CBSE Class 11 Chemistry Redox Reactions Assignment

Question. The oxidation number of Cr in K₂Cr₂O₇ is:
a. +6
b. – 7
c. +2
d. – 2
Answer : A

Question. The oxidation state of nitrogen is highest in:
a. N₃H
b. NH₂OH
c. N₂H₄
d. NH₃
Answer : A

Question. What is the net charge on ferrous ion?
a. + 2
b. + 3
c. + 4
d. + 5
Answer : A

Question. When KMnO₄ is reduced with oxalic acid in acidic solution, the oxidation number of Mn changes from?
a. 7 to 4
b. 6 to 4
c. 7 to 2
d. 4 to 2
Answer : C

Question. The oxidation state of Mn in K₂MnO₄ :
a. + 2
b. + 7
c. – 2
d. + 6
Answer : D

Question. Oxidation number of nitrogen in NaNO₂ is:
a. + 2
b. + 3
c. + 4
d. – 3
Answer : B

Question. Oxidation number of cobalt in K[Co(CO)₄] is:
a. + 1
b. + 3
c. – 1
d. – 3
Answer : C

Question. In which compound, oxidation state of nitrogen is 1?
a. NO
b. N₂O
c. NH₂OH
d. N₂H₄
Answer : B

Question. The oxidation number and covalency of sulphur in the sulphur molecule (S₈) are respectively:
a. 0 and 2
b. 6 and 8
c. 0 and 8
d. 6 and 2
Answer : A

Question. In which of the following compound transition metal has zero oxidation state?
a. CrO₅
b. NH₂NH₂
c. NOClO₄
d. [Fe(CO)₅]
Answer : D

Question. The equivalent weight of phosphoric acid (H₃PO₄) in the reaction NaOH + H₃PO₄ → NaH₂PO₄ + H₂O is:
a. 25
b. 49
c. 59
d. 98
Answer : D

Question. What is the equivalent mass of 4 IO⁻₄ when it is converted into I₂ in acid medium?
a. M/6
b. M/7
c. M/5
d. M/4
Answer : B

Question. The oxidation state of nitrogen in N₃H is:
a. + 1/3
b. + 3
c. −1
d. − 1/3
Answer : D

Question. Sulphur has highest oxidation state in:
a. SO₂
b. H₂ SO₄
c. Na₂ S₂ O₃
d. Na₂ S₄ O₆
Answer : B

Question. An element which never has a positive oxidation number in any of its compounds:
a. Boron
b. Oxygen
c. Chlorine
d. Fluorine
Answer : D

Question. C₂H₂(g) + nO₂ → CO₂ (g) + H₂O(l) In this equation, the ratio of the coefficients of  CO₂ and H₂O is:
a. 1 : 1
b. 2 : 3
c. 3 : 2
d. 1 : 3
Answer : B

Question. Starch paper is used to test for the presence of:
a. Iodine
b. Oxidising agent
c. Iodide ion
d. Reducing agent
Answer : A

Question. In the balanced chemical reaction, IO^-₃ + a_I^- + bH⁺ → cH₂O + dI₂ a, b, c and d respectively correspond to:
a. 5, 6, 3, 3
b. 5, 3, 6, 3
c. 3, 5, 3, 6
d. 5, 6, 5, 5
Answer : A

Question. 2MnO^-₄ + 5H₂O₂ + 6H⁺ → 2Z + 5O₂ + 8H₂O In this reaction Z is:
a. Mn⁺²
b. Mn⁺⁴
c. MnO₂
d. Mn
Answer : A

Question. To prevent rancidification of food material, which of the following is added?
a. Reducing agent
b. Anti-oxidant
c. Oxidising agent
d. None of these
Answer : B

Question. 15 g of KMnO₄ in acidic medium is equal to:
a. 0.09493 moles
b. 0.4446 g equivalents
c. 9.492 L of 0.05 N KMnO₄
d. 10 mL of 0.05 M KMnO₄
Answer : A, C

Question. In the equation H₂S + 2HNO₃ → 2H₂O + 2NO₂ + S
The equivalent weight of hydrogen sulphide is:
a. 16
b. 68
c. 34
d. 17
Answer : D

Question. The metal used in galvanizing of iron is:
a. Pb
b. Zn
c. Al
d. Sn
Answer : B

Question. Which one of the following nitrates will leave behind a metal on strong heating?
a. Ferric nitrate
b. Copper nitrate
c. Manganese nitrate
d. Silver nitrate
Answer : D

Question. The oxidation number of phosphorus vary from
(1) – 3 to + 5
(2) – 1 to + 1
(3) – 3 to + 3
(4) – 5 to + 1
Answer (1)

Question. White P reacts with caustic soda, the products are PH3 and NaH₂ PO ₂. The reaction is an example of
(1) Oxidation
(2) Reduction
(3) Disproportionation
(4) Neutralization
Answer (3)

Question. Oxidation number of oxygen atom in O₃ molecule is
(1) 0
(2) –2
(3) + 2
(4) – ½
Answer (1)

Question. Oxidation number of iodine varies from
(1) – 1 to + 1
(2) – 1 to + 7
(3) + 3 to + 5
(4) – 1 to + 5
Answer (2)

Question.The process in which oxidation number increases is
(1) Reduction
(2) Hydrolysis
(3) Oxidation
(4) Decomposition
Answer (3)

Question. The oxidation number of C in CH₄, CH₃Cl, CH₂Cl₂, CHCl₃and CCl ₄ is respectively
(1) –4, –2, 0, +2, +4
(2) +2, 4, 0, –2, –4
(3) 4, 2, 0, –2, 4
(4) 0, 2, –2, 4, 4
Answer (1)

Question. Chlorine is in +3 oxidation number in
(1) HCl
(2) HClO ₄
(3) ICl
(4) ClF₃
Answer (4)

Question. Given that the oxidation number of sulphur is –2, the equivalent weight of sulphur is
(1) 16
(2) 32
(3) 9
(4) 4
Answer (1)

Question.The brown ring complex compound is formulated as [Fe(H₂O)₅ NO]SO₄ . The oxidation state of iron is
(1) +1
(2) + 2
(3) + 3
(4) + 6
Answer (1)

Question.When iron is added to CuSO₄ solution, copper is precipitated. It is due to
(1) Oxidation of Cu⁺²
(2) Reduction of Cu⁺²
(3) Hydrolysis of CuSO₄
(4) Ionization of CuSO₄
Answer (2)

Question. Which one of the following is the correct match?
(1) Cl_2- only reducing agent
(2) HNO₂- only oxidising agent
(3) HNO₃- both oxidising and reducing agent
(4) SO₂- both oxidising and reducing agent
Answer (4)

Question. Oxidation number of sodium in sodium 3 amalgam is
(1) +1
(2) 0
(3) –1
(4) +2
Answer (2)

Question. In a reaction between zinc and iodine, zinc iodide is formed. Which is being oxidised?
(1) Zinc ions
(2) Iodide ions
(3) Zinc atom
(4) Iodine
Answer (3)

Question.Oxidation number of Cr in CrO ₅ is
(1) + 10
(2) + 6
(3) + 4
(4) + 5
Answer (2)

Question.In which of the following compounds, the oxidation number of iodine is fractional?
(1) IF₇
(2) I₃^–
(3) IF₅
(4) IF₃
Answer (2)

Question. Which of the following compound cannot be oxidised by O₃ ?
(1) KI
(2) KMnO₄
(3) K₂MnO₄
(4) FeSO₄
Answer (2)

Question.The oxidation number of sulphur in H₂S₂O₈ is
(1) + 7
(2) + 6
(3) – 6
(4) + 4
Answer (2)

.

ONE MARK QUESTIONS

Question 1. What do you mean by oxidation number ?
Answer: Oxidation number denotes the degree of oxidation or apparent charge assigned to an atom in a molecule or ion, calculated on the basis of a set of arbitrary rules (usually based on electronegativity values).
In simple words: It is the residual charge an atom appears to have when all bonded atoms are removed according to electronegativity rules.

Exam Tip: Always remember that oxidation number can be positive, negative, zero, or even fractional.

Question 2. Name a compound each in which hydrogen exists in i) +1 ii) -1 oxidation states
Answer: i) In water (\(\text{H}_2\text{O}\)), hydrogen exists in +1 oxidation state.
ii) In sodium hydride (\(\text{NaH}\)), hydrogen exists in -1 oxidation state.
In simple words: Hydrogen shows +1 when combined with non-metals and -1 when combined with active metals (hydrides).

Exam Tip: Metal hydrides like \(\text{NaH}\), \(\text{LiH}\), and \(\text{CaH}_2\) are classic examples where hydrogen exhibits a -1 oxidation state.

Question 3. Define Redox couple.
Answer: A redox couple is defined as together taking part in an oxidation or reduction half-reaction, consisting of an oxidized and a reduced form of the same substance participating in an electron transfer process, represented as Oxidized form / Reduced form (e.g., \(\text{Zn}^{2+} / \text{Zn}\)).
In simple words: A pair of oxidized and reduced species belonging to the same element involved in a half-reaction is called a redox couple.

Question 4. Find the oxidation number of the element underlined (each carries 1 mark)
a) \(\text{Na}_3\underline{\text{V}}\text{O}_4\) b) \(\text{K}_2\underline{\text{Cr}}\text{O}_4\) c) \(\text{C}\text{H}_4\) d) \(\text{S}\text{O}_2\text{Cl}_2\) f) \(\text{N}\text{O}_2\) g) \(\text{Br}\text{F}_3\) h) \(\text{Na}_2\underline{\text{S}}_4\text{O}_6\) i) \(\text{C}\text{H}_2\text{Cl}_2\) j) \(\underline{\text{Cl}}\text{O}_4^-\)
Answer:
a) In \(\text{Na}_3\text{VO}_4\): Let the oxidation number of \(\text{V}\) be \(x\).
\(3(+1) + x + 4(-2) = 0\)
\(3 + x - 8 = 0 \implies x = +5\)
b) In \(\text{K}_2\text{CrO}_4\): Let the oxidation number of \(\text{Cr}\) be \(x\).
\(2(+1) + x + 4(-2) = 0\)
\(2 + x - 8 = 0 \implies x = +6\)
c) In \(\text{CH}_4\): Let the oxidation number of \(\text{C}\) be \(x\).
\(x + 4(+1) = 0 \implies x = -4\)
d) In \(\text{SO}_2\text{Cl}_2\): Let the oxidation number of \(\text{S}\) be \(x\).
\(x + 2(-2) + 2(-1) = 0\)
\(x - 4 - 2 = 0 \implies x = +6\)
f) In \(\text{NO}_2\): Let the oxidation number of \(\text{N}\) be \(x\).
\(x + 2(-2) = 0 \implies x = +4\)
g) In \(\text{BrF}_3\): Let the oxidation number of \(\text{Br}\) be \(x\).
\(x + 3(-1) = 0 \implies x = +3\)
h) In \(\text{Na}_2\text{S}_4\text{O}_6\): Let the oxidation number of \(\text{S}\) be \(x\).
\(2(+1) + 4x + 6(-2) = 0\)
\(2 + 4x - 12 = 0 \implies 4x = 10 \implies x = +2.5\)
i) In \(\text{CH}_2\text{Cl}_2\): Let the oxidation number of \(\text{C}\) be \(x\).
\(x + 2(+1) + 2(-1) = 0\)
\(x + 2 - 2 = 0 \implies x = 0\)
j) In \(\text{ClO}_4^-\): Let the oxidation number of \(\text{Cl}\) be \(x\).
\(x + 4(-2) = -1\)
\(x - 8 = -1 \implies x = +7\)
In simple words: Set up a linear equation equating the sum of all oxidation numbers in a neutral compound to 0 and in an ion to the net charge of the ion.

Exam Tip: Fractional oxidation numbers (like in \(\text{Na}_2\text{S}_4\text{O}_6\)) represent the average oxidation state of all atoms of that element in the molecule.

TWO MARKS QUESTIONS

Question 1. Name a compound each in which oxygen exists in i) +1 ii) -1 iii) +2 iv) -2 oxidation states
Answer:
i) +1: Dioxygen difluoride (\(\text{O}_2\text{F}_2\))
ii) -1: Hydrogen peroxide (\(\text{H}_2\text{O}_2\)) or sodium peroxide (\(\text{Na}_2\text{O}_2\))
iii) +2: Oxygen difluoride (\(\text{OF}_2\))
iv) -2: Water (\(\text{H}_2\text{O}\)) or carbon dioxide (\(\text{CO}_2\))
In simple words: Oxygen shows positive oxidation states only when bonded to fluorine (which is more electronegative than oxygen).

Exam Tip: Fluorine is always -1 in its compounds, which forces oxygen to take positive oxidation states in \(\text{OF}_2\) (+2) and \(\text{O}_2\text{F}_2\) (+1).

Question 2. Define
i) stock notation
ii) Standard electrode potential
Answer:
i) Stock notation: A system introduced by Alfred Stock in which the oxidation number of a metal in a compound is written in Roman numerals in parentheses immediately after the name (or chemical symbol) of the metal.
ii) Standard electrode potential: The potential associated with a half-cell reaction relative to a standard hydrogen electrode (SHE) measured at standard conditions (1 M concentration of ions, 1 atm pressure for gases, and 298 K temperature).
In simple words: Stock notation uses Roman numerals for metal oxidation states, and standard electrode potential measures a half-cell's tendency to gain or lose electrons under standard conditions.

Exam Tip: Examples of stock notation include \(\text{AuCl}_3\) as Gold(III) chloride and \(\text{FeO}\) as Iron(II) oxide.

Question 3. Explain the following by giving example
i) Displacement redox reaction.
ii) Disproportionation reactions
Answer:
i) Displacement redox reaction: A reaction in which an element or ion in a compound is displaced by another element or ion. Example:
\(\text{Zn}(s) + \text{CuSO}_4(aq) \rightarrow \text{ZnSO}_4(aq) + \text{Cu}(s)\)
ii) Disproportionation reaction: A reaction in which an element in one oxidation state is simultaneously oxidized and reduced, yielding two different oxidation states. Example:
\(2\text{H}_2\text{O}_2(aq) \rightarrow 2\text{H}_2\text{O}(l) + \text{O}_2(g)\) (where oxygen goes from -1 to -2 and 0).
In simple words: In displacement, one element replaces another; in disproportionation, a single element undergoes both oxidation and reduction simultaneously.

Exam Tip: For disproportionation, the element must possess at least three oxidation states; it must be in an intermediate oxidation state.

Question 4. Calculate the standard emf of the following cell at 298K using the standard electrode potential. \(\text{Al}(s) \mid \text{Al}^{3+}(aq) \parallel \text{Fe}^{2+}(aq) \mid \text{Fe}(s)\)
Given \(E^\circ_{\text{Al}^{3+}/\text{Al}} = -1.66\text{ V}\) and \(E^\circ_{\text{Fe}^{2+}/\text{Fe}} = -0.44\text{ V}\).
And what will be the cell reaction?
Answer:
Standard EMF of the cell (\(E^\circ_{\text{cell}}\)) = \(E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}\)
Anode (oxidation): \(\text{Al} \rightarrow \text{Al}^{3+} + 3e^-\) (\(E^\circ = -1.66\text{ V}\))
Cathode (reduction): \(\text{Fe}^{2+} + 2e^- \rightarrow \text{Fe}\) (\(E^\circ = -0.44\text{ V}\))
\(E^\circ_{\text{cell}} = (-0.44\text{ V}) - (-1.66\text{ V})\)
\(E^\circ_{\text{cell}} = -0.44 + 1.66 = +1.22\text{ V}\)
Cell Reaction:
Multiplying anode reaction by 2 and cathode reaction by 3 to balance electrons:
Oxidation half-reaction: \(2\text{Al}(s) \rightarrow 2\text{Al}^{3+}(aq) + 6e^-\)
Reduction half-reaction: \(3\text{Fe}^{2+}(aq) + 6e^- \rightarrow 3\text{Fe}(s)\)
Overall Cell Reaction: \(2\text{Al}(s) + 3\text{Fe}^{2+}(aq) \rightarrow 2\text{Al}^{3+}(aq) + 3\text{Fe}(s)\)
In simple words: Subtract the lower standard reduction potential from the higher standard reduction potential to get the cell EMF, and balance electrons to write the overall reaction.

Exam Tip: The metal with the more negative reduction potential acts as the anode.

Question 5. What is salt bridge? What are its functions.
Answer: A salt bridge is a U-shaped tube containing an inert electrolyte (like \(\text{KCl}\), \(\text{KNO}_3\), or \(\text{NH}_4\text{NO}_3\)) in an agar-agar gel that connects the two half-cells of a galvanic cell.
Functions of salt bridge:
1. It completes the electrical circuit by allowing the flow of ions between the two half-cells without direct mixing of solutions.
2. It maintains electrical neutrality in both half-cells by preventing the accumulation of charges.
In simple words: A salt bridge is a U-tube that bridges two half-cells, completing the circuit and keeping solutions electrically neutral.

Exam Tip: The electrolyte used in a salt bridge must have ions whose transport numbers are nearly identical (e.g., \(\text{K}^+\) and \(\text{NO}_3^-\)).

Question 6. Write the cell reaction for the following Galvanic cells:
i) \(\text{Mg}(s) \mid \text{Mg}^{2+}(aq) \parallel \text{Al}^{3+}(aq) \mid \text{Al}(s)\)
ii) \(\text{Zn}(s) \mid \text{Zn}^{2+}(aq) \parallel \text{Ag}^+(aq) \mid \text{Ag}(s)\)
Answer:
i) For cell \(\text{Mg}(s) \mid \text{Mg}^{2+}(aq) \parallel \text{Al}^{3+}(aq) \mid \text{Al}(s)\):
Anode reaction: \(\text{Mg}(s) \rightarrow \text{Mg}^{2+}(aq) + 2e^-\) (multiplied by 3)
Cathode reaction: \(\text{Al}^{3+}(aq) + 3e^- \rightarrow \text{Al}(s)\) (multiplied by 2)
Overall cell reaction: \(3\text{Mg}(s) + 2\text{Al}^{3+}(aq) \rightarrow 3\text{Mg}^{2+}(aq) + 2\text{Al}(s)\)
ii) For cell \(\text{Zn}(s) \mid \text{Zn}^{2+}(aq) \parallel \text{Ag}^+(aq) \mid \text{Ag}(s)\):
Anode reaction: \(\text{Zn}(s) \rightarrow \text{Zn}^{2+}(aq) + 2e^-\)
Cathode reaction: \(\text{Ag}^+(aq) + e^- \rightarrow \text{Ag}(s)\) (multiplied by 2)
Overall cell reaction: \(\text{Zn}(s) + 2\text{Ag}^+(aq) \rightarrow \text{Zn}^{2+}(aq) + 2\text{Ag}(s)\)
In simple words: Write the oxidation half-reaction for the left electrode (anode) and the reduction half-reaction for the right electrode (cathode), then balance the electrons.

Exam Tip: Single vertical lines represent phase boundaries, while double vertical lines represent the salt bridge.

Question 7. Balance the following equations: (each carries 2 marks)
i) \(\text{MnO}_4^- + \text{C}_2\text{H}_2\text{O}_4 \rightarrow \text{Mn}^{2+} + \text{CO}_2\) [acid]
ii) \(\text{MnO}_4^- + \text{Br}^- \rightarrow \text{Mn}^{2+} + \text{Br}_2\) [acid]
iii) \(\text{HNO}_2 + \text{I}^- \rightarrow \text{NO} + \text{I}_2\) [acid]
Answer:
i) \(2\text{MnO}_4^- + 5\text{C}_2\text{H}_2\text{O}_4 + 6\text{H}^+ \rightarrow 2\text{Mn}^{2+} + 10\text{CO}_2 + 8\text{H}_2\text{O}\)
ii) \(2\text{MnO}_4^- + 10\text{Br}^- + 16\text{H}^+ \rightarrow 2\text{Mn}^{2+} + 5\text{Br}_2 + 8\text{H}_2\text{O}\)
iii) \(2\text{HNO}_2 + 2\text{I}^- + 2\text{H}^+ \rightarrow 2\text{NO} + \text{I}_2 + 2\text{H}_2\text{O}\)
In simple words: Balance oxidation and reduction half-reactions separately by adding electrons, then balance oxygen using water and hydrogen using hydrogen ions in acidic medium.

Exam Tip: In acidic medium, add \(\text{H}_2\text{O}\) to balance oxygen atoms and \(\text{H}^+\) ions to balance hydrogen atoms.

THREE MARKS QUESTIONS

Question 1. Identify the redox reactions and classify them.
a. \(2\text{Na} + \text{H}_2 \rightarrow 2\text{NaH}\)
b. \(\text{AgNO}_3 + \text{NaCl} \rightarrow \text{AgCl} + \text{NaNO}_3\)
c. \(\text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2\)
d. \(2\text{HCHO} + \text{NaOH} \rightarrow \text{HCOONa} + \text{CH}_3\text{OH}\)
Answer:
a. \(2\text{Na} + \text{H}_2 \rightarrow 2\text{NaH}\): It is a redox reaction (specifically, a combination redox reaction). Sodium is oxidized from 0 to +1, and hydrogen is reduced from 0 to -1.
b. \(\text{AgNO}_3 + \text{NaCl} \rightarrow \text{AgCl} + \text{NaNO}_3\): Not a redox reaction (it is a double displacement/precipitation reaction with no change in oxidation numbers).
c. \(\text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2\): Not a redox reaction (it is a thermal decomposition reaction with no change in oxidation numbers).
d. \(2\text{HCHO} + \text{NaOH} \rightarrow \text{HCOONa} + \text{CH}_3\text{OH}\): It is a redox reaction (specifically, a disproportionation reaction, known as the Cannizzaro reaction). Carbon is simultaneously oxidized from 0 to +2 (in \(\text{HCOONa}\)) and reduced from 0 to -2 (in \(\text{CH}_3\text{OH}\)).
In simple words: Check for changes in oxidation numbers; reactions where elements change oxidation states are redox, while double displacement and simple thermal decompositions are not.

Exam Tip: Precipitation and neutralization reactions are almost never redox reactions.

Question 2. i) Predict whether the following redox reaction is feasible or not under standard conditions
\(\text{Sn}^{2+}(aq) + \text{Cu}(s) \rightarrow \text{Sn}(s) + \text{Cu}^{2+}(aq)\)
Given that \(E^\circ_{\text{Sn}^{2+}/\text{Sn}} = -0.136\text{ V}\) and \(E^\circ_{\text{Cu}^{2+}/\text{Cu}} = 0.34\text{ V}\)
ii) Differentiate between Activity series and Electrochemical series
Answer:
i) Feasibility prediction:
Anode (oxidation): \(\text{Cu}(s) \rightarrow \text{Cu}^{2+}(aq) + 2e^-\) (\(E^\circ_{\text{Cu}^{2+}/\text{Cu}} = 0.34\text{ V}\))
Cathode (reduction): \(\text{Sn}^{2+}(aq) + 2e^- \rightarrow \text{Sn}(s)\) (\(E^\circ_{\text{Sn}^{2+}/\text{Sn}} = -0.136\text{ V}\))
\(E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}\)
\(E^\circ_{\text{cell}} = (-0.136\text{ V}) - (0.34\text{ V}) = -0.476\text{ V}\)
Since \(E^\circ_{\text{cell}}\) is negative (\(-0.476\text{ V}\)), the given redox reaction is not feasible under standard conditions.
ii) Difference between Activity series and Electrochemical series:
- Activity series arranges metals in order of decreasing chemical reactivity based on displacement reactions and ease of oxidation.
- Electrochemical series arranges elements in order of their standard reduction potentials measured relative to the standard hydrogen electrode (SHE).
In simple words: A reaction is feasible only if the calculated cell EMF is positive.

Exam Tip: A positive \(E^\circ_{\text{cell}}\) implies a negative \(\Delta G^\circ\), which corresponds to a spontaneous reaction.

Question 3. Balance the following equations: (each carries 3 marks)
a. \(\text{HNO}_2 + \text{I}^- \rightarrow \text{NO} + \text{I}_2\) [acid]
b. \(\text{I}_2 + \text{NO}_3^- + \text{H}^+ \rightarrow \text{NO}_2 + \text{IO}_3^-\) [acid]
c. \(\text{Al} + \text{NO}_3^- \rightarrow \text{Al(OH)}_4^- + \text{NH}_3\) [basic]
d. \(\text{CrO}_3^- + \text{H}_2\text{O}_2 \rightarrow \text{CrO}_4^{2-} + \text{H}_2\text{O}\) [basic]
e. \(\text{Fe(OH)}_2 + \text{H}_2\text{O}_2 \rightarrow \text{Fe(OH)}_3 + \text{H}_2\text{O}\) [basic]
f. \(\text{I}^- + \text{IO}_3^- \rightarrow \text{I}_2 + \text{H}_2\text{O}\) [acid]
Answer:
a. \(2\text{HNO}_2 + 2\text{I}^- + 2\text{H}^+ \rightarrow 2\text{NO} + \text{I}_2 + 2\text{H}_2\text{O}\)
b. \(\text{I}_2 + 10\text{NO}_3^- + 8\text{H}^+ \rightarrow 2\text{IO}_3^- + 10\text{NO}_2 + 4\text{H}_2\text{O}\)
c. \(8\text{Al} + 3\text{NO}_3^- + 5\text{OH}^- + 18\text{H}_2\text{O} \rightarrow 8\text{Al(OH)}_4^- + 3\text{NH}_3\)
d. \(2\text{CrO}_3^- + 3\text{H}_2\text{O}_2 + 2\text{OH}^- \rightarrow 2\text{CrO}_4^{2-} + 4\text{H}_2\text{O}\)
e. \(2\text{Fe(OH)}_2 + \text{H}_2\text{O}_2 \rightarrow 2\text{Fe(OH)}_3 + 2\text{H}_2\text{O}\)
f. \(5\text{I}^- + \text{IO}_3^- + 6\text{H}^+ \rightarrow 3\text{I}_2 + 3\text{H}_2\text{O}\)
In simple words: Balance half-reactions by equating electrons, then add \(\text{H}^+\) and \(\text{H}_2\text{O}\) for acidic medium or \(\text{OH}^-\) and \(\text{H}_2\text{O}\) for basic medium.

Exam Tip: In basic medium, balance oxygen atoms by adding twice the number of \(\text{OH}^-\) ions on the side deficient in oxygen, and balance hydrogen by adding \(\text{H}_2\text{O}\) to the opposite side.

VALUE BASED QUESTION (FOUR MARKS)

Question 1. A solution of an electrolyte can be stored in a particular vessel only in case there is no chemical reaction taking place with the material of the vessel. The teacher asked a student Sachin, is it possible to store silver nitrate in copper vessel. Sachin answered ; it is not possible to store silver nitrate in copper vessel & gave his explanation.
i) What explanation did Sachin gave to his teacher.
ii) What is oxidation & reduction according to electronic concept?
iii) What are the values displayed by Sachin?
Answer:
i) Sachin explained that copper is more reactive than silver (having a lower standard reduction potential, \(E^\circ_{\text{Cu}^{2+}/\text{Cu}} = 0.34\text{ V}\) vs \(E^\circ_{\text{Ag}^+/\text{Ag}} = 0.80\text{ V}\)), so copper will displace silver from silver nitrate solution, forming copper nitrate and depositing silver metal. Hence, the container would get corroded and damaged.
ii) According to the electronic concept, oxidation is defined as the loss of electrons, while reduction is defined as the gain of electrons.
iii) The values displayed by Sachin are scientific awareness, analytical thinking, attentiveness in class, and honesty in applying theoretical chemical knowledge to practical situations.
In simple words: Sachin correctly stated that a more reactive metal (copper) cannot store a less reactive metal salt solution (\text{AgNO}_3) because a displacement reaction would occur and destroy the vessel.

Exam Tip: A container can safely store a solution only if the container metal is less reactive than the metal ion present in the solution.