31 Year NEET Previous Year Questions: Redox Reactions - NEET MCQ

The element undergo oxidation itself and reduces others is known as reducing agent.
In this reaction O. N. of Ni Changes from 0 to + 2 and hence Ni acts as a reducing agent.

O.N. of P in H₃PO₃ (phosphorous acid) 3 × 1 + x + 3 × (– 2) = 0 or x = + 3 In orthophosphoric acid (H₃PO₄) O.N. of P is + 5, in hypophosphorous acid (H₃PO₂) it is + 1 while in metaphosphoric acid (HPO₃), it is + 5,

In this reaction, none of the elements undergoes a change in oxidation number or valency.

Let x = oxidation no. of Cr in K₂Cr₂O₇.
∴ (2 × 1) + (2 × x) + 7 (– 2) = 0
or 2 + 2x – 14  = 0 or x = + 6.

Losing of electron is called oxidation,

Carbon has the maximum oxidation state of + 4, therefore carbon dioxide (CO₂) cannot act as a reducing agent.

(not redox reaction)

here oxidation of N₂ & reduction of O₂ is taking place

(not redox reaction)

Pyrophosphoric acid H₄P₂O₇  
Let oxidation state of phosphorus is x(4 × 1 + (– 2) × 7 + 2 x) = 0
∴ 2x = 10   or  x = +5

operate eq. (1) × 4 + eq. (2) × 1

Oxidation number of a compound must be 0.
Using the values for A, B and C in the four options we find that A3(BC₄)₂ is the answer.
Check : (+2)3 + [(+5)+4(–2)]2 = 6 + (5–8)2 = 0

Zinc gives H₂ gas with dil H₂SO₄/HCl but not with HNO₃ because in HNO₃, NO₃^– ion is reduced and give NH₄NO₃, N₂O, NO and NO₂ (based upon the concentration of HNO₃)

Zn is on the top position of hydrogen in electrochemical series. So Zn displaces H₂ from dilute H₂SO₄ and HCl with liberation of H₂.

The correct answer is option B
Oxidation state of S₂​O₄²⁻​
2(x) + 4(−2) = −2
 2x = 8 − 2
 2x = 6
  x = 3
Oxidation state of SO₃²⁻​
x + 3(−2) = −2
 x = 6 − 2
 x = 4
Oxidation state of S₂​O₆²⁻​
2(x) + 6(−2) = −2
 2x = 12 − 2
 2x = 10
   x = 5
So the oxidation state of sulphur in the anions S₂​O₄²⁻​, S₂​O₄²⁻​ and S2​O62−​ follows the order.S₂​O₄²⁻​ < SO₃²⁻​ < S₂​O₆²⁻​.
 

(i) Sum of oxidation states of all atoms = charge of ion.
(ii) oxidation number of oxygen = -2
Let the oxidation state of P in PO₄^3- is x.
 PO₄^3- 

x + 4 (-2) = - 3
x-8 = - 3
x = +5

Let the oxidation state of S in SO₄^2- is y
y + 4(-2) = -2
y-8 = - 2
y = +6

Let the oxidation state of Cr in Cr₂O₇^2- is z.
2 x z+7(-2) = -2
2z-14 = - 2
z=+6
Hence, oxidation state of P, S and Cr are +5, +6 and +6

The correct answer is Option B.

The reaction of chlorine gas with hot and concentrated sodium hydroxide solution is

3Cl2 ​+ 6NaOH ⟶ NaClO3 ​+ 5NaCl + 3H2​O

 Oxidation number of Cl is 0 in Cl2, −1 in NaCl and +5 in NaClO3

​So, the oxidation number of chlorine changes from Zero to -1 and Zero to +5.

i.e. maximum change in oxidation number is observed in Cl (+5 to –1).

Higher the value of reduction potential higher will be the oxidising power whereas the lower the value of reduction potential higher will be the reducing power.