathrm{H_2O} + \mathrm{xe^-} \rightarrow \mathrm{Cl^-} + 2\mathrm{OH^-} \)
Answer: The value of x is 2 in the given reaction :
\( \mathrm{ClO^-} + \mathrm{H_2O} + 2\mathrm{e^-} \rightarrow \mathrm{Cl^-} + 2\mathrm{OH^-} \)
This reaction shows a gain of two electrons, meaning the chlorine atom is reduced. Balancing the electrons is crucial for accurate chemical equations.
In simple words: In this chemical equation, 'x' stands for the number of electrons. Chlorine goes from a +1 oxidation state in \( \mathrm{ClO^-} \) to -1 in \( \mathrm{Cl^-} \), so it gains 2 electrons, which means x is 2.

🎯 Exam Tip: To find 'x' (the number of electrons), compare the oxidation state of the changing element on both sides of the equation. If the oxidation state decreases, electrons are gained; if it increases, electrons are lost.

 

Question 9. Identify the reducing agent in the following reaction: \( 5\mathrm{H_2O_2}+\mathrm{Br_2} \rightarrow 2\mathrm{HBrO_3} +4\mathrm{H_2O} \)
Answer: In the given reaction, Bromine \( (\mathrm{Br_2}) \) acts as the reducing agent. This is because bromine's oxidation state increases from 0 in \( \mathrm{Br_2} \) to +5 in \( \mathrm{HBrO_3} \) (it gets oxidized). Since a reducing agent is a substance that causes another substance to be reduced (by being oxidized itself), \( \mathrm{Br_2} \) is the reducing agent. Meanwhile, hydrogen peroxide \( (\mathrm{H_2O_2}) \) is reduced (oxygen's oxidation state changes from -1 in \( \mathrm{H_2O_2} \) to -2 in \( \mathrm{H_2O} \)), making \( \mathrm{H_2O_2} \) the oxidizing agent. The overall reaction shows how one substance donates electrons while another accepts them.
In simple words: In this reaction, bromine \( (\mathrm{Br_2}) \) is the reducing agent. It causes hydrogen peroxide to gain electrons (be reduced) while bromine itself loses electrons (gets oxidized).

🎯 Exam Tip: A reducing agent is always oxidized in a redox reaction, meaning its oxidation number increases. An oxidizing agent is always reduced, meaning its oxidation number decreases.

 

Question 10. In the given structure the electronegativity of Y is more then Z. Find the oxidation Number of Z. X X-Y-Z-Y-X X
Answer: In this structure, the oxidation number of Z is zero. This happens because Z is bonded to two Y atoms. Since the electronegativity of Y is greater than Z, the bonds between Y and Z will be polarized such that Z effectively loses electrons to Y. However, Z is also bonded to two X atoms (assuming X is less electronegative or has no formal charge contribution to Z). If the contributions from all surrounding atoms balance out due to Z being in the center and Y atoms balancing each other in terms of electron pull, or if Z is connected to X which is of the same electronegativity, it may result in zero. Assuming X is also less electronegative or the entire molecule is neutral and symmetrical around Z, Z would not have a net charge if all bonds cancel out or it's a pure elemental bond equivalent.
In simple words: Even though Y pulls electrons from Z, Z also connects to other atoms. If all these pulls and pushes balance out perfectly, like in a symmetric structure, Z ends up with an oxidation number of zero.

🎯 Exam Tip: When calculating oxidation numbers in a complex structure, remember to consider the electronegativity of all surrounding atoms and how it influences electron distribution in each bond. If the central atom is bonded symmetrically to atoms of similar electronegativity, its oxidation state can often be zero.

 

Question 12. Arrange the following in increasing order of oxidation states of Mn, \( \mathrm{K_2MnO_4} \), \( \mathrm{MnO_2} \), \( \mathrm{KMnO_4} \)
Answer: To find the oxidation state of Mn in each compound:
1. For \( \mathrm{K_2MnO_4} \): Potassium (K) is +1, Oxygen (O) is -2.
\( 2(+1) + x + 4(-2) = 0 \)
\( 2 + x - 8 = 0 \)
\( x = 8 - 2 \)
\( x = +6 \)
So, oxidation state of Mn in \( \mathrm{K_2MnO_4} \) is +6.

2. For \( \mathrm{MnO_2} \): Oxygen (O) is -2.
\( x + 2(-2) = 0 \)
\( x - 4 = 0 \)
\( x = +4 \)
So, oxidation state of Mn in \( \mathrm{MnO_2} \) is +4.

3. For \( \mathrm{KMnO_4} \): Potassium (K) is +1, Oxygen (O) is -2.
\( 1(+1) + x + 4(-2) = 0 \)
\( 1 + x - 8 = 0 \)
\( x - 7 = 0 \)
\( x = +7 \)
So, oxidation state of Mn in \( \mathrm{KMnO_4} \) is +7.

Therefore, the increasing order of oxidation states of Mn is:
\( \mathrm{MnO_2} < \mathrm{K_2MnO_4} < \mathrm{KMnO_4} \)
This shows that manganese can exhibit a range of oxidation states, often depending on the compound it forms.
In simple words: First, find the oxidation number for manganese in each chemical. For \( \mathrm{MnO_2} \), it's +4. For \( \mathrm{K_2MnO_4} \), it's +6. For \( \mathrm{KMnO_4} \), it's +7. So, when arranged from smallest to largest, the order is \( \mathrm{MnO_2} \), then \( \mathrm{K_2MnO_4} \), then \( \mathrm{KMnO_4} \).

🎯 Exam Tip: When calculating oxidation states, always assign known values first (like +1 for alkali metals, -2 for oxygen in most compounds, -1 for halogens unless bonded to more electronegative atoms) and then solve for the unknown element.

 

Question 13. What do you mean by redox reaction ?
Answer: A redox reaction is a chemical reaction where both oxidation and reduction processes happen simultaneously. In these reactions, one chemical species loses electrons (oxidation) while another chemical species gains those electrons (reduction). This electron transfer is fundamental to many biological and industrial processes, such as cellular respiration and corrosion.
In simple words: A redox reaction is a special kind of chemical reaction where one substance loses electrons (called oxidation) and another substance gains those electrons (called reduction) at the very same time.

🎯 Exam Tip: For full marks, define both oxidation (loss of electrons, increase in oxidation number) and reduction (gain of electrons, decrease in oxidation number) within your answer to fully explain a redox reaction.

 

Question 15. Calculate the oxidation number of Fe in \( \mathrm{K_4[Fe(CN)_6]} \).
Answer: To calculate the oxidation number of Fe in \( \mathrm{K_4[Fe(CN)_6]} \):
Potassium (K) has an oxidation number of +1.
The cyanide ion \( (\mathrm{CN^-}) \) has a charge of -1.
Let 'x' be the oxidation number of Fe.
The sum of oxidation numbers in a neutral compound is zero.
\( 4(+1) + x + 6(-1) = 0 \)
\( 4 + x - 6 = 0 \)
\( x - 2 = 0 \)
\( x = +2 \)
Therefore, the oxidation number of Fe in \( \mathrm{K_4[Fe(CN)_6]} \) is +2. This complex ion is an important compound in chemistry and shows iron in a common oxidation state.
In simple words: For \( \mathrm{K_4[Fe(CN)_6]} \), potassium is +1 and cyanide \( (\mathrm{CN^-}) \) is -1. When you add up all the charges and set them to zero, you find that iron (Fe) has an oxidation number of +2.

🎯 Exam Tip: When dealing with complex ions, remember to consider the overall charge of the complex and the known charges of the ligands (like \( \mathrm{CN^-} \)).

 

Question 16. What is the maximum oxidation state of sulphur ?
Answer: The maximum oxidation state of sulfur is +6. Sulfur can reach this oxidation state when it is bonded to highly electronegative elements like oxygen or fluorine, such as in sulfuric acid \( (\mathrm{H_2SO_4}) \) or sulfur hexafluoride \( (\mathrm{SF_6}) \). This ability comes from sulfur being a Group 16 element, meaning it can use all its valence electrons in bonding to achieve stability.
In simple words: Sulfur can have an oxidation state as high as +6. This happens when it joins with other atoms that pull electrons very strongly, like in sulfuric acid.

🎯 Exam Tip: For main group elements, the maximum positive oxidation state is usually equal to their group number (e.g., Group 16 for sulfur means +6), and the minimum is Group Number - 8 (e.g., 16-8 = -2 for sulfur).

 

RBSE Class 11 Chemistry Chapter 8 Short Answer Type Questions

 

Question 17. How to balance a reaction taking place in basic medium by ion electron method ? Explain.
Answer: To balance a reaction in a basic medium using the ion-electron method, follow these steps:
Step I: Write down the unbalanced skeletal equation and assign oxidation numbers to all elements.
Step II: Identify which species are being oxidized and which are being reduced based on changes in oxidation numbers.
Step III: Separate the overall equation into two half-reactions: one for oxidation and one for reduction.
Step IV: Balance each half-reaction individually using these steps:

• First, balance all atoms except oxygen and hydrogen.
• Next, balance oxygen atoms by adding \( \mathrm{H_2O} \) molecules to the side that needs oxygen.
• Then, balance hydrogen atoms by adding \( \mathrm{H^+} \) ions to the side that needs hydrogen.
• Since the reaction is in a basic medium, add \( \mathrm{OH^-} \) ions to both sides of the equation, equal to the number of \( \mathrm{H^+} \) ions present. Combine \( \mathrm{H^+} \) and \( \mathrm{OH^-} \) to form \( \mathrm{H_2O} \).
• Finally, balance the charges by adding electrons \( (\mathrm{e^-}) \) to the more positive side.

Step V: Multiply each balanced half-reaction by an appropriate integer so that the number of electrons lost in the oxidation half-reaction equals the number of electrons gained in the reduction half-reaction.
Step VI: Add the two balanced half-reactions together and cancel out any common species (like electrons, water molecules, or \( \mathrm{OH^-} \) ions) present on both sides. This method helps to systematically balance even complex redox reactions.
In simple words: To balance a reaction in a basic solution, you first split it into two parts: one where electrons are lost (oxidation) and one where electrons are gained (reduction). Then, for each part, you balance the atoms that are not oxygen or hydrogen, then balance oxygen using water, then balance hydrogen using \( \mathrm{H^+} \). Because it's basic, you add \( \mathrm{OH^-} \) to both sides for every \( \mathrm{H^+} \) and turn them into water. Finally, balance the charges with electrons and make sure both parts have the same number of electrons before adding them back together.

🎯 Exam Tip: The key difference for balancing in basic medium versus acidic medium is the final step of adding \( \mathrm{OH^-} \) ions to neutralize any \( \mathrm{H^+} \) ions and form water molecules, ensuring the reaction environment is basic.

 

Question 18. Explain the difference between oxidation state and valency of an element.
Answer:

Understanding the distinction between these two concepts is crucial for describing chemical behavior and reactions. Valency describes how many bonds an atom can form, while oxidation state describes the electron shift in those bonds.
In simple words: Oxidation state tells you the charge an atom would have if all its bonds were completely ionic, and it can be positive, negative, or zero. Valency, on the other hand, is about how many bonds an atom can make with others, and it is always a positive whole number.

🎯 Exam Tip: Always remember that valency is a simple combining capacity, while oxidation state is a more complex concept reflecting electron distribution in bonds, especially useful in redox reactions.

 

Question 19. \( 2\mathrm{KI} + \mathrm{Cl_2} \rightarrow 2\mathrm{KCl} + \mathrm{I_2} \), Identify the reducing agent in the reaction.
Answer: For the reaction: \( 2\mathrm{KI} + \mathrm{Cl_2} \rightarrow 2\mathrm{KCl} + \mathrm{I_2} \)
Let's find the oxidation states of the elements:
Oxidation state of I in KI: K is +1, so \( 1 + x = 0 \Rightarrow x = -1 \).
Oxidation state of I in \( \mathrm{I_2} \): Since it's an element, the oxidation state is 0.

Oxidation state of Cl in \( \mathrm{Cl_2} \): Since it's an element, the oxidation state is 0.
Oxidation state of Cl in KCl: K is +1, so \( 1 + x = 0 \Rightarrow x = -1 \).

From these calculations:
Iodine (I) changes from -1 to 0. This is an increase in oxidation state, so Iodine is oxidized.
Chlorine (Cl) changes from 0 to -1. This is a decrease in oxidation state, so Chlorine is reduced.

Since the reducing agent is the species that gets oxidized (loses electrons), the reducing agent in this reaction is KI (specifically, the iodide ion, \( \mathrm{I^-} \)). It causes the reduction of \( \mathrm{Cl_2} \) to \( \mathrm{Cl^-} \).
In simple words: In this reaction, iodide \( (\mathrm{I^-}) \) in KI is the reducing agent. This is because iodide loses electrons and changes its oxidation state from -1 to 0, meaning it gets oxidized, which in turn causes chlorine to be reduced.

🎯 Exam Tip: To correctly identify the reducing agent, pinpoint the substance whose oxidation state increases during the reaction. That substance is giving away electrons and thus causing reduction in another species.

 

Question 20. Calculate the oxidation number :
(i) Mo in \( \mathrm{(NH_4)_2MoO_4} \)
(ii) Ni in \( \mathrm{[Ni (CN)_4]^{2-}} \)
Answer:
(i) To calculate the oxidation number of Mo in \( \mathrm{(NH_4)_2MoO_4} \):
The ammonium ion \( (\mathrm{NH_4^+}) \) has a charge of +1. Oxygen (O) has an oxidation number of -2.
Let 'x' be the oxidation number of Mo.
\( 2(+1) + x + 4(-2) = 0 \)
\( 2 + x - 8 = 0 \)
\( x - 6 = 0 \)
\( x = +6 \)
The oxidation number of Mo in \( \mathrm{(NH_4)_2MoO_4} \) is +6. This high oxidation state is common for transition metals when bonded to oxygen.

(ii) To calculate the oxidation number of Ni in \( \mathrm{[Ni(CN)_4]^{2-}} \):
The cyanide ion \( (\mathrm{CN^-}) \) has a charge of -1.
Let 'x' be the oxidation number of Ni.
The overall charge of the complex ion is -2.
\( x + 4(-1) = -2 \)
\( x - 4 = -2 \)
\( x = -2 + 4 \)
\( x = +2 \)
The oxidation number of Ni in \( \mathrm{[Ni(CN)_4]^{2-}} \) is +2. This is a common oxidation state for nickel in coordination compounds.
In simple words: (i) For Mo in \( \mathrm{(NH_4)_2MoO_4} \), ammonium \( (\mathrm{NH_4^+}) \) is +1 and oxygen is -2. Adding these up shows Mo is +6. (ii) For Ni in \( \mathrm{[Ni(CN)_4]^{2-}} \), cyanide \( (\mathrm{CN^-}) \) is -1, and the whole ion has a -2 charge. This means Ni is +2.

🎯 Exam Tip: Always remember the charges of common polyatomic ions (like \( \mathrm{NH_4^+} \) and \( \mathrm{CN^-} \)) and the overall charge of the compound or complex ion when calculating oxidation numbers.

 

Question 21. Arrange the following metals in the order in which they displace each other from the solution of their salts - Cr, Cu, Mg, Zn, Fe, Al.
Answer: The ability of a metal to displace another metal from its salt solution depends on its reactivity, which is related to its standard electrode potential. A metal with a more negative standard electrode potential is more active and can displace metals with less negative (or positive) standard electrode potentials. These more active metals are also stronger reducing agents.

Here are the standard electrode potentials \( (\mathrm{E^\circ}) \) for the given metals:
Magnesium (Mg): -2.36 V
Aluminum (Al): -1.66 V
Zinc (Zn): -0.76 V
Chromium (Cr): -0.74 V
Iron (Fe): -0.44 V
Copper (Cu): +0.34 V

Arranging them from most active (most negative \( \mathrm{E^\circ} \)) to least active (most positive \( \mathrm{E^\circ} \)) gives the order in which they can displace each other from their salt solutions:
Mg > Al > Zn > Cr > Fe > Cu

Therefore, the order of metals in which they displace each other from the solution of their salts, from most active to least active, is:
Mg, Al, Zn, Cr, Fe, Cu
(More active) (Least active)
This order is crucial for predicting the outcome of displacement reactions and understanding the reactivity series of metals.
In simple words: The more negative a metal's standard electrode potential, the more active it is and the better it can push out other metals from their salts. Based on their values, Magnesium is the most active, followed by Aluminum, Zinc, Chromium, Iron, and then Copper as the least active among these.

🎯 Exam Tip: A metal with a more negative standard electrode potential is a stronger reducing agent and will displace any metal with a less negative or positive standard electrode potential from its salt solution.

 

Question 23. Define oxidation and reduction :
(a) On the basis of electronic concept
(b) On the basic of oxidation number.
Answer:
(a) On the basis of the electronic concept:

Oxidation: Oxidation is defined as a chemical process where an atom, ion, or molecule loses one or more electrons. For example, when sodium metal loses an electron to become a sodium ion: \( \mathrm{Na} \rightarrow \mathrm{Na^+} + \mathrm{e^-} \). This loss of electrons is key to many chemical reactions.

Reduction: Reduction is defined as a chemical process where an atom, ion, or molecule gains one or more electrons. For example, when a chlorine atom gains an electron to become a chloride ion: \( \mathrm{Cl} + \mathrm{e^-} \rightarrow \mathrm{Cl^-} \). This gain of electrons often results in a decrease in the charge of the species.

(b) On the basis of oxidation number:

Oxidation: Oxidation is a process in which the oxidation number of an element in an atom, molecule, or ion increases. For example, in the reaction \( \mathrm{Fe^{2+}} \rightarrow \mathrm{Fe^{3+}} + \mathrm{e^-} \), the oxidation number of iron increases from +2 to +3, indicating oxidation.

Reduction: Reduction is a process in which the oxidation number of an element in an atom, molecule, or ion decreases. For example, in the reaction \( \mathrm{Cu^{2+}} + 2\mathrm{e^-} \rightarrow \mathrm{Cu} \), the oxidation number of copper decreases from +2 to 0, indicating reduction.
Both definitions help us understand how electrons are transferred during chemical changes.
In simple words: (a) By electrons, oxidation means losing electrons, and reduction means gaining electrons. (b) By oxidation numbers, oxidation means the number goes up, and reduction means the number goes down.

🎯 Exam Tip: Always provide both the electronic definition (loss/gain of electrons) and the oxidation number definition (increase/decrease in oxidation number) for full understanding of oxidation and reduction.

 

Question 24. Identify oxidising agent and reducing agent in the following reactions :
\( \mathrm{H_2O_2} + \mathrm{O_3} \rightarrow \mathrm{H_2O} + 2\mathrm{O_2} \)
\( 2\mathrm{Na_2S_2O_3} + \mathrm{I_2} \rightarrow \mathrm{Na_2S_4O_6} + 2\mathrm{NaI} \)
Answer:
Let's analyze the first reaction: \( \mathrm{H_2O_2} + \mathrm{O_3} \rightarrow \mathrm{H_2O} + 2\mathrm{O_2} \)
Oxidation states:
In \( \mathrm{H_2O_2} \), Oxygen is -1.
In \( \mathrm{O_3} \), Oxygen is 0.
In \( \mathrm{H_2O} \), Oxygen is -2.
In \( \mathrm{O_2} \), Oxygen is 0.

Here, \( \mathrm{H_2O_2} \) undergoes both oxidation (Oxygen goes from -1 to 0 in \( \mathrm{O_2} \)) and reduction (Oxygen goes from -1 to -2 in \( \mathrm{H_2O} \)). This means \( \mathrm{H_2O_2} \) acts as both an oxidizing agent and a reducing agent. Ozone \( (\mathrm{O_3}) \) acts as an oxidizing agent, as it oxidizes \( \mathrm{H_2O_2} \) to \( \mathrm{O_2} \). The decomposition of \( \mathrm{H_2O_2} \) itself is a redox reaction where it dismutates.

Let's analyze the second reaction: \( 2\mathrm{Na_2S_2O_3} + \mathrm{I_2} \rightarrow \mathrm{Na_2S_4O_6} + 2\mathrm{NaI} \)
Oxidation states:
In \( \mathrm{Na_2S_2O_3} \) (sodium thiosulfate), Sulfur has an average oxidation state of +2.
In \( \mathrm{I_2} \), Iodine is 0.
In \( \mathrm{Na_2S_4O_6} \) (sodium tetrathionate), Sulfur has an average oxidation state of +2.5.
In \( \mathrm{NaI} \), Iodine is -1.

Here, Iodine \( (\mathrm{I_2}) \) changes from 0 to -1, which is a decrease in oxidation state, so \( \mathrm{I_2} \) is reduced. Therefore, \( \mathrm{I_2} \) is the **oxidizing agent**.
Sulfur in \( \mathrm{Na_2S_2O_3} \) changes from +2 to +2.5, which is an increase in oxidation state, so \( \mathrm{Na_2S_2O_3} \) is oxidized. Therefore, \( \mathrm{Na_2S_2O_3} \) is the **reducing agent**. This reaction is a classic example of an iodometric titration.
In simple words: In the first reaction, \( \mathrm{H_2O_2} \) both gains and loses electrons, so it acts as both the oxidizer and the reducer, while \( \mathrm{O_3} \) is an oxidizer. In the second reaction, \( \mathrm{I_2} \) gains electrons, making it the oxidizing agent, and \( \mathrm{Na_2S_2O_3} \) loses electrons, making it the reducing agent.

🎯 Exam Tip: For reactions where a single substance acts as both an oxidizing and reducing agent (disproportionation), make sure to show how its oxidation state changes in two different ways to reflect both processes.

 

Question 25. When an iron rod is dipped in silver nitrate solution, it became silverish. Explain the reason.
Answer: When an iron rod is dipped into a silver nitrate \( (\mathrm{AgNO_3}) \) solution, the iron rod becomes silverish due to a displacement reaction. This occurs because iron (Fe) is a more reactive metal than silver (Ag). We can determine this by comparing their standard electrode potentials:

Standard electrode potential of iron (\( \mathrm{Fe^{2+}/Fe} \)): \( -\mathrm{0.44V} \)
Standard electrode potential of silver (\( \mathrm{Ag^+/Ag} \)): \( +\mathrm{0.80V} \)

Since iron has a more negative standard electrode potential than silver, iron has a stronger tendency to lose electrons (get oxidized) compared to silver. Consequently, iron metal will displace silver ions \( (\mathrm{Ag^+}) \) from the solution. The iron atoms on the rod will oxidize to \( \mathrm{Fe^{2+}} \) ions and dissolve into the solution, while the silver ions \( (\mathrm{Ag^+}) \) in the solution will gain electrons (get reduced) and deposit as solid silver metal onto the iron rod, giving it a silverish appearance. This process demonstrates a fundamental principle of electrochemical reactivity.
The reaction is:
\( \mathrm{Fe} + 2\mathrm{AgNO_3} \rightarrow \mathrm{Fe(NO_3)_2} + 2\mathrm{Ag} \)
In simple words: An iron rod turns silver because iron is more reactive than silver. When put in silver nitrate, the iron takes electrons from the silver ions, making the iron dissolve a bit and pure silver metal stick to the rod.

🎯 Exam Tip: Remember that a metal with a more negative (or less positive) standard electrode potential will displace a metal with a more positive (or less negative) standard electrode potential from its salt solution.

 

Question 26. What is the oxidation number of oxygen in \( \mathrm{KO_2} \) ?
Answer: To find the oxidation number of oxygen in \( \mathrm{KO_2} \):
Potassium (K) is an alkali metal, so its oxidation number is always +1.
Let the oxidation number of oxygen be 'x'.
In \( \mathrm{KO_2} \), there are one potassium atom and two oxygen atoms.
The sum of oxidation numbers in a neutral compound is zero.
\( (+1) + 2x = 0 \)
\( 2x = -1 \)
\( x = -\frac{1}{2} \)
So, the oxidation number of oxygen in \( \mathrm{KO_2} \) is \( -\frac{1}{2} \). This unusual fractional oxidation state indicates that \( \mathrm{KO_2} \) contains the superoxide ion, \( \mathrm{O_2^-} \), where the two oxygen atoms share a total charge of -1.
In simple words: In the compound \( \mathrm{KO_2} \), potassium always has a +1 charge. Since the whole compound must be neutral, the two oxygen atoms together must have a total charge of -1. This means each oxygen atom has an oxidation number of \( -\frac{1}{2} \).

🎯 Exam Tip: While oxygen usually has an oxidation state of -2, remember to look out for exceptions like peroxides \( (\mathrm{O_2^{2-}} \text{, -1}) \), superoxides \( (\mathrm{O_2^-} \text{, -}\frac{1}{2}) \), and compounds with fluorine \( (\mathrm{OF_2} \text{, +2}) \).

 

Question 27. During test of nitrate, a complex \( \mathrm{[Fe (H_2O)_5 (NO)(SO_4)]} \) is formed. What the oxidation number of Fe in this complex?
Answer: To find the oxidation number of Fe in the complex \( \mathrm{[Fe (H_2O)_5 (NO)(SO_4)]} \):
Water \( (\mathrm{H_2O}) \) is a neutral ligand, so its contribution to the charge is 0.
Nitrosyl \( (\mathrm{NO}) \) can be present as \( \mathrm{NO^+} \), \( \mathrm{NO} \), or \( \mathrm{NO^-} \). In most brown ring complexes (like this one formed in the nitrate test), it acts as a \( \mathrm{NO^+} \) ligand, contributing +1 charge.
Sulfate \( (\mathrm{SO_4}) \) is a polyatomic ion with a charge of -2 \( (\mathrm{SO_4^{2-}}) \).
Let 'x' be the oxidation number of Fe.
The complex itself is neutral.
So, \( x + 5(0) + (+1) + (-2) = 0 \)
\( x + 0 + 1 - 2 = 0 \)
\( x - 1 = 0 \)
\( x = +1 \)
Therefore, the oxidation number of Fe in this complex is +1. This complex, known as the brown ring complex, is a characteristic product of the nitrate test.
In simple words: In the complex \( \mathrm{[Fe (H_2O)_5 (NO)(SO_4)]} \), water has no charge, the nitrosyl group \( (\mathrm{NO}) \) acts as +1, and sulfate \( (\mathrm{SO_4}) \) is -2. Adding these up to zero (since the complex is neutral) means that the iron (Fe) has an oxidation number of +1.

🎯 Exam Tip: For coordination complexes, remember to identify the charge of each ligand. Water is neutral, while ligands like \( \mathrm{NO} \) can sometimes be treated as \( \mathrm{NO^+} \) depending on the context, especially in the brown ring complex.

 

Question 28. Arrange the following in increasing order of oxidation states of iodine : \( \mathrm{I_2, HI, HIO_4, ICI} \)
Answer: To arrange the given compounds in increasing order of oxidation states of iodine, we first need to calculate the oxidation state of iodine in each compound:

1. In \( \mathrm{I_2} \): As an elemental substance, the oxidation state of Iodine is 0.

2. In \( \mathrm{HI} \): Hydrogen (H) usually has an oxidation state of +1. Since the molecule is neutral, let 'x' be the oxidation state of I.
\( (+1) + x = 0 \Rightarrow x = -1 \).
So, the oxidation state of I in HI is -1.

3. In \( \mathrm{HIO_4} \) (Periodic acid): Hydrogen (H) is +1, and Oxygen (O) is -2. Let 'x' be the oxidation state of I.
\( (+1) + x + 4(-2) = 0 \)
\( 1 + x - 8 = 0 \)
\( x - 7 = 0 \Rightarrow x = +7 \).
So, the oxidation state of I in \( \mathrm{HIO_4} \) is +7.

4. In \( \mathrm{ICl} \): Chlorine (Cl) is more electronegative than Iodine, so Cl will have an oxidation state of -1. Let 'x' be the oxidation state of I.
\( x + (-1) = 0 \Rightarrow x = +1 \).
So, the oxidation state of I in ICl is +1.

Now, arranging these in increasing order of oxidation states of iodine:
-1 (HI) < 0 (\( \mathrm{I_2} \)) < +1 (ICl) < +7 (\( \mathrm{HIO_4} \))

Thus, the increasing order is:
\( \mathrm{HI} < \mathrm{I_2} < \mathrm{ICl} < \mathrm{HIO_4} \)
This demonstrates the wide range of oxidation states that iodine, a halogen, can exhibit.
In simple words: First, find the oxidation number for iodine in each chemical: \( \mathrm{HI} \) has -1, \( \mathrm{I_2} \) has 0, \( \mathrm{ICl} \) has +1, and \( \mathrm{HIO_4} \) has +7. Then, put them in order from the smallest number to the largest.

🎯 Exam Tip: When determining oxidation states for halogens like iodine, always compare their electronegativity with the atoms they are bonded to. For example, in \( \mathrm{ICl} \), chlorine is more electronegative, so it takes the -1 oxidation state, forcing iodine to be +1.

 

RBSE Class 11 Chemistry Chapter 8 Long Answer Type Questions

 

Question 29. What is oxidation number ? How oxidation and reduction can be identified on the basis of change in oxidation number ? Write the step of balancing equation by oxidation number method.
Answer:
**Oxidation Number:**
The oxidation number (or oxidation state) of an element in an atom, ion, or molecule is the hypothetical charge that atom would have if all its bonds to other atoms were ionic. It represents the number of electrons gained or lost by an atom compared to its elemental state. This number helps to track electron transfers in chemical reactions.

**Identification of Oxidation and Reduction by Oxidation Number:**
Oxidation and reduction can be easily identified by observing changes in the oxidation number of an element during a reaction:

• **Oxidation:** If the oxidation number of an element increases during a reaction, the element is said to be oxidized. This indicates a loss of electrons.
• **Reduction:** If the oxidation number of an element decreases during a reaction, the element is said to be reduced. This indicates a gain of electrons.

**Steps for Balancing Equations by Oxidation Number Method:**
The oxidation number method provides a systematic way to balance redox reactions. Here are the steps:
Step I: Write the unbalanced skeletal equation and assign oxidation numbers to all elements in the reactants and products. This helps in identifying the changes.
Step II: Identify the elements that undergo a change in oxidation number. Mark these changes clearly.
Step III: Calculate the total increase in oxidation number for the element that is oxidized and the total decrease in oxidation number for the element that is reduced. If more than one atom of the same element is involved in the change, multiply the change per atom by the number of atoms.
Step IV: Multiply the oxidizing agent and reducing agent in the equation by suitable coefficients so that the total increase in oxidation number equals the total decrease in oxidation number. This ensures electron balance.
Step V: Balance all other atoms (like those of hydrogen and oxygen) by inspection. For acidic medium, balance oxygen by adding \( \mathrm{H_2O} \) and hydrogen by adding \( \mathrm{H^+} \). For basic medium, balance oxygen by adding \( \mathrm{H_2O} \) and hydrogen by adding \( \mathrm{OH^-} \) (and then combining \( \mathrm{H^+} \) and \( \mathrm{OH^-} \) to form \( \mathrm{H_2O} \)).
Step VI: Finally, check if the atoms and charges are balanced on both sides of the equation. This ensures the law of conservation of mass and charge is followed.
This method is particularly useful for complex redox reactions where simple inspection balancing is difficult.
In simple words: An oxidation number tells us the charge an atom would have if it took or gave electrons completely. If this number goes up, it's oxidation; if it goes down, it's reduction. To balance equations, you first find these numbers, then figure out which atoms are gaining or losing electrons. You then adjust the number of molecules so that the total electrons lost equal the total electrons gained. Finally, you balance all other atoms, especially hydrogen and oxygen, adding water or \( \mathrm{H^+} \) (for acid) or \( \mathrm{OH^-} \) (for base) as needed.

🎯 Exam Tip: The most critical step in the oxidation number method is Step IV: ensuring that the total increase in oxidation number (oxidation) exactly matches the total decrease in oxidation number (reduction). This balances the electron transfer.

 

Question 30. What is standard electrode potential ? What is the importance of standard electrode potential in chemical reactions? Explain with example.
Answer:
**Standard Electrode Potential (\( \mathrm{E^\circ} \)):**
The standard electrode potential is defined as the potential difference between an electrode and its solution when the electrode is in equilibrium with its ions at standard conditions. These standard conditions are: 1 M concentration for all ions, 1 atmospheric pressure (or 1 bar for gases), and a temperature of 298 K (\( 25^\circ\mathrm{C} \)). It is measured relative to a standard hydrogen electrode (SHE), which is arbitrarily assigned an electrode potential of 0.00 V. This potential indicates the tendency of a species to gain or lose electrons.

**Importance of Standard Electrode Potential in Chemical Reactions:**
Standard electrode potentials are very important in chemistry because they allow us to predict and understand various aspects of chemical reactions, especially redox reactions:
1. **Predicting Reactivity of Metals:** Metals with more negative standard electrode potentials are more active (stronger reducing agents). They have a greater tendency to lose electrons and get oxidized. For example, zinc \( (\mathrm{E^\circ = -0.76 V}) \) is more active than copper \( (\mathrm{E^\circ = +0.34 V}) \).

2. **Predicting Displacement Reactions:** A metal with a more negative standard electrode potential can displace a metal with a less negative (or positive) standard electrode potential from its salt solution. For example, iron can displace silver from silver nitrate solution because \( \mathrm{E^\circ_{Fe^{2+}/Fe}} \) (-0.44 V) is less than \( \mathrm{E^\circ_{Ag^+/Ag}} \) (+0.80 V). This means the more active metal reacts.

3. **Identifying Good Reducing Agents:** Metals at the top of the activity series (those with highly negative \( \mathrm{E^\circ} \) values), like alkali metals and alkaline earth metals, are excellent reducing agents. Their cations cannot be easily reduced by chemical means, so they are often obtained by electrolytic reduction. For these elements, the value of \( \mathrm{E^\circ} \) will be very negative.

4. **Hydrogen Displacement from Acids:** Metals placed above hydrogen in the activity series (i.e., metals with negative standard electrode potentials) can reduce hydrogen ions \( (\mathrm{H^+}) \) from acids to form hydrogen gas \( (\mathrm{H_2}) \). For instance, zinc reacts with acids to liberate hydrogen gas:
\( \mathrm{Zn} + 2\mathrm{H^+} \rightarrow \mathrm{Zn^{2+}} + \mathrm{H_2} \)
\( \mathrm{Mg} + 2\mathrm{H^+} \rightarrow \mathrm{Mg^{2+}} + \mathrm{H_2} \)

5. **Displacement of Anions by Non-metals:** Non-metals higher in the reactivity series (stronger oxidizing agents) can displace anions of less reactive non-metals from their solutions. For example, chlorine \( (\mathrm{Cl_2}) \) can displace bromide ions \( (\mathrm{Br^-}) \) from a bromide solution:
\( \mathrm{Cl_2} + 2\mathrm{Br^-} \rightarrow 2\mathrm{Cl^-} + \mathrm{Br_2} \)

6. **Identifying Good Oxidizing Agents:** Elements that are lower in the activity series or have highly positive standard electrode potentials have a greater tendency to accept electrons. These are good oxidizing agents. For example, their cations can oxidize \( \mathrm{H_2} \) to \( \mathrm{H^+} \) ions:
\( 2\mathrm{Au^+} + 3\mathrm{H_2} \rightarrow 2\mathrm{Au} + 6\mathrm{H^+} \)

7. **Precipitation of Precious Metals:** More electronegative metals (stronger reducing agents) can precipitate precious metals like gold and silver from their salt solutions. For example, zinc can precipitate silver from a complex silver cyanide solution:
\( 2\mathrm{Na[Ag(CN)_2]} + \mathrm{Zn} \rightarrow \mathrm{Na_2[Zn(CN)_4]} + 2\mathrm{Ag} \downarrow \)
Standard electrode potentials are vital tools for understanding the driving force of redox reactions and designing electrochemical cells.
In simple words: Standard electrode potential tells us how easily a substance will gain or lose electrons under normal conditions. It's important because it helps us guess how reactive metals are, which metals can push out others from solutions, which chemicals are good at taking electrons, and if a metal will make hydrogen gas from an acid. For example, a metal with a very low number can easily give away its electrons and displace other metals.

🎯 Exam Tip: When explaining the importance of standard electrode potential, focus on its predictive power for reactivity, displacement reactions, and identifying oxidizing/reducing agents. Providing concrete examples for each point strengthens your answer.

 

Question 31. Balance the following reaction by oxidation number method and also identify oxidising agent.
(i) \( \mathrm{H_2S} + \mathrm{MnO_4^-} \rightarrow \mathrm{S} + \mathrm{Mn^{2+}} + \mathrm{H_2O} \) (Acidic medium)
(ii) \( \mathrm{Cl_2O_7} + \mathrm{H_2O_2} \rightarrow \mathrm{ClO_2^-} + \mathrm{O_2(gas)} \) (Basic medium)
Answer:
(i) Balancing \( \mathrm{H_2S} + \mathrm{MnO_4^-} \rightarrow \mathrm{S} + \mathrm{Mn^{2+}} + \mathrm{H_2O} \) (Acidic medium):
**Step I: Write unbalanced equation with oxidation numbers.**
\( \overset{-2}{\mathrm{H_2S}} + \overset{+7}{\mathrm{MnO_4^-}} \rightarrow \overset{0}{\mathrm{S}} + \overset{+2}{\mathrm{Mn^{2+}}} + \mathrm{H_2O} \)

**Step II: Identify elements changing oxidation number.**
Sulfur (S): from -2 to 0 (oxidation, increase by 2)
Manganese (Mn): from +7 to +2 (reduction, decrease by 5)

**Step III: Write half-reactions (conceptual) and balance electron change.**
Oxidation: \( \mathrm{H_2S} \rightarrow \mathrm{S} \) (Increase by 2 per S atom)
Reduction: \( \mathrm{MnO_4^-} \rightarrow \mathrm{Mn^{2+}} \) (Decrease by 5 per Mn atom)

**Step IV: Equalize total increase and decrease in oxidation numbers.**
Multiply oxidation half by 5 and reduction half by 2:
Total increase: \( 5 \times 2 = 10 \)
Total decrease: \( 2 \times 5 = 10 \)
\( 5\mathrm{H_2S} + 2\mathrm{MnO_4^-} \rightarrow 5\mathrm{S} + 2\mathrm{Mn^{2+}} \)

**Step V: Balance other atoms (O and H) for acidic medium.**
Balance Oxygen: \( 2\mathrm{MnO_4^-} \) has 8 oxygen atoms on the left. Add 8 \( \mathrm{H_2O} \) to the right.
\( 5\mathrm{H_2S} + 2\mathrm{MnO_4^-} \rightarrow 5\mathrm{S} + 2\mathrm{Mn^{2+}} + 8\mathrm{H_2O} \)
Balance Hydrogen: Left side has \( 5 \times 2 = 10 \) H from \( \mathrm{H_2S} \). Right side has \( 8 \times 2 = 16 \) H from \( \mathrm{H_2O} \). Add \( 6\mathrm{H^+} \) to the left side.
\( 5\mathrm{H_2S} + 2\mathrm{MnO_4^-} + 6\mathrm{H^+} \rightarrow 5\mathrm{S} + 2\mathrm{Mn^{2+}} + 8\mathrm{H_2O} \)

**Step VI: Check balance of atoms and charges.**
Atoms: S=5, Mn=2, O=8, H=16 on both sides. (Balanced)
Charge: Left side \( (2 \times -1) + (6 \times +1) = -2 + 6 = +4 \)
Right side \( (2 \times +2) = +4 \) (Balanced)
The balanced equation is: \( 5\mathrm{H_2S} + 2\mathrm{MnO_4^-} + 6\mathrm{H^+} \rightarrow 5\mathrm{S} + 2\mathrm{Mn^{2+}} + 8\mathrm{H_2O} \)
The oxidizing agent is \( \mathrm{MnO_4^-} \) because it gets reduced.

(ii) Balancing \( \mathrm{Cl_2O_7} + \mathrm{H_2O_2} \rightarrow \mathrm{ClO_2^-} + \mathrm{O_2(g)} \) (Basic medium):
**Step I: Write unbalanced equation with oxidation numbers.**
\( \overset{+7}{\mathrm{Cl_2O_7}} + \overset{-1}{\mathrm{H_2O_2}} \rightarrow \overset{+3}{\mathrm{ClO_2^-}} + \overset{0}{\mathrm{O_2(g)}} \)

**Step II: Identify elements changing oxidation number.**
Chlorine (Cl): from +7 (in \( \mathrm{Cl_2O_7} \)) to +3 (in \( \mathrm{ClO_2^-} \)). Decrease by 4 per Cl atom. Since there are 2 Cl atoms in \( \mathrm{Cl_2O_7} \), total decrease is \( 2 \times 4 = 8 \). (Reduction)
Oxygen (O): from -1 (in \( \mathrm{H_2O_2} \)) to 0 (in \( \mathrm{O_2} \)). Increase by 1 per O atom. Since there are 2 O atoms in \( \mathrm{H_2O_2} \), total increase is \( 2 \times 1 = 2 \). (Oxidation)

**Step III: Write half-reactions (conceptual) and balance electron change.**
Reduction: \( \mathrm{Cl_2O_7} \rightarrow 2\mathrm{ClO_2^-} \) (Decrease by 8 electrons)
Oxidation: \( \mathrm{H_2O_2} \rightarrow \mathrm{O_2} \) (Increase by 2 electrons)

**Step IV: Equalize total increase and decrease in oxidation numbers.**
Multiply oxidation half by 4 (to get 8 electrons) and reduction half by 1:
\( \mathrm{Cl_2O_7} + 4\mathrm{H_2O_2} \rightarrow 2\mathrm{ClO_2^-} + 4\mathrm{O_2} \)

**Step V: Balance other atoms (O and H) for basic medium.**
Balance Oxygen: Left side: 7 (from \( \mathrm{Cl_2O_7} \)) + 8 (from \( 4\mathrm{H_2O_2} \)) = 15 O. Right side: 4 (from \( 2\mathrm{ClO_2^-} \)) + 8 (from \( 4\mathrm{O_2} \)) = 12 O. Add 3 \( \mathrm{H_2O} \) to the right side.
\( \mathrm{Cl_2O_7} + 4\mathrm{H_2O_2} \rightarrow 2\mathrm{ClO_2^-} + 4\mathrm{O_2} + 3\mathrm{H_2O} \)
Balance Hydrogen: Left side: \( 4 \times 2 = 8 \) H from \( \mathrm{H_2O_2} \). Right side: \( 3 \times 2 = 6 \) H from \( \mathrm{H_2O} \). Add \( 2\mathrm{H^+} \) to the right side.
\( \mathrm{Cl_2O_7} + 4\mathrm{H_2O_2} \rightarrow 2\mathrm{ClO_2^-} + 4\mathrm{O_2} + 3\mathrm{H_2O} + 2\mathrm{H^+} \)
Since it's basic medium, add \( 2\mathrm{OH^-} \) to both sides:
\( \mathrm{Cl_2O_7} + 4\mathrm{H_2O_2} + 2\mathrm{OH^-} \rightarrow 2\mathrm{ClO_2^-} + 4\mathrm{O_2} + 3\mathrm{H_2O} + 2\mathrm{H^+} + 2\mathrm{OH^-} \)
Combine \( 2\mathrm{H^+} + 2\mathrm{OH^-} \) to \( 2\mathrm{H_2O} \):
\( \mathrm{Cl_2O_7} + 4\mathrm{H_2O_2} + 2\mathrm{OH^-} \rightarrow 2\mathrm{ClO_2^-} + 4\mathrm{O_2} + 3\mathrm{H_2O} + 2\mathrm{H_2O} \)
Simplify water:
\( \mathrm{Cl_2O_7} + 4\mathrm{H_2O_2} + 2\mathrm{OH^-} \rightarrow 2\mathrm{ClO_2^-} + 4\mathrm{O_2} + 5\mathrm{H_2O} \)

**Step VI: Check balance of atoms and charges.**
Atoms: Cl=2, O=15, H=8 on both sides. (Balanced)
Charge: Left side \( (2 \times -1) = -2 \)
Right side \( (2 \times -1) = -2 \) (Balanced)
The balanced equation is: \( \mathrm{Cl_2O_7} + 4\mathrm{H_2O_2} + 2\mathrm{OH^-} \rightarrow 2\mathrm{ClO_2^-} + 4\mathrm{O_2} + 5\mathrm{H_2O} \)
The oxidizing agent is \( \mathrm{Cl_2O_7} \) because it gets reduced.
In simple words: For reaction (i) in acid, sulfur goes from -2 to 0 (oxidation), and manganese goes from +7 to +2 (reduction). We balance the electron changes and then add \( \mathrm{H_2O} \) and \( \mathrm{H^+} \) to balance oxygen and hydrogen. \( \mathrm{MnO_4^-} \) is the oxidizer. For reaction (ii) in base, chlorine goes from +7 to +3 (reduction), and oxygen in hydrogen peroxide goes from -1 to 0 (oxidation). We balance electrons, then balance oxygen with water, and hydrogen with \( \mathrm{H^+} \), then change \( \mathrm{H^+} \) to \( \mathrm{OH^-} \) to fit the basic conditions. \( \mathrm{Cl_2O_7} \) is the oxidizer.

🎯 Exam Tip: When balancing redox reactions in basic medium, a common mistake is forgetting to convert \( \mathrm{H^+} \) ions to \( \mathrm{H_2O} \) by adding an equal number of \( \mathrm{OH^-} \) ions to both sides of the equation.

 

Question 32. Write the step of balancing equation by Ion Electron method and balance the following reactions :
(i) \( \mathrm{Al} + \mathrm{NO_3^-} \rightarrow \mathrm{Al(OH)_4^-} + \mathrm{NH_3} \) (Basic)
(ii) \( \mathrm{MnO_4^-} + \mathrm{Br^-} \rightarrow \mathrm{Mn^{2+}} + \mathrm{Br_2} \) (Acidic)
(iii) \( \mathrm{Cr_2O_7^{2-}} + \mathrm{Fe^{2+}} \rightarrow \mathrm{Cr^{3+}} + \mathrm{Fe^{3+}} \) (Acidic)
Answer:
**Steps for Balancing Equations by Ion-Electron (Half-Reaction) Method:**
1. **Write the skeletal equation:** Write down the unbalanced reaction.
2. **Assign oxidation numbers:** Determine the oxidation states of all atoms to identify which elements are oxidized and reduced.
3. **Separate into half-reactions:** Divide the overall reaction into two half-reactions, one for oxidation and one for reduction.
4. **Balance atoms (excluding O and H):** Balance all atoms in each half-reaction except oxygen and hydrogen.
5. **Balance oxygen atoms:** For reactions in acidic medium, add \( \mathrm{H_2O} \) molecules to the side deficient in oxygen. For basic medium, add \( \mathrm{H_2O} \) to the side deficient in oxygen, then add \( \mathrm{OH^-} \) to the opposite side (twice the number of \( \mathrm{H_2O} \) added) to balance hydrogen.
6. **Balance hydrogen atoms:** For reactions in acidic medium, add \( \mathrm{H^+} \) ions to the side deficient in hydrogen. For basic medium, add \( \mathrm{H_2O} \) to the side deficient in hydrogen, then add \( \mathrm{OH^-} \) to the opposite side (equal to the number of \( \mathrm{H_2O} \) added) to balance hydrogen.
7. **Balance charges:** Add electrons (\( \mathrm{e^-} \)) to the side with the more positive charge to balance the charges in each half-reaction.
8. **Equalize electrons:** Multiply each half-reaction by an appropriate integer so that the number of electrons lost in the oxidation half-reaction equals the number of electrons gained in the reduction half-reaction.
9. **Combine half-reactions:** Add the two balanced half-reactions together and cancel out identical species (electrons, \( \mathrm{H_2O} \), \( \mathrm{H^+} \), or \( \mathrm{OH^-} \)) on opposite sides of the equation.
10. **Verify:** Check that the atoms and charges are balanced on both sides of the final equation.

(i) Balancing \( \mathrm{Al} + \mathrm{NO_3^-} \rightarrow \mathrm{Al(OH)_4^-} + \mathrm{NH_3} \) (Basic medium):
**Oxidation half-reaction:** \( \mathrm{Al} \rightarrow \mathrm{Al(OH)_4^-} \)
1. Balance Al: Already balanced.
2. Balance O: Add \( 4\mathrm{H_2O} \) to the left: \( \mathrm{Al} + 4\mathrm{H_2O} \rightarrow \mathrm{Al(OH)_4^-} \)
3. Balance H: Add \( 4\mathrm{H^+} \) to the right: \( \mathrm{Al} + 4\mathrm{H_2O} \rightarrow \mathrm{Al(OH)_4^-} + 4\mathrm{H^+} \)
4. Convert to basic medium: Add \( 4\mathrm{OH^-} \) to both sides and combine \( \mathrm{H^+} \) and \( \mathrm{OH^-} \).
\( \mathrm{Al} + 4\mathrm{H_2O} + 4\mathrm{OH^-} \rightarrow \mathrm{Al(OH)_4^-} + 4\mathrm{H_2O} \)
Simplify \( \mathrm{H_2O} \): \( \mathrm{Al} + 4\mathrm{OH^-} \rightarrow \mathrm{Al(OH)_4^-} \)
5. Balance charge: Left is -4, right is -1. Add 3 electrons to the right:
\( \mathrm{Al} + 4\mathrm{OH^-} \rightarrow \mathrm{Al(OH)_4^-} + 3\mathrm{e^-} \) (Equation 1)

**Reduction half-reaction:** \( \mathrm{NO_3^-} \rightarrow \mathrm{NH_3} \)
1. Balance N: Already balanced.
2. Balance O: Add \( 3\mathrm{H_2O} \) to the right: \( \mathrm{NO_3^-} \rightarrow \mathrm{NH_3} + 3\mathrm{H_2O} \)
3. Balance H: Add \( 9\mathrm{H^+} \) to the left: \( \mathrm{NO_3^-} + 9\mathrm{H^+} \rightarrow \mathrm{NH_3} + 3\mathrm{H_2O} \)
4. Convert to basic medium: Add \( 9\mathrm{OH^-} \) to both sides and combine \( \mathrm{H^+} \) and \( \mathrm{OH^-} \).
\( \mathrm{NO_3^-} + 9\mathrm{H_2O} \rightarrow \mathrm{NH_3} + 3\mathrm{H_2O} + 9\mathrm{OH^-} \)
Simplify \( \mathrm{H_2O} \): \( \mathrm{NO_3^-} + 6\mathrm{H_2O} \rightarrow \mathrm{NH_3} + 9\mathrm{OH^-} \)
5. Balance charge: Left is -1. Right is -9. Add 8 electrons to the left:
\( \mathrm{NO_3^-} + 6\mathrm{H_2O} + 8\mathrm{e^-} \rightarrow \mathrm{NH_3} + 9\mathrm{OH^-} \) (Equation 2)

**Combine half-reactions:**
Multiply Equation 1 by 8 and Equation 2 by 3 to equalize electrons (24 electrons).
\( 8(\mathrm{Al} + 4\mathrm{OH^-} \rightarrow \mathrm{Al(OH)_4^-} + 3\mathrm{e^-}) \Rightarrow 8\mathrm{Al} + 32\mathrm{OH^-} \rightarrow 8\mathrm{Al(OH)_4^-} + 24\mathrm{e^-} \)
\( 3(\mathrm{NO_3^-} + 6\mathrm{H_2O} + 8\mathrm{e^-} \rightarrow \mathrm{NH_3} + 9\mathrm{OH^-}) \Rightarrow 3\mathrm{NO_3^-} + 18\mathrm{H_2O} + 24\mathrm{e^-} \rightarrow 3\mathrm{NH_3} + 27\mathrm{OH^-} \)
Add the two multiplied equations:
\( 8\mathrm{Al} + 32\mathrm{OH^-} + 3\mathrm{NO_3^-} + 18\mathrm{H_2O} + 24\mathrm{e^-} \rightarrow 8\mathrm{Al(OH)_4^-} + 24\mathrm{e^-} + 3\mathrm{NH_3} + 27\mathrm{OH^-} \)
Cancel electrons and simplify \( \mathrm{OH^-} \):
\( 8\mathrm{Al} + 5\mathrm{OH^-} + 3\mathrm{NO_3^-} + 18\mathrm{H_2O} \rightarrow 8\mathrm{Al(OH)_4^-} + 3\mathrm{NH_3} \)
This is the balanced equation.

(ii) Balancing \( \mathrm{MnO_4^-} + \mathrm{Br^-} \rightarrow \mathrm{Mn^{2+}} + \mathrm{Br_2} \) (Acidic medium):
**Oxidation half-reaction:** \( \mathrm{Br^-} \rightarrow \mathrm{Br_2} \)
1. Balance Br: \( 2\mathrm{Br^-} \rightarrow \mathrm{Br_2} \)
2. Balance charge: \( 2\mathrm{Br^-} \rightarrow \mathrm{Br_2} + 2\mathrm{e^-} \) (Equation 1)

**Reduction half-reaction:** \( \mathrm{MnO_4^-} \rightarrow \mathrm{Mn^{2+}} \)
1. Balance Mn: Already balanced.
2. Balance O: Add \( 4\mathrm{H_2O} \) to the right: \( \mathrm{MnO_4^-} \rightarrow \mathrm{Mn^{2+}} + 4\mathrm{H_2O} \)
3. Balance H: Add \( 8\mathrm{H^+} \) to the left: \( \mathrm{MnO_4^-} + 8\mathrm{H^+} \rightarrow \mathrm{Mn^{2+}} + 4\mathrm{H_2O} \)
4. Balance charge: Left is \( (-1) + 8(+1) = +7 \). Right is +2. Add 5 electrons to the left:
\( \mathrm{MnO_4^-} + 8\mathrm{H^+} + 5\mathrm{e^-} \rightarrow \mathrm{Mn^{2+}} + 4\mathrm{H_2O} \) (Equation 2)

**Combine half-reactions:**
Multiply Equation 1 by 5 and Equation 2 by 2 to equalize electrons (10 electrons).
\( 5(2\mathrm{Br^-} \rightarrow \mathrm{Br_2} + 2\mathrm{e^-}) \Rightarrow 10\mathrm{Br^-} \rightarrow 5\mathrm{Br_2} + 10\mathrm{e^-} \)
\( 2(\mathrm{MnO_4^-} + 8\mathrm{H^+} + 5\mathrm{e^-} \rightarrow \mathrm{Mn^{2+}} + 4\mathrm{H_2O}) \Rightarrow 2\mathrm{MnO_4^-} + 16\mathrm{H^+} + 10\mathrm{e^-} \rightarrow 2\mathrm{Mn^{2+}} + 8\mathrm{H_2O} \)
Add the two multiplied equations:
\( 10\mathrm{Br^-} + 2\mathrm{MnO_4^-} + 16\mathrm{H^+} + 10\mathrm{e^-} \rightarrow 5\mathrm{Br_2} + 10\mathrm{e^-} + 2\mathrm{Mn^{2+}} + 8\mathrm{H_2O} \)
Cancel electrons:
\( 10\mathrm{Br^-} + 2\mathrm{MnO_4^-} + 16\mathrm{H^+} \rightarrow 5\mathrm{Br_2} + 2\mathrm{Mn^{2+}} + 8\mathrm{H_2O} \)
This is the balanced equation.

(iii) Balancing \( \mathrm{Cr_2O_7^{2-}} + \mathrm{Fe^{2+}} \rightarrow \mathrm{Cr^{3+}} + \mathrm{Fe^{3+}} \) (Acidic medium):
**Oxidation half-reaction:** \( \mathrm{Fe^{2+}} \rightarrow \mathrm{Fe^{3+}} \)
1. Balance Fe: Already balanced.
2. Balance charge: \( \mathrm{Fe^{2+}} \rightarrow \mathrm{Fe^{3+}} + \mathrm{e^-} \) (Equation 1)

**Reduction half-reaction:** \( \mathrm{Cr_2O_7^{2-}} \rightarrow \mathrm{Cr^{3+}} \)
1. Balance Cr: \( \mathrm{Cr_2O_7^{2-}} \rightarrow 2\mathrm{Cr^{3+}} \)
2. Balance O: Add \( 7\mathrm{H_2O} \) to the right: \( \mathrm{Cr_2O_7^{2-}} \rightarrow 2\mathrm{Cr^{3+}} + 7\mathrm{H_2O} \)
3. Balance H: Add \( 14\mathrm{H^+} \) to the left: \( \mathrm{Cr_2O_7^{2-}} + 14\mathrm{H^+} \rightarrow 2\mathrm{Cr^{3+}} + 7\mathrm{H_2O} \)
4. Balance charge: Left is \( (-2) + 14(+1) = +12 \). Right is \( 2(+3) = +6 \). Add 6 electrons to the left:
\( \mathrm{Cr_2O_7^{2-}} + 14\mathrm{H^+} + 6\mathrm{e^-} \rightarrow 2\mathrm{Cr^{3+}} + 7\mathrm{H_2O} \) (Equation 2)

**Combine half-reactions:**
Multiply Equation 1 by 6 and Equation 2 by 1 to equalize electrons (6 electrons).
\( 6(\mathrm{Fe^{2+}} \rightarrow \mathrm{Fe^{3+}} + \mathrm{e^-}) \Rightarrow 6\mathrm{Fe^{2+}} \rightarrow 6\mathrm{Fe^{3+}} + 6\mathrm{e^-} \)
\( 1(\mathrm{Cr_2O_7^{2-}} + 14\mathrm{H^+} + 6\mathrm{e^-} \rightarrow 2\mathrm{Cr^{3+}} + 7\mathrm{H_2O}) \Rightarrow \mathrm{Cr_2O_7^{2-}} + 14\mathrm{H^+} + 6\mathrm{e^-} \rightarrow 2\mathrm{Cr^{3+}} + 7\mathrm{H_2O} \)
Add the two multiplied equations:
\( 6\mathrm{Fe^{2+}} + \mathrm{Cr_2O_7^{2-}} + 14\mathrm{H^+} + 6\mathrm{e^-} \rightarrow 6\mathrm{Fe^{3+}} + 6\mathrm{e^-} + 2\mathrm{Cr^{3+}} + 7\mathrm{H_2O} \)
Cancel electrons:
\( 6\mathrm{Fe^{2+}} + \mathrm{Cr_2O_7^{2-}} + 14\mathrm{H^+} \rightarrow 6\mathrm{Fe^{3+}} + 2\mathrm{Cr^{3+}} + 7\mathrm{H_2O} \)
This is the balanced equation.
In simple words: First, you split the main chemical reaction into two smaller parts: one where atoms lose electrons (oxidation) and one where atoms gain electrons (reduction). Then, for each part, you balance all atoms except oxygen and hydrogen. After that, you balance oxygen by adding water \( (\mathrm{H_2O}) \) and hydrogen by adding \( \mathrm{H^+} \) (for acid) or \( \mathrm{OH^-} \) (for base). Next, balance the electrical charges by adding electrons. Make sure both parts have the same number of electrons by multiplying them. Finally, put the two parts back together and remove anything that appears on both sides.

🎯 Exam Tip: The most crucial aspect of the ion-electron method is to correctly identify and balance the electron transfer in each half-reaction. Any error in electron count will lead to an unbalanced overall equation.

 

Question 33. Draw the diagram of Galvanic cell which represents the following reaction : \( Zn + 2Ag^{+} \rightarrow Zn^{2+}+ 2 Ag \) Answer the following question :
(a) Give equation for the reaction taking place at each electrode.
(b) In which direction electrons in external circuit flows ?
(c) Identify anode and cathode.
(d) If \( E^\circ Zn^{2+}/Zn = -0.76 \) volt and \( E^\circ Ag^{+}/ Ag = +0.80 \) volt, then whether \( E^\circ = 1.56 \) volt ?

Answer:
(a) Reaction taking place at anode: \( Zn \rightarrow Zn^{2+} + 2e^{-} \)
Reaction taking place at cathode: \( 2Ag^{+} + 2e^{-} \rightarrow 2Ag \)
(b) Electrons in the external circuit flow from the anode to the cathode.
(c) Zinc (Zn) acts as the anode, and Silver (Ag) acts as the cathode. These are the two points where the electrical current enters and leaves.
(d) To calculate the standard cell potential \( E^\circ_{cell} \), we use the formula: \( E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} \).
Given \( E^\circ_{Zn^{2+}/Zn} = -0.76 \) V (anode) and \( E^\circ_{Ag^{+}/Ag} = +0.80 \) V (cathode).
So, \( E^\circ_{cell} = (+0.80 \, V) - (-0.76 \, V) \)
\( E^\circ_{cell} = 0.80 \, V + 0.76 \, V \)
\( E^\circ_{cell} = 1.56 \, V \)
Yes, the calculated standard cell potential \( E^\circ_{cell} \) is \( 1.56 \) V.
In simple words: The diagram shows how a Galvanic cell uses zinc and silver to make electricity. Zinc gives away electrons (oxidation) and silver takes them (reduction). Electrons move from zinc to silver through the wire. The total voltage that this cell makes is \( 1.56 \) volts.

🎯 Exam Tip: Always remember that the anode is where oxidation occurs (loss of electrons) and the cathode is where reduction occurs (gain of electrons). Electron flow is always from anode to cathode in the external circuit.

 

Question 34. What is electro chemical series ? The electrode potential of elements A, B, C and D are \( +0.79, -0.74, +1.08 \) and \( -0.31 \) volts. Arrange them in increasing order of reactivity. To store \( 1 \) m HCl, which one aluminum or silver is suitable to use ? \( (E^\circ Al^{3+}/Al = -1.66V \) and \( E^\circ Ag/Ag = +0.80V) \)
Answer:
Electro chemical series: The electrochemical series is an arrangement of elements in order of decreasing reduction potential values. It helps us understand how reactive different metals are and predict reaction outcomes.

Increasing order of reactivity of these elements: The reactivity of a metal is inversely proportional to its standard reduction potential. A more negative potential means a more active metal.
Comparing the given values: \( C (+1.08 \, V) > A (+0.79 \, V) > D (-0.31 \, V) > B (-0.74 \, V) \).
So, the order of reactivity from least active to most active is: \( C < A < D < B \).
To store \( 1 \) M HCl: We need a metal that will not react with the acid. A metal will react with HCl if its standard reduction potential is more negative than that of hydrogen (which is \( 0.00 \) V). This means a more active metal will displace hydrogen from HCl.
Given \( E^\circ_{Al^{3+}/Al} = -1.66 \) V and \( E^\circ_{Ag^{+}/Ag} = +0.80 \) V.
Aluminum has a very negative potential (\( -1.66 \) V), making it highly reactive and a strong reducing agent. It would react with HCl, displacing hydrogen. Silver, with a positive potential (\( +0.80 \) V), is less reactive than hydrogen and will not react with HCl. Therefore, silver is suitable to store \( 1 \) M HCl. In general, metals above hydrogen in the activity series react with acids, while those below do not.
In simple words: The electrochemical series lists how easily elements gain electrons. Elements with more negative values are more reactive and can push out other metals from their solutions. To store acid like HCl, we need a metal that does not react with it. Silver is a good choice because it is less reactive than hydrogen, so it will not react with the acid.

🎯 Exam Tip: Remember, the more negative the standard electrode potential, the stronger the reducing agent and the more reactive the metal. For storage, choose a less reactive metal than the substance it will hold.

 

Question 35. Balance the following reaction using Ion electron method :
(i) \( MnO_4^{-} + SO_3^{2-} \rightarrow Mn^{2+} + SO_4^{2-} \)
(ii) \( Cr_2O_7^{2-} + H^{+} + I^{-} \rightarrow Cr^{3+}+ H_2O +I_2 \) (Acidic medium)
(iii) \( Cl_2O_7 + H_2O_2 \rightarrow ClO_2^{-} + O_2 + H^{+} \) (Basic medium)
(iv) \( P_4 + OH^{-} \rightarrow PH_3 + HPO_2^{-} \)

Answer:
(i) Balancing \( MnO_4^{-} + SO_3^{2-} \rightarrow Mn^{2+} + SO_4^{2-} \)
First half reaction: \( MnO_4^{-} \rightarrow Mn^{2+} \) (Reduction)
(a) Balance Mn atoms: Already balanced.
(b) Balance O atoms by adding \( H_2O \) to the right side: \( MnO_4^{-} \rightarrow Mn^{2+} + 4H_2O \)
(c) Balance H atoms by adding \( H^{+} \) to the left side: \( MnO_4^{-} + 8H^{+} \rightarrow Mn^{2+} + 4H_2O \)
(d) Balance charge by adding electrons to the left side: \( MnO_4^{-} + 8H^{+} + 5e^{-} \rightarrow Mn^{2+} + 4H_2O \) ...(1)

Second half reaction: \( SO_3^{2-} \rightarrow SO_4^{2-} \) (Oxidation)
(a) Balance S atoms: Already balanced.
(b) Balance O atoms by adding \( H_2O \) to the left side: \( SO_3^{2-} + H_2O \rightarrow SO_4^{2-} \)
(c) Balance H atoms by adding \( H^{+} \) to the right side: \( SO_3^{2-} + H_2O \rightarrow SO_4^{2-} + 2H^{+} \)
(d) Balance charge by adding electrons to the right side: \( SO_3^{2-} + H_2O \rightarrow SO_4^{2-} + 2H^{+} + 2e^{-} \) ...(2)

Multiply equation (1) by \( 2 \) and equation (2) by \( 5 \) to equalize electrons:
\( 2[MnO_4^{-} + 8H^{+} + 5e^{-} \rightarrow Mn^{2+} + 4H_2O] \)
\( 5[SO_3^{2-} + H_2O \rightarrow SO_4^{2-} + 2H^{+} + 2e^{-}] \)

This gives:
\( 2MnO_4^{-} + 16H^{+} + 10e^{-} \rightarrow 2Mn^{2+} + 8H_2O \)
\( 5SO_3^{2-} + 5H_2O \rightarrow 5SO_4^{2-} + 10H^{+} + 10e^{-} \)

Adding these two equations and cancelling common terms (like \( 10e^{-} \), \( 10H^{+} \), \( 5H_2O \)):
\( 2MnO_4^{-} + 5SO_3^{2-} + (16-10)H^{+} + (5-8)H_2O \rightarrow 2Mn^{2+} + 5SO_4^{2-} \)
\( 2MnO_4^{-} + 5SO_3^{2-} + 6H^{+} \rightarrow 2Mn^{2+} + 5SO_4^{2-} + 3H_2O \)
This is the balanced equation. This method helps visualize the electron transfer in different parts of a reaction.

(ii) Balancing \( Cr_2O_7^{2-} + H^{+} + I^{-} \rightarrow Cr^{3+}+ H_2O +I_2 \) (Acidic medium)
First half reaction: \( Cr_2O_7^{2-} \rightarrow Cr^{3+} \) (Reduction)
(a) Balance Cr atoms: \( Cr_2O_7^{2-} \rightarrow 2Cr^{3+} \)
(b) Balance O atoms by adding \( H_2O \) to the right side: \( Cr_2O_7^{2-} \rightarrow 2Cr^{3+} + 7H_2O \)
(c) Balance H atoms by adding \( H^{+} \) to the left side: \( Cr_2O_7^{2-} + 14H^{+} \rightarrow 2Cr^{3+} + 7H_2O \)
(d) Balance charge by adding electrons to the left side: \( Cr_2O_7^{2-} + 14H^{+} + 6e^{-} \rightarrow 2Cr^{3+} + 7H_2O \) ...(1)

Second half reaction: \( I^{-} \rightarrow I_2 \) (Oxidation)
(a) Balance I atoms: \( 2I^{-} \rightarrow I_2 \)
(b) Balance charge by adding electrons to the right side: \( 2I^{-} \rightarrow I_2 + 2e^{-} \) ...(2)

Multiply equation (1) by \( 1 \) and equation (2) by \( 3 \) to equalize electrons:
\( 1[Cr_2O_7^{2-} + 14H^{+} + 6e^{-} \rightarrow 2Cr^{3+} + 7H_2O] \)
\( 3[2I^{-} \rightarrow I_2 + 2e^{-}] \)

This gives:
\( Cr_2O_7^{2-} + 14H^{+} + 6e^{-} \rightarrow 2Cr^{3+} + 7H_2O \)
\( 6I^{-} \rightarrow 3I_2 + 6e^{-} \)

Adding these two equations and cancelling common terms:
\( Cr_2O_7^{2-} + 14H^{+} + 6I^{-} \rightarrow 2Cr^{3+} + 3I_2 + 7H_2O \)
This is the balanced equation. Ion-electron method systematically balances mass and charge.

(iii) Balancing \( Cl_2O_7 + H_2O_2 \rightarrow ClO_2^{-} + O_2 + H^{+} \) (Basic medium)
First half reaction: \( Cl_2O_7 \rightarrow ClO_2^{-} \) (Reduction)
(a) Balance Cl atoms: \( Cl_2O_7 \rightarrow 2ClO_2^{-} \)
(b) Balance O atoms by adding \( H_2O \) to the left side: \( Cl_2O_7 + 6H_2O \rightarrow 2ClO_2^{-} \)
(c) Balance H atoms by adding \( H^{+} \) to the right side: \( Cl_2O_7 + 6H_2O \rightarrow 2ClO_2^{-} + 12H^{+} \)
(d) Since it's a basic medium, add \( OH^{-} \) to both sides equal to \( H^{+} \) ions:
\( Cl_2O_7 + 6H_2O + 12OH^{-} \rightarrow 2ClO_2^{-} + 12H_2O \)
Simplify \( H_2O \): \( Cl_2O_7 + 12OH^{-} \rightarrow 2ClO_2^{-} + 6H_2O \)
(e) Balance charge by adding electrons to the left side: \( Cl_2O_7 + 12OH^{-} + 10e^{-} \rightarrow 2ClO_2^{-} + 6H_2O \) ...(1)

Second half reaction: \( H_2O_2 \rightarrow O_2 \) (Oxidation)
(a) Balance O atoms: Already balanced.
(b) Balance H atoms by adding \( H^{+} \) to the right side: \( H_2O_2 \rightarrow O_2 + 2H^{+} \)
(c) Since it's a basic medium, add \( OH^{-} \) to both sides equal to \( H^{+} \) ions:
\( H_2O_2 + 2OH^{-} \rightarrow O_2 + 2H_2O \)
(d) Balance charge by adding electrons to the right side: \( H_2O_2 + 2OH^{-} \rightarrow O_2 + 2H_2O + 2e^{-} \) ...(2)

Multiply equation (1) by \( 1 \) and equation (2) by \( 5 \) to equalize electrons:
\( 1[Cl_2O_7 + 12OH^{-} + 10e^{-} \rightarrow 2ClO_2^{-} + 6H_2O] \)
\( 5[H_2O_2 + 2OH^{-} \rightarrow O_2 + 2H_2O + 2e^{-}] \)

This gives:
\( Cl_2O_7 + 12OH^{-} + 10e^{-} \rightarrow 2ClO_2^{-} + 6H_2O \)
\( 5H_2O_2 + 10OH^{-} \rightarrow 5O_2 + 10H_2O + 10e^{-} \)

Adding these two equations and cancelling common terms (like \( 10e^{-} \), \( 10OH^{-} \), \( 6H_2O \)):
\( Cl_2O_7 + 5H_2O_2 + (12-10)OH^{-} \rightarrow 2ClO_2^{-} + 5O_2 + (10-6)H_2O \)
\( Cl_2O_7 + 5H_2O_2 + 2OH^{-} \rightarrow 2ClO_2^{-} + 5O_2 + 4H_2O \)
This is the balanced equation. Understanding medium-specific balancing is crucial for accuracy.

(iv) Balancing \( P_4 + OH^{-} \rightarrow PH_3 + HPO_2^{-} \)
First half reaction: \( P_4 \rightarrow PH_3 \) (Reduction)
(a) Balance P atoms: \( P_4 \rightarrow 4PH_3 \)
(b) Balance H atoms by adding \( H_2O \) to the left side and \( OH^{-} \) to the right side (basic medium):
\( P_4 + 12H_2O \rightarrow 4PH_3 + 12OH^{-} \)
(c) Balance charge by adding electrons to the left side: \( P_4 + 12H_2O + 12e^{-} \rightarrow 4PH_3 + 12OH^{-} \) ...(1)

Second half reaction: \( P_4 \rightarrow HPO_2^{-} \) (Oxidation)
(a) Balance P atoms: \( P_4 \rightarrow 4HPO_2^{-} \)
(b) Balance O atoms by adding \( H_2O \) to the left side: \( P_4 + 8H_2O \rightarrow 4HPO_2^{-} \)
(c) Balance H atoms by adding \( H^{+} \) to the right side and then convert to basic medium with \( OH^{-} \):
\( P_4 + 8H_2O \rightarrow 4HPO_2^{-} + 12H^{+} \)
Add \( 12OH^{-} \) to both sides: \( P_4 + 8H_2O + 12OH^{-} \rightarrow 4HPO_2^{-} + 12H_2O \)
Simplify \( H_2O \): \( P_4 + 12OH^{-} \rightarrow 4HPO_2^{-} + 4H_2O \)
(d) Balance charge by adding electrons to the right side: \( P_4 + 12OH^{-} \rightarrow 4HPO_2^{-} + 4H_2O + 8e^{-} \) ...(2)

Multiply equation (1) by \( 2 \) and equation (2) by \( 3 \) to equalize electrons:
\( 2[P_4 + 12H_2O + 12e^{-} \rightarrow 4PH_3 + 12OH^{-}] \)
\( 3[P_4 + 12OH^{-} \rightarrow 4HPO_2^{-} + 4H_2O + 8e^{-}] \)

This gives:
\( 2P_4 + 24H_2O + 24e^{-} \rightarrow 8PH_3 + 24OH^{-} \)
\( 3P_4 + 36OH^{-} \rightarrow 12HPO_2^{-} + 12H_2O + 24e^{-} \)

Adding these two equations and cancelling common terms:
\( (2+3)P_4 + (24-12)H_2O + (36-24)OH^{-} \rightarrow 8PH_3 + 12HPO_2^{-} \)
\( 5P_4 + 12H_2O + 12OH^{-} \rightarrow 8PH_3 + 12HPO_2^{-} \)
This is the balanced equation. Balancing complex redox reactions in basic media requires careful step-by-step application of these rules.
In simple words: Balancing reactions means making sure the number of each type of atom and the total charge are the same on both sides. The ion-electron method breaks the reaction into two parts, one where electrons are lost (oxidation) and one where electrons are gained (reduction). By balancing atoms and charges in each part, then putting them back together, we get a complete and correct reaction. Always double check oxygen, hydrogen, and total charge.

🎯 Exam Tip: When balancing in basic medium, first balance as if in acidic medium, then add \( OH^{-} \) ions to both sides equal to the number of \( H^{+} \) ions. \( H^{+} \) and \( OH^{-} \) on the same side combine to form \( H_2O \). Simplify water molecules afterward.

 

Question 36. Balance the following reaction using Ion Electron method,
(i) \( \mathrm{NO}_{3}^{-} + H^{+} + I^{-} \rightarrow NO+ H_2O + I_2 \)
(ii) \( CrO_4^{2-} + SO_3^{2-} + OH^{-} \rightarrow CrO_2^{2-} + \mathrm{SO}_{4}^{2-} \)
(iii) \( \mathrm{MnO}_{4}^{-} + Fe_3O_4 + OH^{-} \rightarrow Fe_2O_3 + MnO_2 \)
(iv) \( P_4 + OH^{-} \rightarrow PH_3 + \mathrm{HPO}_{2}^{-} \)

Answer:
(i) Balancing \( \mathrm{NO}_{3}^{-} + H^{+} + I^{-} \rightarrow NO+ H_2O + I_2 \)
First half reaction: \( NO_3^{-} \rightarrow NO \) (Reduction)
(a) Balance N atoms: Already balanced.
(b) Balance O atoms by adding \( H_2O \) to the right side: \( NO_3^{-} \rightarrow NO + 2H_2O \)
(c) Balance H atoms by adding \( H^{+} \) to the left side: \( NO_3^{-} + 4H^{+} \rightarrow NO + 2H_2O \)
(d) Balance charge by adding electrons to the left side: \( NO_3^{-} + 4H^{+} + 3e^{-} \rightarrow NO + 2H_2O \) ...(1)

Second half reaction: \( I^{-} \rightarrow I_2 \) (Oxidation)
(a) Balance I atoms: \( 2I^{-} \rightarrow I_2 \)
(b) Balance charge by adding electrons to the right side: \( 2I^{-} \rightarrow I_2 + 2e^{-} \) ...(2)

Multiply equation (1) by \( 2 \) and equation (2) by \( 3 \) to equalize electrons:
\( 2[NO_3^{-} + 4H^{+} + 3e^{-} \rightarrow NO + 2H_2O] \)
\( 3[2I^{-} \rightarrow I_2 + 2e^{-}] \)

This gives:
\( 2NO_3^{-} + 8H^{+} + 6e^{-} \rightarrow 2NO + 4H_2O \)
\( 6I^{-} \rightarrow 3I_2 + 6e^{-} \)

Adding these two equations and cancelling common terms:
\( 2NO_3^{-} + 8H^{+} + 6I^{-} \rightarrow 2NO + 3I_2 + 4H_2O \)
This is the balanced equation. The method ensures conservation of mass and charge.

(ii) Balancing \( CrO_4^{2-} + SO_3^{2-} + OH^{-} \rightarrow CrO_2^{2-} + SO_4^{2-} \)
First half reaction: \( CrO_4^{2-} \rightarrow CrO_2^{2-} \) (Reduction)
(a) Balance Cr atoms: Already balanced.
(b) Balance O atoms by adding \( H_2O \) to the right side: \( CrO_4^{2-} \rightarrow CrO_2^{2-} + 2H_2O \)
(c) Balance H atoms by adding \( H^{+} \) to the left side: \( CrO_4^{2-} + 4H^{+} \rightarrow CrO_2^{2-} + 2H_2O \)
(d) Since it's a basic medium (indicated by \( OH^{-} \) in reactants), add \( OH^{-} \) to both sides equal to \( H^{+} \):
\( CrO_4^{2-} + 4H_2O \rightarrow CrO_2^{2-} + 2H_2O + 4OH^{-} \)
Simplify \( H_2O \): \( CrO_4^{2-} + 2H_2O \rightarrow CrO_2^{2-} + 4OH^{-} \)
(e) Balance charge by adding electrons to the left side: \( CrO_4^{2-} + 2H_2O + 2e^{-} \rightarrow CrO_2^{2-} + 4OH^{-} \) ...(1)

Second half reaction: \( SO_3^{2-} \rightarrow SO_4^{2-} \) (Oxidation)
(a) Balance S atoms: Already balanced.
(b) Balance O atoms by adding \( H_2O \) to the left side: \( SO_3^{2-} + H_2O \rightarrow SO_4^{2-} \)
(c) Balance H atoms by adding \( H^{+} \) to the right side: \( SO_3^{2-} + H_2O \rightarrow SO_4^{2-} + 2H^{+} \)
(d) Since it's a basic medium, add \( OH^{-} \) to both sides equal to \( H^{+} \):
\( SO_3^{2-} + H_2O + 2OH^{-} \rightarrow SO_4^{2-} + 2H_2O \)
Simplify \( H_2O \): \( SO_3^{2-} + 2OH^{-} \rightarrow SO_4^{2-} + H_2O \)
(e) Balance charge by adding electrons to the right side: \( SO_3^{2-} + 2OH^{-} \rightarrow SO_4^{2-} + H_2O + 2e^{-} \) ...(2)

The number of electrons is already equal. Add the two half-reactions:
\( CrO_4^{2-} + 2H_2O + 2e^{-} + SO_3^{2-} + 2OH^{-} \rightarrow CrO_2^{2-} + 4OH^{-} + SO_4^{2-} + H_2O + 2e^{-} \)
Cancel common terms (\( 2e^{-} \), \( H_2O \), \( 2OH^{-} \)):
\( CrO_4^{2-} + SO_3^{2-} + H_2O \rightarrow CrO_2^{2-} + SO_4^{2-} + 2OH^{-} \)
This is the balanced equation. Understanding basic medium rules is important for these types of reactions.

(iii) Balancing \( \mathrm{MnO}_{4}^{-} + Fe_3O_4 + OH^{-} \rightarrow Fe_2O_3 + MnO_2 \)
First half reaction: \( MnO_4^{-} \rightarrow MnO_2 \) (Reduction)
(a) Balance Mn atoms: Already balanced.
(b) Balance O atoms by adding \( H_2O \) to the right side: \( MnO_4^{-} \rightarrow MnO_2 + 2H_2O \)
(c) Balance H atoms by adding \( H^{+} \) to the left side: \( MnO_4^{-} + 4H^{+} \rightarrow MnO_2 + 2H_2O \)
(d) Since it's a basic medium, add \( OH^{-} \) to both sides equal to \( H^{+} \):
\( MnO_4^{-} + 4H_2O \rightarrow MnO_2 + 2H_2O + 4OH^{-} \)
Simplify \( H_2O \): \( MnO_4^{-} + 2H_2O \rightarrow MnO_2 + 4OH^{-} \)
(e) Balance charge by adding electrons to the left side: \( MnO_4^{-} + 2H_2O + 3e^{-} \rightarrow MnO_2 + 4OH^{-} \) ...(1)

Second half reaction: \( Fe_3O_4 \rightarrow Fe_2O_3 \) (Oxidation)
(a) Balance Fe atoms: \( 2Fe_3O_4 \rightarrow 3Fe_2O_3 \)
(b) Balance O atoms by adding \( H_2O \) to the left side: \( 2Fe_3O_4 + H_2O \rightarrow 3Fe_2O_3 \)
(c) Balance H atoms by adding \( H^{+} \) to the right side: \( 2Fe_3O_4 + H_2O \rightarrow 3Fe_2O_3 + 2H^{+} \)
(d) Since it's a basic medium, add \( OH^{-} \) to both sides equal to \( H^{+} \):
\( 2Fe_3O_4 + H_2O + 2OH^{-} \rightarrow 3Fe_2O_3 + 2H_2O \)
Simplify \( H_2O \): \( 2Fe_3O_4 + 2OH^{-} \rightarrow 3Fe_2O_3 + H_2O \)
(e) Balance charge by adding electrons to the right side: \( 2Fe_3O_4 + 2OH^{-} \rightarrow 3Fe_2O_3 + H_2O + 2e^{-} \) ...(2)

Multiply equation (1) by \( 2 \) and equation (2) by \( 3 \) to equalize electrons:
\( 2[MnO_4^{-} + 2H_2O + 3e^{-} \rightarrow MnO_2 + 4OH^{-}] \)
\( 3[2Fe_3O_4 + 2OH^{-} \rightarrow 3Fe_2O_3 + H_2O + 2e^{-}] \)

This gives:
\( 2MnO_4^{-} + 4H_2O + 6e^{-} \rightarrow 2MnO_2 + 8OH^{-} \)
\( 6Fe_3O_4 + 6OH^{-} \rightarrow 9Fe_2O_3 + 3H_2O + 6e^{-} \)

Adding these two equations and cancelling common terms:
\( 2MnO_4^{-} + 6Fe_3O_4 + (4-3)H_2O + (6-8)OH^{-} \rightarrow 2MnO_2 + 9Fe_2O_3 \)
\( 2MnO_4^{-} + 6Fe_3O_4 + H_2O \rightarrow 2MnO_2 + 9Fe_2O_3 + 2OH^{-} \)
This is the balanced equation. Oxidation states change during this type of reaction.

(iv) Balancing \( P_4 + OH^{-} \rightarrow PH_3 + HPO_2^{-} \)
First half reaction: \( P_4 \rightarrow PH_3 \) (Reduction)
(a) Balance P atoms: \( P_4 \rightarrow 4PH_3 \)
(b) Balance H atoms by adding \( H_2O \) to the left side and \( OH^{-} \) to the right side (basic medium):
\( P_4 + 12H_2O \rightarrow 4PH_3 + 12OH^{-} \)
(c) Balance charge by adding electrons to the left side: \( P_4 + 12H_2O + 12e^{-} \rightarrow 4PH_3 + 12OH^{-} \) ...(1)

Second half reaction: \( P_4 \rightarrow HPO_2^{-} \) (Oxidation)
(a) Balance P atoms: \( P_4 \rightarrow 4HPO_2^{-} \)
(b) Balance O atoms by adding \( H_2O \) to the left side: \( P_4 + 8H_2O \rightarrow 4HPO_2^{-} \)
(c) Balance H atoms by adding \( H^{+} \) to the right side and then convert to basic medium with \( OH^{-} \):
\( P_4 + 8H_2O \rightarrow 4HPO_2^{-} + 12H^{+} \)
Add \( 12OH^{-} \) to both sides: \( P_4 + 8H_2O + 12OH^{-} \rightarrow 4HPO_2^{-} + 12H_2O \)
Simplify \( H_2O \): \( P_4 + 12OH^{-} \rightarrow 4HPO_2^{-} + 4H_2O \)
(d) Balance charge by adding electrons to the right side: \( P_4 + 12OH^{-} \rightarrow 4HPO_2^{-} + 4H_2O + 8e^{-} \) ...(2)

Multiply equation (1) by \( 2 \) and equation (2) by \( 3 \) to equalize electrons:
\( 2[P_4 + 12H_2O + 12e^{-} \rightarrow 4PH_3 + 12OH^{-}] \)
\( 3[P_4 + 12OH^{-} \rightarrow 4HPO_2^{-} + 4H_2O + 8e^{-}] \)

This gives:
\( 2P_4 + 24H_2O + 24e^{-} \rightarrow 8PH_3 + 24OH^{-} \)
\( 3P_4 + 36OH^{-} \rightarrow 12HPO_2^{-} + 12H_2O + 24e^{-} \)

Adding these two equations and cancelling common terms:
\( (2+3)P_4 + (24-12)H_2O + (36-24)OH^{-} \rightarrow 8PH_3 + 12HPO_2^{-} \)
\( 5P_4 + 12H_2O + 12OH^{-} \rightarrow 8PH_3 + 12HPO_2^{-} \)
This is the balanced equation. This is an example of a disproportionation reaction, where phosphorus is both oxidized and reduced.
In simple words: When balancing these reactions, we break them into two smaller reactions: one where atoms lose electrons (oxidation) and one where atoms gain electrons (reduction). We adjust atoms like oxygen and hydrogen, and then balance the charges using electrons. Finally, we combine the two balanced parts to get the full balanced reaction.

🎯 Exam Tip: For balancing redox reactions, carefully identify the species undergoing oxidation and reduction. Remember that in basic medium, \( OH^{-} \) ions are used to balance \( H^{+} \) ions by forming \( H_2O \).

 

Question 37. Balance the following reactions using Ion Electron method :
(i) \( MnO_4^{-} + H_2O_2 \rightarrow MnO_2 + O_2 + OH^{-} \)
(ii) \( As O_3^{3-} + H_2O + I_2 \rightarrow ASO_4^{3-} + H^{+} + I^{-} \)
(iii) \( Cl_2O_7 + H_2O_2 \rightarrow CIO_2^{-} + O_2 + H^{+} \)
(iv) \( Cr_2O_7^{2-}+SO_2 \rightarrow Cr^{3+} + SO_4^{2-} \)

Answer:
(i) Balancing \( MnO_4^{-} + H_2O_2 \rightarrow MnO_2 + O_2 + OH^{-} \)
First half reaction: \( MnO_4^{-} \rightarrow MnO_2 \) (Reduction)
(a) Balance Mn atoms: Already balanced.
(b) Balance O atoms by adding \( H_2O \) to the right side: \( MnO_4^{-} \rightarrow MnO_2 + 2H_2O \)
(c) Balance H atoms by adding \( H^{+} \) to the left side: \( MnO_4^{-} + 4H^{+} \rightarrow MnO_2 + 2H_2O \)
(d) The reaction is in basic medium, so add \( OH^{-} \) to both sides equal to \( H^{+} \):
\( MnO_4^{-} + 4H_2O \rightarrow MnO_2 + 2H_2O + 4OH^{-} \)
Simplify \( H_2O \): \( MnO_4^{-} + 2H_2O \rightarrow MnO_2 + 4OH^{-} \)
(e) Balance charge by adding electrons to the left side: \( MnO_4^{-} + 2H_2O + 3e^{-} \rightarrow MnO_2 + 4OH^{-} \) ...(1)

Second half reaction: \( H_2O_2 \rightarrow O_2 \) (Oxidation)
(a) Balance O atoms: Already balanced.
(b) Balance H atoms by adding \( H^{+} \) to the right side: \( H_2O_2 \rightarrow O_2 + 2H^{+} \)
(c) The reaction is in basic medium, so add \( OH^{-} \) to both sides equal to \( H^{+} \):
\( H_2O_2 + 2OH^{-} \rightarrow O_2 + 2H_2O \)
(d) Balance charge by adding electrons to the right side: \( H_2O_2 + 2OH^{-} \rightarrow O_2 + 2H_2O + 2e^{-} \) ...(2)

Multiply equation (1) by \( 2 \) and equation (2) by \( 3 \) to equalize electrons:
\( 2[MnO_4^{-} + 2H_2O + 3e^{-} \rightarrow MnO_2 + 4OH^{-}] \)
\( 3[H_2O_2 + 2OH^{-} \rightarrow O_2 + 2H_2O + 2e^{-}] \)

This gives:
\( 2MnO_4^{-} + 4H_2O + 6e^{-} \rightarrow 2MnO_2 + 8OH^{-} \)
\( 3H_2O_2 + 6OH^{-} \rightarrow 3O_2 + 6H_2O + 6e^{-} \)

Adding these two equations and cancelling common terms:
\( 2MnO_4^{-} + 3H_2O_2 + (4-6)H_2O + (6-8)OH^{-} \rightarrow 2MnO_2 + 3O_2 \)
\( 2MnO_4^{-} + 3H_2O_2 \rightarrow 2MnO_2 + 3O_2 + 2OH^{-} + 2H_2O \)
The final equation is \( 2MnO_4^{-} + 3H_2O_2 \rightarrow 2MnO_2 + 3O_2 + 2OH^{-} + 2H_2O \). This demonstrates how hydrogen peroxide can act as a reducing agent.

(ii) Balancing \( AsO_3^{3-} + H_2O + I_2 \rightarrow AsO_4^{3-} + H^{+} + I^{-} \)
First half reaction: \( AsO_3^{3-} \rightarrow AsO_4^{3-} \) (Oxidation)
(a) Balance As atoms: Already balanced.
(b) Balance O atoms by adding \( H_2O \) to the left side: \( AsO_3^{3-} + H_2O \rightarrow AsO_4^{3-} \)
(c) Balance H atoms by adding \( H^{+} \) to the right side: \( AsO_3^{3-} + H_2O \rightarrow AsO_4^{3-} + 2H^{+} \)
(d) Balance charge by adding electrons to the right side: \( AsO_3^{3-} + H_2O \rightarrow AsO_4^{3-} + 2H^{+} + 2e^{-} \) ...(1)

Second half reaction: \( I_2 \rightarrow I^{-} \) (Reduction)
(a) Balance I atoms: \( I_2 \rightarrow 2I^{-} \)
(b) Balance charge by adding electrons to the left side: \( I_2 + 2e^{-} \rightarrow 2I^{-} \) ...(2)

The number of electrons is already equal. Add the two half-reactions:
\( AsO_3^{3-} + H_2O + I_2 + 2e^{-} \rightarrow AsO_4^{3-} + 2H^{+} + 2e^{-} + 2I^{-} \)
Cancel common terms (\( 2e^{-} \)):
\( AsO_3^{3-} + H_2O + I_2 \rightarrow AsO_4^{3-} + 2H^{+} + 2I^{-} \)
This is the balanced equation. Redox reactions often involve changes in multiple oxidation states.

(iii) Balancing \( Cl_2O_7 + H_2O_2 \rightarrow ClO_2^{-} + O_2 + H^{+} \) (Acidic medium is implied by \( H^{+} \) in products)
First half reaction: \( Cl_2O_7 \rightarrow ClO_2^{-} \) (Reduction)
(a) Balance Cl atoms: \( Cl_2O_7 \rightarrow 2ClO_2^{-} \)
(b) Balance O atoms by adding \( H_2O \) to the left side: \( Cl_2O_7 + 3H_2O \rightarrow 2ClO_2^{-} \)
(c) Balance H atoms by adding \( H^{+} \) to the right side: \( Cl_2O_7 + 3H_2O \rightarrow 2ClO_2^{-} + 6H^{+} \)
(d) Balance charge by adding electrons to the left side: \( Cl_2O_7 + 3H_2O + 8e^{-} \rightarrow 2ClO_2^{-} + 6H^{+} \) ...(1)

Second half reaction: \( H_2O_2 \rightarrow O_2 \) (Oxidation)
(a) Balance O atoms: Already balanced.
(b) Balance H atoms by adding \( H^{+} \) to the right side: \( H_2O_2 \rightarrow O_2 + 2H^{+} \)
(c) Balance charge by adding electrons to the right side: \( H_2O_2 \rightarrow O_2 + 2H^{+} + 2e^{-} \) ...(2)

Multiply equation (1) by \( 1 \) and equation (2) by \( 4 \) to equalize electrons:
\( 1[Cl_2O_7 + 3H_2O + 8e^{-} \rightarrow 2ClO_2^{-} + 6H^{+}] \)
\( 4[H_2O_2 \rightarrow O_2 + 2H^{+} + 2e^{-}] \)

This gives:
\( Cl_2O_7 + 3H_2O + 8e^{-} \rightarrow 2ClO_2^{-} + 6H^{+} \)
\( 4H_2O_2 \rightarrow 4O_2 + 8H^{+} + 8e^{-} \)

Adding these two equations and cancelling common terms (like \( 8e^{-} \), \( 6H^{+} \)):
\( Cl_2O_7 + 3H_2O + 4H_2O_2 \rightarrow 2ClO_2^{-} + (8-6)H^{+} + 4O_2 \)
\( Cl_2O_7 + 3H_2O + 4H_2O_2 \rightarrow 2ClO_2^{-} + 2H^{+} + 4O_2 \)
This is the balanced equation. Hydrogen peroxide can sometimes act as a reducing agent, especially when reacting with strong oxidizers.

(iv) Balancing \( Cr_2O_7^{2-}+SO_2 \rightarrow Cr^{3+} + SO_4^{2-} \)
First half reaction: \( Cr_2O_7^{2-} \rightarrow Cr^{3+} \) (Reduction)
(a) Balance Cr atoms: \( Cr_2O_7^{2-} \rightarrow 2Cr^{3+} \)
(b) Balance O atoms by adding \( H_2O \) to the right side: \( Cr_2O_7^{2-} \rightarrow 2Cr^{3+} + 7H_2O \)
(c) Balance H atoms by adding \( H^{+} \) to the left side: \( Cr_2O_7^{2-} + 14H^{+} \rightarrow 2Cr^{3+} + 7H_2O \)
(d) Balance charge by adding electrons to the left side: \( Cr_2O_7^{2-} + 14H^{+} + 6e^{-} \rightarrow 2Cr^{3+} + 7H_2O \) ...(1)

Second half reaction: \( SO_2 \rightarrow SO_4^{2-} \) (Oxidation)
(a) Balance S atoms: Already balanced.
(b) Balance O atoms by adding \( H_2O \) to the left side: \( SO_2 + 2H_2O \rightarrow SO_4^{2-} \)
(c) Balance H atoms by adding \( H^{+} \) to the right side: \( SO_2 + 2H_2O \rightarrow SO_4^{2-} + 4H^{+} \)
(d) Balance charge by adding electrons to the right side: \( SO_2 + 2H_2O \rightarrow SO_4^{2-} + 4H^{+} + 2e^{-} \) ...(2)

Multiply equation (1) by \( 1 \) and equation (2) by \( 3 \) to equalize electrons:
\( 1[Cr_2O_7^{2-} + 14H^{+} + 6e^{-} \rightarrow 2Cr^{3+} + 7H_2O] \)
\( 3[SO_2 + 2H_2O \rightarrow SO_4^{2-} + 4H^{+} + 2e^{-}] \)

This gives:
\( Cr_2O_7^{2-} + 14H^{+} + 6e^{-} \rightarrow 2Cr^{3+} + 7H_2O \)
\( 3SO_2 + 6H_2O \rightarrow 3SO_4^{2-} + 12H^{+} + 6e^{-} \)

Adding these two equations and cancelling common terms (like \( 6e^{-} \), \( 12H^{+} \), \( 6H_2O \)):
\( Cr_2O_7^{2-} + 3SO_2 + (14-12)H^{+} + (6-7)H_2O \rightarrow 2Cr^{3+} + 3SO_4^{2-} \)
\( Cr_2O_7^{2-} + 3SO_2 + 2H^{+} \rightarrow 2Cr^{3+} + 3SO_4^{2-} + H_2O \)
This is the balanced equation. This reaction is common in environmental chemistry where sulfur dioxide is oxidized.
In simple words: For each reaction, we separate it into two parts: one where atoms lose electrons (oxidation) and one where atoms gain electrons (reduction). We then add water and hydrogen ions (or hydroxide ions for basic reactions) to balance atoms, and electrons to balance the charges. Finally, we combine the two parts to get the full balanced reaction.

🎯 Exam Tip: When balancing redox reactions in acidic media, use \( H^{+} \) and \( H_2O \) to balance hydrogen and oxygen atoms. In basic media, use \( OH^{-} \) and \( H_2O \). Always ensure the total charge on both sides of the final equation is equal.

 

Question 38. Calculate oxidation number of the following compounds :
(i) Fe in \( FeSO_4 \)
(ii) Fe in \( Fe(CO)_5 \)
(iii) P in \( H_3PO_3 \)
(iv) S in \( H_2S_2O_7 \)
(v) C in \( C_{12}H_{22}O_{11} \)
(vi) P in \( NaH_2PO_2 \)

Answer:
(i) Fe in \( FeSO_4 \)
Let the oxidation number of Fe be \( x \).
In \( SO_4^{2-} \), the sulfate ion, sulfur typically has an oxidation number of \( +6 \), and oxygen is \( -2 \). So, \( S + 4(-2) = -2 \implies S = +6 \). Therefore, the sulfate ion has a charge of \( -2 \).
Since \( FeSO_4 \) is a neutral compound, the sum of oxidation numbers is zero.
\( x + (-2) = 0 \)
\( x = +2 \)
Therefore, the oxidation number of Fe in \( FeSO_4 \) is \( +2 \). The iron atom gives up two electrons in this compound.

(ii) Fe in \( Fe(CO)_5 \)
Let the oxidation number of Fe be \( x \).
Carbonyl (CO) is a neutral ligand, so its oxidation number is \( 0 \).
Since \( Fe(CO)_5 \) is a neutral compound, the sum of oxidation numbers is zero.
\( x + 5(0) = 0 \)
\( x = 0 \)
Therefore, the oxidation number of Fe in \( Fe(CO)_5 \) is \( 0 \). This is common for metals in carbonyl complexes.

(iii) P in \( H_3PO_3 \)
Let the oxidation number of P be \( x \).
Hydrogen (H) usually has an oxidation number of \( +1 \).
Oxygen (O) usually has an oxidation number of \( -2 \).
Since \( H_3PO_3 \) is a neutral compound, the sum of oxidation numbers is zero.
\( 3(+1) + x + 3(-2) = 0 \)
\( 3 + x - 6 = 0 \)
\( x - 3 = 0 \)
\( x = +3 \)
Therefore, the oxidation number of P in \( H_3PO_3 \) is \( +3 \). Phosphorous can exhibit various oxidation states.

(iv) S in \( H_2S_2O_7 \)
Let the oxidation number of S be \( x \).
Hydrogen (H) has an oxidation number of \( +1 \).
Oxygen (O) has an oxidation number of \( -2 \).
Since \( H_2S_2O_7 \) is a neutral compound, the sum of oxidation numbers is zero.
\( 2(+1) + 2x + 7(-2) = 0 \)
\( 2 + 2x - 14 = 0 \)
\( 2x - 12 = 0 \)
\( 2x = 12 \)
\( x = \frac{12}{2} = 6 \)
Therefore, the oxidation number of S in \( H_2S_2O_7 \) is \( +6 \). This compound is also known as pyrosulfuric acid.

(v) C in \( C_{12}H_{22}O_{11} \)
Let the oxidation number of C be \( x \).
Hydrogen (H) has an oxidation number of \( +1 \).
Oxygen (O) has an oxidation number of \( -2 \).
Since \( C_{12}H_{22}O_{11} \) is a neutral compound, the sum of oxidation numbers is zero.
\( 12x + 22(+1) + 11(-2) = 0 \)
\( 12x + 22 - 22 = 0 \)
\( 12x + 0 = 0 \)
\( 12x = 0 \)
\( x = 0 \)
Therefore, the oxidation number of C in \( C_{12}H_{22}O_{11} \) is \( 0 \). This compound is sucrose, a common sugar.

(vi) P in \( NaH_2PO_2 \)
Let the oxidation number of P be \( x \).
Sodium (Na) is an alkali metal, so its oxidation number is \( +1 \).
Hydrogen (H) has an oxidation number of \( +1 \).
Oxygen (O) has an oxidation number of \( -2 \).
Since \( NaH_2PO_2 \) is a neutral compound, the sum of oxidation numbers is zero.
\( 1(+1) + 2(+1) + x + 2(-2) = 0 \)
\( 1 + 2 + x - 4 = 0 \)
\( x - 1 = 0 \)
\( x = +1 \)
Therefore, the oxidation number of P in \( NaH_2PO_2 \) is \( +1 \). This compound is sodium hypophosphite.
In simple words: To find the oxidation number, we assign common numbers to elements like hydrogen (+1) and oxygen (-2). Then, we use algebra to find the unknown oxidation number of the element we are interested in. For neutral compounds, all oxidation numbers add up to zero. For ions, they add up to the ion's charge.

🎯 Exam Tip: Remember the common oxidation states of elements (e.g., alkali metals +1, alkaline earth metals +2, oxygen -2, hydrogen +1, halogens -1) as a starting point. Exceptions exist, like peroxides for oxygen or metal hydrides for hydrogen.

 

Question 39. Identify oxidising and reducing agent in the following:
(i) \( 3I_2 + NaOH \rightarrow NaIO_3 + 5NaI + 3H_2O \)
(ii) \( AlCl_3 + 3K \rightarrow Al+ 3KCl \)
(iii) \( SO_2 + 2H_2 S \rightarrow 3S + 2H_2O \)
(iv) \( SnCl_2 + 2FeCl_3 \rightarrow SnCl_4 + 2FeCl_2 \)
(v) \( H_2 O_2 + H_2 O_2 \rightarrow 2H_2O+ O_2 \)

Answer:
(i) \( 3I_2 + NaOH \rightarrow NaIO_3 + 5NaI + 3H_2O \)
Let's find the oxidation states:
In \( I_2 \): I is \( 0 \)
In \( NaIO_3 \): Na is \( +1 \), O is \( -2 \). So, \( +1 + x + 3(-2) = 0 \implies x = +5 \). (Iodine is oxidized)
In \( NaI \): Na is \( +1 \). So, \( +1 + x = 0 \implies x = -1 \). (Iodine is reduced)
In this reaction, iodine (\( I_2 \)) is both oxidized to \( NaIO_3 \) and reduced to \( NaI \). Therefore, \( I_2 \) acts as both an oxidising agent and a reducing agent. This is an example of a disproportionation reaction. The \( NaOH \) acts as a medium and does not change its oxidation state.
Oxidizing agent: \( I_2 \)
Reducing agent: \( I_2 \)

(ii) \( AlCl_3 + 3K \rightarrow Al+ 3KCl \)
Let's find the oxidation states:
In \( AlCl_3 \): Cl is \( -1 \). So, \( x + 3(-1) = 0 \implies x = +3 \).
In \( K \): K is \( 0 \).
In \( Al \): Al is \( 0 \). (Aluminum is reduced)
In \( KCl \): K is \( +1 \), Cl is \( -1 \). (Potassium is oxidized)
In this reaction, potassium (K) reduces \( AlCl_3 \) to \( Al \) by losing electrons, so K is the reducing agent. \( AlCl_3 \) oxidizes K to \( KCl \) by gaining electrons, so \( AlCl_3 \) is the oxidising agent. This is a single displacement reaction.
Oxidizing agent: \( AlCl_3 \)
Reducing agent: \( K \)

(iii) \( SO_2 + 2H_2 S \rightarrow 3S + 2H_2O \)
Let's find the oxidation states:
In \( SO_2 \): O is \( -2 \). So, \( x + 2(-2) = 0 \implies x = +4 \).
In \( H_2S \): H is \( +1 \). So, \( 2(+1) + x = 0 \implies x = -2 \).
In \( S \) (product): S is \( 0 \).
\( S \) in \( SO_2 \) goes from \( +4 \) to \( 0 \) (reduced). So, \( SO_2 \) is the oxidising agent.
\( S \) in \( H_2S \) goes from \( -2 \) to \( 0 \) (oxidized). So, \( H_2S \) is the reducing agent.
In this reaction, sulfur dioxide (\( SO_2 \)) oxidizes hydrogen sulfide (\( H_2S \)) to sulfur, while \( H_2S \) reduces \( SO_2 \) to sulfur. Both sulfur-containing compounds participate in the redox process.
Oxidizing agent: \( SO_2 \)
Reducing agent: \( H_2S \)

(iv) \( SnCl_2 + 2FeCl_3 \rightarrow SnCl_4 + 2FeCl_2 \)
Let's find the oxidation states:
In \( SnCl_2 \): Cl is \( -1 \). So, \( x + 2(-1) = 0 \implies x = +2 \).
In \( FeCl_3 \): Cl is \( -1 \). So, \( x + 3(-1) = 0 \implies x = +3 \).
In \( SnCl_4 \): Cl is \( -1 \). So, \( x + 4(-1) = 0 \implies x = +4 \). (Tin is oxidized)
In \( FeCl_2 \): Cl is \( -1 \). So, \( x + 2(-1) = 0 \implies x = +2 \). (Iron is reduced)
In this reaction, \( FeCl_3 \) oxidizes \( SnCl_2 \) to \( SnCl_4 \) (Fe goes from \( +3 \) to \( +2 \), Sn goes from \( +2 \) to \( +4 \)). Thus, \( FeCl_3 \) is the oxidising agent. \( SnCl_2 \) reduces \( FeCl_3 \) to \( FeCl_2 \). So, \( SnCl_2 \) is the reducing agent. This reaction is commonly used in titrations.
Oxidizing agent: \( FeCl_3 \)
Reducing agent: \( SnCl_2 \)

(v) \( H_2 O_2 + H_2 O_2 \rightarrow 2H_2O+ O_2 \)
Let's find the oxidation states:
In \( H_2O_2 \): H is \( +1 \), O is \( -1 \) (in peroxide).
In \( H_2O \): H is \( +1 \), O is \( -2 \).
In \( O_2 \): O is \( 0 \).
In this reaction, one molecule of \( H_2O_2 \) is oxidized (O goes from \( -1 \) to \( 0 \) in \( O_2 \)), and another molecule of \( H_2O_2 \) is reduced (O goes from \( -1 \) to \( -2 \) in \( H_2O \)). Therefore, \( H_2O_2 \) acts as both an oxidising agent and a reducing agent. This is another disproportionation reaction.
Oxidizing agent: \( H_2O_2 \)
Reducing agent: \( H_2O_2 \)
In simple words: An oxidising agent helps another substance lose electrons (get oxidized) and itself gains electrons (gets reduced). A reducing agent helps another substance gain electrons (get reduced) and itself loses electrons (gets oxidized). We look at how the oxidation numbers change to tell which is which.

🎯 Exam Tip: To identify oxidising and reducing agents, first determine the oxidation numbers of all elements in reactants and products. An element whose oxidation number increases is oxidized, and the compound containing it is the reducing agent. An element whose oxidation number decreases is reduced, and the compound containing it is the oxidising agent.

 

Question 40. In which of the following reaction, \( H_2O_2 \) acts as oxidising agent and in which it acts as reducing agent?
(i) \( H_2O_2 \) oxidises KI to \( I_2 \)
(ii) \( H_2O_2 \) reduces \( Cl_2 \) to HCl

Answer:
Hydrogen peroxide (\( H_2O_2 \)) is a unique compound because the oxidation state of oxygen in it is \( -1 \), which is an intermediate state between \( 0 \) (in \( O_2 \)) and \( -2 \) (in \( H_2O \)). This allows \( H_2O_2 \) to either be oxidized to \( O_2 \) or be reduced to \( H_2O \), meaning it can act as both an oxidising agent and a reducing agent depending on the reaction.

(i) When \( H_2O_2 \) acts as an oxidising agent (oxidation state of oxygen in \( H_2O_2 \) changes from \( -1 \) to \( -2 \) in \( H_2O \)):
Example: \( H_2O_2 \) oxidizes KI to \( I_2 \)
The reaction is: \( 2KI + H_2O_2 \rightarrow 2KOH + I_2 \)
Oxidation states:
In \( H_2O_2 \), O is \( -1 \). In \( H_2O \), O is \( -2 \). So \( H_2O_2 \) is reduced.
In \( KI \), I is \( -1 \). In \( I_2 \), I is \( 0 \). So \( KI \) is oxidized.
Since \( H_2O_2 \) is reduced, it acts as an oxidising agent in this reaction. It causes the iodide ions to lose electrons.

(ii) When \( H_2O_2 \) acts as a reducing agent (oxidation state of oxygen in \( H_2O_2 \) changes from \( -1 \) to \( 0 \) in \( O_2 \)):
Example: \( H_2O_2 \) reduces \( Cl_2 \) to HCl
The reaction is: \( Cl_2 + H_2O_2 \rightarrow 2HCl + O_2 \)
Oxidation states:
In \( Cl_2 \), Cl is \( 0 \). In \( HCl \), Cl is \( -1 \). So \( Cl_2 \) is reduced.
In \( H_2O_2 \), O is \( -1 \). In \( O_2 \), O is \( 0 \). So \( H_2O_2 \) is oxidized.
Since \( H_2O_2 \) is oxidized, it acts as a reducing agent in this reaction. It causes chlorine to gain electrons.
In simple words: Hydrogen peroxide is special because its oxygen atom can either gain an electron (making it an oxidising agent) or lose an electron (making it a reducing agent). When it makes other things lose electrons (like turning KI into \( I_2 \)), it's an oxidising agent. When it makes other things gain electrons (like turning \( Cl_2 \) into HCl), it's a reducing agent.

🎯 Exam Tip: Remember that hydrogen peroxide contains oxygen in the \( -1 \) oxidation state, allowing it to undergo both oxidation (to \( O_2 \), state \( 0 \)) and reduction (to \( H_2O \), state \( -2 \)). Identify the change in oxidation state of oxygen in \( H_2O_2 \) to determine its role in a reaction.

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