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Previous Year Questions of Oxidation, Reduction and Electrochemistry

Previous Year Questions of Oxidation, Reduction and Electrochemistry

Q 2354134954

Which one of the following elements will not react with dilute HCl to produce `H_2`?
NDA Paper 2 2016
(A)

Hg

(B)

AI

(C)

Mg

(D)

Fe

Solution:

Hg will not react with dilute HCl to produce `H_2` because in reactivity series, Hg is less reactive than the given, i.e. the hydrogen which has a stronger attraction to the Cl than the mercury.
Correct Answer is `=>` (A) Hg
Q 2109401318

Which one of the following is a reduction reaction?
NDA Paper 2 2016
(A)

`2Mg(s)+ O_( 2) (g) -> 2MgO(s)`

(B)

`S(s)+ O_( 2) (g) -> SO_(2) (g)`

(C)

`2HgO(s) overset (Heat) rightarrow 2 Hg(l) + O_( 2) (g)`

(D)

`Mg(s) + S (s) -> MgS (s)`

Solution:

`2HgO(s) overset (Heat) rightarrow 2 Hg(l) + O_( 2) (g)`

Above reaction is a reduction reaction. In this

reaction, the removal of oxygen from mercuric oxide

occurs.
Correct Answer is `=>` (C) `2HgO(s) overset (Heat) rightarrow 2 Hg(l) + O_( 2) (g)`
Q 1618656509

In the reaction `ZnO + C -> Zn +CO, 'C'` acts as
NDA Paper 2 2015
(A)

an acid

(B)

a base

(C)

an oxidising agent

(D)

a reducing agent

Solution:

Removal of oxygen `(O)` is 'reduction' and its addition is 'oxidation'. Since here `Zn^(+2)` is reduced to `Zn` by `C` (i.e. `C` causes its reduction) so `C` acts as reducing agent.
Correct Answer is `=>` (D) a reducing agent
Q 1710434310

What is the oxidising agent in the following equation?

`HAsO_2(aq) + Sn^(2+) (aq) + H^(+) (aq) ->`

`As` `(s) + Sn^(4+) (aq) + H_2O (l)`
NDA Paper 2 2014
(A)

`HAsO_2(aq)`

(B)

`Sn^(2+) (aq)`

(C)

`H^(+) (aq)`

(D)

`Sn^(4+) (aq)`

Solution:

In the given reaction, `HAsO_2 (aq)` is the

oxidising agent since the oxidation number of `As`

decreases from `+ 3` to `0`.
Correct Answer is `=>` (A) `HAsO_2(aq)`
Q 1783767647

Which of the following is correct regarding the reaction of fluorine with water?

`2F_2 (g) + 2H_2 O(l) -> 4H^+ (aq) +4F^(-)
(aq) + O_2(g)`
NDA Paper 2 2014
(A)

Fluorine is oxidised to `F^-`

(B)

Water is oxidised to `O_2`

(C)

Water is reduced to `H^+`

(D)

Oxidation state of fluorine does not change

Solution:

In the given reaction, `HAsO_2 (aq)` is the

oxidising agent since the oxidation number of `As`

decreases from `+ 3` to `0`.
Correct Answer is `=>` (A) Fluorine is oxidised to `F^-`
Q 2336191072

A gas is evolved when a piece of zinc metal is placed in dilute sulphuric acid (`H_2SO_4` ). What is the gas?
NDA Paper 2 2013
(A)

Hydrogen

(B)

Oxygen

(C)

Water vapour

(D)

Sulphur dioxide

Solution:

Hydrogen gas is evolved when a piece of zinc metal is placed in dilute sulphuric acid (`H_2SO_4`). The chemical reaction is
`Zn + H_2SO_4 -> ZnSO_4 + H_2` .
Correct Answer is `=>` (A) Hydrogen
Q 2317045889

Which one among the following is an electrochemical cell that cannot be charged?
NDA Paper 2 2012
(A)

Electrolytic cell

(B)

Storage cell

(C)

Primary cell

(D)

Fuel cell

Solution:

A primary cell is a battery that is designed to be used once and discharged, and not recharged with electricity and reused like a secondary cell (rechargeable battery). In general, the electrochemical reaction occurring in the cell is not reversible, rendering the cell unrechargeable. As a primary cell is used, chemical reactions in the battery use up the chemicals that generate the power; when they are gone, the battery stops producing electricity and is useless.
Correct Answer is `=>` (C) Primary cell
Q 2377856786

In `KMnO_4` molecule, the oxidation states of the elements potassium (K), manganese (Mn) and oxygen (O) are respectively
NDA Paper 2 2012
(A)

+1, +5, -2

(B)

+1, +7, -2

(C)

0, 0, 0

(D)

+1, +7, 0

Solution:

Potassium pennanganate is an inorganic chemical compound with the formula `KMnO_4` . In potassium permanganate, the oxidation states of the elements potassium (K) is + 1 , manganese (Mn) + 7, and oxygen (O) is- 2.
Correct Answer is `=>` (B) +1, +7, -2
Q 2357056884

The following equation is an example of a redox reaction, in which `Cl`, is the oxidising agent and `FeBr_3` is the reducing agent `2FeBr_3 (aq) + 3 Cl_2 (g) -> 2FeCl_3 (aq) + 3 Br_2 (l)` Which one among the following statements is incorrect for this redox reaction?
NDA Paper 2 2012
(A)

Oxidising agents are themselves reduced

(B)

Reducing agents gain or appear to gain electrons

(C)

Reducing agents are themselves oxidised

(D)

Oxidising agents oxidise other substances

Solution:

The substance in which reduction occurs is called oxidising agent, while the substance in which oxidation occurs is called reducing agent. Thus the oxidising agent is the substance which gains electrons, while the reducing agent is the substance which loses electron.
Correct Answer is `=>` (B) Reducing agents gain or appear to gain electrons
Q 2377367286

Which of the following statements regarding oxidation and reduction are correct?

1. In oxidation, loss of electron takes place whereas in reduction, gain of electron takes place.
2. In oxidation, gain of electron takes place whereas in reduction, loss of electron takes place.
3. Oxidising agent decreases the oxidation number but reducing agent increases the oxidation number.
4. oxidising agent increases the oxidation number but reducing agent reduces the oxidation number.

Select the correct answer using the codes given below
NDA Paper 2 2012
(A)

1 and 3

(B)

2 and 4

(C)

2 and 3

(D)

1 and 4

Solution:

Oxidation is gain of oxygen and reduction is loss of oxygen.
Correct Answer is `=>` (B) 2 and 4
Q 2317478380

In oxidation,
1. Hydrogen is displaced from a substance.
2. An electropositive element is added to or proportion of electropositive element increase in a substance.
Select the correct answer using the codes given below
NDA Paper 2 2011
(A)

Only 1

(B)

Only 2

(C)

Both 1 and 2

(D)

Neither 1 nor 2

Solution:

Oxidation is a process which involves
1. addition of oxygen or any other electronegative element.
2. removal of hydrogen or any other electropositive element.
3. loss of electrons.
4. increase in oxidation number of its atom.
Correct Answer is `=>` (A) Only 1
Q 2337878782

When a copper rod is dipped in aqueous silver nitrate solution, the colour of the solution changes to blue. This is because
NDA Paper 2 2011
(A)

Cu is more easily reduced than Ag

(B)

Ag is more easily reduced than Cu

(C)

Nitrate ion acts as an oxidising agent

(D)

Nitrate ion acts as a reducing agent

Solution:

According to electrochemical series the reduction potential of Ag ( + 0.80 V) is more than that of Cu ( + 0.34 V), hence silver has greater tendency to get reduced in comparison to copper.
`undersettext[Silver Nitrate (Colourless Solution)](2AgNO_3(aq)) + Cu(s) -> undersettext[Copper nitrate (blue solution)](Cu(NO_3)_2 (aq)) + 2Ag (s)`
Correct Answer is `=>` (B) Ag is more easily reduced than Cu
Q 2367878785

Iron nails are dipped into blue copper sulphate solution. After sometime iron nails are
NDA Paper 2 2011
(A)

dissolved and blue colour is discharged

(B)

dissolved but blue colour is not discharged

(C)

not dissolved and blue colour is not discharged

(D)

not dissolved but blue colour is discharged

Solution:

According to reactivity series of metals, iron is more reactive than copper, hence it displaces copper from copper sulphate solution.

`underset(Iron)(Fe) + underset(Blue)(CusO_4) -> underset(Green)(FeSO_4) + underset(Red-Brown)(Cu)`

So, after sometime blue colour of the solution slowly fades as copper is removed from the copper (II) sulphate solution. This copper is deposited over iron nail.
Correct Answer is `=>` (D) not dissolved but blue colour is discharged
Q 2377078886

Silverware turns black after a period of time due to formation of
NDA Paper 2 2011
(A)

nitrate coating on silver

(B)

sulphide coating on silver

(C)

chloride coating on silver

(D)

oxide coating on silver

Solution:

Silver in contact with `H_2S` of air forms silver sulphide which is black in colour. Hence, silverware turns black after a period of time.

`2Ag + H_2S -> underset(Black)(Ag_2S) + 2H^+`
Correct Answer is `=>` (B) sulphide coating on silver
Q 2327180981

An oxidising agent is a substance which
NDA Paper 2 2010
(A)

increases the oxidation number of an element in a given substance

(B)

decreases the oxidation number of an element in a given substance

(C)

is oxidised itself in an oxidation-reduction reaction

(D)

loses electrons in an oxidation-reduction reaction

Solution:

An oxidising agent is a substance

(i) which oxidises some other species and is itself reduced
(ii) which gains one or more electrons i.e., as electron-acceptor.
(iii) which increases the oxidation number of an element in a
given substance or the oxidation number of whose atom or
atoms decreases.

`underset(+1 -4 -2)(H_2SO_3)+ underset(+1-2)(2H_2S)-> underset(0)(3S)+underset(+1-2)(3H_2O)`
Correct Answer is `=>` (A) increases the oxidation number of an element in a given substance
Q 2307391288

In the reaction `4 Fe+3O_2-> 4 Fe^(3+)+6 O^(2-)`
Which of the following statements is incorrect?
NDA Paper 2 2010
(A)

It is a redox reaction

(B)

Metallic iron acts as a reducing agent

(C)

`O_2` acts as an oxidising agent

(D)

Metallic iron is reduced to `Fe^( 3+)`

Solution:

`overset(0)(4Fe)+overset(0)(3O_2)-> 4Fe^(3+)+6O^(2+)`

This reaction is a redox reaction. In this reaction. iron acts as a reducing agent while `O_2`, acts as an oxidising agent, hence metallic `Fe` is oxidised to `Fe^( 3+)`' and `O_2` is reduced to `O^(2-)`.
Correct Answer is `=>` (D) Metallic iron is reduced to `Fe^( 3+)`
Q 2348512403

Which one among the following metals is more reactive than hydrogen?
NDA Paper 2 2010
(A)

Mercury

(B)

Copper

(C)

Silver

(D)

Tin

Solution:

In electrochemical series (activity series), tin is present above hydrogen while all other given metals are present below it. Thus, Sn (tin) is more reactive than hydrogen. The order of reactivity is as
`Sn > H > Cu > Hg > Ag`
Correct Answer is `=>` (D) Tin
Q 2338612502

Which one of the following elements will replace hydrogen from acids to form salts?
NDA Paper 2 2010
(A)

Sulphur (S)

(B)

Silicon (Si)

(C)

Zinc (Zn)

(D)

Phosphorus (P)

Solution:

The metals that are more reactive than hydrogen have a high tendency to donate electrons. These electrons are accepted by the `H^ +` ions of acid and thus, H + get reduced to `H_2` gas. Thus, such metals (like Zn which is moderately active) form salt when react with acids and also evolve hydrogen gas. Non-metals like S, Si, P, however, do not have a tendency to donate electrons, thus they do not evolve hydrogen with acids.

`Zn+H_2SO_4-> ZnSO_4+H_2`

Non-metals + Acid -> No reaction
Correct Answer is `=>` (C) Zinc (Zn)
Q 2388612507

Which one among the following is the correct order of reactivity of the elements'?
NDA Paper 2 2010
(A)

Cu > Mg > Zn > Na

(B)

Na > Zn > Mg > Cu

(C)

Cu > Zn > Mg > Na

(D)

Na > Mg > Zn > Cu

Solution:

The metal that occupy higher position in the activity series is more reactive as compared to that which occupy lower position. Thus, the order of reactivity of the given metals is `Na > Mg > Zn > Cu`
Correct Answer is `=>` (D) Na > Mg > Zn > Cu
Q 2307223188

Aluminium is more reactive than iron but aluminium is less easily corroded than iron, because.
NDA Paper 2 2009
(A)

oxygen forms a protective oxide layer

(B)

it is a noble metal

(C)

iron undergoes reaction easily with water

(D)

iron forms ions

Solution:

Aluminium is more reactive than iron but aluminium is
Jess easily corroded than iron because the action of moist air or
oxygen on aluminium metal forms a thin layer of aluminium oxide
over the aluminium metal. This thin aluminium oxide layer is very
tough and protects aluminium objects from further corrosion.
Correct Answer is `=>` (A) oxygen forms a protective oxide layer
Q 2377167086

Which one of the following is the correct order of oxidation number of iodine (`I`) in `I_2 ,HI, HIO_ 4` and `ICl`?
NDA Paper 2 2008
(A)

`HIO_4 < HI < ICI < I_2`

(B)

`HI < I_2 < ICI < HIO_4`

(C)

`I_2 < ICI < HIO_4 < HI`

(D)

`ICI < HIO_ 4 < HI < I`

Solution:

Let the oxidation number of iodine in each compound
is `x`.
The oxidation state of any element in free state is `0`. i.e., oxidation
number of `I_2 = 0`

`HI`

` +1 +x= 0`

`x= -1`

`HIO_2`

`1 +x+ (-2) xx 4 = 0`

`x= +7`

`ICI`

`x -1 = 0`

`x= +1`

Hence, the order of oxidation states of iodine is

`HI < I_2 < ICI < HIO_4`
Correct Answer is `=>` (B) `HI < I_2 < ICI < HIO_4`
Q 2337856782

Which one of the following is an element which never exhibits positive oxidation state in any of its compounds?
NDA Paper 2 2008
(A)

Oxygen

(B)

Chlorine

(C)

Fluorine

(D)

Carbon

Solution:

Fluorine is the most electromotive element in whole periodic table. Therefore, it never exhibits positive oxidation state. It always exhibit -1 oxidation state.
Correct Answer is `=>` (C) Fluorine
Q 2317867789

Which one of the following elements cannot displace hydrogen gas from a dilute acid?
NDA Paper 2 2008
(A)

Zinc

(B)

Copper

(C)

Magnesium

(D)

Iron

Solution:

In electrochemical series, copper is present below hydrogen, (i.e., have high reduction potential than `H _2` ). so it cannot displace hydrogen from the dilute acids. While zinc. magnesium and iron are present above hydrogen. So, they can displace hydrogen from the dilute acids.

`Zn + 2HCl -> ZnCl_2 +· H_2 uparrow`

`Mg + 2HCl -> MgCl_2 + H_2 uparrow`

`Fe + 2HCl -> FeCl_2 + H_2 uparrow`'

`Cu + HCl ->` No reaction
Correct Answer is `=>` (B) Copper
Q 2327778681

What is the correct order of oxidation states of sulphur in anions `SO_3^(2-), S_2O_4^(2-)` and `S_2 O_6^(2-)`?
NDA Paper 2 2007
(A)

`S_2O_4^(2-) < SO_3^(2-) < S_2 O_6^(2-)`

(B)

`S_2O_6^(2-) < S_2O_4^(2-) < S_O_3^(2-)`

(C)

`SO_3^(2-) < S_2O_4^(2-) < S_2O_6^(2-)`

(D)

`S_2O_4^(2-) < S_2O_6^(2-) < SO_3^(2-)`

Solution:

Let the oxidation state of sulphur is `x` in each
compound.

`SO_3^(2-)`

`x + 3(-2) = -2`

`x- 6 = -2`

`x=+4`

`S_2O_4^(2-)`

`2x + 4(-2) = -2`

`2x- 8 = -2`

`x=+3`

`S_2O_6^(2-)`

`2x + 6(-2) = -2`

`2x -12 = -2`

`x=+5`

Hence, the increasing order of oxidation state of '`S`' is as

`S_2O_4^(2-) < SO_3^(2-) < S_2 O_6^(2-)`
Correct Answer is `=>` (A) `S_2O_4^(2-) < SO_3^(2-) < S_2 O_6^(2-)`
 
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