NCERT MOST IMPORTANT QUESTIONS CLASS – 11| CHEMISTRY IMPORTANT QUESTIONS PART-2 | CHAPTER – 1 | REDOX REACTION | EDUGROWN |

In This Post we are  providing Chapter-1 REDOX REACTION NCERT MOST IMPORTANT QUESTIONS for Class 11 CHEMISTRY which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These MCQS  can be really helpful in the preparation of Board exams and will provide you with a brief knowledge of the chapter.

NCERT MOST IMPORTANT QUESTIONS ON REDOX REACTION

Question 1.
What are redox reactions? Give an example.

Answer:
Redox reaction is a reaction in which oxidation and reduction take place simultaneously, e.g.
Zn (s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s)

Question 2.
Define oxidation and reduction in terms of electrons.

Answer:
Oxidation involves loss and reduction involves the gain of electrons.

Question 3.
How many grams of K₂Cr₂O₇ is required to oxidize Fe²⁺ present in 15.2 gm of FeSO₄ to Fe³⁺ if the reaction is carried out in an acidic medium.

Answer:
The balanced chemical equation for the redox reaction is
K₂Cr₂O₇ + 6FeSO₄ + 7H₂SO₄ → K₂SO₄ + Cr₂(SO₄)₃ + 7H₂O
from the balanced equation, it is clear that 6 moles of FeSO₄ = 1

a mole of K₂Cr₂O₇ or 6 × 152 gm of FeSO₄ are oxidized by
K₂Cr₂O₇ = 294 gm
or
15.2 gm of FeSO₄ are oxidized by K₂Cr₂O₇

Question 4.
15.0 cm3 of 0.12 M KMnO₄ solution are required to oxidize 20 ml of FeSO₄ solution in an acidic medium. What is the concentration of FeSO₄ solution?

Answer:
The balanced chemical equation for the redox reaction is
2KMnO₄ + 10FeSO₄ + 8H₂SO₄ → K₂SO₄ + 2MnSO₄ + 5Fe₂(SO₄)₃ + 8 H₂O

Applying molarity equation to the above redox reaction

Question 5.
16.6 gm of pure KI was dissolved in water and the solution was made up to one liter, cm³ of this solution was acidified with 20 cm³ of 2 MHCl the resulting solution required 10 cm³ of decinormal KIO₃ for complete oxidation of I- ions to ICl. Find out the value of v.

Answer:
The chemical equation for the redox reaction is
IO- + 2I^– + 6HCl → 3ICl + 3Cl^– + 3H₂O

Molarity of KI solution = 16.6166 = 0.1 m

Applying molarity equation
0.1×v2 (KI) = 10×0.11(KIO3)
v = 20 cm³

Question 6.
Calculate the cone of hypo (Na₂S₂O₃ 5H₂O) solution in g dm-3 if 10.0 of this solution decolorized 15 ml of M/40 iodine solution.

Answer:
The balanced equation for the redox reaction is
2S₂O₃^2- + I₂ → 2I^– + S₄O₆^2-
from the balanced equation, it is evident that
2 moles of Na₂S₂O₃ = 1 mole of I₂

Applying the molarity equation we have,

Thus, the molarity of the hypo solution = 3/40 M
mol. mass of Na₂S₂O₃.5H₂O = 248 g mol^-1
cone, of Na₂S₂O₃.5H₂O = 248×340
= 18.6 gdm^-3

Question 7.
How many millimoles of potassium dichromate is required to oxidize 24 cm³ of 0.5 M mohr’s salt solution in an acidic medium.

Answer:
No. of millimoles of K₂Cr₂O₇ present in 24 cm³ of 0.5 m solution = 24 × 0.5 = 12
The balanced chemical equation for the redox reaction is
K₂Cr₂O₇ + 6(NH₄)₂SO₄.FeSO₄.6H₂O + 7H₂SO₄ → K₂SO₄ + 6(NH₄)₂SO₄ + 3Fe₂(SO₄)₃ + Cr₂(SO₄)₃ + 43H₂O from the balanced equations.

6 moles mohr’s salt are oxidised by K₂Cr₂O₇ = 1 moles
∴ 12 millimoles of mohr’s salt will be oxidised by
K₂Cr₂O₇ = 16 × 12 = 2 millimoies.

Question 8.
2.48 gm of Na₂S₂O₃.xH₂O was dissolved per liter of the solution. 20 cm³ of this solution required 10 cm³ of 0.01 M iodine solution. Find out the value of x?

Answer:
The balanced equation for the redox reaction is
2Na₂S₂O₃ + I2 → Na₂S₄O₆ + 2NaI
Let the molarity of Na₂S₂O₃ .xH₂O solution = M₁.

Applying molarity equation to the above redox reaction, we have
M1×202(Na2S2O3) = 10×.011(I2)
∴ M₁ = 0.01 M

mol wt. of Na₂S₂O₃.xH₂O
= 2 × 23 + 2 × 32 + 3 × 16 + x × 18
= 158 + 18x

Amount of Na₂S₂O₃.xH₂O present per litre
= (158 + 18x) × 0.01 g

But the actual amount dissolved = 2.48 g. equating these values, we have
(158 + 18x) × 0.01 = 2.48
or
x = 5.

Question 9.
The half cell reactions with their oxidation potentials are:
Pb(s) → Pb²⁺(aq) + 2e^–, E°_oxi = + 0.13 v.
Ag(s) → Ag⁺(aq) + e^–, E°_oxi = – 0.80 v.
Write the cell reaction and calculate its EMF.

Answer:
The equations are as
Pb²⁺(aq) + 2e^– → Pb(s), E°_oxi = – 0.13 v
Ag⁺(aq) + e^– → Ag(s), (E° = + 0.80 v) …(2)

To obtain the equation for the cell reaction, multiply equation (2) by 2 and subtract equation (1) from it we get
Pb(s) + 2Ag⁺(aq) → Pb²⁺(aq) + 2Ag(s)
E°_cell = + 0.80 – ( – 0.13) = + 0.93 v

Question 10.
Predict whether zinc or silver reacts with 1 M H₂SO₄ to give out hydrogen or not. Given that the standard potential of zinc & silver are – 0.76 v & + 0.80 v respectively.

Answer:
(a) To predict the reaction of zinc with H₂SO₄ If Zn reacts, the following reactions should take place.
Zn + H₂SO₄ → ZnSO₄ + H₂
i.e. Zn + 2H⁺ → Zn²⁺ + H₂

By conventions, the cell will be represented as-
Zn / Zn²⁺ ∥ H⁺ / H₂

standard EMF of the cell,
E°cell = E°_H⁺/H2 – E°_zn²⁺/zn
= 0 – (- 0.76) = + 0.76 v

Thus the EMF of the cells comes out to be positive. Hence the reaction takes place.

(b) To predict the reaction, of silver with H₂SO₄.
If Ag reacts, the following reactions should take place.
2Ag + H₂SO₄ → Ag₂SO₄ + H₂
2Ag + 2H⁺ → 2Ag⁺ + H₂

By convention, the cell may be written as
Ag / Ag⁺ ∥ H⁺ / H₂
E°_cell = E°_H⁺,H2 – E°_Ag⁺,Ag
= 0 – 0.80 = – 0.80 v.
The EMF of the cell is negative
Hence, this reaction does not take place.