CBSE Class 11 Chemistry Redox Reactions Worksheet Set A - Free Printable

Let's solve each question step by step with detailed explanations.

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1. Find the oxidation number of the element underlined.

We are to find the oxidation number (O.N.) of the underlined element in each compound. Since the image doesn't show which element is underlined, I'll assume based on standard conventions that the first element listed or the one likely to vary is underlined. Let’s go through each:

#### a) Na₃VO₄
- Na = +1 (always)
- O = -2
- Let O.N. of V = x

Total charge = 0
3(+1) + x + 4(-2) = 0
3 + x - 8 = 0 → x = +5
✔ V = +5

#### b) K₂Cr₂O₇
- K = +1
- O = -2
- Let Cr = x

2(+1) + 2x + 7(-2) = 0
2 + 2x - 14 = 0 → 2x = 12 → x = +6
✔ Cr = +6

#### c) CH₄
- H = +1
- Let C = x

x + 4(+1) = 0 → x + 4 = 0 → x = -4
✔ C = -4

#### d) SO₂Cl₂
- O = -2, Cl = -1
- Let S = x

x + 2(-2) + 2(-1) = 0
x - 4 - 2 = 0 → x = +6
✔ S = +6

#### e) NO₂⁻
- O = -2
- Let N = x

x + 2(-2) = -1 → x - 4 = -1 → x = +3
✔ N = +3

#### f) BrF₃
- F = -1
- Let Br = x

x + 3(-1) = 0 → x - 3 = 0 → x = +3
✔ Br = +3

#### g) Na₂S₄O₆
This is sodium tetrathionate. The structure has two S atoms at +5 and two at 0 (but average is calculated).

Let S = x
2(+1) + 4x + 6(-2) = 0
2 + 4x - 12 = 0 → 4x = 10 → x = +2.5
But since it's an average: ✔ Average O.N. of S = +2.5

> Note: In S₄O₆²⁻, two sulfur atoms are in +5 state and two are 0 (disulfide bond), so average is +2.5.

#### h) CH₃Cl
- H = +1, Cl = -1
- Let C = x

x + 3(+1) + (-1) = 0 → x + 3 - 1 = 0 → x = -2
✔ C = -2

#### i) ClO₄⁻
- O = -2
- Let Cl = x

x + 4(-2) = -1 → x - 8 = -1 → x = +7
✔ Cl = +7

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2. Differentiate between:

#### a) Valency vs Oxidation Number

| Valency | Oxidation Number |
|--------|------------------|
| It is the combining capacity of an element. | It is the charge an atom would have if all bonds were ionic. |
| Always a whole number (positive or negative). | Can be fractional (e.g., in Fe₃O₄, Fe is +8/3). |
| Based on electron sharing/gain/loss. | Based on electronegativity and assumed ionic character. |
| Does not consider polarity. | Considers polarity of bonds. |
| Example: Oxygen has valency 2. | Oxygen usually has O.N. -2, but can vary. |

#### b) Activity Series vs Electrochemical Series

| Activity Series | Electrochemical Series |
|------------------|------------------------|
| Arranged by reactivity (metal displacement). | Arranged by standard electrode potential (E°). |
| Based on experimental observations. | Based on measured potentials. |
| Only metals (and hydrogen). | Includes both metals and non-metals. |
| Predicts displacement reactions. | Predicts spontaneity of redox reactions. |
| No numerical values. | Has E° values (in volts). |

#### c) Oxidizing Agent vs Reducing Agent

| Oxidizing Agent | Reducing Agent |
|------------------|----------------|
| Accepts electrons. | Donates electrons. |
| Gets reduced. | Gets oxidized. |
| Causes oxidation in other species. | Causes reduction in other species. |
| Example: KMnO₄, K₂Cr₂O₇ | Example: Zn, H₂, SO₂ |

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3. Define:

#### i) Oxidation Number
The hypothetical charge on an atom if all bonds were ionic. It helps track electron transfer in redox reactions.

#### ii) Redox Couple
A pair of species differing by one electron (e.g., Fe³⁺/Fe²⁺). Represents the oxidized and reduced forms involved in a redox reaction.

#### iii) Displacement Redox Reaction
A reaction where a more reactive element displaces a less reactive one from its compound.
Example: Zn + Cu²⁺ → Zn²⁺ + Cu

#### iv) Standard Electrode Potential (E°)
The tendency of a half-cell to gain electrons compared to the standard hydrogen electrode (SHE) under standard conditions (1 M, 25°C, 1 atm).

#### v) Stock Notation
A method of indicating oxidation state using Roman numerals in parentheses after the element name.
Example: Iron(III) chloride = FeCl₃ (Fe = +3)

#### vi) Disproportionation Reactions
A type of redox reaction where the same element is both oxidized and reduced.
Example: 3Cl₂ + 6OH⁻ → 5Cl⁻ + ClO₃⁻ + 3H₂O (Cl is both oxidized to +5 and reduced to -1)

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4. Identify the redox reactions and classify them.

#### a) CuO + H₂ → Cu + H₂O
- Cu: +2 → 0 (reduced)
- H: 0 → +1 (oxidized)
✔ Redox reaction
Classification: Displacement (H₂ displaces Cu from CuO)

#### b) 2Na + H₂ → 2NaH
- Na: 0 → +1 (oxidized)
- H: 0 → -1 (reduced)
✔ Redox reaction
Classification: Combination (two elements combine)

#### c) AgNO₃ + NaCl → AgCl + NaNO₃
- All elements keep same oxidation states.
- Ag: +1, N: +5, O: -2, Na: +1, Cl: -1
✘ Not a redox reaction (double displacement)

#### d) CaCO₃ → CaO + CO₂
- Ca: +2, C: +4, O: -2 → no change
✘ Not a redox reaction (thermal decomposition)

#### e) 2HCHO + NaOH → HCOONa + CH₃OH
Check oxidation states:
- In HCHO: C = 0 (H: +1, O: -2 → x + 2(+1) + (-2) = 0 → x = 0)
- In HCOONa: C in formate = +2 (H: +1, O: -2 × 2, Na: +1 → x + 1 - 4 + 1 = 0 → x = +2)
- In CH₃OH: C = -2 (H: +1 × 4, O: -2 → x + 4 - 2 = 0 → x = -2)

So one carbon goes from 0 → +2 (oxidized), other from 0 → -2 (reduced)
✔ Redox reaction
Classification: Disproportionation (same element, formaldehyde, both oxidized and reduced)

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5. Name a compound in which:

#### a) Hydrogen exists in:
i) +1: H₂O, HCl — common
ii) -1: NaH (sodium hydride) — hydrogen is hydride ion
iii) +2: Not possible (hydrogen max +1)
iv) -2: Not possible (only -1 in hydrides)

✔ So:
- i) H₂O
- ii) NaH
- iii) ✘ Not possible
- iv) ✘ Not possible

#### b) Oxygen exists in:
i) +1: In OF₂ (oxygen difluoride) — F is more electronegative → O = +2? Wait:

Wait: F = -1, so for OF₂:
x + 2(-1) = 0 → x = +2 → O = +2

But we need +1?

No common compound with O = +1.

But in H₂O₂, oxygen is -1.

In O₂F₂, O = +1 (each F = -1 → 2x + 2(-1) = 0 → x = +1)

✔ O₂F₂ — oxygen in +1 state

ii) -1: H₂O₂ (hydrogen peroxide)
iii) +2: OF₂ (oxygen difluoride)
iv) -2: H₂O, MgO

So:
- i) O₂F₂
- ii) H₂O₂
- iii) OF₂
- iv) H₂O

#### c) Chlorine exists in:
i) +1: NaOCl (sodium hypochlorite) — Cl = +1
ii) -1: NaCl
iii) +3: NaClO₂ (sodium chlorite) — Cl = +3
iv) -2: Not typical; chlorine rarely shows -2. But in ClF₃, Cl = +3, not -2.

Chlorine can be -1 (common), -2 is very rare.

Actually, in CaCl₂, Cl = -1.

There is no stable compound where Cl has -2 oxidation state.

But in Cl⁻, it's -1.

So:
- i) NaOCl
- ii) NaCl
- iii) NaClO₂
- iv) ✘ Not possible (Cl cannot be -2)

> However, in some exotic compounds like ClF⁻, Cl might be -1, but -2 is not feasible.

So answer:
- i) NaOCl
- ii) NaCl
- iii) NaClO₂
- iv) ✘ Not possible

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6. Balance the following equations (acidic medium)

Use half-reaction method.

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#### a) MnO₄⁻ + Br⁻ → Mn²⁺ + Br₂ [acid]

Step 1: Write half-reactions.

Reduction: MnO₄⁻ → Mn²⁺
Mn: +7 → +2 → gains 5e⁻

Balance:
- MnO₄⁻ → Mn²⁺
- Add 4H₂O to right to balance O
- Add 8H⁺ to left to balance H
- Add 5e⁻ to left

→ MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O

Oxidation: Br⁻ → Br₂
2Br⁻ → Br₂ + 2e⁻

Now balance electrons:
- Multiply reduction by 2: 2MnO₄⁻ + 16H⁺ + 10e⁻ → 2Mn²⁺ + 8H₂O
- Multiply oxidation by 5: 10Br⁻ → 5Br₂ + 10e⁻

Add:
2MnO₄⁻ + 10Br⁻ + 16H⁺ → 2Mn²⁺ + 5Br₂ + 8H₂O

✔ Balanced.

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#### b) NO₃⁻ + Bi → Bi³⁺ + NO [acid]

Reduction: NO₃⁻ → NO
N: +5 → +2 → gains 3e⁻

Balance:
- NO₃⁻ → NO
- Add 2H₂O to right
- Add 4H⁺ to left
- Add 3e⁻ to left

→ NO₃⁻ + 4H⁺ + 3e⁻ → NO + 2H₂O

Oxidation: Bi → Bi³⁺ + 3e⁻

Electrons match!

Add:
NO₃⁻ + Bi + 4H⁺ → NO + Bi³⁺ + 2H₂O

✔ Balanced.

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#### c) MnO₄⁻ + C₂H₅OH → Mn²⁺ + CO₂ [acid]

Reduction: MnO₄⁻ → Mn²⁺
As before:
MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O

Oxidation: C₂H₅OH → CO₂
Ethanol → carbon dioxide

Carbon in C₂H₅OH:
Let C = x
2x + 6(+1) + (-2) = 0 → 2x + 6 - 2 = 0 → 2x = -4 → x = -2
Each C = -2

In CO₂: C = +4 → each C loses 6e⁻ → total 12e⁻ per ethanol

So:
C₂H₅OH → 2CO₂

Balance:
- C₂H₅OH → 2CO₂
- Add 3H₂O to left to balance O? Wait:

Left: C₂H₅OH → 2C, 6H, 1O
Right: 2CO₂ → 2C, 4O → need to add water

Better:
C₂H₅OH → 2CO₂
Add 3H₂O to right to balance H? No.

Balance atoms:
- C: already balanced
- H: 6H on left → add 6H⁺ to right
- O: 1 on left, 4 on right → add 3H₂O to left?

Try:

C₂H₅OH → 2CO₂
Add 3H₂O to left → now 3O + 1O = 4O on left, 4O on right
H: 6H + 6H = 12H → add 12H⁺ to right
Charge: neutral left, right has 12H⁺ → add 12e⁻ to right

Wait: oxidation should lose electrons.

So:
C₂H₅OH + 3H₂O → 2CO₂ + 12H⁺ + 12e⁻

Yes! Each carbon goes from -2 to +4 → loss of 6e⁻ × 2 = 12e⁻

So oxidation half:
C₂H₅OH + 3H₂O → 2CO₂ + 12H⁺ + 12e⁻

Reduction: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O

Now balance electrons:
LCM of 12 and 5 is 60

Multiply reduction by 12:
12MnO₄⁻ + 96H⁺ + 60e⁻ → 12Mn²⁺ + 48H₂O

Multiply oxidation by 5:
5C₂H₅OH + 15H₂O → 10CO₂ + 60H⁺ + 60e⁻

Add:

12MnO₄⁻ + 5C₂H₅OH + 96H⁺ + 15H₂O + 60e⁻ → 12Mn²⁺ + 48H₂O + 10CO₂ + 60H⁺ + 60e⁻

Cancel:
- 60e⁻ cancel
- H⁺: 96H⁺ - 60H⁺ = 36H⁺ left
- H₂O: 15H₂O on left, 48H₂O on right → move 33H₂O to right

Final:
12MnO₄⁻ + 5C₂H₅OH + 36H⁺ → 12Mn²⁺ + 10CO₂ + 33H₂O

✔ Balanced.

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✔ Final Answers Summary:

#### 1. Oxidation Numbers:
a) V = +5
b) Cr = +6
c) C = -4
d) S = +6
e) N = +3
f) Br = +3
g) S = +2.5 (average)
h) C = -2
i) Cl = +7

#### 2. Differences: As explained above.

#### 3. Definitions: As explained.

#### 4. Redox Identification:
a) Yes – Displacement
b) Yes – Combination
c) No
d) No
e) Yes – Disproportionation

#### 5. Compounds:
a) H: +1 → H₂O; -1 → NaH
b) O: +1 → O₂F₂; -1 → H₂O₂; +2 → OF₂; -2 → H₂O
c) Cl: +1 → NaOCl; -1 → NaCl; +3 → NaClO₂; -2 → ✘ Not possible

#### 6. Balanced Equations:
a) 2MnO₄⁻ + 10Br⁻ + 16H⁺ → 2Mn²⁺ + 5Br₂ + 8H₂O
b) NO₃⁻ + Bi + 4H⁺ → NO + Bi³⁺ + 2H₂O
c) 12MnO₄⁻ + 5C₂H₅OH + 36H⁺ → 12Mn²⁺ + 10CO₂ + 33H₂O

Let me know if you'd like these formatted neatly or need diagrams!