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Redox Reaction Part-01

Redox Reaction DPP-01

1. Oxidation is defined as –
(1) Gain of electrons
(2) Decrease in positive valency
(3) Loss of electrons
(4) Addition of electropositive element

2. Reduction is defined as –
(1) Increase in positive valency
(2) Gain of electrons
(3) Loss of protons
(4) Decrease in negative valency

3. In the reaction MnO4– + SO3–2 + H+ ⎯⎯→ SO4–2 + Mn2+ + H2O

(1) MnO4– and H+ both are reduced
(2) MnO4– is reduced and H+ is oxidised
(3) MnO4– is reduced and SO32– is oxidised
(4) MnO4– is oxidised and SO32– is reduced

4. Which of the following reactions do not involve oxidation-reduction ?

(1) 2Rb + 2H2O ⎯⎯→ 2RbOH + H2
(2) 2CuI2 ⎯⎯→ 2CuI + I2
(3) NH4Cl + NaOH ⎯⎯→ NaCl + NH3 + H2O
(4) 3Mg + N2 ⎯⎯→ Mg3N2

5. Choose the redox reaction from the following–

(1) Cu + 2H2SO4 ⎯⎯→ CuSO4 + SO2 + 2H2O
(2) BaCl2 + H2SO4 ⎯⎯→ BaSO4 + 2HCl
(3) 2NaOH + H2SO4 ⎯⎯→ Na2SO4 + 2H2O
(4) KNO3 + H2SO4 ⎯⎯→ 2HNO3 + K2SO4

6. Which of the following is not a redox reaction ?

(1) MnO4– ⎯⎯→ MnO2 + O2
(2) Cl2 + H2O ⎯⎯→ HCl + HClO
(3) 2CrO42– + 2H+ ⎯⎯→ Cr2O72– + H2O
(4) MnO4– + 8H+ + 5Ag ⎯⎯→ Mn+2 + 4H2O + 5Ag+

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Redox Reaction Part-01

Answer Key
Question 1 2 3 4 5 6
Answer 3 2 3 3 1 3

SOLUTIONS DPP-01

1. According to modern theory of oxidation, the atom/ions which looses electron are oxidised.

2. According to modern theory of reduction, the atoms/ions which gain electrons are reduced.

3.

+7 +4 +6
MnO4− S O3−2 S O4−2

+7
Here, MnO4− gains 5e–and gets reduced to Mn+2 and SO3–2 loses 2e– and oxidized to SO4–2.

−3 +1 −1 +1 −2 +1 +1 −1 −3 +1 +1 −2
4. NH4 Cl + NaOH ⎯⎯
→ NaCl + NH3 + H2 O
In the above reaction as we can see that there is no change in oxidation state of any atoms/groups.

0 +6 +2 +4
5. Cu + 2H2 S O4 ⎯⎯
→ C uSO4 + S O2 + 2H2O
+4
Cu gets oxidized to Cu+2 and S gets reduced from SO4–2 to S O2 .

6.

CrO42− Cr2O72−

As there is no change in oxidation state, it is not a redox reaction.

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Redox Reaction Part-03

Redox Reaction DPP-02

1. In which of the following compounds, the oxidation state of I-atom is highest ?

(1) KI3
(2) KIO4
(3) KIO3
(4) IF5

2. The oxidation number of phosphorus in Ba(H2PO2)2 is –

(1) +3
(2) +2
(3) +1
(4) –1

3. Oxidation number of Ni in Ni(CO)4 is –

(1) 0
(2) 4
(3) 8
(4) 2

4. Positive oxidation state of an element indicates that it is –

(1) Elementry form
(2) Oxidised
(3) Reduced
(4) Only reductant

5. Predict the highest and lowest oxidation state of (a) Ti and (b) Tl in combined state.
(1) a[0, +3] b[0, +2]
(2) a[+3, 0] b[+4, 0]
(3) a[+4, 0] b[+4, 0]
(4) a[+4, +2] b[+3, +1]

6. The oxidation state of oxygen atom in potassium superoxide is –

(1) Zero
(2) –1/2
(3) –1
(4) –2

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Redox Reaction Part-03

7. Sulphur has highest oxidation state in

(1) SO2
(2) H2SO4
(3) Na2S2O3
(4) Na2S4O6

8. Oxidation number of oxygen in O2 molecule is

(1) + 1
(2) 0
(3) + 2
(4) – 2

9. Oxidation number of N in (NH4)2SO4 is

(1) – 1 / 3
(2) – 1
(3) + 1
(4) – 3

10. The oxidation number of Pt in [Pt(C2H4)Cl3]– is

(1) + 1
(2) + 2
(3) + 3
(4) + 4

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Redox Reaction Part-03

Answer Key
Question 1 2 3 4 5 6 7 8 9 10
Answer 2 3 1 2 4 2 2 2 4 2

SOLUTIONS DPP-02

1. In KIO4 oxidation state of I is +7

ON of K + ON of I + 4 × ON of O = 0
(+1) × 1 + (x) × 1 + 4 × (–2) = 0
⇒ x = +7

2. Ba(H2PO2)2 ⇒ Ba+2 + 2H2PO2–1

ON of H × 2 + ON of P × 1 + ON of O × 2 = Total charge
(+1) × 2 + (x) × 1 + (–2) × 2 = –1 ⇒ x = +1

3. Since, carbonyl [CO] group is neutral ligand thus, ON of Ni in Ni(CO)4 is zero.

4. When any element is oxidized, generally it gains positive charge on it.

5. Ti22 → 1s2 2s2 2p6 3s2 3p6 4s2 3d2 Highest → +4 Lowest → + 2
Tl81 → [Xe]4f 5d 6s 6p
14 10 2 1
Highest → +3 Lowest → + 1

6. KO2 ⇒ ON of K × 1 + ON of O × 2 = 0
(+1) × 1 + x × 2 = 0
1
x=–
2


7. SO2 = +4

H2SO4 = +6

Na 2S2O3 = +2

5
Na 2S4O6 = + .
2

8. Each molecule always show zero oxidation state in elemental state.

9. (NH4 )2 SO4 ⇌ 2NH4+ + SO4− 2

x + 4 = +1; x = 1 – 4 = –3.

10. ON of Pt + ON of C2H4 + ON of Cl × 3 = Total charge

(x) × 1 + (0) × 1 + (–1) × 3 = – 1 ⇒ x = +2

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Redox Reaction Part-04

Redox Reaction DPP-03

1. Which of the following behaves as both oxidising and reducing agents

(1) H2SO4
(2) SO2
(3) H2S
(4) HNO3

2. The reaction H2S + H2O2 → 2H2O + S shows

(1) Oxidizing action of H2O2
(2) Reducing action of H2O2
(3) Alkaline nature of H2O2
(4) Acidic nature of H2O2

3. In the reaction
C2O24− + MnO4− + H+ →Mn2+ + CO2 + H2O
the reductant is
(1) C2O24−
(2) MnO4−
2+
(3) Mn
+
(4) H

4. Identify the correct statement about H2O2

(1) It acts as reducing agent only
(2) It acts as both oxidising and reducing agent
(3) It is neither an oxidiser nor reducer
(4) It acts as oxidising agent only

5. Which of the following can work as oxidising agent

(1) O2
(2) MnO4–
(3) I2
(4) All of these

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Redox Reaction Part-04

Answer Key
Question 1 2 3 4 5
Answer 2 1 1 2 4

SOLUTIONS DPP-03

1. In SO2 oxidation state of S is +4. Thus, it can gain as well as donate its electron. Hence, it behaves as both
oxidising and reducing agent.

−2 −1 −2 0
2. H2 S + H2 O2 ⎯⎯
→ 2H2 O + S
H2O2 oxidises sulphur from –2 to 0.

3. C2O24− + MnO4− + H+ → Mn2+ + CO2 + H2O .

In this reaction C2O24− act as a reducing agent, as it converts Mn+7 to Mn+2.

4. In H2O2 oxidation state of O is –1 thus, it can gain as well as donate its electron. Hence, it acts as both oxidising
and reducing agent.

5. All of these acts as an oxidising agent.

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Redox Reaction Part-07

Redox Reaction DPP-04

1. In the half reaction : 2ClO3– ⎯⎯→ Cl2

(1) 5 electrons are gained
(2) 5 electrons are liberated
(3) 10 electrons are gained
(4) 10 electrons are liberated

2. The number of electrons required to balance the following equation –

NO3– + 4H+ + e– ⎯⎯→ 2H2O + NO are –
(1) 5
(2) 4
(3) 3
(4) 2

3. Which of the following equations is a balanced one ?

(1) 5BiO3– + 22H+ + Mn2+ ⎯⎯→ 5Bi3+ + 7H2O + MnO4–
(2) 5BiO3– + 14H+ + 2Mn2+ ⎯⎯→ 5Bi3+ + 7H2O + 2MnO4–
(3) 2BiO3– + 4H+ + Mn2+ ⎯⎯→ 2Bi3+ + 2H2O + MnO4–
(4) 6BiO3– + 12H+ + 3Mn2+ ⎯⎯→ 6Bi3+ + 6H2O + 3MnO4–

4. C2H6 (g) + nO2 → CO2 (g) + H2O(l)

In this equation, the ratio of the coefficients of CO2 and H2O are
(1) 1 : 1
(2) 2 : 3
(3) 3 : 2
(4) 1 : 3

5. 2MnO4− + 5H2O2 + 6H+ → 2 Z + 5O2 + 8H2O . In this reaction Z is

(1) Mn+2
(2) Mn+4
(3) MnO2
(4) Mn

6. In the balanced chemical reaction,

IO3− + a I− + b H+ → c H2O + d I2
a, b, c and d respectively correspond to
(1) 5, 6, 3, 3
(2) 5, 3, 6, 3
(3) 3, 5, 3, 6
(4) 5, 6, 5, 5

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Redox Reaction Part-07

Answer Key
Question 1 2 3 4 5 6
Answer 3 3 2 2 1 1

SOLUTIONS DPP-04

+5 0
1. 2ClO3− ⎯⎯
→ Cl2
2 × 5 = 10, electrons are gained in this half cell reaction.

2. NO3– + 4H+ + xe– → 2H2O + NO

3e– will balance the following equation.

3. 5BiO3– + 14H+ + 2Mn2+ ⎯⎯→ 5Bi3+ + 7H2O + 2MnO4–

4. The balanced chemical equation is

2C2H6 + 7O2 → 4CO2 + 6H2O.
Ratio of the coefficients of CO2 and H2O is 4 : 6 or 2 : 3.

5. 2MnO–4 + 5H2O2 + 6H+ → 2Mn2+ + 5O2 + 8H2O

6. IO3– + aI– + bH+ → cH2O + dI2

Step 1 : I–1 → I2 (oxidation)
IO3– → I2 (reduction)
Step 2 : 2IO3– + 12H+ → I2 + 6H2O
Step 3 : 2IO3– + 12H+ + 10e → I2 + 6H2O
2I– → I2 + 2e
Step 4 : 2IO3– + 12H+ + 10e– → I2 + 6H2O
[2I– → I2 + 2e]5
Step 5 : 2IO3– + 10I– + 12H+ → 6I2 + 6H2O
IO3– + 5I– + 6H+ → 3I2 + 3H2O
On comparing, a = 5, b = 6, c = 3, d = 3

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Redox Reaction Part-10

Redox Reaction DPP-05

1. Molecular weight of KMnO4 in acidic medium and neutral medium will be respectively –
(1) 7 × equivalent weight and 2 × equivalent weight
(2) 5 × equivalent weight and 3 × equivalent weight
(3) 4 × equivalent weight and 5 × equivalent weight
(4) 2 × equivalent weight and 4 × equivalent weight

2. In acidic medium, equivalent weight of K2Cr2O7 (Molecular weight = M) is –

(1) M/3
(2) M/4
(3) M/6
(4) M/2

3. In alkaline condition KMnO4 reacts as 2KMnO4 + 2KOH → 2K 2MnO4 + H2O + O . The equivalent weight of
KMnO4 would be (Atomic mass of K = 39, Mn = 55, O = 16)
(1) 158.0
(2) 79.0
(3) 52.7
(4) 31.6

4. What is ‘A’ in the following reaction 2Fe3+(aq) + Sn2+(aq) → 2Fe2+(aq) + A

(1) Sn3+(aq)
(2) Sn4+(aq)
(3) Sn2+(aq)
(4) Sn

5. The number of moles of KMnO4 reduced by one mole of KI in alkaline medium is:
(1) One fifth
(2) five
(3) One
(4) Two
6. How many moles of K2Cr2O7 can be reduced by 1 mole of Sn2+
(1) 1/3
(2) 1/6
(3) 2/3
(4) 1

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Redox Reaction Part-10

7. What is the equivalent mass of IO4− when it is converted into I2 in acidic medium
(1) M/6
(2) M/7
(3) M/5
(4) M/4

8. For decolourization of 1 mole of KMnO4, the moles of H2O2 required is

(1) 1/2
(2) 3/2
(3) 5/2
(4) 7/2

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Redox Reaction Part-10

Answer Key
Question 1 2 3 4 5 6 7 8
Answer 2 3 1 2 4 1 2 3

SOLUTIONS DPP-05

+7
1. In acidic medium, MnO4− ⎯⎯
→Mn +2 5e– change
+7 +4
Neutral medium, MnO4− ⎯⎯
→MnO2 3e– change
Molecular weight = n factor × equivalent weight
5 × equivalent weight and 3 × equivalent weight

2. n factor = 6
Molecular weight
Equivalent weight =
6

3. e− + Mn7+ → Mn6+
M
E= .
1

4.

2Fe3+ + Sn2+ → 2Fe2+ + Sn4+

5. In alkaline medium
2KMnO4 + KI + H2O → 2MnO2 + 2KOH + KIO3 .

6. Cr2O72− + 14H+ + 6e− → 2Cr3+ + 7H2O

(Sn2+ → Sn4+ + 2e− ) 3

Cr2O72− + 14H+ + 3Sn2+ → 3Sn4+ + 2Cr3+ + 7H2O

It is clear from this equation that 3 moles of Sn2+ reduce one mole of Cr2O72− , hence 1 mol. of Sn2+ will reduce
1
moles of Cr2O72− .
3

7. Equivalent mass
Molecular weight
=
Change in oxidation number per mole
Suppose molecular weight is M
Oxidation number of I2 in IO4− in
Acidic medium i.e., I × (–8) + 1e– = +7
So eq. wt. = M/7.

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Redox Reaction Part-10

8. →K2SO4 + 2MnSO4 + 3H2O + 5O

2KMnO4 + 3H2SO4 ⎯⎯
5H2O2 + 5O ⎯⎯
→5H2O + 5O2
→K2SO4 + 2MnSO4 + 8H2O + 5O2
2KMnO4 + 3H2SO4 + 5H2O2 ⎯⎯

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