CBSE Class 11 Chemistry Redox Reaction Worksheet

Question. 2CuI → Cu +CuI₂ the reaction is:
a. Redox
b. Neutralisation
c. Oxidation
d. Reduction
Answer : A

Question. H₂S reacts with halogens, the halogens:
a. Form sulphur halides
b. Are oxidised
c. Are reduced
d. None of these
Answer : C

Question. H₂O₂ reduces K₄Fe(CN)₆ :
a. In neutral solution
b. In acidic solution
c. In non-polar solvent
d. In alkaline solution
Answer : B

Question. Which halide is not oxidised by MnO₂ ?
a. F
b. Cl
c. Br
d. I
Answer : A

Question. Oxidation number of N in HNO₃ is:
a. – 3.5
b. + 3.5
c. – 3, +5
d. + 5
Answer : D

Question. The oxidation number of Mn in MnO⁻¹₄ is:
a. + 7
b. – 5
c. + 6
d. + 5
Answer : A

Question. The compound(s) behaving both oxidizing as well as reducing agent is/are:
a. H₂O₂
b. SO₂
c. HNO₂
d. HNO₃
Answer : A, B, D

Question. When Fe²⁺ changes to Fe³⁺ in a reaction?
a. It loses an electron
b. It gains an electron
c. It loses a proton
d. It gains a proton
Answer : A

Question. In acid solution, the reaction MnO^-₄ → Mn²⁺ involves:
a. Oxidation by 3 electrons
b. Reduction by 3 electrons
c. Oxidation by 5 electrons
d. Reduction by 5 electrons
Answer : D

Question. When iron or zinc is added to CuSO₄ solution, copper is precipitated. It is due to?
a. Oxidation of 2 Cu⁺²
b. Reduction of 2 Cu⁺²
c. Hydrolysis of 4 CuSO₄
d. Ionization of 4 CuSO₄
Answer : B

Question. The reaction H₂S₂+H₂O₂ → 2H₂O+S shows:
a. Oxidizing action of H₂O₂
b. Reducing action of H₂O₂
c. Alkaline nature of H₂O₂
d. Acidic nature of H₂O₂
Answer : A

Question. Which of the following is not a reducing agent?
a. NaNO₂
b. NaNO₃
c. HI
d. SnCl₂
Answer : B

Question. The compound in which the same element possesses two different oxidation states is/are:
a. bleaching powder
b. hypo
c. perdisulphuric acid
d. potassium permanganate
Answer : A, B, C

Question. Which of the following cannot work as oxidising agent?
a. O₂
b. KMnO₄
c. I₂
d. None of these
Answer : C

Assertion and Reason

Note: Read the Assertion (A) and Reason (R) carefully to mark the correct option out of the options given below:
a. If both assertion and reason are true and the reason is the correct explanation of the assertion.
b. If both assertion and reason are true but reason is not the correct explanation of the assertion.
c. If assertion is true but reason is false.
d. If the assertion and reason both are false.
e. If assertion is false but reason is true.

Question. Assertion: The oxidation numbers are artificial; they are useful as a ‘book-keeping’ device of electrons in reactions.
Reason: The oxidation numbers do not usually represent real charges on atoms, they are simply conventions that indicate what the maximum charge could possibly be on an atom in a molecule.
Answer : A

Question. Assertion: H₂ SO₄ cannot act as reducing agent.
Reason: Sulphur cannot increase its oxidation number beyond + 6.
Answer : A

Question. Assertion: Equivalent weight of NH₃ in the reaction N₂ → NH₃ is 17/3 while that of N₂ is 28/6.
Reason: Equivalent weight = Molecular weight/number of e¯ lost or gained
Answer : A

Question. Assertion: SO₂ and Cl₂ both are bleaching agents.
Reason: Both are reducing agents.
Answer : C

Question. Assertion: Fluorine exists only in −1 oxidation state.
Reason: Fluorine has 2s² 2p⁵ configuration.
Answer : B

Question. Assertion: Stannous chloride is a powerful oxidising agent which oxidises mercuric chloride to mercury.
Reason: Stannous chloride gives grey precipitate with mercuric chloride, but stannic chloride does not do so
Answer : C

Question. Assertion: HClO₄ is a stronger acid than HClO₃.
Reason: Oxidation state of Cl in HClO₄ is +VII and in HClO₃ + V.
Answer : B

Question. Assertion: In a reaction Zn(s) + CuSO₄ (aq) → ZnSO₄ (aq) + Cu(s), Zn is a reductant but itself get oxidized.
Reason: In a redox reaction, oxidant is reduced by accepting electrons and reductant is oxidized by losing electrons.
Answer : A

Question. Assertion: Oxidation number of carbon in CH₂O is zero.
Reason: CH₂O formaldehyde, is a covalent compound.
Answer : B

Question. Assertion: A reaction between Fe and 2 I occurs, but a reaction between Fe²⁺ and I⁻ does not occur.
Reason: Fe is a better reducing agent than I .
Answer : A

Comprehension Based

Paragraph–I

Consider the following unbalanced redox reaction: H₂O + AX+ BY → HA + OY + X₂B
The oxidation number of X is −2, and neither X nor water is involved in the redox process.

Question. The element(s) undergoing oxidation is/are:
a. A
b. B
c. Y
d. B or Y or both
Answer : D

Question. The possible oxidation state of B and Y in BY are, respectively:
a. +1, −1
b. +2, − 2
c. +3, − 3
d. All of these
Answer : D

Question. If the above reaction is balanced with smallest whole number coefficients, the sum of the stoichiometric coefficients of all the compounds is:
a. 9
b. 8
c. 7
d. 6
Answer : B

Paragraph–II

Oxidation reaction involves loss of electrons, and reduction reaction involves gain of electrons. The reaction in which a species disproportionate into two oxidation states (lower and higher) is called disproportionation reaction

Question. Which of the following statements is wrong?
a. A acidified K₂Cr₂O₇ paper on being exposed to SO₂ turns green.
b. Mercuric chloride and stannous chloride cannot exist as such.
c. Iron turning on addition to CuSO₄ solution decolourises the blue colour.
d. [CuI₄]²⁻ is formed but [CuCl₄]²⁻ is not.
Answer : D

Question. Which of the following statements is wrong?
a. Accidified KMnO₄ solution decolourieses on the addition of sodium oxalate.
b. In the reaction between Br₂ and CsI, Br₂ is an oxidising agent and CsI is a reducing agent.
c. In the reaction 2K₂S₂O₃ + I₂ → 2KI + K₂S₄O₆₊ , the change in the oxidation number of S is 0.5
d. C has the same oxidation number in both CH₄ and CO₂
Answer : D

Question. Which of the following statements is correct?
a. An element in the lowest oxidation state acts only as a reducing agent.
b. An element in the highest oxidation sate acts only as a reducing agent.
c. The oxidation number of V in Rb₄K(HV₁₀O₂₈) is +4.
d. The oxidation number and valency of Hg in calomel is +1
Answer : A

Question. Which of the following statements is wrong?
a. The algebraic sum of the oxidation numbers of all atoms in an ion is zero
b. The oxidation number is an arbitrary number. It can have positive, negative, zero, or fractional values.
c. When a negative ion changes to neutral species, the process is oxidation.
d. The oxidation number of phosphorous can very form –3 to +5.
Answer : A

Question. Which of the following is not a disproportionation reaction?
a. OH^- + Br₂ → Br^- + BrO^-₃ 
b. Cu₂O + 2H⁺  → Cu + Cu²⁺ +  H₂O
c. (CN)- → CO₃^2- + NO^-₃ 
d. (CN)₂ + 2OH → CN^- + CNO^- + H₂O
Answer : C

Integer

Question. Find out the value of n in MnO^-₄ + 8H⁺ + ne → Mn⁺² + 4H₂O
Answer : 5

Question. One mole of N₂H₄ loses 10 mole electrons to form a new compound Y. Assuming that all the N₂ appears in new compound, what is oxidation state of N in Y? There is no change in oxidation state of H.
Answer : 3

Question. The equivalent weight of HNO₃ (molecular weight = 63) in the following reaction is 3Cu + 8HNO₃ → 3Cu(NO₃)₂ + 2NO + 4H₂O
Answer : 84

Question. In the following reaction xZn + yHNO₃ (dil) → aZn(NO₃)₃ + bH₂O + cNH₄NO₃
What is the sum of the coefficients (a + b + c) ?
Answer : 8

Question. CN⁻ ion is oxidized by a powerful oxidising agent of NO⁻₃ and CO₂ or CO₃²⁻ depending on the acidity of the reaction mixture CN⁻ → CO₂ + NO⁻₃ + H⁺ + ne⁻
What is the number (n) of electrons involved in the process, divided by 10?
Answer : 3

Question 1. Name the oxidiser and reducer in the following reaction:- \(\text{SnCl}_2 + 2\text{FeCl}_3 \rightarrow \text{SnCl}_4 + 2\text{FeCl}_2\)
Answer: Oxidiser (Oxidizing agent): \(\text{FeCl}_3\) (or \(\text{Fe}^{3+}\))
Reducer (Reducing agent): \(\text{SnCl}_2\) (or \(\text{Sn}^{2+}\))
Explanation: In this reaction, the oxidation state of tin (\(\text{Sn}\)) increases from +2 in \(\text{SnCl}_2\) to +4 in \(\text{SnCl}_4\), meaning \(\text{SnCl}_2\) undergoes oxidation and acts as the reducing agent. The oxidation state of iron (\(\text{Fe}\)) decreases from +3 in \(\text{FeCl}_3\) to +2 in \(\text{FeCl}_2\), meaning \(\text{FeCl}_3\) undergoes reduction and acts as the oxidizing agent.
In simple words: The substance that loses electrons gets oxidized (reducer), and the one that gains electrons gets reduced (oxidiser).

Exam Tip: Always assign oxidation numbers to identify which species is oxidized (increase in oxidation number) and which is reduced (decrease in oxidation number).

Question 2. Define oxidation number and calculate the oxidation number of Cr in \(\text{K}_2\text{Cr}_2\text{O}_7\).
Answer: Oxidation number is defined as the residual charge that an atom appears to have when all other atoms in the molecule are removed according to set electronegativity guidelines.
Calculation of oxidation number of \(\text{Cr}\) in \(\text{K}_2\text{Cr}_2\text{O}_7\):
Let the oxidation number of \(\text{Cr}\) be \(x\).
The oxidation number of potassium (\(\text{K}\)) is +1.
The oxidation number of oxygen (\(\text{O}\)) is -2.
Sum of oxidation states in a neutral compound = 0
\(2(+1) + 2(x) + 7(-2) = 0\)
\(2 + 2x - 14 = 0\)
\(2x - 12 = 0\)
\(2x = 12\)
\(x = +6\)
Therefore, the oxidation number of \(\text{Cr}\) in \(\text{K}_2\text{Cr}_2\text{O}_7\) is +6.
In simple words: Oxidation number is the apparent charge on an atom, calculated by balancing the known charges of other atoms in the formula.

Exam Tip: Remember standard oxidation numbers: Group 1 elements are always +1, fluorine is -1, and oxygen is usually -2 (except in peroxides and fluorides).

Question 3. What is the usual oxidation state of oxygen? In which type of compounds oxygen shows an oxidation no. of -1 and +2?
Answer: The usual oxidation state of oxygen is -2.
Oxygen shows an oxidation number of -1 in peroxides (e.g., \(\text{H}_2\text{O}_2\), \(\text{Na}_2\text{O}_2\)).
Oxygen shows an oxidation number of +2 in oxygen difluoride (\(\text{OF}_2\)), because fluorine is more electronegative than oxygen.
In simple words: Oxygen is normally -2, but it becomes -1 in peroxides and +2 when bonded to fluorine.

Exam Tip: Electronegativity dictates sign: since fluorine is the most electronegative element, oxygen takes a positive oxidation state only when combined with fluorine.

Question 4. What is meant by half reaction?
Answer: A half reaction is either the oxidation half or the reduction half of a complete redox reaction. It explicitly shows the electrons lost during oxidation or gained during reduction.
In simple words: A redox reaction split into two separate equations—one showing loss of electrons and the other showing gain of electrons.

Exam Tip: In any balanced redox equation, the total number of electrons lost in the oxidation half reaction must equal the total number of electrons gained in the reduction half reaction.

Question 5. Balance the following equation in the acidic medium by oxidation number method.
\(\text{MnO}_4^{-}(\text{aq}) + \text{C}_2\text{H}_2\text{O}_4(\text{aq}) + \text{H}^{+} \rightarrow \text{Mn}^{2+}(\text{aq}) + \text{CO}_2(\text{g}) + \text{H}_2\text{O}(\text{l})\)
Answer: Step 1: Assign oxidation numbers to the atoms undergoing change.
In \(\text{MnO}_4^{-}\), \(\text{Mn}\) is +7. In \(\text{Mn}^{2+}\), \(\text{Mn}\) is +2. (Change = 5 electrons per \(\text{Mn}\))
In \(\text{C}_2\text{H}_2\text{O}_4\), each \(\text{C}\) is +3. In \(\text{CO}_2\), \(\text{C}\) is +4. (Total change = 2 carbons \(\times 1\) = 2 electrons per molecule)
Step 2: Equalize the increase and decrease in oxidation numbers by cross-multiplying.
Multiply \(\text{MnO}_4^{-}\) by 2 and \(\text{C}_2\text{H}_2\text{O}_4\) by 5:
\(2\text{MnO}_4^{-} + 5\text{C}_2\text{H}_2\text{O}_4 \rightarrow 2\text{Mn}^{2+} + 10\text{CO}_2\)
Step 3: Balance oxygen atoms by adding water (\(\text{H}_2\text{O}\)) to the side deficient in oxygen.
Reactant oxygen = \(2(4) + 5(4) = 8 + 20 = 28\)
Product oxygen (in \(\text{CO}_2\)) = \(10(2) = 20\)
Deficiency on product side = \(28 - 20 = 8\). Add \(8\text{H}_2\text{O}\) to the products.
\(2\text{MnO}_4^{-} + 5\text{C}_2\text{H}_2\text{O}_4 \rightarrow 2\text{Mn}^{2+} + 10\text{CO}_2 + 8\text{H}_2\text{O}\)
Step 4: Balance hydrogen atoms in acidic medium by adding \(\text{H}^{+}\).
Product hydrogen = \(8(2) = 16\)
Reactant hydrogen (in \(\text{C}_2\text{H}_2\text{O}_4\)) = \(5(2) = 10\)
Deficiency on reactant side = \(16 - 10 = 6\). Add \(6\text{H}^{+}\) to the reactants.
Final balanced equation:
\(2\text{MnO}_4^{-}(\text{aq}) + 5\text{C}_2\text{H}_2\text{O}_4(\text{aq}) + 6\text{H}^{+} \rightarrow 2\text{Mn}^{2+}(\text{aq}) + 10\text{CO}_2(\text{g}) + 8\text{H}_2\text{O}(\text{l})\)
In simple words: Balance changes in oxidation numbers first, then balance oxygen with water and hydrogen with hydrogen ions.

Exam Tip: Always double check that both atom balance and total ionic charge balance are conserved on both sides of the final equation.

Question 6. Write the half reactions for the following Redox reaction:
(a) \(2\text{Fe}^{3+}(\text{aq}) + 2\text{I}^{-}(\text{aq}) \rightarrow 2\text{Fe}^{2+}(\text{aq}) + \text{I}_2(\text{aq})\)
(b) \(\text{Zn}(\text{s}) + 2\text{H}^{+}(\text{aq}) \rightarrow \text{Zn}^{2+}(\text{aq}) + \text{H}_2(\text{g})\)
Answer:
(a) For reaction (a):
Oxidation half reaction: \(2\text{I}^{-}(\text{aq}) \rightarrow \text{I}_2(\text{aq}) + 2\text{e}^{-}\)
Reduction half reaction: \(2\text{Fe}^{3+}(\text{aq}) + 2\text{e}^{-} \rightarrow 2\text{Fe}^{2+}(\text{aq})\)
(b) For reaction (b):
Oxidation half reaction: \(\text{Zn}(\text{s}) \rightarrow \text{Zn}^{2+}(\text{aq}) + 2\text{e}^{-}\)
Reduction half reaction: \(2\text{H}^{+}(\text{aq}) + 2\text{e}^{-} \rightarrow \text{H}_2(\text{g})\)
In simple words: Separate the overall equation into the oxidation part (loss of electrons) and the reduction part (gain of electrons).

Exam Tip: Electrons must always appear on the product side for oxidation half reactions and on the reactant side for reduction half reactions.

Question 7. Identify the strongest and the weakest reducing agents from the following metals: Zn, Cu, Ag, Na
Answer: Strongest reducing agent: Sodium (\(\text{Na}\))
Weakest reducing agent: Silver (\(\text{Ag}\))
Explanation: Reducing ability depends on the ease of oxidation (tendency to lose electrons), which corresponds to having a more negative standard reduction potential. Sodium has the most negative reduction potential, making it the easiest to oxidize and thus the strongest reducing agent. Silver has a positive standard reduction potential, making it the most difficult to oxidize and thus the weakest reducing agent.
In simple words: Highly reactive alkali metals like sodium give up electrons very easily (strongest reducing agents), whereas noble metals like silver hold onto their electrons tightly (weakest reducing agents).

Exam Tip: The electrochemical series order of reducing strength is the reverse of standard reduction potentials: lower (more negative) reduction potential = stronger reducing agent.

Question 8. The standard reduction potential for \(\text{Cu}^{2+}/\text{Cu}\) is +0.34 V. Calculate the standard reduction potential at \(\text{pH} = 14\) for the above couple. \(K_{sp}\) of \(\text{Cu}(\text{OH})_2\) is \(1.0 \times 10^{-19}\).
Answer: Given data:
\(E^{\circ}_{\text{Cu}^{2+}/\text{Cu}} = +0.34\text{ V}\)
\(\text{pH} = 14 \Rightarrow [\text{OH}^{-}] = 10^{-\text{pH}} = 10^{-14}\text{ M}\)
\(K_{sp} \text{ of } \text{Cu}(\text{OH})_2 = 1.0 \times 10^{-19}\)
The solubility product expression for \(\text{Cu}(\text{OH})_2\) is:
\(K_{sp} = [\text{Cu}^{2+}][\text{OH}^{-}]^2\)
Substitute the known values to find \([\text{Cu}^{2+}]\):
\(1.0 \times 10^{-19} = [\text{Cu}^{2+}](10^{-14})^2\)
\(1.0 \times 10^{-19} = [\text{Cu}^{2+}] \times 10^{-28}\)
\([\text{Cu}^{2+}] = \frac{1.0 \times 10^{-19}}{10^{-28}} = 1.0 \times 10^9\text{ M}\)
Now, apply the Nernst equation for the half-cell reaction \(\text{Cu}^{2+} + 2\text{e}^{-} \rightarrow \text{Cu}\):
\(E = E^{\circ} - \frac{0.0591}{n} \log\left(\frac{1}{[\text{Cu}^{2+}]}\right)\)
Here \(n = 2\):
\(E = 0.34 - \frac{0.0591}{2} \log\left(\frac{1}{1.0 \times 10^9}\right)\)
\(E = 0.34 - \frac{0.0591}{2} \log(10^{-9})\)
\(E = 0.34 - \frac{0.0591}{2} (-9)\)
\(E = 0.34 + 0.02955 \times 9\)
\(E = 0.34 + 0.26595 = +0.606\text{ V}\)
In simple words: Use the solubility product to find the copper ion concentration at pH 14, then plug it into the Nernst equation to find the new reduction potential.

Exam Tip: Changes in pH affect the concentration of hydroxide ions, which in turn alter metal ion concentrations through precipitation equilibria handled via Nernst equation.

Question 9. Calculate pH of the following half cell \(\text{Pt}, \text{H}_2 / \text{H}_2\text{SO}_4\). The oxidation potential is +0.3 V.
Answer: Given data:
Oxidation potential (\(E_{\text{ox}}\)) = +0.3 V
Standard oxidation potential of hydrogen electrode (\(E^{\circ}_{\text{ox}}\)) = 0.0 V
The oxidation half-cell reaction is:
\(\text{H}_2(\text{g}) \rightarrow 2\text{H}^{+}(\text{aq}) + 2\text{e}^{-}\)
Assume standard pressure of \(\text{H}_2\) gas is 1 atm (\(P_{\text{H}_2} = 1\text{ bar}\)).
Apply the Nernst equation for oxidation potential:
\(E_{\text{ox}} = E^{\circ}_{\text{ox}} - \frac{0.0591}{n} \log\left(\frac{[\text{H}^{+}]^2}{P_{\text{H}_2}}\right)\)
Substitute the values (\(n = 2\), \(E^{\circ}_{\text{ox}} = 0\)):
\(0.3 = 0 - \frac{0.0591}{2} \log([\text{H}^{+}]^2)\)
\(0.3 = -0.0591 \log[\text{H}^{+}]\)
Since \(\text{pH} = -\log[\text{H}^{+}]\):
\(0.3 = 0.0591 \times \text{pH}\)
\(\text{pH} = \frac{0.3}{0.0591} \approx 5.08\)
In simple words: Use the Nernst equation relating cell potential to hydrogen ion concentration, then convert it directly into pH.

Exam Tip: Remember that pH is defined as \(-\log[\text{H}^{+}]\); be careful with negative signs when simplifying logarithmic terms in the Nernst equation.