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All questions of Redox Reactions for NEET Exam

Oxidation numbers of P in PO43 , of S in SO42– and that of Cr in Cr2O72–  are respectively [2009]
  • a)
    + 3, + 6 and + 5
  • b)
    + 5, + 3 and + 6
  • c)
    – 3, + 6 and + 6
  • d)
    + 5, + 6 and + 6
Correct answer is option 'D'. Can you explain this answer?

Hansa Sharma answered
(i) Sum of oxidation states of all atoms = charge of ion.
(ii) oxidation number of oxygen = -2
Let the oxidation state of P in PO43- is x.
 PO43- 

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

Let the oxidation state of S in SO42- is y
y + 4(-2) = -2
y-8 = - 2
y = +6

Let the oxidation state of Cr in Cr2O72- 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

Which of the following involves a redox reaction?
  • a)
    Reaction of H2SO4 with NaOH
  • b)
    Production of ozone from oxygen in the atmosphere by lightning
  • c)
    Production of nitrogen oxides from nitrogen and oxygen in the atmosphere by lightning
  • d)
    Evaporation of water
Correct answer is option 'C'. Can you explain this answer?

Rocky Handsome answered
•The 1st reaction is acid base reaction which does not involve either oxidation or reduction.

•O3 formation from O2 does not involve either oxidation or reduction.

•Nitrogen oxides from N2 and O2 involves oxidation of nitrogen and reduction of oxygen.

For example, N2 + O2---> NO2.
• Evaporation of H2O is a physical change and is not a chemical change.

Hence answer is C.

The oxidation number of chromium in potassium dichromate is [1995]
  • a)
    + 6
  • b)
    – 5
  • c)
    – 2
  • d)
    + 2
Correct answer is option 'A'. Can you explain this answer?

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

Phosphorus has the oxidation state of + 3 in
  • a)
    Phosphorous acid [1994]
  • b)
    Orthophosphoric acid
  • c)
    Hypophosphorous acid
  • d)
    Metaphosphoric acid.
Correct answer is option 'A'. Can you explain this answer?

Sushant Goyal answered
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,

The oxidation number of phosphorus in pyrophosphoric acid is [1999]
  • a)
    +3
  • b)
    +1
  • c)
    +4
  • d)
    +5
Correct answer is option 'D'. Can you explain this answer?

The oxidation number of an element in a compound is a measure of the charge that element would have if all the bonding electrons were assigned to the more electronegative atom. Oxidation numbers are used to track the flow of electrons in chemical reactions and to determine the potential for redox reactions.

In pyrophosphoric acid (H4P2O7), there are two phosphorus atoms. To determine the oxidation number of phosphorus in this compound, we can use the following steps:

1. Assign oxidation numbers to the other elements in the compound:
- Hydrogen (H) is usually assigned an oxidation number of +1.
- Oxygen (O) is usually assigned an oxidation number of -2.

2. Determine the overall charge of the compound:
- The overall charge of pyrophosphoric acid is 0, since it is a neutral compound.

3. Apply the oxidation number rules to determine the oxidation number of phosphorus:
- The sum of all oxidation numbers in a compound must equal the overall charge of the compound.
- Since there are two phosphorus atoms in pyrophosphoric acid, let's assign the oxidation number of one of the phosphorus atoms as "x".

4. Calculate the oxidation number of phosphorus:
- The sum of the oxidation numbers of all the elements in the compound must equal zero. Therefore, we can write the equation:
(4)(+1) + 2(x) + (7)(-2) = 0
4 + 2x - 14 = 0
2x - 10 = 0
2x = 10
x = 5

Therefore, the oxidation number of phosphorus in pyrophosphoric acid is +5 or 5. Thus, the correct answer is option 'D'.

The oxidation states of sulphur in the anions SO32–, S2O42– and S2O62– follow the order [2003]
  • a)
  • b)
  • c)
  • d)
Correct answer is option 'B'. Can you explain this answer?

Suresh Reddy answered
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−​.
 

The oxide, which cannot act as a reducing agent, is[1995]
  • a)
    NO2
  • b)
    SO2
  • c)
    CO2
  • d)
    ClO2
Correct answer is option 'C'. Can you explain this answer?

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

Zn gives H2 gas with H2SO4 and HCl but not with HNO3 because [2002]
  • a)
    Zn acts as an oxidising agent when it reacts with HNO3
  • b)
    HNO3 is weaker acid than H2SO4 and HCl
  • c)
    In electroch emical series, Zn is above hydrogen
  • d)
    is reduced in preference to hydronium ion
Correct answer is option 'D'. Can you explain this answer?

Zinc is or.r the top position of hydrogen in electrochemical series. So Zn displaces H2 from dilute H2SO4 and HCl with iiberation of H2.

On the other hand HNO3 is an oxidising agent. Hydrogen obtained in this reaction is converted into H2O.

Which substan ce serves as reducing agent in the following reaction ? [1994]
  • a)
    H2O
  • b)
    Ni
  • c)
    H+
  • d)
Correct answer is option 'B'. Can you explain this answer?

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.

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