Maharashtra Board Class 11 Chemistry Chapter 6 Redox Reactions Solutions

Redox Reactions Class 11 Exercise Question Answers Solutions Maharashtra Board

Class 11 Chemistry Chapter 6 Exercise Solutions Maharashtra Board

Chemistry Class 11 Chapter 6 Exercise Solutions

Exercise 1. Choose The Most Correct Option

Question A.Oxidction numbers of Cl atoms marked as Clª and Clb in CaOCl2 (bleaching powder) are
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक कैल्शियम ऑक्सीक्लोराइड (विरंजक चूर्ण) अणु की संरचना को दर्शाता है। इसमें एक कैल्शियम आयन (Ca²⁺) और दो क्लोराइड आयन हैं, जिनमें से एक ऑक्सीजन से जुड़ा है (O-Cl⁻), और दूसरा सीधे कैल्शियम से जुड़ा है (Cl⁻)।
(a) zero in each
(b) -1 in Clª and +1 in Clb
(c) +1 in Clª and -1 in Clb
(d) 1 in each
Answer: (b) -1 in Clª and +1 in Clb
In simple words: In bleaching powder, the two chlorine atoms have different oxidation states: one is -1 and the other is +1, due to their bonding arrangements.

🎯 Exam Tip: Remember that in mixed halides like bleaching powder, the oxidation states of identical atoms (like Cl) can differ based on their chemical environment and bonding. Determine oxidation numbers by considering electronegativity and formal charges.

Question B.Which of the following is not an example of redox reacton ?
(a) CuO + H2 \(\rightarrow\) Cu + H2O
(b) Fe2O3 + 3CO2 \(\rightarrow\) 2Fe + 3CO2
(c) 2K + F2 \(\rightarrow\) 2KF
(d) BaCl2 + H2SO4 \(\rightarrow\) BaSO4 + 2HCl
Answer: (d) BaCl2 + H2SO4 \(\rightarrow\) BaSO4 + 2HCl
In simple words: A redox reaction involves a change in oxidation states, but in the reaction BaCl2 + H2SO4 \(\rightarrow\) BaSO4 + 2HCl, no elements change their oxidation states, making it a double displacement reaction, not a redox one.

🎯 Exam Tip: To identify a non-redox reaction, check if the oxidation states of all elements remain unchanged from reactants to products. Double displacement reactions are typically non-redox.

Question C.A compound contains atoms of three elements A, B and C. If the oxidation state of A is +2, B is +5 and that of C is -2, the compound is possibly represented by
(a) A2(BC3)2
(b) A3(BC4)2
(c) A3(B4C)2
(d) ABC2
Answer: (b) A3(BC4)2
In simple words: To find the correct formula, we need to balance the total positive and negative oxidation states; for A3(BC4)2, the total charge is balanced with A having +2, B having +5, and C having -2.

🎯 Exam Tip: When determining the formula of a compound from oxidation states, ensure that the sum of the oxidation states of all atoms in the formula equals zero for a neutral compound.

Question D.The coefficients p, q, r, s in the reaction \(pCr_2O_7^{2-} + q Fe^{2+} \rightarrow r Cr^{3+} + s Fe^{3+} + H_2O\) respectively are :
(a) 1, 2, 6, 6
(b) 6, 1, 2, 4
(c) 1, 6, 2, 6
(d) 1, 2, 4, 6
Answer: (c) 1, 6, 2, 6
In simple words: Balancing this redox reaction requires finding the correct stoichiometric coefficients (p, q, r, s) that ensure both atoms and charges are conserved on both sides of the equation.

🎯 Exam Tip: For balancing redox reactions, use the oxidation number method or half-reaction method systematically to find the correct coefficients. Practice is key for these types of questions.

Question E.For the following redox reactions, find the correct statement. \(Sn^{2+} + 2Fe^{3+} \rightarrow Sn^{4+} + 2Fe^{2+}\)
(a) \(Sn^{2+}\) is undergoing oxidation
(b) \(Fe^{3+}\) is undergoing oxidation
(c) It is not a redox reaction
(d) Both \(Sn^{2+}\) and \(Fe^{3+}\) are oxidised
Answer: (a) \(Sn^{2+}\) is undergoing oxidation
In simple words: In this reaction, the oxidation state of tin (Sn) increases from +2 to +4, which means \(Sn^{2+}\) is losing electrons and therefore undergoing oxidation.

🎯 Exam Tip: Oxidation is defined as an increase in oxidation state (loss of electrons), while reduction is a decrease in oxidation state (gain of electrons). Identify the changes in oxidation numbers to determine which species is oxidized or reduced.

Question F.Oxidation number of carbon in H2CO3 is
(a) +1
(b) +2
(c) +3
(d) +4
Answer: (d) +4
In simple words: In carbonic acid (H2CO3), assuming hydrogen is +1 and oxygen is -2, the oxidation state of carbon calculates to +4 to make the overall molecule neutral.

🎯 Exam Tip: To calculate the oxidation number of an element in a compound, use the known oxidation states of common elements (like H=+1, O=-2) and ensure the sum equals the charge of the molecule or ion.

Question G.Which is the correct stock notation for magenese dioxide ?
(a) Mn(I)O2
(b) Mn(II)O2
(c) Mn(III)O2
(d) Mn(IV)O2
Answer: (d) Mn(IV)O2
In simple words: In manganese dioxide (MnO2), oxygen typically has an oxidation state of -2; therefore, manganese must have an oxidation state of +4 for the compound to be neutral.

🎯 Exam Tip: Stock notation uses Roman numerals in parentheses to indicate the oxidation state of the metal element, especially for metals that can exist in multiple oxidation states.

Question I.Oxidation number of oxygen in superoxide is
(a) -2
(b) -1
(c) \(-\frac{1}{2}\)
(d) 0
Answer: (c) \(-\frac{1}{2}\)
In simple words: In a superoxide ion (O2⁻), since the overall charge is -1 and there are two oxygen atoms, each oxygen atom shares that charge, resulting in an oxidation state of \(-\frac{1}{2}\).

🎯 Exam Tip: Remember the unusual oxidation states of oxygen: -2 in most oxides, -1 in peroxides (e.g., H2O2), \(-\frac{1}{2}\) in superoxides (e.g., KO2), and positive in compounds with fluorine (e.g., OF2).

Question J.Which of the following halogens does always show oxidation state -1 ?
(a) F
(b) Cl
(c) Br
(d) I
Answer: (a) F
In simple words: Fluorine (F) is the most electronegative element and almost always exists with an oxidation state of -1 in its compounds, as it strongly attracts electrons.

🎯 Exam Tip: Fluorine's exceptionally high electronegativity means it will always pull electrons towards itself, resulting in a consistent -1 oxidation state in all its compounds.

Question K.The process SO2 \(\rightarrow\) S2Cl2 is
(a) Reduction
(b) Oxidation
(c) Neither oxidation nor reduction
(d) Oxidation and reduction.
Answer: (a) Reduction
In simple words: In SO2, sulfur has an oxidation state of +4, but in S2Cl2, sulfur's oxidation state is +1; this decrease in oxidation state indicates reduction.

🎯 Exam Tip: To classify a process as oxidation or reduction, calculate the oxidation state of the key element on both sides of the reaction. A decrease in oxidation state signifies reduction.

Exercise 2. Write The Formula For The Following Compounds :

A. Mercury(II) chloride
B. Thallium(I) sulphate
C. Tin(IV) oxide
D. Chromium(III) oxide
Answer:
(i) HgCl2
(ii) Tl2SO4
(iii) SnO2
(iv) Cr2O3
In simple words: The chemical formulas are determined by balancing the charges of the given ions, using the Roman numeral to indicate the metal's oxidation state.

🎯 Exam Tip: When writing chemical formulas from names, use the Roman numeral to determine the cation's charge, then balance it with the anion's charge to achieve a neutral compound.

Exercise 3. Answer The Following Questions

Question A.In which chemical reaction does carbon exibit variation of oxidation state from -4 to +4? Write balanced chemical reaction.
Answer:In combustion of methane, carbon exhibits variation from -4 to +4. The reaction is as follows: CH4 + 2O2 \(\rightarrow\) CO2 + 2H2O In CH4, the oxidation state of carbon is -4 while in CO2, the oxidation state of carbon is +4.
In simple words: Methane combustion is the reaction where carbon changes its oxidation state from -4 in CH4 to +4 in CO2, showcasing a wide range of oxidation states for carbon.

🎯 Exam Tip: Combustion reactions often involve significant changes in oxidation states, making them prime examples of redox processes. Ensure equations are balanced for both mass and charge.

Question B.In which reaction does nitrogen exhibit variation of oxidation state from -3 to +5 ?
Answer:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह दो चरणों में नाइट्रोजन के ऑक्सीकरण को दर्शाता है। पहले चरण में, अमोनिया (NH3) ऑक्सीजन (O2) की उपस्थिति में प्लैटिनम/रोडियम उत्प्रेरक (Pt/Rh gauge catalyst) पर 500 K और 9 बार पर नाइट्रिक ऑक्साइड (NO) और पानी (H2O) बनाता है। नाइट्रोजन का ऑक्सीकरण अवस्था -3 से +2 तक बदलता है। दूसरे चरण में, नाइट्रिक ऑक्साइड (NO) ऑक्सीजन (O2) के साथ प्रतिक्रिया करके नाइट्रोजन डाइऑक्साइड (NO2) बनाता है, और फिर नाइट्रोजन डाइऑक्साइड (NO2) पानी (H2O) के साथ प्रतिक्रिया करके नाइट्रिक एसिड (HNO3) और नाइट्रिक ऑक्साइड (NO) बनाता है, जहाँ नाइट्रोजन का ऑक्सीकरण अवस्था +4 से +5 तक बदलता है।
\[\begin{array}{rcl} -3 & & +2 \\ 4NH_{3(g)} + 5O_{2(g)} & \xrightarrow{\text{Pt/Rh gauge catalyst}} & 4NO_{(g)} + 6H_2O_{(g)} \\ \text{Ammonia Oxygen} & & \text{Nitric oxide} \end{array}\] \[\begin{array}{rcl} +4 & & +5 \\ 2NO_{(g)} + O_{2(g)} & \rightarrow & 2NO_{2(g)} \\ \text{Nitric oxide Oxygen} & & \text{Nitrogen dioxide} \\ 3NO_{2(g)} + H_2O_{(l)} & \rightarrow & 2HNO_{3(l)} + NO_{(g)} \\ \text{Nitrogen Water} & & \text{Nitric Nitric} \\ \text{dioxide} & & \text{acid oxide} \end{array}\]
In simple words: Nitrogen shows an oxidation state change from -3 to +5 in reactions like the catalytic oxidation of ammonia to nitric acid, involving several intermediate nitrogen oxides.

🎯 Exam Tip: Reactions involving nitrogen compounds often exhibit a wide range of oxidation states due to nitrogen's ability to gain or lose varying numbers of electrons. Carefully calculate oxidation states at each step.

Question C.Calculate the oxidation number of underlined atoms.
a. H2SO4
b. HNO3
c. H3PO3
d. K2C2O4
e. H2S4O6
f. Cr2O7\(^{2-}\)
g. NaH2PO4
Answer:
(i) H2SO4 Oxidation number of H = +1 Oxidation number of O = -2 H2SO4 is a neutral molecule.
\[ \therefore \text{ Sum of the oxidation numbers of all atoms of } H_2SO_4 = 0 \]
\[ \therefore 2 \times (\text{Oxidation number of H}) + (\text{Oxidation number of S}) + 4 \times (\text{Oxidation number of O}) = 0 \]
\[ \therefore 2 \times (+1) + (\text{Oxidation number of S}) + 4 \times (-2) = 0 \]
\[ \therefore \text{ Oxidation number of S} + 2 - 8 = 0 \]
\[ \therefore \text{ Oxidation number of S in } H_2SO_4 = +6 \]
(ii) HNO3 Oxidation number of H = +1 Oxidation number of O = -2 HNO3 is a neutral molecule.
\[ \therefore \text{ Sum of the oxidation numbers of all atoms of } HNO_3 = 0 \]
\[ \therefore (\text{Oxidation number of H}) + (\text{Oxidation number of N}) + 3 \times (\text{Oxidation number of O}) = 0 \]
\[ \therefore (+1) + (\text{Oxidation number of N}) + 3 \times (-2) = 0 \]
\[ \therefore \text{ Oxidation number of N} + 1 - 6 = 0 \]
\[ \therefore \text{ Oxidation number of N in } HNO_3 = +5 \]
(iii) H3PO3 Oxidation number of O = -2 Oxidation number of H = +1 H3PO3 is a neutral molecule.
\[ \therefore \text{ Sum of the oxidation numbers of all atoms} = 0 \]
\[ \therefore 3 \times (\text{Oxidation number of H}) + (\text{Oxidation number of P}) + 3 \times (\text{Oxidation number of O}) = 0 \]
\[ \therefore 3 \times (+1) + (\text{Oxidation number of P}) + 3 \times (-2) = 0 \]
\[ \therefore \text{ Oxidation number of P} + 3 - 6 = 0 \]
Oxidation number of P in H3PO3 = +3
(iv) K2C2O4 Oxidation number of K = +1 Oxidation number of O = -2 K2C2O4 is a neutral molecule.
\[ \therefore \text{ Sum of the oxidation number of all atoms} = 0 \]
\[ \therefore 2 \times (\text{Oxidation number of K}) + 2 \times (\text{Oxidation number of C}) + 4 \times (\text{Oxidation number of O}) = 0 \]
\[ \therefore 2 \times (+1) + 2 \times (\text{Oxidation number of C}) + 4 \times (-2) = 0 \]
\[ \therefore 2 \times (\text{Oxidation number of C}) + 2 - 8 = 0 \]
\[ \therefore 2 \times (\text{Oxidation number of C}) = +6 \]
\[ \therefore \text{ Oxidation number of C} = \frac{+6}{2} = +3 \]
\[ \therefore \text{ Oxidation number of C in } K_2C_2O_4 = +3 \]
(v) H2S4O6 Oxidation number of H = +1 Oxidation number of O = -2 H2S4O6 is a neutral molecule.
\[ \therefore \text{ Sum of the oxidation numbers of all atoms} = 0 \]
\[ \therefore 2 \times (\text{Oxidation number of H}) + 4 \times (\text{Oxidation number of S}) + 6 \times (\text{Oxidation number of O}) = 0 \]
\[ \therefore 2 \times (+1) + 4 \times (\text{Oxidation number of S}) + 6 \times (-2) = 0 \]
\[ \therefore 4 \times (\text{Oxidation number of S}) + 2 - 12 = 0 \]
\[ \therefore 4 \times (\text{Oxidation number of S}) = +10 \]
\[ \therefore \text{ Oxidation number of S} = \frac{+10}{4} = +2.5 \]
\[ \therefore \text{ Oxidation number of S in } H_2S_4O_6 = +2.5 \]
(vi) Cr2O7\(^{2-}\) Oxidation of O = -2 Cr2O7\(^{2-}\) is an ionic species.
\[ \therefore \text{ Sum of the oxidation numbers of all atoms} = -2 \]
\[ \therefore 2 \times (\text{Oxidation number of Cr}) + 7 \times (\text{Oxidation number of O}) = -2 \]
\[ \therefore 2 \times (\text{Oxidation number of Cr}) + 7 \times (-2) = -2 \]
\[ \therefore 2 \times (\text{Oxidation number of Cr}) - 14 = -2 \]
\[ \therefore 2 \times (\text{Oxidation number of Cr}) = -2 + 14 \]
\[ \therefore \text{ Oxidation number of Cr} = \frac{+12}{2} = +6 \]
\[ \therefore \text{ Oxidation number of Cr in } Cr_2O_7^{2-} = +6 \]
(vii) NaH2PO4 Oxidation number of Na = +1 Oxidation number of H = +1 Oxidation number of O = -2 NaH2PO4 is a neutral molecule Sum of the oxidation numbers of all atoms = 0
\[ (\text{Oxidation number of Na}) + 2 \times (\text{Oxidation number of H}) + (\text{Oxidation number of P}) + 4 \times (\text{Oxidation number of O}) = 0 \]
\[ (+1) + 2 \times (+1) + (\text{Oxidation number of P}) + 4 \times (-2) = 0 \]
\[ (\text{Oxidation number of P}) + 3 - 8 = 0 \]
Oxidation number of P in NaH2PO4 = +5
In simple words: Oxidation numbers are calculated by assigning standard values to known elements and then solving for the unknown element, ensuring the total charge of the molecule or ion is balanced.

🎯 Exam Tip: Pay close attention to the charge of the overall species (neutral molecule or polyatomic ion) when calculating oxidation numbers. Remember that the sum of oxidation numbers must equal the charge of the species.

Question D.Justify that the following reactions are redox reaction; identify the species oxidized/reduced, which acts as an oxidant and which act as a reductant.
a. 2Cu2O(s) + Cu2S(s) \(\rightarrow\) 6Cu(s) + SO2(g)
b. HF(aq) + OH\(^-\)(aq) \(\rightarrow\) H2O(l) + F\(^-\)(aq)
c. I2(aq) + 2 S2O3\(^{2-}\)(aq) \(\rightarrow\) S4O6\(^{2-}\)(aq) + 2I\(^-\)(aq)
Answer:
(i) 2Cu2O(s) + Cu2S(s) \(\rightarrow\) 6Cu(s) + SO2(g)
a. Write oxidation number of all the atoms of reactants and products.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह 2Cu2O(s) + Cu2S(s) \(\rightarrow\) 6Cu(s) + SO2(g) अभिक्रिया में परमाणुओं की ऑक्सीकरण अवस्थाओं में परिवर्तन को दर्शाता है। बाईं ओर, Cu2O में Cu का ऑक्सीकरण अवस्था +1 है और O का -2 है। Cu2S में Cu का +1 और S का -2 है। दाईं ओर, Cu(s) में Cu का 0 है और SO2 में S का +4 और O का -2 है।
\[\begin{array}{cc} 2Cu_2O_{(s)} + Cu_2S_{(s)} & \rightarrow 6Cu_{(s)} + SO_{2(g)} \\ \uparrow\uparrow \quad \uparrow\uparrow & \quad \uparrow \quad \uparrow\uparrow \\ +1\, -2 \quad +1\, -2 & \quad 0 \quad +4\, -2 \end{array}\]
b. Identify the species that undergoes change in oxidation number.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह Cu2O और Cu2S में Cu और S की ऑक्सीकरण अवस्थाओं में परिवर्तन को दर्शाता है। Cu2O और Cu2S में Cu की ऑक्सीकरण अवस्था +1 से घटकर 6Cu में 0 हो जाती है, जो इलेक्ट्रॉन के लाभ (Gain of e⁻) को दर्शाता है। Cu2S में S की ऑक्सीकरण अवस्था -2 से बढ़कर SO2 में +4 हो जाती है, जो इलेक्ट्रॉन की हानि (Loss of e⁻) को दर्शाता है।
\[\begin{array}{rcl} \quad \text{Gain of e}^- \\ \quad \downarrow \\ 2Cu_2O_{(s)} + Cu_2S_{(s)} & \rightarrow 6Cu_{(s)} + SO_{2(g)} \\ \quad \uparrow \\ \quad \text{Loss of e}^- \end{array}\]
c. The oxidation number of S increases from -2 to +4 and that of Cu decreases from +1 to 0. Because oxidation number of one species increases and that of the other decreases, the reaction is a redox reaction.
d. The oxidation number of S increases by loss of electrons and therefore, S is a reducing agent and it itself is oxidised. On the other hand, the oxidation number of Cu decreases by gain of electrons and therefore, Cu is an oxidising agent and itself is reduced.
Result:
1. The given reaction is a redox reaction.
2. Oxidant/oxidising agents (Reduced species): Cu2O/ Cu2S
3. Reductant/reducing agent (Oxidised species): Cu2S
[Note: Cu in both Cu2O and Cu2S undergoes reduction. Hence, both Cu2O and Cu2S can be termed as oxidising agents in the given reaction.]
In simple words: This reaction is a redox reaction because sulfur's oxidation state increases (oxidation) and copper's oxidation state decreases (reduction). Copper oxides/sulfides act as oxidants, while the sulfide acts as a reductant.

🎯 Exam Tip: For identifying redox components, explicitly write down the initial and final oxidation states for each element. A species that gets oxidized is the reducing agent, and a species that gets reduced is the oxidizing agent.

(ii) HF(aq) + OH\(^-\)(aq) \(\rightarrow\) H2O(l) + F\(^-\)(aq)
a. Write oxidation number of all the atoms of reactants and products.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह HF(aq) + OH⁻(aq) \(\rightarrow\) H2O(l) + F⁻(aq) अभिक्रिया में परमाणुओं की ऑक्सीकरण अवस्थाओं को दर्शाता है। HF में H का +1 और F का -1 है। OH⁻ में O का -2 और H का +1 है। H2O में H का +1 और O का -2 है। F⁻ में F का -1 है।
\[\begin{array}{cc} HF_{(aq)} + OH^-_{(aq)} & \rightarrow H_2O_{(l)} + F^-_{(aq)} \\ \uparrow\uparrow \quad \uparrow\uparrow & \quad \uparrow\uparrow\uparrow \quad \uparrow \\ +1\, -1 \quad -2\, +1 & \quad +1\, -2 \quad -1 \end{array}\]
b. Since, the oxidation numbers of all the species remain same, this is NOT a redox reaction. Result: The given reaction is NOT a redox reaction.
In simple words: In the reaction between hydrofluoric acid and hydroxide, all elements maintain their initial oxidation states, indicating no electron transfer, so it is not a redox reaction but an acid-base neutralization.

🎯 Exam Tip: Neutralization reactions (acid-base reactions) are generally not redox reactions unless one of the participating ions can undergo a change in its oxidation state. Always verify oxidation states.

(iii) I2(aq) + 2 S2O3\(^{2-}\)(aq) \(\rightarrow\) S4O6\(^{2-}\)(aq) + 2I\(^-\)(aq)
a. Write oxidation number of all the atoms of reactants and products.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह I2(aq) + 2 S2O3\(^{2-}\)(aq) \(\rightarrow\) S4O6\(^{2-}\)(aq) + 2I\(^-\)(aq) अभिक्रिया में परमाणुओं की ऑक्सीकरण अवस्थाओं में परिवर्तन को दर्शाता है। बाईं ओर, I2 में I का ऑक्सीकरण अवस्था 0 है। S2O3\(^{2-}\) में S का +2 और O का -2 है। दाईं ओर, S4O6\(^{2-}\) में S का +2.5 और O का -2 है। 2I\(^-\) में I का -1 है।
\[\begin{array}{cc} I_{2(aq)} + 2S_2O_3^{2-}{}_{(aq)} & \rightarrow S_4O_6^{2-}{}_{(aq)} + 2I^-_{(aq)} \\ \uparrow \quad \uparrow\uparrow & \quad \uparrow\uparrow \quad \uparrow \\ 0 \quad +2\, -2 & \quad +2.5\, -2 \quad -1 \end{array}\]
b. Identify the species that undergoes change in oxidation number.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह I2 और S2O3\(^{2-}\) में I और S की ऑक्सीकरण अवस्थाओं में परिवर्तन को दर्शाता है। I2 में I की ऑक्सीकरण अवस्था 0 से घटकर 2I⁻ में -1 हो जाती है, जो इलेक्ट्रॉन के लाभ (Gain of e⁻) को दर्शाता है। S2O3\(^{2-}\) में S की ऑक्सीकरण अवस्था +2 से बढ़कर S4O6\(^{2-}\) में +2.5 हो जाती है, जो इलेक्ट्रॉन की हानि (Loss of e⁻) को दर्शाता है।
\[\begin{array}{rcl} \quad \text{Gain of e}^- \\ \quad \downarrow \\ I_{2(aq)} + 2S_2O_3^{2-}{}_{(aq)} & \rightarrow S_4O_6^{2-}{}_{(aq)} + 2I^-_{(aq)} \\ \quad \uparrow \\ \quad \text{Loss of e}^- \end{array}\]
c. The oxidation number of S increases from +2 to +2.5 and that of I decreases from 0 to -1. Because oxidation number of one species increases and that of the other decreases, the reaction is a redox reaction.
d. The oxidation number of S increases by loss of electrons and therefore, S is a reducing agent and it itself is oxidised. On the other hand, the oxidation number of I decreases by gain of electrons and therefore, I is an oxidising agent and itself is reduced.
Result:
1. The given reaction is a redox reaction.
2. Oxidant/oxidising agent (Reduced species): I2
3. Reductant/reducing agent (Oxidised species): S2O3\(^{2-}\)
In simple words: This reaction is a redox process because iodine's oxidation state decreases (reduction) while sulfur's oxidation state increases (oxidation); I2 acts as the oxidant, and S2O3\(^{2-}\) acts as the reductant.

🎯 Exam Tip: When dealing with polyatomic ions like S2O3\(^{2-}\) and S4O6\(^{2-}\), calculate the average oxidation state of sulfur. Even a fractional change indicates a redox process.

Question E.What is oxidation? Which one of the following pairs of species is in its oxidized state ?
a. Mg / Mg2+
b. Cu / Cu2+
c. O2/O2-
d. Cl2 / Cl\(^-\)
Answer:
a. Mg / Mg2+ Here, Mg loses two electrons to form Mg2+ ion. Mg(s) \(\rightarrow\) Mg\(^{2+}\)(aq) + 2e- Hence, Mg / Mg2+ is an oxidized state.
b. Cu/Cu2+ Here, Cu loses two electrons to form Cu2+ ion. Cu(s) \(\rightarrow\) Cu\(^{2+}\)(aq) + 2e- Hence, Cu/Cu2+ is in an oxidized state.
c. O2 / O2- Here, each O gains two electrons to form O\(^{2-}\) ion. O2(g) + 4e\(^-\) \(\rightarrow\) 2O\(^{2-}\)(aq) Hence, O2 / O2- is in a reduced state.
d. Cl2 / Cl\(^-\) Here, each Cl gains one electron to form Cl\(^-\) ion. Cl2(g) + 2e\(^-\) \(\rightarrow\) 2Cl\(^-\)(aq) Hence, Cl2 / Cl\(^-\) is in a reduced state.
In simple words: Oxidation is the loss of electrons or an increase in oxidation state. Pairs like Mg/Mg\(^{2+}\) and Cu/Cu\(^{2+}\) show the oxidized state because the metal atoms have lost electrons to become cations.

🎯 Exam Tip: A species is in its oxidized state if its oxidation number has increased (indicating electron loss) compared to its elemental or initial state. Look for positive charges on metals as a common indicator.

Question F.Justify the following reaction as redox reaction. 2 Na(s) + S(s) \(\rightarrow\) Na2S(s) Find out the oxidizing and reducing agents.
Answer:
i. Redox reaction can be described as electron transfer as shown below: 2Na(s) + S(s) \(\rightarrow\) 2Na\(^+\) + S\(^{2-}\)
ii. Charge development suggests that each sodium atom loses one electron to form Na\(^+\) and sulphur atom gains two electrons to form S\(^{2-}\). This can be represented as follows:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह 2Na(s) + S(s) \(\rightarrow\) 2Na⁺ + S²⁻ अभिक्रिया में इलेक्ट्रॉन हस्तांतरण को दर्शाता है। सोडियम (Na) इलेक्ट्रॉन खो देता है (Loss of e⁻) और Na⁺ बन जाता है, जबकि सल्फर (S) इलेक्ट्रॉन प्राप्त कर लेता है (Gain of e⁻) और S²⁻ बन जाता है।
\[\begin{array}{c} \text{Loss of e}^- \\ \downarrow \\ 2Na_{(s)} + S_{(s)} \rightarrow 2Na^+ + S^{2-} \\ \uparrow \\ \text{Gain of e}^- \end{array}\]
iii. When Na is oxidised to Na2S, the neutral Na atom loses electrons to form Na\(^+\) in Na2S while the elemental sulphur gains electrons and forms S\(^{2-}\) in Na2S.
iv. Each of the above steps represents a half reaction which involves electron transfer (loss or gain).
v. Sum of these two half reactions or the overall reaction is a redox reaction.
vi. Oxidising agent is an electron acceptor and hence, S is an oxidising agent. Reducing agent is an electron donor and hence, Na is a reducing agent.
In simple words: The reaction between sodium and sulfur is a redox reaction because sodium loses electrons (oxidized, acting as reducing agent) and sulfur gains electrons (reduced, acting as oxidizing agent).

🎯 Exam Tip: Always clearly identify the electron donor (reducing agent) and electron acceptor (oxidizing agent) by tracking the change in oxidation states or explicit electron transfer. This demonstrates full understanding of the redox concept.

Question G.Provide the stock notation for the following compounds: HAuCl4, Tl2O, FeO, Fe2O3, MnO and CuO.
Answer:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह विभिन्न यौगिकों के लिए स्टॉक नोटेशन को दर्शाता है, जहाँ धातु परमाणु की ऑक्सीकरण अवस्था को रोमन अंकों में इंगित किया जाता है। HAuCl4 में Au की ऑक्सीकरण अवस्था +3 है, इसलिए इसे HAu(III)Cl4 लिखा जाता है। Tl2O में Tl की ऑक्सीकरण अवस्था +1 है, इसलिए इसे Tl2(I)O लिखा जाता है। FeO में Fe की ऑक्सीकरण अवस्था +2 है, इसलिए इसे Fe(II)O लिखा जाता है। Fe2O3 में Fe की ऑक्सीकरण अवस्था +3 है, इसलिए इसे Fe2(III)O3 लिखा जाता है। MnO में Mn की ऑक्सीकरण अवस्था +2 है, इसलिए इसे Mn(II)O लिखा जाता है। CuO में Cu की ऑक्सीकरण अवस्था +2 है, इसलिए इसे Cu(II)O लिखा जाता है।
(i) \(\begin{array}{c} +1 +3 -1 \\ HAuCl_4 \end{array} \implies HAu(III)Cl_4\)
(ii) \(\begin{array}{c} +1 -2 \\ Tl_2O \end{array} \implies Tl_2(I)O\)
(iii) \(\begin{array}{c} +2 -2 \\ FeO \end{array} \implies Fe(II)O\)
(iv) \(\begin{array}{c} +3 -2 \\ Fe_2O_3 \end{array} \implies Fe_2(III)O_3\)
(v) \(\begin{array}{c} +2 -2 \\ MnO \end{array} \implies Mn(II)O\)
(vi) \(\begin{array}{c} +2 -2 \\ CuO \end{array} \implies Cu(II)O\)
In simple words: Stock notation uses Roman numerals in parentheses to indicate the oxidation state of the central metal atom in compounds, ensuring clarity for metals with variable oxidation states.

🎯 Exam Tip: To determine the stock notation, first calculate the oxidation state of the metal using the known oxidation states of other elements (like H, O, Cl), then express it in Roman numerals. For anions, write their full name or common abbreviation.

Question H.Assign oxidation number to each atom in the following species.
a. Cr(OH)4\(^-\)
b. Na2S2O3
c. H3BO3
Answer:
(i) Cr(OH)4\(^-\) Oxidation number of O = -2 Oxidation number of H = +1 Cr(OH)4\(^-\) is an ionic species.
\[ \therefore \text{ Sum of the oxidation numbers of all atoms} = -1 \]
\[ \therefore \text{ Oxidation number of Cr} + 4 \times (\text{Oxidation number of O}) + 4 \times (\text{Oxidation number of H}) = -1 \]
\[ \therefore \text{ Oxidation number of Cr} + 4 \times (-2) + 4 \times (+1) = -1 \]
\[ \therefore \text{ Oxidation number of Cr} - 8 + 4 = -1 \]
\[ \therefore \text{ Oxidation number of Cr} - 4 = -1 \]
\[ \therefore \text{ Oxidation number of Cr} = -1 + 4 \]
\[ \therefore \text{ Oxidation number of Cr in } Cr(OH)_4^- = +3 \]
(ii) Na2S2O3 Oxidation number of Na = +1 Oxidation number of O = -2 Na2S2O3 is a neutral molecule.
\[ \therefore \text{ Sum of the oxidation numbers of all atoms} = 0 \]
\[ \therefore 2 \times (\text{Oxidation number of Na}) + 2 \times (\text{Oxidation number of S}) + 3 \times (\text{Oxidation number of O}) = 0 \]
\[ \therefore 2 \times (+1) + 2 \times (\text{Oxidation number of S}) + 3 \times (-2) = 0 \]
\[ \therefore 2 \times (\text{Oxidation number of S}) + 2 - 6 = 0 \]
\[ \therefore 2 \times (\text{Oxidation number of S}) = +4 \]
\[ \therefore \text{ Oxidation number of S} = \frac{+4}{2} = +2 \]
\[ \therefore \text{ Oxidation number of S in } Na_2S_2O_3 = +2 \]
(iii) H3BO3 Oxidation number of H = +1 Oxidation number of O = -2 H3BO3 is a neutral molecule.
\[ \therefore \text{ Sum of the oxidation numbers of all atoms} = 0 \]
\[ \therefore 3 \times (\text{Oxidation number of H}) + (\text{Oxidation number of B}) + 3 \times (\text{Oxidation number of O}) = 0 \]
\[ \therefore 3 \times (+1) + (\text{Oxidation number of B}) + 3 \times (-2) = 0 \]
\[ \therefore \text{ Oxidation number of B} + 3 - 6 = 0 \]
\[ \therefore \text{ Oxidation number of B in } H_3BO_3 = +3 \]
In simple words: The oxidation number for each underlined atom is determined by setting the sum of all oxidation states in the compound or ion equal to its overall charge, using known values for common elements.

🎯 Exam Tip: For polyatomic ions, remember to equate the sum of oxidation numbers to the net charge of the ion, not zero. For neutral compounds, the sum is always zero.

Question I.Which of the following redox couple is stronger oxidizing agent ?
a. Cl2 (Eº = 1.36 V) and Br2 (E° = 1.09 V)
b. MnO4\(^-\) (E0 = 1.51 V) and Cr2O7\(^{2-}\) (E° = 1.33 V)
Answer:
a. Cl2 has a larger positive value of Eº than Br2. Thus, Cl2 is a stronger oxidizing agent than Br2.
b. MnO4\(^-\) has larger positive value of Eº than Cr2O7\(^{2-}\). Thus, MnO4\(^-\) is stronger oxidizing agent than Cr2O7\(^{2-}\).
In simple words: A stronger oxidizing agent corresponds to a higher positive standard electrode potential (E°), as it has a greater tendency to be reduced and thus oxidize other species.

🎯 Exam Tip: Higher positive standard reduction potential (E°) values indicate a stronger oxidizing agent. Conversely, lower (more negative) E° values indicate a stronger reducing agent.

Question J.Which of the following redox couple is stronger reducing agent ?
a. Li (E° = -3.05 V) and Mg(E° = -2.36 V)
b. Zn(E° = -0.76 V) and Fe(E° = -0.44 V)
Answer:
a. Li has a larger negative value of Eº than Mg. Thus, Li is a stronger reducing agent than Mg.
b. Zn has a larger negative value of Eº than Fe. Thus, Zn is a stronger reducing agent than Fe.
In simple words: A stronger reducing agent is indicated by a more negative standard electrode potential (E°), meaning it has a greater tendency to be oxidized and thus reduce other species.

🎯 Exam Tip: A substance with a more negative standard reduction potential (or a more positive standard oxidation potential) is a stronger reducing agent because it readily loses electrons.

Exercise 4. Balance The Reactions/Equations :

Question A.Balance the following reactions by oxidation number method
a. Cr2O7\(^{2-}\)(aq) + SO3\(^{2-}\)(aq) \(\rightarrow\) Cr\(^{3+}\)(aq) + SO4\(^{2-}\)(aq) (acidic)
b. MnO4\(^-\)(aq) + Br\(^-\)(aq) \(\rightarrow\) MnO2(s) + BrO3\(^-\)(aq) (basic)
c. H2SO4 (aq) + C (s) \(\rightarrow\) CO2 (g) + SO2 (g) + H2O (l) (acidic)
d. Bi (OH)3 (g) + Sn(OH)3\(^-\)(aq) \(\rightarrow\) Bi (s) + Sn (OH)6\(^{2-}\)(aq) (basic)
Answer:
(i) Cr2O7\(^{2-}\)(aq) + SO3\(^{2-}\)(aq) \(\rightarrow\) Cr\(^{3+}\)(aq) + SO4\(^{2-}\)(aq) (acidic)
Step 1: Write skeletal equation and balance the elements other than O and H. Cr2O7\(^{2-}\)(aq) + SO3\(^{2-}\)(aq) \(\rightarrow\) 2Cr\(^{3+}\)(aq) + SO4\(^{2-}\)(aq)
Step 2: Assign oxidation number to Cr and S. Calculate the increase and decrease in the oxidation number and make them equal.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह Cr2O7\(^{2-}\) + SO3\(^{2-}\) \(\rightarrow\) 2Cr\(^{3+}\) + SO4\(^{2-}\) अभिक्रिया में क्रोमियम और सल्फर की ऑक्सीकरण अवस्थाओं में परिवर्तन को दर्शाता है। Cr2O7\(^{2-}\) में Cr की ऑक्सीकरण अवस्था +6 है, जो 2Cr\(^{3+}\) में +3 तक घट जाती है (प्रति परमाणु 3 की कमी, कुल 6 की कमी)। SO3\(^{2-}\) में S की ऑक्सीकरण अवस्था +4 है, जो SO4\(^{2-}\) में +6 तक बढ़ जाती है (प्रति परमाणु 2 की वृद्धि)।
\[\begin{array}{cc} Cr_2O_7^{2-}{}_{(aq)} + SO_3^{2-}{}_{(aq)} & \rightarrow 2Cr^{3+}{}_{(aq)} + SO_4^{2-}{}_{(aq)} \\ \uparrow\uparrow \quad \uparrow\uparrow & \quad \uparrow \quad \uparrow \\ +6\, -2 \quad +4\, -2 & \quad +3 \quad +6 \end{array}\] Increase in oxidation number: \(\begin{array}{rcl} SO_3^{2-} & \rightarrow & SO_4^{2-} \\ \uparrow & & \uparrow \\ +4 & & +6 \end{array}\) (Increase per atom = 2) Decrease in oxidation number: \(\begin{array}{rcl} Cr_2O_7^{2-} & \rightarrow & 2Cr^{3+} \\ \uparrow & & \uparrow \\ +6 & & +3 \end{array}\) (Decrease per atom = 3) To make the net increase and decrease equal, we must take 3 atoms of S and 2 atoms of Cr. (There are already 2 Cr atoms.)
Step 3: Balance 'O' atoms by adding 4H2O to the right-hand side. Cr2O7\(^{2-}\)(aq) + 3SO3\(^{2-}\)(aq) \(\rightarrow\) 2Cr\(^{3+}\)(aq) + 3SO4\(^{2-}\)(aq) + 4H2O(l)
Step 4: The medium is acidic. To make the charges and hydrogen atoms on the two sides equal, add 8H\(^+\) on the left-hand side. Cr2O7\(^{2-}\)(aq) + 3SO3\(^{2-}\)(aq) + 8H\(^+\)(aq) \(\rightarrow\) 2Cr\(^{3+}\)(aq) + 3SO4\(^{2-}\)(aq) + 4H2O(l)
Step 5: Check two sides for balance of atoms and charges. Hence, balanced equation: Cr2O7\(^{2-}\)(aq) + 3SO3\(^{2-}\)(aq) + 8H\(^+\)(aq) \(\rightarrow\) 2Cr\(^{3+}\)(aq) + 3SO4\(^{2-}\)(aq) + 4H2O(l)
In simple words: To balance this acidic redox reaction, we first balance atoms other than O and H, then equate the total increase and decrease in oxidation numbers by adjusting coefficients, and finally balance oxygen with water and hydrogen with H\(^+\) ions.

🎯 Exam Tip: The oxidation number method for balancing reactions in acidic medium involves 5 steps: skeletal equation, oxidation numbers, balancing O (with H2O), balancing H (with H\(^+\)), and finally verifying charges and atoms.

(ii) MnO4\(^-\)(aq) + Br\(^-\)(aq) \(\rightarrow\) MnO2(s) + BrO3\(^-\)(aq) (basic)
Step 1: Write skeletal equation and balance the elements other than O and H. MnO4\(^-\)(aq) + Br\(^-\)(aq) \(\rightarrow\) MnO2(s) + BrO3\(^-\)(aq)
Step 2: Assign oxidation number to Mn and Br. Calculate the increase and decrease in the oxidation number and make them equal.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह MnO4\(^-\) + Br\(^-\) \(\rightarrow\) MnO2 + BrO3\(^-\) अभिक्रिया में मैंगनीज और ब्रोमीन की ऑक्सीकरण अवस्थाओं में परिवर्तन को दर्शाता है। MnO4\(^-\) में Mn की ऑक्सीकरण अवस्था +7 है, जो MnO2 में +4 तक घट जाती है (3 की कमी)। Br\(^-\) में Br की ऑक्सीकरण अवस्था -1 है, जो BrO3\(^-\) में +5 तक बढ़ जाती है (6 की वृद्धि)।
\[\begin{array}{cc} MnO_4^-{}_{(aq)} + Br^-{}_{(aq)} & \rightarrow MnO_{2(s)} + BrO_3^-{}_{(aq)} \\ \uparrow\uparrow \quad \uparrow & \quad \uparrow \quad \uparrow \\ +7\, -2 \quad -1 & \quad +4 \quad +5 \end{array}\] Increase in oxidation number: \(\begin{array}{rcl} Br^- & \rightarrow & BrO_3^- \\ \uparrow & & \uparrow \\ -1 & & +5 \end{array}\) (Increase per atom = 6) Decrease in oxidation number: \(\begin{array}{rcl} MnO_4^- & \rightarrow & MnO_2 \\ \uparrow & & \uparrow \\ +7 & & +4 \end{array}\) (Decrease per atom = 3) To make the net increase and decrease equal, we must take 2 atoms of Mn. 2MnO4\(^-\)(aq) + Br\(^-\)(aq) \(\rightarrow\) 2MnO2(s) + BrO3\(^-\)(aq)
Step 3: Balance 'O' atoms by adding H2O to the right-hand side. 2MnO4\(^-\)(aq) + Br\(^-\)(aq) \(\rightarrow\) 2MnO2(s) + BrO3\(^-\)(aq) + H2O(l)
Step 4: The medium is basic. To make the charges and hydrogen atoms on the two sides equal, add 2H\(^+\) on the left-hand side. 2MnO4\(^-\)(aq) + Br\(^-\)(aq) + 2H\(^+\)(aq) \(\rightarrow\) 2MnO2(s) + BrO3\(^-\)(aq) + H2O(l) Add OH\(^-\) ions equal to the number of H\(^+\) ions on both sides of the equation. 2MnO4\(^-\)(aq) + Br\(^-\)(aq) + 2H\(^+\)(aq) + 2OH\(^-\)(aq) \(\rightarrow\) 2MnO2(s) + BrO3\(^-\)(aq) + H2O(l) + 2OH\(^-\)(aq) The H\(^+\) and OH\(^-\) ions appearing on the same side of the reaction are combined to give H2O molecules. 2MnO4\(^-\)(aq) + Br\(^-\)(aq) + 2H2O(l) \(\rightarrow\) 2MnO2(s) + BrO3\(^-\)(aq) + H2O(l) + 2OH\(^-\)(aq) Simplify H2O: 2MnO4\(^-\)(aq) + Br\(^-\)(aq) + H2O(l) \(\rightarrow\) 2MnO2(s) + BrO3\(^-\)(aq) + 2OH\(^-\)(aq)
Step 5: Check two sides for balance of atoms and charges. Hence, balanced equation: 2MnO4\(^-\)(aq) + Br\(^-\)(aq) + H2O(l) \(\rightarrow\) 2MnO2(s) + BrO3\(^-\)(aq) + 2OH\(^-\)(aq)
In simple words: Balancing this basic redox reaction involves equating changes in oxidation numbers for Mn and Br, then adding water to balance oxygen, and finally adding H\(^+\) and OH\(^-\) to balance hydrogen and overall charge, simplifying any H2O formations.

🎯 Exam Tip: When balancing in basic medium, remember to add H\(^+\) ions first, then add an equal number of OH\(^-\) ions to both sides, and combine H\(^+\) and OH\(^-\) to form H2O. Finally, simplify the H2O molecules.

iii. H2SO4(aq) + C(s) → CO2(g) + SO2(g) + H2O(l) (acidic)
Step 1: Write skeletal equation and balance the elements other than O and H.
\( \text{H}_2\text{SO}_4(\text{aq}) + \text{C}(\text{s}) \implies \text{CO}_2(\text{g}) + \text{SO}_2(\text{g}) + \text{H}_2\text{O}(\text{l}) \)
Step 2: Assign oxidation number to S and C. Calculate the increase and decrease in the oxidation number and make them equal.
\( \text{H}_2\text{SO}_4(\text{aq}) + \text{C}(\text{s}) \implies \text{CO}_2(\text{g}) + \text{SO}_2(\text{g}) + \text{H}_2\text{O}(\text{l}) \)

ℹ️ चित्र व्याख्या (Diagram Explanation): इस रासायनिक समीकरण के नीचे परमाणुओं के ऑक्सीकरण अंक दिए गए हैं। हाइड्रोजन का +1, सल्फर का +6, और ऑक्सीजन का -2 है। कार्बन का ऑक्सीकरण अंक 0 से +4 में बदलता है, जबकि सल्फर का +6 से +4 में बदलता है।
In simple words: This section shows how to balance a redox reaction in acidic medium by tracking the changes in oxidation numbers of sulfur and carbon. The goal is to make the total increase in oxidation number equal to the total decrease by adjusting coefficients.

🎯 Exam Tip: Accurately identifying oxidation states for all atoms in reactants and products is a critical first step for balancing redox reactions.

Increase in oxidation number: C → CO2
(Increase per atom = 4)
Decrease in oxidation number: H2SO4 → SO2
(Decrease per atom = 2)
To make the net increase and decrease equal, we must take 2 atoms of S.
\( 2\text{H}_2\text{SO}_4(\text{aq}) + \text{C}(\text{s}) \implies \text{CO}_2(\text{g}) + 2\text{SO}_2(\text{g}) + \text{H}_2\text{O}(\text{l}) \)
Step 3: Balance 'O' atoms by adding H2O to the right-hand side.
\( 2\text{H}_2\text{SO}_4(\text{aq}) + \text{C}(\text{s}) \implies \text{CO}_2(\text{g}) + 2\text{SO}_2(\text{g}) + \text{H}_2\text{O}(\text{l}) + \text{H}_2\text{O}(\text{l}) \)
Step 4: The medium is acidic. There is no charge on either side. Hydrogen atoms are equal on both side.
\( 2\text{H}_2\text{SO}_4(\text{aq}) + \text{C}(\text{s}) \implies \text{CO}_2(\text{g}) + 2\text{SO}_2(\text{g}) + \text{H}_2\text{O}(\text{l}) \)
Step 5: Check two sides for balance of atoms and charges.
Hence, balanced equation: \( 2\text{H}_2\text{SO}_4(\text{aq}) + \text{C}(\text{s}) \implies \text{CO}_2(\text{g}) + 2\text{SO}_2(\text{g}) + \text{H}_2\text{O}(\text{l}) \)

In simple words: After determining the oxidation changes, balance the overall reaction by ensuring the number of atoms for each element and the total charge are equal on both sides of the equation.

🎯 Exam Tip: Always double-check atom count and total charge on both sides of the equation in the final step to ensure proper balancing.

iv. Bi(OH)3(s) + Sn(OH)3-(aq) → Bi(s) + Sn(OH)62-(aq) (basic)
Step 1: Write skeletal equation and balance the elements other than O and H.
\( \text{Bi}(\text{OH})_3(\text{s}) + \text{Sn}(\text{OH})_3^-(\text{aq}) \implies \text{Bi}(\text{s}) + \text{Sn}(\text{OH})_6^{2-}(\text{aq}) \)
Step 2: Assign oxidation numbers to Bi and Sn. Calculate the increase and decrease in the oxidation number and make them equal.
\( \text{Bi}(\text{OH})_3(\text{s}) + \text{Sn}(\text{OH})_3^-(\text{aq}) \implies \text{Bi}(\text{s}) + \text{Sn}(\text{OH})_6^{2-}(\text{aq}) \)

ℹ️ चित्र व्याख्या (Diagram Explanation): यहाँ बिस्मथ का ऑक्सीकरण अंक +3 से 0 तक घटता है, जो अपचयन को दर्शाता है। टिन का ऑक्सीकरण अंक +2 से +4 तक बढ़ता है, जो ऑक्सीकरण को दर्शाता है। ये परिवर्तन रेडॉक्स प्रतिक्रिया की पुष्टि करते हैं।
In simple words: This part details balancing a redox reaction in a basic medium, specifically involving bismuth and tin compounds, by equalizing the changes in their oxidation numbers.

🎯 Exam Tip: Remember that balancing redox reactions in basic media often involves adding OH- ions and H2O to balance hydrogen and oxygen atoms.

Increase in oxidation number: Sn(OH)3- → Sn(OH)62-
(Increase per atom = 2)
Decrease in oxidation number: Bi(OH)3 → Bi
(Decrease per atom = 3)
To make the net increase and decrease equal, we must take 3 atoms of Sn and 2 atoms of Bi.
\( 2\text{Bi}(\text{OH})_3(\text{s}) + 3\text{Sn}(\text{OH})_3^-(\text{aq}) \implies 2\text{Bi}(\text{s}) + 3\text{Sn}(\text{OH})_6^{2-}(\text{aq}) \)
Step 3: Balance 'O' atoms by adding 3H2O to the left-hand side.
\( 2\text{Bi}(\text{OH})_3(\text{s}) + 3\text{Sn}(\text{OH})_3^-(\text{aq}) + 3\text{H}_2\text{O}(\text{l}) \implies 2\text{Bi}(\text{s}) + 3\text{Sn}(\text{OH})_6^{2-}(\text{aq}) \)
Step 4: The medium is basic. To make hydrogen atoms on the two sides equal, add 3H+ on the right-hand side.
\( 2\text{Bi}(\text{OH})_3(\text{s}) + 3\text{Sn}(\text{OH})_3^-(\text{aq}) + 3\text{H}_2\text{O}(\text{l}) \implies 2\text{Bi}(\text{s}) + 3\text{Sn}(\text{OH})_6^{2-}(\text{aq}) + 3\text{H}^+(\text{aq}) \)
Add OH- ions equal to the number of H+ ions on both sides of the equation.
\( 2\text{Bi}(\text{OH})_3(\text{s}) + 3\text{Sn}(\text{OH})_3^-(\text{aq}) + 3\text{H}_2\text{O}(\text{l}) + 3\text{OH}^-(\text{aq}) \implies 2\text{Bi}(\text{s}) + 3\text{Sn}(\text{OH})_6^{2-}(\text{aq}) + 3\text{H}^+(\text{aq}) + 3\text{OH}^-(\text{aq}) \)
The H+ and OH- ions appearing on the same side of the reaction are combined to give H2O molecules.
\( 2\text{Bi}(\text{OH})_3(\text{s}) + 3\text{Sn}(\text{OH})_3^-(\text{aq}) + 3\text{H}_2\text{O}(\text{l}) + 3\text{OH}^-(\text{aq}) \implies 4\text{Bi}(\text{s}) + 3\text{Sn}(\text{OH})_6^{2-}(\text{aq}) + 3\text{H}_2\text{O}(\text{l}) \)
\( 2\text{Bi}(\text{OH})_3(\text{s}) + 3\text{Sn}(\text{OH})_3^-(\text{aq}) + 3\text{OH}^-(\text{aq}) \implies 2\text{Bi}(\text{s}) + 3\text{Sn}(\text{OH})_6^{2-}(\text{aq}) \)
Step 5: Check two sides for balance of atoms and charges.
Hence, balanced equation: \( 2\text{Bi}(\text{OH})_3(\text{s}) + 3\text{Sn}(\text{OH})_3^-(\text{aq}) + 3\text{OH}^-(\text{aq}) \implies 2\text{Bi}(\text{s}) + 3\text{Sn}(\text{OH})_6^{2-}(\text{aq}) \)

In simple words: This process systematically balances the redox reaction by ensuring that the number of electrons lost during oxidation equals the number gained during reduction, and then adjusting water and hydroxide ions for basic conditions.

🎯 Exam Tip: Balancing H and O atoms correctly in basic media requires careful addition of H2O and OH- ions; always cancel out H+ and OH- forming H2O if they appear on the same side.

Question B. Balance the following redox equation by half reaction method
a. H2C2O4 (aq) + MnO4- (aq) → CO2(g) + Mn2+ (aq) (acidic)
b. Bi (OH)3 (s) +SnO22- (aq) → SnO32- (aq) + Bi (s) (basic)
Answer:
i. \( \text{H}_2\text{C}_2\text{O}_4(\text{aq}) + \text{MnO}_4^-(\text{aq}) \implies \text{CO}_2(\text{g}) + \text{Mn}^{2+}(\text{aq}) \)
Step 1: Write unbalanced equation for the redox reaction. Assign oxidation number to all the atoms in reactants and products. Divide the equation into two half equations.
\( \text{H}_2\text{C}_2\text{O}_4(\text{aq}) + \text{MnO}_4^-(\text{aq}) \implies \text{CO}_2(\text{g}) + \text{Mn}^{2+}(\text{aq}) \)

ℹ️ चित्र व्याख्या (Diagram Explanation): इस आरेख में, H2C2O4 में कार्बन का ऑक्सीकरण अंक +3 से CO2 में +4 तक बढ़ता है, जो इलेक्ट्रॉन हानि (ऑक्सीकरण) को दर्शाता है। वहीं, MnO4- में मैंगनीज का ऑक्सीकरण अंक +7 से Mn2+ में +2 तक घटता है, जो इलेक्ट्रॉन लाभ (अपचयन) को दर्शाता है।
In simple words: This problem asks to balance two redox reactions using the half-reaction method, one in acidic and one in basic medium, by breaking them down into oxidation and reduction steps.

🎯 Exam Tip: The half-reaction method systematically separates the oxidation and reduction processes, making complex reactions easier to balance, especially in different media.

Oxidation half reaction: \( \text{H}_2\text{C}_2\text{O}_4(\text{aq}) \implies \text{CO}_2(\text{g}) \)
Reduction half reaction: \( \text{MnO}_4^-(\text{aq}) \implies \text{Mn}^{2+}(\text{aq}) \)
Step 2: Balance the atoms except O and H in each half equation. Balance half equation for O atoms by adding 4H2O to the right side of reduction half equation.
Oxidation: \( \text{H}_2\text{C}_2\text{O}_4(\text{aq}) \implies 2\text{CO}_2(\text{g}) \)
Reduction: \( \text{MnO}_4^-(\text{aq}) \implies \text{Mn}^{2+}(\text{aq}) + 4\text{H}_2\text{O}(\text{l}) \)
Step 3: Balance H atoms by adding H+ ions to the side with less H. Hence, add 2H+ ions to the right side of oxidation half equation and 8H+ ions to the left side of reduction half equation.
Oxidation: \( \text{H}_2\text{C}_2\text{O}_4(\text{aq}) \implies 2\text{CO}_2(\text{g}) + 2\text{H}^+(\text{aq}) \)
Reduction: \( \text{MnO}_4^-(\text{aq}) + 8\text{H}^+(\text{aq}) \implies \text{Mn}^{2+}(\text{aq}) + 4\text{H}_2\text{O}(\text{l}) \)
Step 4: Now add 2 electrons to the right side of oxidation half equation and 5 electrons to the left side of reduction half equation to balance the charges.
Oxidation: \( \text{H}_2\text{C}_2\text{O}_4(\text{aq}) \implies 2\text{CO}_2(\text{g}) + 2\text{H}^+(\text{aq}) + 2\text{e}^- \)
Reduction: \( \text{MnO}_4^-(\text{aq}) + 8\text{H}^+(\text{aq}) + 5\text{e}^- \implies \text{Mn}^{2+}(\text{aq}) + 4\text{H}_2\text{O}(\text{l}) \)
Step 5: Multiply oxidation half equation by 5 and reduction half equation by 2 to equalize number of electrons in two half equations. Then add two half equation.
Oxidation: \( 5\text{H}_2\text{C}_2\text{O}_4(\text{aq}) \implies 10\text{CO}_2(\text{g}) + 10\text{H}^+(\text{aq}) + 10\text{e}^- \)
Reduction: \( 2\text{MnO}_4^-(\text{aq}) + 16\text{H}^+(\text{aq}) + 10\text{e}^- \implies 2\text{Mn}^{2+}(\text{aq}) + 8\text{H}_2\text{O}(\text{l}) \)
Add two half equations:
\( 5\text{H}_2\text{C}_2\text{O}_4(\text{aq}) + 2\text{MnO}_4^-(\text{aq}) + 6\text{H}^+(\text{aq}) \implies 10\text{CO}_2(\text{g}) + 2\text{Mn}^{2+}(\text{aq}) + 8\text{H}_2\text{O}(\text{l}) \)
The equation is balanced in terms of number of atoms and the charges.
Hence, balanced equation: \( 5\text{H}_2\text{C}_2\text{O}_4(\text{aq}) + 2\text{MnO}_4^-(\text{aq}) + 6\text{H}^+(\text{aq}) \implies 10\text{CO}_2(\text{g}) + 2\text{Mn}^{2+}(\text{aq}) + 8\text{H}_2\text{O}(\text{l}) \)

In simple words: This detailed process balances the acidic redox reaction by first separating it into oxidation and reduction half-reactions, then balancing atoms (O with H2O, H with H+), charges with electrons, and finally combining them after multiplying to equalize electrons.

🎯 Exam Tip: Pay close attention to the medium (acidic or basic) as it dictates whether you use H+ or OH- and H2O to balance hydrogen and oxygen atoms.

ii. Bi(OH)3(s) + SnO22-(aq) → SnO32-(aq) + Bi(s)
Step 1: Write unbalanced equation for the redox reaction. Assign oxidation number to all the atoms in reactants and products. Divide the equation into two half equations.
\( \text{Bi}(\text{OH})_3(\text{s}) + \text{SnO}_2^{2-}(\text{aq}) \implies \text{SnO}_3^{2-}(\text{aq}) + \text{Bi}(\text{s}) \)

ℹ️ चित्र व्याख्या (Diagram Explanation): बिस्मथ हाइड्रोक्साइड में बिस्मथ का ऑक्सीकरण अंक +3 है, जो उत्पाद बिस्मथ धातु में 0 हो जाता है, जिससे इलेक्ट्रॉन लाभ (अपचयन) होता है। स्टैनिट आयन में टिन का ऑक्सीकरण अंक +2 है, जो स्टैनेट आयन में +4 हो जाता है, जिससे इलेक्ट्रॉन हानि (ऑक्सीकरण) होती है।
In simple words: This problem involves balancing a redox reaction in a basic medium by identifying oxidation number changes and then separating the reaction into oxidation and reduction half-reactions.

🎯 Exam Tip: When setting up half-reactions, ensure all atoms involved in the oxidation or reduction are correctly represented before balancing oxygen and hydrogen.

Oxidation half reaction: \( \text{SnO}_2^{2-}(\text{aq}) \implies \text{SnO}_3^{2-}(\text{aq}) \)
Reduction half reaction: \( \text{Bi}(\text{OH})_3(\text{s}) \implies \text{Bi}(\text{s}) \)
Step 2: Balance half equations for O atoms by adding H2O to the side with less O atoms. Add 1H2O to left side of oxidation half equation and 3H2O to the right side of reduction half equation.
Oxidation: \( \text{SnO}_2^{2-}(\text{aq}) + \text{H}_2\text{O}(\text{l}) \implies \text{SnO}_3^{2-}(\text{aq}) \)
Reduction: \( \text{Bi}(\text{OH})_3(\text{s}) \implies \text{Bi}(\text{s}) + 3\text{H}_2\text{O}(\text{l}) \)
Step 3: Balance H atoms by adding H+ ions to the side with less H. Hence, add 2H+ ions to the right side of oxidation half equation and 3H+ ions to the left side of reduction half equation.
Oxidation: \( \text{SnO}_2^{2-}(\text{aq}) + \text{H}_2\text{O}(\text{l}) \implies \text{SnO}_3^{2-}(\text{aq}) + 2\text{H}^+(\text{aq}) \)
Reduction: \( \text{Bi}(\text{OH})_3(\text{s}) + 3\text{H}^+(\text{aq}) \implies \text{Bi}(\text{s}) + 3\text{H}_2\text{O}(\text{l}) \)
Step 4: Now add 2 electrons to the right side of oxidation half equation and 3 electrons to the left side of reduction half equation to balance the charges.
Oxidation: \( \text{SnO}_2^{2-}(\text{aq}) + \text{H}_2\text{O}(\text{l}) \implies \text{SnO}_3^{2-}(\text{aq}) + 2\text{H}^+(\text{aq}) + 2\text{e}^- \)
Reduction: \( \text{Bi}(\text{OH})_3(\text{s}) + 3\text{H}^+(\text{aq}) + 3\text{e}^- \implies \text{Bi}(\text{s}) + 3\text{H}_2\text{O}(\text{l}) \)
Step 5: Multiply oxidation half equation by 3 reduction half equation by 2 to equalize number of electrons in two half equations. Then add two half equation.
Oxidation: \( 3\text{SnO}_2^{2-}(\text{aq}) + 3\text{H}_2\text{O}(\text{l}) \implies 3\text{SnO}_3^{2-}(\text{aq}) + 6\text{H}^+(\text{aq}) + 6\text{e}^- \)
Reduction: \( 2\text{Bi}(\text{OH})_3(\text{s}) + 6\text{H}^+(\text{aq}) + 6\text{e}^- \implies 2\text{Bi}(\text{s}) + 6\text{H}_2\text{O}(\text{l}) \)
Add two half equations:
\( 2\text{Bi}(\text{OH})_3(\text{s}) + 3\text{SnO}_2^{2-}(\text{aq}) \implies 3\text{SnO}_3^{2-}(\text{aq}) + 2\text{Bi}(\text{s}) + 3\text{H}_2\text{O}(\text{l}) \)
Reaction occurs in basic medium. However, H+ ions cancel out and the reaction is balanced. Hence, no need to add OH- ions. The equation is balanced in terms of number of atoms and the charges.
Hence, balanced equation:
\( 2\text{Bi}(\text{OH})_3(\text{s}) + 3\text{SnO}_2^{2-}(\text{aq}) \implies 3\text{SnO}_3^{2-}(\text{aq}) + 2\text{Bi}(\text{s}) + 3\text{H}_2\text{O}(\text{l}) \)

In simple words: This segment outlines the step-by-step balancing of a redox reaction in a basic medium, ensuring atomic and charge balance through the addition of water, hydrogen ions (which then cancel out), and electrons.

🎯 Exam Tip: When H+ and OH- appear on opposite sides, they form H2O; if on the same side, cancel out the common species. Ensure the final equation reflects the correct medium.

Question 5. Complete the following table :
Assign oxidation number to the underlined species and write Stock notation of compound

🎯 Exam Tip: Stock notation is crucial for clearly communicating the oxidation state of a metal in its compound, especially when the metal can exhibit multiple oxidation states.

11th Chemistry Digest Chapter 6 Redox Reactions Intext Questions and Answers

Can you tell? (Textbook Page No. 81)

Question i. Why does cut apple turn brown when exposed to air?
Answer: Cut apple turns brown when exposed to air because polyphenols are released. These polyphenols undergo oxidation in the presence of air and impart brown colour.
In simple words: Apples turn brown because chemicals called polyphenols react with oxygen in the air, a process called oxidation, creating brown-colored compounds.

🎯 Exam Tip: This is a common example of an oxidation reaction visible in everyday life, often used to introduce redox concepts.

Question ii. Why does old car bumper change colour?
Answer: Car bumper is made of iron which undergoes rusting over a period of time. Hence, old car bumper changes colour.
In simple words: Car bumpers, if made of iron, change color over time due to rusting, which is an oxidation reaction where iron reacts with oxygen and moisture to form iron oxides.

🎯 Exam Tip: Rusting is a prime example of corrosion, a redox process where metals slowly react with their environment.

Question iii. Why do new batteries become useless after some days?
Answer: Batteries generate electricity by redox reactions. Once the chemicals taking part in redox reaction are used up, the battery cannot generate power. Hence, new batteries become useless after some days.
In simple words: Batteries stop working when the chemicals inside them, which power the redox reactions producing electricity, are fully consumed.

🎯 Exam Tip: Understanding that battery function relies on consumable reactants in redox reactions helps explain their finite lifespan.

Can you recall? (Textbook Page No. 81)

Question i. What is combustion reaction?
Answer: Combustion is a process in which a substance combines with oxygen.
In simple words: Combustion is a fast chemical reaction where a substance reacts with oxygen, usually producing heat and light.

🎯 Exam Tip: Emphasize that combustion is fundamentally an oxidation reaction, often vigorous and exothermic.

Question ii. Write an equation for combustion of methane.
Answer: Combustion of methane: \( \text{CH}_4 + 2\text{O}_2 \implies \text{CO}_2 + 2\text{H}_2\text{O} + \text{Heat} + \text{Light} \)
In simple words: Methane burning with oxygen produces carbon dioxide, water, heat, and light.

🎯 Exam Tip: Balancing combustion equations correctly is important, ensuring the conservation of mass for all elements.

Question iii. What is the driving force behind reactions of elements?
Answer: The ability of element to combine with other element or the ability of element to replace other element in compound is the driving force behind the reactions. This may involve formation of precipitates, formation of water, release of gas, etc.
In simple words: Elements react to become more stable, either by combining with other elements or displacing them, often resulting in energy changes, precipitate formation, or gas release.

🎯 Exam Tip: Relate the driving force to achieving a more stable electron configuration, which is a core concept in chemical reactivity.

Try this. (Textbook Page No. 82)

Question 1. Complete the following table of displacement reactions. Identify oxidising and reducing agents involved.

🎯 Exam Tip: For displacement reactions, remember that the element undergoing oxidation is the reducing agent, and the element undergoing reduction is the oxidising agent.

Try this (Textbook Page No. 88)

Question 1. Classify the following unbalanced half equations as oxidation and reduction.

🎯 Exam Tip: To correctly classify a half-reaction, determine the oxidation state of the key element on both sides of the arrow; an increase means oxidation, and a decrease means reduction.

11th Std Chemistry Questions And Answers:

• Some Basic Concepts of Chemistry Class 11 Chemistry Questions And Answers
• Introduction to Analytical Chemistry Class 11 Chemistry Questions And Answers
• Basic Analytical Techniques Class 11 Chemistry Questions And Answers
• Structure of Atom Class 11 Chemistry Questions And Answers
• Chemical Bonding Class 11 Chemistry Questions And Answers
• Redox Reactions Class 11 Chemistry Questions And Answers
• Modern Periodic Table Class 11 Chemistry Questions And Answers
• Elements of Group 1 and 2 Class 11 Chemistry Questions And Answers