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All questions of Redox Reactions for NEET Exam

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Which is chlorate (I) ion?
a)
b)
c)
d)
Correct answer is option 'B'. Can you explain this answer?

Pooja Shah answered
  • ClO3- : A very reactive inorganic anion.
  • The term chlorate can also be used to describe any compound containing the chlorate ion, normally chlorate salts. 
  • Example: Potassium chlorate, KClO3

Which of the following is not an example of redox reaction?
  • a)
    BaCl2 + H2SO4 ⎯→ BaSO4 + 2HCl
  • b)
    Fe2O3 + 3CO ⎯→ 2Fe + 3CO2
  • c)
    2K + F2 ⎯→ 2KF
  • d)
    CuO + H2 ⎯→ Cu + H2O
Correct answer is option 'A'. Can you explain this answer?

Raghav Bansal answered
a) BaCl2 + H2SO4 → BaSO4 + 2HCl is not a redox reaction, as there is no change in the oxidation state of any element.
It is an example of double displacement reactions.

The oxidation half reaction for following reaction is
Fe2+(aq) + Cr2O72-(aq) → Fe3+ (aq) + Cr3+(aq)
  • a)
    Fe3+(aq) → Fe2+ (aq)
  • b)
    Cr2O72-(aq) → Cr3+(aq)
  • c)
    Cr3+(aq) → Cr2O72-(aq)
  • d)
    Fe2+ (aq) → Fe3+(aq)
Correct answer is option 'D'. Can you explain this answer?

Geetika Shah answered
Oxidation half reaction for a reaction is that reaction which gives us the reactant and product formed after the oxidation of the reactant. In this case Fe+2 oxidizes itself to Fe+3 and so the oxidation of Fe+2 is oxidation half reaction. Option d correct.

Oxidation numbers of P in PO4−3, of S in SO42− and that of Cr in Cr2O72− are respectively,
  • a)
     +5, +6 and +6
  • b)
    +3, +6 and +5
  • c)
    +5, +3 and +6               
  • d)
    -3, +6 and +6
Correct answer is option 'A'. Can you explain this answer?

Raghav Bansal answered
The correct answer is option A
(I) xPO43−​ ⇒ x + 4 × (−2) = −3
                ⇒x = −3 + 8 = +5
                ⇒x = +5
Oxidation number of P = +5
(II) xSO42−​ ⇒ x + 4 × (−2) = −2
                ⇒x = −2 + 8
                  ⇒x = +6
Oxidation number of S=+6
(III) xCr2​O72− ​    ⇒2x + 7 × (−2) = −2
                        ⇒2x =−2+14
                        ⇒2x=12
                        ⇒x= 12/2​ = +6

Oxidation number of 1/2 is assigned to oxygen atom in
  • a)
    superoxides
  • b)
    when oxygen is bonded to fluorine
  • c)
    when oxygen is bonded to metals
  • d)
    peroxidesperoxides
Correct answer is option 'A'. Can you explain this answer?

Arka Desai answered
The oxidation number is a concept used in chemistry to keep track of the distribution of electrons in a compound or molecule. It is a measure of the charge that an atom would have if all the shared electrons were assigned to the more electronegative atom in a bond.

In the case of oxygen, its most common oxidation number is -2. However, in certain compounds, such as superoxides and peroxides, the oxidation number of oxygen deviates from -2.

Oxidation number of 1/2 is assigned to the oxygen atom in superoxides. Superoxides are a class of compounds that contain the superoxide ion, O2-. In this ion, each oxygen atom has an oxidation number of -1/2. This is because the oxygen-oxygen bond in the superoxide ion is a single bond with a bond order of 1/2. Therefore, each oxygen atom is assigned an oxidation number of -1/2 to account for the distribution of electrons in the bond.

In the case of peroxides, such as hydrogen peroxide (H2O2), the oxidation number of oxygen is -1. In peroxides, the oxygen-oxygen bond is a single bond with a bond order of 1. Each oxygen atom is assigned an oxidation number of -1 to account for the distribution of electrons in the bond.

When oxygen is bonded to fluorine, the oxidation number of oxygen is -1. Fluorine is the most electronegative element, and therefore, it attracts the shared electrons in the bond more strongly than oxygen. As a result, oxygen is assigned an oxidation number of -1 to account for the unequal distribution of electrons in the bond.

When oxygen is bonded to metals, the oxidation number of oxygen is typically -2. However, there are some exceptions to this rule, such as in certain metal peroxides or superoxides, where the oxidation number of oxygen deviates from -2.

In summary, the oxidation number of 1/2 is assigned to the oxygen atom in superoxides, where the oxygen-oxygen bond is a single bond with a bond order of 1/2. In other compounds, such as peroxides, fluorides, and most metal oxides, the oxidation number of oxygen is typically -1 or -2.

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.

Select the set of compounds with oxidation-reduction duality.
  • a)
    Cl2, H3PO4, HCHO, HNO2
  • b)
    Cl2, H3PO3, C6H5CHO, H2O2
  • c)
    Br2, H3PO2, CH3CHO, H3PO4
  • d)
    CrO2Cl2, KMnO4, SO3, CO2
Correct answer is option 'B'. Can you explain this answer?

Neha Sharma answered
Com pounds having oxidising and reducing nature in given reaction are said to have oxidation-reduction duality. Such compounds are said to undergo disproportionation reaction.



(This is called Cannizzaro reaction.)

Note Such compounds have O.N. of the affected atoms intermediate of oxidation part and reduction part

 In this method, the two half equations are balanced separately and then added together to give balanced equation
  • a)
    Reluctant method
  • b)
    Oxidizing agent method
  • c)
    Reducing agent method
  • d)
    Half reaction method
Correct answer is option 'D'. Can you explain this answer?

Kavita Joshi answered
In the ion-electron method (also called the half-reaction method), the redox equation is separated into two half-equations - one for oxidation and one for reduction. Each of these half-reactions is balanced separately and then combined to give the balanced redox equation.

Intensity of blue colour increases gradually when _________________
  • a)
    copper rod is dipped in silver nitrate solution
  • b)
    silver rod is dipped in copper nitrate solution
  • c)
    zinc rod is dipped in silver solution
  • d)
    copper rod is dipped in zinc rod solution
Correct answer is option 'A'. Can you explain this answer?

EduRev NEET answered
When a copper rod is dipped in silver nitrate solution, a redox reaction occurs between Copper and an aqueous solution of silver nitrate.
  • So the intensity of blue colour increases gradually as silver deposits on the rod.

The oxidation number of oxygen in most compounds is
  • a)
    -3
  • b)
    1
  • c)
    4
  • d)
    -2.0
Correct answer is option 'D'. Can you explain this answer?

Niharika Nair answered
Oxidation number of oxygen in most compounds is -2.

Explanation:
- Oxidation number is the number assigned to an atom to indicate its degree of oxidation or loss/gain of electrons.
- Oxygen is a highly electronegative element, meaning it has a strong tendency to attract electrons.
- In most compounds, oxygen has an oxidation number of -2 because it tends to gain electrons to achieve a stable octet configuration (8 valence electrons).
- For example, in water (H2O), each hydrogen atom has an oxidation number of +1 and the oxygen atom has an oxidation number of -2, which balances out the charge to zero.
- There are some exceptions to this rule, such as in peroxides where oxygen has an oxidation number of -1, and in compounds with more electronegative elements where oxygen may have a positive oxidation number.
- Overall, the oxidation number of oxygen in most compounds is -2.

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.

When a zinc rod is kept in a copper nitrate solution what happens?
  • a)
    zinc is deposited on copper
  • b)
    copper is deposited in the beaker
  • c)
    zinc is deposited in the beaker
  • d)
    copper is deposited on zinc
Correct answer is option 'D'. Can you explain this answer?

EduRev NEET answered
When zinc is placed in copper nitrate solution the intensity of the blue colour is produced and copper iron is deposited on zinc.
  • This is a Redox reaction between zinc and an aqueous solution of copper nitrate occurring in a beaker.

The process in which the strength of an unknown solution is calculated using a known standard solution.
  • a)
    Titration
  • b)
    Oxidation
  • c)
    Reduction
  • d)
    None of these
Correct answer is option 'A'. Can you explain this answer?

Nandini Iyer answered
A titration is a technique used to work out the concentration of an unknown solution based on its chemical reaction with a solution of known concentration. The process usually involves adding the known solution (the titrant) to a known quantity of the unknown solution (the analyte) until the reaction is complete.

The correct order of N-compounds in its decreasing order of oxidation states is
  • a)
    HNO3, NO, N2, NH4Cl
  • b)
    HNO3, NO, NH4Cl, N2
  • c)
    HNO3, NH4Cl, NO, N2
  • d)
    NH4Cl, N2, NO, HN03
Correct answer is option 'A'. Can you explain this answer?

Infinity Academy answered
To determine the decreasing order of oxidation states of nitrogen in the given compounds, we need to find the oxidation state of nitrogen in each compound:
  1. HNO3 (Nitric acid): Oxidation state of nitrogen: +5
  2. NO (Nitric oxide): Oxidation state of nitrogen: +2
  3. N2 (Dinitrogen): Oxidation state of nitrogen: 0
  4. NH4Cl (Ammonium chloride): Oxidation state of nitrogen: -3
Now, let's arrange these compounds in decreasing order of oxidation states:
  • HNO3: +5
  • NO: +2
  • N2: 0
  • NH4Cl: -3
So, the correct order in decreasing oxidation state is:
HNO3, NO, N2, NH4Cl

Which of the following is true as per metal activity series?
  • a)
    Zn>Cu>Ag
  • b)
    Zn<Cu<Ag
  • c)
    Zn>Ag>Cu
  • d)
    Zn<Ag<Cu
Correct answer is option 'A'. Can you explain this answer?

EduRev NEET answered
Metal activity series or electrochemical series is a series in the decreasing order of metals which are active during a chemical reaction comparatively with each other.
  • Here, Zinc’s activity is greater than Copper’s activity and Copper’s activity is greater than that of silver.

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?

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

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?

Keerthana Datta answered
Explanation:

Reducing Agent:
A reducing agent is a substance that donates electrons to another chemical species in a redox reaction. It itself gets oxidized in the process.

Oxides:
Oxides are compounds that contain oxygen as an anion.

Identifying the Correct Answer:
To determine which oxide cannot act as a reducing agent, we need to consider the oxidation states of the elements present in each oxide. In general, nonmetals tend to gain electrons (get reduced) and act as oxidizing agents, while metals tend to lose electrons (get oxidized) and act as reducing agents.

Oxides Given:
a) NO2 - Nitrogen has an oxidation state of +4 in NO2, and it can be reduced to lower oxidation states, making it a potential reducing agent.
b) SO2 - Sulfur has an oxidation state of +4 in SO2, and it can be reduced to lower oxidation states, making it a potential reducing agent.
c) CO2 - Carbon has an oxidation state of +4 in CO2, and it cannot be further reduced as it already has the lowest common oxidation state for carbon, making it unable to act as a reducing agent.
d) ClO2 - Chlorine has an oxidation state of +4 in ClO2, and it can be reduced to lower oxidation states, making it a potential reducing agent.

Correct Answer:
The oxide that cannot act as a reducing agent is CO2 because carbon already has the lowest common oxidation state in this compound and cannot be further reduced.

The oxidation number of Cl in Cl2O7 is
  • a)
    + 3
  • b)
    + 5
  • c)
    + 7
  • d)
    – 7
Correct answer is option 'C'. Can you explain this answer?

answered
Cl show different oxidation state as -1 to +7 due to vacant d orbital.
As oxygen is more electronegative than Cl.
Oxygen size is small hence its more electronegative and show -2 oxidation states.
2x + 7 (-2) = 0
x = +7

In the reaction of metallic cobalt placed in nickel sulphate solution, therein is a competition for release of electrons At equilibrium, chemical tests reveal that both Ni+2 (aq) and Co+2 (aq) are present at moderate concentrations. The result is that:
  • a)
    Only one reactant and one product is greatly favoured.
  • b)
    Only [Co(s) and Ni+2 (aq)] are favoured
  • c)
    Only Co+2 (aq) and Ni (s)] are favoured
  • d)
    neither the reactants nor the products [are greatly favoured.
Correct answer is option 'D'. Can you explain this answer?

Arya Reddy answered
The reaction of metallic cobalt in a nickel sulfate solution involves a competition for the release of electrons. This means that the cobalt metal can react with the nickel ions in the solution, or the nickel can deposit on the cobalt metal.
At equilibrium, the reaction has balanced out with no net change in the concentration of the reactants and products. The fact that both Ni+2 (aq) and Co+2 (aq) are present at moderate concentrations at equilibrium signifies that neither forward nor reverse reactions are greatly favoured.
  • A: This option is incorrect because both reactants and products are present in moderate concentrations, indicating that neither is greatly favoured.
  • B: This statement is not correct either. Even though Co (s) and Ni+2 (aq) are part of the reaction, the fact that Co+2 (aq) is also present at moderate concentrations shows that they are not the only favoured species.
  • C: This option is also incorrect. Even though Co+2 (aq) and Ni (s) are part of the reaction, the fact that Ni+2 (aq) is also present at moderate concentrations shows that they are not the only favoured species.
  • D: This is the correct answer. When a reaction is at equilibrium, it means that the rate of the forward reaction equals the rate of the reverse reaction. Therefore, neither the reactants nor the products are greatly favoured. In other words, the concentrations of the reactants and products remain constant over time


Coefficient x, y and z are respectively 
  • a)
    a
  • b)
    b
  • c)
    c
  • d)
    d
Correct answer is option 'C'. Can you explain this answer?

Hansa Sharma answered

Cross -multiply by change in oxidation number and balance H by H+ ions.

Thus, x = 2, y = 5 and z = 6 

the oxidation number of sulphur in S8,S2F2,H2S respectively are
  • a)
    0,+1 and -2
  • b)
    +2, +1 and -2 
  • c)
    0, +1 and +2
  • d)
    -2, +1 and -2
Correct answer is option 'A'. Can you explain this answer?

Sparsh Datta answered
(i) Oxidation state of element in its free state is zero.
(ii) Sum of oxidation states of all atoms in compound is zero.
O.N of S in S8=0;    O.N of S in S2F2=+1
O.N of S in H2S=-2

Which of the following atom has been assigned only single oxidation number?
  • a)
    H
  • b)
    O
  • c)
    N
  • d)
    F
Correct answer is option 'D'. Can you explain this answer?

Amrita Kumar answered
Fluorine is the most electronegative element. It can gain one electron
F+ e- → F
-
Oxidation number = -1

Due to very high (IE), it cannot lose electron.
(a) HH- (-1), (H+ (+1)
(b) O -2 in oxide
-1 in peroxide
+2 in OF2
(c) N - 3 in N2O3, NH3
+3 in HNO2,
+5 in HNO3,
+ 1 in N2O

In balancing the half-reaction, CN- → CNO-
The number of electrons that must be added is 
  • a)
    0
  • b)
    4 on ther ight
  • c)
    4 on the left
  • d)
    2 on the right
Correct answer is option 'D'. Can you explain this answer?

EduRev NEET answered
Okay, here’s the short version without symbols:
  • In CN⁻, carbon = +2
  • In CNO⁻, carbon = +4
  • Change: +2 to +4 (oxidation, loss of 2 electrons)
Half-reaction:
CN⁻ → CNO⁻ + 2e⁻

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