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Tricks of Oxidation, Reduction and Electrochemistry For NDA

Oxidation and Oxidising Agent

Q 2858056804

The reaction which takes place at the anode is

(A)

Oxidation

(B)

Reduction

(C)

dissociation

(D)

ionisation

Solution:

Correct Answer is `=>` (A) Oxidation
Q 2828156901

Which of the following act as an oxidising agent .

(A)

`HNO_3`

(B)

`K_2Cr_2O_7`

(C)

`KMnO_4`

(D)

All of these

Solution:

Correct Answer is `=>` (D) All of these
Q 2858156904

The process on `text()_(28)^(56)Fe^(2+) → text()_(28)^(56)Fe^(3+)` is appropriate classified as

(A)

ionisation

(B)

oxidation

(C)

reduction

(D)

nuclear reaction

Solution:

Correct Answer is `=>` (B) oxidation
Q 2838867702

Which of the statements is/are true?
I. The process of oxidation leads to gain of electrons.
II. The process of oxidation leads to loss of electrons.
III. The process of reduction leads to gain to electrons.
IV. The process of reduction leads to loss of electrons.
Select the correct answer from the codes given below.

(A)

I and IV

(B)

II and Ill

(C)

Only I

(D)

Only IV

Solution:

Correct Answer is `=>` (B) II and Ill
Q 2818167000

Among `Cl^- , Br^- ` and `I^-` the correct order for being oxidised to dihalogen is

(A)

`I^(-) > Cl^(-) > Br^(-)`

(B)

`Cl^(-) > Br^(-) > I^(-)`

(C)

`I^(-) > Br^(-) > Cl^(-)`

(D)

`Br^(-) > I^(-) > Cl^(-)`

Solution:

Correct Answer is `=>` (C) `I^(-) > Br^(-) > Cl^(-)`
Q 2858167004

The process in which oxidation number of element increases is

(A)

Oxidation

(B)

reduction

(C)

autoxidation

(D)

None of these

Solution:

Correct Answer is `=>` (A) Oxidation
Q 2818667509

An oxidising agent is a substance which

(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:

Correct Answer is `=>` (A) increases the oxidation number of an element in a given substance
Q 2818378209

In oxidation
I. Hydrogen is displaced from a substance.
II. An electropositive element is added to or proportion of electropositive element increase in a substance.
Select the correct answer using the codes given above.

(A)

only I

(B)

only II

(C)

Both I and II

(D)

Neither I nor II

Solution:

Correct Answer is `=>` (A) only I

Reduction and Reducing Agent

Q 2344780653

In the following chemical reaction,

` Ag_(2)O + H_(2)O + 2e^(-) -> 2Ag + 2OH^(-)`
BITSAT Mock
(A)

water is oxidised

(B)

silver is oxidised

(C)

silver is reduced

(D)

it is not a redox reaction

Solution:

`Ag_(2)^(+1)O^(-2) + H^(+1) + 2e^(-) -> 2 Ag^0 + 2OH^(-)`

`2Ag^+ + 2e^(-) -> 2Ag^0` (reduction is gain of electron).
Correct Answer is `=>` (C) silver is reduced
Q 2868056805

The strongest reducing agent is

(A)

`HNO_3`

(B)

`H_2S`

(C)

`H_2SO_3`

(D)

`SnCl_2`

Solution:

`Ag_(2)^(+1)O^(-2) + H^(+1) + 2e^(-) -> 2 Ag^0 + 2OH^(-)`

`2Ag^+ + 2e^(-) -> 2Ag^0` (reduction is gain of electron).
Correct Answer is `=>` (B) `H_2S`
Q 2888056807

Which one of the following is not a reducing agent?

(A)

`NaNO_2`

(B)

`NaNO_3`

(C)

`HI`

(D)

`SnCl_2`

Solution:

`Ag_(2)^(+1)O^(-2) + H^(+1) + 2e^(-) -> 2 Ag^0 + 2OH^(-)`

`2Ag^+ + 2e^(-) -> 2Ag^0` (reduction is gain of electron).
Correct Answer is `=>` (B) `NaNO_3`
Q 2838156902

Which of the following is best reducing agent?

(A)

`Na`

(B)

`Ca`

(C)

`Sr`

(D)

`Ba`

Solution:

`Ag_(2)^(+1)O^(-2) + H^(+1) + 2e^(-) -> 2 Ag^0 + 2OH^(-)`

`2Ag^+ + 2e^(-) -> 2Ag^0` (reduction is gain of electron).
Correct Answer is `=>` (D) `Ba`
Q 2848156903

Which is strongest reducing agent?

(A)

`F^-`

(B)

`Cl^-`

(C)

`Br^-`

(D)

`F`

Solution:

`Ag_(2)^(+1)O^(-2) + H^(+1) + 2e^(-) -> 2 Ag^0 + 2OH^(-)`

`2Ag^+ + 2e^(-) -> 2Ag^0` (reduction is gain of electron).
Correct Answer is `=>` (D) `F`
Q 2888156907

In the reaction, `4Fe+3O_2 → 4Fe^(3+) +6O^(2-)` Which of the following statements is incorrect?

(A)

It is redox reaction

(B)

Metallic iron is a reducing agent

(C)

`Fe^(3+)` is an oxidising agent

(D)

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

Solution:

`Ag_(2)^(+1)O^(-2) + H^(+1) + 2e^(-) -> 2 Ag^0 + 2OH^(-)`

`2Ag^+ + 2e^(-) -> 2Ag^0` (reduction is gain of electron).
Correct Answer is `=>` (D) Metallic iron is reduced to `Fe^(3+)`
Q 2838167002

Among the following chemical compounds, which one of the following has the highest reducing power?

(A)

`AsH_3`

(B)

`BiH_3`

(C)

`NH_3`

(D)

`PH_3`

Solution:

`Ag_(2)^(+1)O^(-2) + H^(+1) + 2e^(-) -> 2 Ag^0 + 2OH^(-)`

`2Ag^+ + 2e^(-) -> 2Ag^0` (reduction is gain of electron).
Correct Answer is `=>` (B) `BiH_3`
Q 2818856709

A reducing agent is a substance which can

(A)

accept electrons

(B)

donate electrons

(C)

accept protons

(D)

donate protons

Solution:

`Ag_(2)^(+1)O^(-2) + H^(+1) + 2e^(-) -> 2 Ag^0 + 2OH^(-)`

`2Ag^+ + 2e^(-) -> 2Ag^0` (reduction is gain of electron).
Correct Answer is `=>` (B) donate electrons

Redox Reactions

Q 1716456370

Which of the following is not an example of redox reaction?
NCERT Exemplar
(A)

`CuO + H_2 rightarrow Cu + H_2O`

(B)

`Fe_2O_3 + 3CO rightarrow2Fe+ 3CO_2`

(C)

`2K+ F_2rightarrow 2KF`

(D)

`BaCl_2+ H_SO_4 rightarrow BaSO_4 + 2HCl`

Solution:

Following are the examples of redox reaction
(a) `CuO + H_2 rightarrow Cu + H_2O`

(b) `Fe_2O_3 + 3CO rightarrow2Fe+ 3CO_2`


(c) `2K+ F_2rightarrow 2KF`

Option (d) is not an example of redox reaction
Correct Answer is `=>` (D) `BaCl_2+ H_SO_4 rightarrow BaSO_4 + 2HCl`
Q 2888667507

In the reaction, `3Br_2+6CO_3^(2-) +3H_2O → 5Br^-+BrO_3^- +6 HCO_3^-`

(A)

bromine is oxidised and carbonate is reduced

(B)

bromine is the both reduced and oxidised

(C)

bromine is neither reduced nor oxidised

(D)

bromine is reduced and water is oxidised

Solution:

Following are the examples of redox reaction
(a) `CuO + H_2 rightarrow Cu + H_2O`

(b) `Fe_2O_3 + 3CO rightarrow2Fe+ 3CO_2`


(c) `2K+ F_2rightarrow 2KF`

Option (d) is not an example of redox reaction
Correct Answer is `=>` (B) bromine is the both reduced and oxidised
Q 2818767600

When a copper rod is dipped in aqueous silver nitrate solution, the colour of the solution changes to blue. This is because

(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:

Following are the examples of redox reaction
(a) `CuO + H_2 rightarrow Cu + H_2O`

(b) `Fe_2O_3 + 3CO rightarrow2Fe+ 3CO_2`


(c) `2K+ F_2rightarrow 2KF`

Option (d) is not an example of redox reaction
Correct Answer is `=>` (B) Ag is more easily reduced than Cu
Q 1756078874

Identify the correct statements with reference to the given reaction

`P_4 + 3OH^- + 3H_2O -> PH_3 + 3H_2PO_2^-`
NCERT Exemplar

(This question may have multiple correct answers)

(A) Phosphorus is undergoing reduction only
(B) Phosphorus is undergoing oxidation only
(C) Phosphorus is undergoing oxidation as well as reduction
(D) Hydrogen is undergoing neither oxidation nor reduction
Solution:

Write the O.N. of each element above its symbol, then

`overset(0)(P_4) + 3overset(-2,+1)(OH^-) + 3overset(+1,-2)(H_2O) -> overset(-3,+1)( PH_3) + 3overset(+1,+1)(H_2PO_2^-)`


In this reaction, O.N. of P increases from 0 in `P_4` to+ 1 in `H_2PO_2^(-)` and decreases to- 3 in `PH_3` ,
therefore, P undergoes both oxidation as well as reduction. Thus, options (a) and (b) are
wrong and option (c) is correct
Further, O.N. of H remains +1 in all the compounds, i.e., H neither undergoes oxidation nor
reduction. Thus, option (d) is correct
Correct Answer is `=>` (A)
Q 2848667503

Consider the following equation, `Cr_2O_7^(2-) (aq) + Br^(-) (aq) + H^+ (aq) → Br_2 (l) + Cr^(3+) (aq) + H_2O(l)`. ln this reaction which one of the following is a reducing agent?

(A)

`Br_2`

(B)

`Br^-`

(C)

`H^+`

(D)

`Cr_2O_7^(2-)`

Solution:

Write the O.N. of each element above its symbol, then

`overset(0)(P_4) + 3overset(-2,+1)(OH^-) + 3overset(+1,-2)(H_2O) -> overset(-3,+1)( PH_3) + 3overset(+1,+1)(H_2PO_2^-)`


In this reaction, O.N. of P increases from 0 in `P_4` to+ 1 in `H_2PO_2^(-)` and decreases to- 3 in `PH_3` ,
therefore, P undergoes both oxidation as well as reduction. Thus, options (a) and (b) are
wrong and option (c) is correct
Further, O.N. of H remains +1 in all the compounds, i.e., H neither undergoes oxidation nor
reduction. Thus, option (d) is correct
Correct Answer is `=>` (B) `Br^-`
Q 2878678506

Following questions consist of two statement I and statement II. Examine both the statements carefully and match the correct choise according to the codes given below.
Statement I : When hydrogen gas is passed over heated metallic oxide it removes the oxygen from it.
Statement II : Hydrogen is an oxidising agent.

(A)

Both the statements are true and Statement II is the correct explanation of Statement I

(B)

Both the statements are true but Statement II is not the correct explanation of Statement I.

(C)

Statement I is true, but Statement II is false

(D)

Statement I is false, but Statement II is true.

Solution:

Write the O.N. of each element above its symbol, then

`overset(0)(P_4) + 3overset(-2,+1)(OH^-) + 3overset(+1,-2)(H_2O) -> overset(-3,+1)( PH_3) + 3overset(+1,+1)(H_2PO_2^-)`


In this reaction, O.N. of P increases from 0 in `P_4` to+ 1 in `H_2PO_2^(-)` and decreases to- 3 in `PH_3` ,
therefore, P undergoes both oxidation as well as reduction. Thus, options (a) and (b) are
wrong and option (c) is correct
Further, O.N. of H remains +1 in all the compounds, i.e., H neither undergoes oxidation nor
reduction. Thus, option (d) is correct
Correct Answer is `=>` (C) Statement I is true, but Statement II is false
Q 1716578479

Identify the correct statement (s) in relation to the following reaction

`Zn+ 2HCl -> ZnCl_2 + H_2`
NCERT Exemplar

(This question may have multiple correct answers)

(A) Zinc is acting as an oxidant
(B) Chlorine is acting as a reductant
(C) Hydrogen ion is acting as an oxidant
(D) Zinc is acting as a reductant
Solution:

Writing the oxidation number of each element above its symbol, so that

`overset(0)(Zn) + 2overset(+1,-1)(HCl) -> overset(+2,-1)( ZnCl_2) +overset(0)(H_2)`

(a) The oxidation number of `Zn` increases from `0` in `Zn` to `+2` in `ZnCl_2` therefore, `Zn` acts as
a reductant. Thus, option (a) is incorrect

(b) The oxidation number of chlorine does not change, therefore, it neither acts as a
reductant nor an oxidant. Therefore, option (b) is incorrect.

(c) The oxidation number of hydrogen decreases from +1 in W to 0 in `H_2` , therefore, `H^+`
acts as an oxidant. Thus, option (c) is correct.

(d) As explained in option (a), Zn acts as reductant, therefore, it cannot act as an oxidant.
Thus, option (d) is correct.
Correct Answer is `=>` (C)

Oxidation Number

Q 2018145009

Match the items in Column I with relevant items in Column II.









Column IColumn II
(A) Ions having positive charge (1) +7
(B) The sum of oxidation number of
all atoms in a neutral molecule 		(2) -1
(C) Oxidation number of hydrogen ion `(H^(+))` (3) +1
(D) Oxidation number of fluorine in NaF (4) 0
(E) Ions having negative charge (5) Cation
`quad` `quad` (6) Anion

NCERT Exemplar
Solution:

`A. -> (5) B.-> (4) C.-> (3) D.-> (2) E.-> (6)`
A. Ions having positive charge - Cation
B. The sum of oxidation number of all atoms in a neutral molecule- Zero
C. Oxidation number of hydrogen ion `(H^(+)) - +1
D. Oxidation number of fluorine in NaF - -1
E. Ions having negative charge- Anion
Q 2888167007

Oxidation state of `Ni` in `Ni (CO)_4` is

(A)

`+2`

(B)

`0`

(C)

`+4`

(D)

`-8`

Solution:

`A. -> (5) B.-> (4) C.-> (3) D.-> (2) E.-> (6)`
A. Ions having positive charge - Cation
B. The sum of oxidation number of all atoms in a neutral molecule- Zero
C. Oxidation number of hydrogen ion `(H^(+)) - +1
D. Oxidation number of fluorine in NaF - -1
E. Ions having negative charge- Anion
Correct Answer is `=>` (B) `0`
Q 2838367202

Oxidation number of oxygen in ozone

(A)

`0`

(B)

`-2`

(C)

`+2`

(D)

`-6`

Solution:

`A. -> (5) B.-> (4) C.-> (3) D.-> (2) E.-> (6)`
A. Ions having positive charge - Cation
B. The sum of oxidation number of all atoms in a neutral molecule- Zero
C. Oxidation number of hydrogen ion `(H^(+)) - +1
D. Oxidation number of fluorine in NaF - -1
E. Ions having negative charge- Anion
Correct Answer is `=>` (A) `0`
Q 2868467305

Oxidation number of oxygen in hydrogen peroxide is

(A)

-1

(B)

+1

(C)

-2

(D)

None of these

Solution:

`A. -> (5) B.-> (4) C.-> (3) D.-> (2) E.-> (6)`
A. Ions having positive charge - Cation
B. The sum of oxidation number of all atoms in a neutral molecule- Zero
C. Oxidation number of hydrogen ion `(H^(+)) - +1
D. Oxidation number of fluorine in NaF - -1
E. Ions having negative charge- Anion
Correct Answer is `=>` (A) -1
Q 2858367204

Oxygen has an oxidation state +2 in

(A)

`H_2O_2`

(B)

`OF_2`

(C)

`H_2O`

(D)

`SO_2`

Solution:

`A. -> (5) B.-> (4) C.-> (3) D.-> (2) E.-> (6)`
A. Ions having positive charge - Cation
B. The sum of oxidation number of all atoms in a neutral molecule- Zero
C. Oxidation number of hydrogen ion `(H^(+)) - +1
D. Oxidation number of fluorine in NaF - -1
E. Ions having negative charge- Anion
Correct Answer is `=>` (B) `OF_2`
Q 1726867771

Which of the following arrangements represent increasing oxidation
number of the central atom ?
NCERT Exemplar
(A)

`CrO_2^(-) ; ClO_3^(-); CrO_4^(2-);. MnO_4^(-)`

(B)

`ClO_3^(-), CrO_4^(2-); MnO_4^(-), CrO_2^(-)`

(C)

`CrO_2^(-), ClO_3^(-), MnO_4^(-), CrO_4^(2-)`

(D)

`CrO_4^(2-), MnO_4^(-), CrO_2^(-). ClO_3^(-)`

Solution:

Writing the oxidation number `(O.N.)` of `Cr, Cl` and `Mn` on each species in the four set of
ions, then,

(a) `overset(+3)(CrO_2^(-)), overset(+5)(ClO_3^(-)), overset(+6)(CrO_4^(-)) , overset(+7)(MnO_4^(-))`
(b) `overset(+5)(ClO_3^(-)), overset(+6)(CrO_4^(2-)), overset(+7)(MnO_4^(-)) :, overset(+3)(CrO_2^(-))`
(c) `overset(+3)(CrO_2^(-)), overset(+5)(ClO_3^(-)), overset(+7)(MnO_4^(-)); overset(+6)(CrO_4^(-))`
(d) `overset(+6)(CrO_4^(2-)), overset(+7)(MnO_4^(-)), overset(+3)(CrO_2^(-)). overset(+5)(ClO_3^(3-))`
Only in the arrangement (a), the O.N. of central atom increases from left to right,
therefore, option (a) is correct.
Correct Answer is `=>` (A) `CrO_2^(-) ; ClO_3^(-); CrO_4^(2-);. MnO_4^(-)`
Q 2878367206

Oxidation state of oxygen atom in potassium super oxide is

(A)

`-1/2`

(B)

`-1`

(C)

`-2`

(D)

`0`

Solution:

Writing the oxidation number `(O.N.)` of `Cr, Cl` and `Mn` on each species in the four set of
ions, then,

(a) `overset(+3)(CrO_2^(-)), overset(+5)(ClO_3^(-)), overset(+6)(CrO_4^(-)) , overset(+7)(MnO_4^(-))`
(b) `overset(+5)(ClO_3^(-)), overset(+6)(CrO_4^(2-)), overset(+7)(MnO_4^(-)) :, overset(+3)(CrO_2^(-))`
(c) `overset(+3)(CrO_2^(-)), overset(+5)(ClO_3^(-)), overset(+7)(MnO_4^(-)); overset(+6)(CrO_4^(-))`
(d) `overset(+6)(CrO_4^(2-)), overset(+7)(MnO_4^(-)), overset(+3)(CrO_2^(-)). overset(+5)(ClO_3^(3-))`
Only in the arrangement (a), the O.N. of central atom increases from left to right,
therefore, option (a) is correct.
Correct Answer is `=>` (A) `-1/2`
Q 2808367208

ln haemoglobin the iron is in oxidation state of

(A)

+2

(B)

+1

(C)

+3

(D)

+4

Solution:

Writing the oxidation number `(O.N.)` of `Cr, Cl` and `Mn` on each species in the four set of
ions, then,

(a) `overset(+3)(CrO_2^(-)), overset(+5)(ClO_3^(-)), overset(+6)(CrO_4^(-)) , overset(+7)(MnO_4^(-))`
(b) `overset(+5)(ClO_3^(-)), overset(+6)(CrO_4^(2-)), overset(+7)(MnO_4^(-)) :, overset(+3)(CrO_2^(-))`
(c) `overset(+3)(CrO_2^(-)), overset(+5)(ClO_3^(-)), overset(+7)(MnO_4^(-)); overset(+6)(CrO_4^(-))`
(d) `overset(+6)(CrO_4^(2-)), overset(+7)(MnO_4^(-)), overset(+3)(CrO_2^(-)). overset(+5)(ClO_3^(3-))`
Only in the arrangement (a), the O.N. of central atom increases from left to right,
therefore, option (a) is correct.
Correct Answer is `=>` (A) +2
Q 2808667508

The correct oxidation states of `N` in `NH_4NO_3` molecule are

(A)

-3 and +5

(B)

+3 and -5

(C)

-3 and +3

(D)

-5 and +5

Solution:

Writing the oxidation number `(O.N.)` of `Cr, Cl` and `Mn` on each species in the four set of
ions, then,

(a) `overset(+3)(CrO_2^(-)), overset(+5)(ClO_3^(-)), overset(+6)(CrO_4^(-)) , overset(+7)(MnO_4^(-))`
(b) `overset(+5)(ClO_3^(-)), overset(+6)(CrO_4^(2-)), overset(+7)(MnO_4^(-)) :, overset(+3)(CrO_2^(-))`
(c) `overset(+3)(CrO_2^(-)), overset(+5)(ClO_3^(-)), overset(+7)(MnO_4^(-)); overset(+6)(CrO_4^(-))`
(d) `overset(+6)(CrO_4^(2-)), overset(+7)(MnO_4^(-)), overset(+3)(CrO_2^(-)). overset(+5)(ClO_3^(3-))`
Only in the arrangement (a), the O.N. of central atom increases from left to right,
therefore, option (a) is correct.
Correct Answer is `=>` (A) -3 and +5
Q 2058145004

Match Column I with Column II for the oxidation states of the central
atoms.









Column IColumn II
(A) `Cr_2O_7^(2-)` (P) `+3`
(B) `MnO_4^(-)` (Q) `+4`
(C) `VO_3^(-)` (R) `+5`
(D) `FeF_6^(3-)` (S) `+6`
`quad` `quad` (T) `+7`

NCERT Exemplar
Solution:

`A.-> (4) B.-> (5) C.-> (3) D. -> (1)`

Suppose that x be the oxidation states of central atoms.
A. Oxidation number of Cr in `Cr_2O_7^(2-)`
`2x+7(-2)=-2`
`2x -14 = -2`
`2x = +12`
`x=+6`
B. Oxidation number of Mn in `MnO_^(-)`
`x+ 4 (-2) = 1`
`x-8=-1`
`x=+7`
C. Oxidation number of V in `VO_3^(-)`
`x+ 3 (-2) = -1`
`x-6=-1`
`x=+5`
D. Oxidation number of Fe in FeFt
or
`x+ 6(-1)=- 3`
`x-6=-3`
`x=+3`
Q 2888467307

In the reaction of potassium permanganate in acidic medium (mol wt =158.04) with ferrous ammonium sulphate (mol. wt. = 392.14). the change in the oxidation state of manganese in potassium permanganate is

(A)

+5 to -2

(B)

+6 to +2

(C)

+7 to +2

(D)

+7 to +3

Solution:

`A.-> (4) B.-> (5) C.-> (3) D. -> (1)`

Suppose that x be the oxidation states of central atoms.
A. Oxidation number of Cr in `Cr_2O_7^(2-)`
`2x+7(-2)=-2`
`2x -14 = -2`
`2x = +12`
`x=+6`
B. Oxidation number of Mn in `MnO_^(-)`
`x+ 4 (-2) = 1`
`x-8=-1`
`x=+7`
C. Oxidation number of V in `VO_3^(-)`
`x+ 3 (-2) = -1`
`x-6=-1`
`x=+5`
D. Oxidation number of Fe in FeFt
or
`x+ 6(-1)=- 3`
`x-6=-3`
`x=+3`
Correct Answer is `=>` (C) +7 to +2
Q 2868667505

A compound contains atom `X , Y , Z` The oxidation number of `X` is `+2 , Y ` is `+5` and `Z` is `-2` therefore, a possible formula of the compound is

(A)

`XY_1Z_2`

(B)

`X_2 (YZ_3)_2`

(C)

`X_3 (YZ_4)_2`

(D)

`X_3 (Y_4 Z)_2`

Solution:

`A.-> (4) B.-> (5) C.-> (3) D. -> (1)`

Suppose that x be the oxidation states of central atoms.
A. Oxidation number of Cr in `Cr_2O_7^(2-)`
`2x+7(-2)=-2`
`2x -14 = -2`
`2x = +12`
`x=+6`
B. Oxidation number of Mn in `MnO_^(-)`
`x+ 4 (-2) = 1`
`x-8=-1`
`x=+7`
C. Oxidation number of V in `VO_3^(-)`
`x+ 3 (-2) = -1`
`x-6=-1`
`x=+5`
D. Oxidation number of Fe in FeFt
or
`x+ 6(-1)=- 3`
`x-6=-3`
`x=+3`
Correct Answer is `=>` (C) `X_3 (YZ_4)_2`
Q 1776467376

The oxidation number of an element in a compound is evaluated on the
basis of certain rules. Which of the following is incorrect in this respect?
NCERT Exemplar
(A)

The oxidation number of hydrogen is always+ 1

(B)

The algebraic sum of all the oxidation numbers in a compound is zero

(C)

An element in the free or the uncombined state bears oxidation number zero

(D)

In all its compounds, the oxidation number of fluorine is -1

Solution:

Oxidation number of hydrogen is always +1 is a wrong rule since, it is +1 in hydrogen
halides, -1 in hydrides and zero in `H_2` molecule.
All the other three statements (mb), (c) and (d) are correct.
Correct Answer is `=>` (A) The oxidation number of hydrogen is always+ 1
Q 2828467301

Oxidation number of sulphur in perdisulphuric acid is

(A)

+8

(B)

-6

(C)

+6

(D)

+4

Solution:

Oxidation number of hydrogen is always +1 is a wrong rule since, it is +1 in hydrogen
halides, -1 in hydrides and zero in `H_2` molecule.
All the other three statements (mb), (c) and (d) are correct.
Correct Answer is `=>` (C) +6
Q 2858467304

The oxidation number of iron in `Fe_3O_4` is

(A)

+2

(B)

+3

(C)

`8/3`

(D)

`2/3`

Solution:

Oxidation number of hydrogen is always +1 is a wrong rule since, it is +1 in hydrogen
halides, -1 in hydrides and zero in `H_2` molecule.
All the other three statements (mb), (c) and (d) are correct.
Correct Answer is `=>` (C) `8/3`
Q 2848567403

Which of the following is correct?

(A)

Oxidation state of bromine varies from -1 to +6

(B)

Among `IO_4^- , ClO_4^-` and `BrO_4^-` the last one `(BrO_4^-)` is the strongest oxidising agent

(C)

`HNO_3` is an oxidising agent

(D)

All of the above

Solution:

Oxidation number of hydrogen is always +1 is a wrong rule since, it is +1 in hydrogen
halides, -1 in hydrides and zero in `H_2` molecule.
All the other three statements (mb), (c) and (d) are correct.
Correct Answer is `=>` (D) All of the above
Q 2878678506

Following questions consist of two statement I and statement II. Examine both the statements carefully and match the correct choise according to the codes given below.
Statement I : Fluorine exhibits the same oxidation state in all its components.
Statement II : Fluorine atom has no d-orbital in the valence shell.

(A)

Both the statements are true and Statement II is the correct explanation of Statement I

(B)

Both the statements are true but Statement II is not the correct explanation of Statement I.

(C)

Statement I is true, but Statement II is false

(D)

Statement I is false, but Statement II is true.

Solution:

Oxidation number of hydrogen is always +1 is a wrong rule since, it is +1 in hydrogen
halides, -1 in hydrides and zero in `H_2` molecule.
All the other three statements (mb), (c) and (d) are correct.
Correct Answer is `=>` (A) Both the statements are true and Statement II is the correct explanation of Statement I
Q 2828678501

Match List I and List II and select the correct answer by using the codes given below the lists







List I (The atom for which the oxidation state is to the found ) List II(Oxidation state)
(A) Oxygen in `BaO_2` (1) `-1`
(B) Sulphur in `S_2O_6^(2-)` (2) `0`
(C) Carbon in `C_(12) H_(22) O_(11)` (3) `+7`
(D) Manganese in `MnO_4^-` (4) `-2`
(5) `5`


(A)

`A →1 , B → 4, C → 2 , D → 3`

(B)

`A →1 , B → 5, C → 2 , D → 3`

(C)

`A →4 , B → 1, C → 2 , D → 3`

(D)

`A →4 , B → 5, C → 3 , D → 1`

Solution:

Oxidation number of hydrogen is always +1 is a wrong rule since, it is +1 in hydrogen
halides, -1 in hydrides and zero in `H_2` molecule.
All the other three statements (mb), (c) and (d) are correct.
Correct Answer is `=>` (B) `A →1 , B → 5, C → 2 , D → 3`

Electrochemistry

Q 2878567406

The correct statement is

(A)

zinc cannot displace `Cu` from `CuSO_4`

(B)

`Cu` cannot displace `Fe` from `FeSO_4`

(C)

`Zn` metal displaces `H_2` gas from `H_2SO_4`

(D)

All of the above

Solution:

Oxidation number of hydrogen is always +1 is a wrong rule since, it is +1 in hydrogen
halides, -1 in hydrides and zero in `H_2` molecule.
All the other three statements (mb), (c) and (d) are correct.
Correct Answer is `=>` (D) All of the above
Q 2828767601

Iron nails are dipped into blue copper sulphate solution. After some time iron nails are

(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:

Oxidation number of hydrogen is always +1 is a wrong rule since, it is +1 in hydrogen
halides, -1 in hydrides and zero in `H_2` molecule.
All the other three statements (mb), (c) and (d) are correct.
Correct Answer is `=>` (A) dissolved and blue colour is discharged
Q 2858767604

Which one among the following is the correct order of reactivity of the elements?

(A)

`Cu > Mg > Zn > Na`

(B)

`Na > Zn > Mg > Cu`

(C)

`Cu > Zn > Mg > Na`

(D)

`Na > Mg > Zn > Cu`

Solution:

Oxidation number of hydrogen is always +1 is a wrong rule since, it is +1 in hydrogen
halides, -1 in hydrides and zero in `H_2` molecule.
All the other three statements (mb), (c) and (d) are correct.
Correct Answer is `=>` (D) `Na > Mg > Zn > Cu`
Q 2818567409

Which of the following statement(s) is/are false for electrolytic cell?

(A)

External source of voltage is applied to carry the chemical reaction

(B)

These cells are mainly used in laboratory and chemical industry

(C)

These cells consist of two copper strips dipped in and aqueous solution of `CuSO_4`

(D)

None of the above

Solution:

Oxidation number of hydrogen is always +1 is a wrong rule since, it is +1 in hydrogen
halides, -1 in hydrides and zero in `H_2` molecule.
All the other three statements (mb), (c) and (d) are correct.
Correct Answer is `=>` (D) None of the above
Q 2838767602

Which one among the following metals is more reactive than hydrogen?

(A)

Mercury

(B)

Copper

(C)

Silver

(D)

Tin

Solution:

Oxidation number of hydrogen is always +1 is a wrong rule since, it is +1 in hydrogen
halides, -1 in hydrides and zero in `H_2` molecule.
All the other three statements (mb), (c) and (d) are correct.
Correct Answer is `=>` (D) Tin

Battery, Electrochemistry and Corrosion

Q 2818667500

Mark t.be correct statement(s) for the corrosion

(A)

It fastly coast the surfaces of metallic object

(B)

Its examples are - Rusting of iron tamishing of silver

(C)

II does not cause damage to buildings, ship made metals especially that are made up of iron

(D)

All of the above

Solution:

Oxidation number of hydrogen is always +1 is a wrong rule since, it is +1 in hydrogen
halides, -1 in hydrides and zero in `H_2` molecule.
All the other three statements (mb), (c) and (d) are correct.
Correct Answer is `=>` (B) Its examples are - Rusting of iron tamishing of silver
Q 2818767609

Silver ware turns black after a period of time due to formation of

(A)

nitrate coating on silver

(B)

sulphide coating on silver

(C)

chloride coating on silver

(D)

oxide coating on silver

Solution:

Oxidation number of hydrogen is always +1 is a wrong rule since, it is +1 in hydrogen
halides, -1 in hydrides and zero in `H_2` molecule.
All the other three statements (mb), (c) and (d) are correct.
Correct Answer is `=>` (B) sulphide coating on silver
Q 2858867704

Which of the following cause the rusting of iron?
I. Oxidation
II. Reduction
III. Chemical reaction with oxygen.
IV. Chemical reaction with `CO_2`.
Select the correct answer from the codes given above.

(A)

I and II

(B)

I and III

(C)

II and III

(D)

III and IV

Solution:

Oxidation number of hydrogen is always +1 is a wrong rule since, it is +1 in hydrogen
halides, -1 in hydrides and zero in `H_2` molecule.
All the other three statements (mb), (c) and (d) are correct.
Correct Answer is `=>` (B) I and III
Q 2529156911

The process of zinc-plating on iron sheet is known as
BCECE Mains 2015
(A)

annealing

(B)

roasting

(C)

galvanisation

(D)

smelting

Solution:

The process of zinc plating on iron sheet is known

as galvanisation.
Correct Answer is `=>` (C) galvanisation
Q 2888767607

Aluminium is more reactive than iron but aluminium is less easily corroded than iron, because

(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:

The process of zinc plating on iron sheet is known

as galvanisation.
Correct Answer is `=>` (A) oxygen forms a protective oxide layer
Q 2818778600

The standard electrode potential values of the element A, Band C are 0.68 V, -2.0 V and -0.50 V respectively. The order of their
reducing power is

(A)

`A > B > C`

(B)

`A < C > B`

(C)

`C > B > A`

(D)

`B > C > A`

Solution:

The process of zinc plating on iron sheet is known

as galvanisation.
Q 2838578402

Match the term;; of Column I and Column II and choose the correct option fron the codes given below.







Column IColumn II
(A) Lectanche cell (1) Cell reaction `2H_2O+O_2 → 2H_2O`
(B) Ni - Cd- Cell (2) Does not involve any
ion in solution and is
used in hearing aids .
(C) Fuel cell (3) Rechargeable
(D) Mercury cell (4) Reaction at anode `Zn → Zn^(2+) +2 e^-`
(5) Converts energy of combustion in to electrical energy


(A)

`A →1 , B → 2, C → 3 , 4 , D → 5`

(B)

`A → 5 , B → 2 , C → 1, 3 , D → 4`

(C)

`A → 2 , B → 3 , C → 1, 5 , D → 4`

(D)

`A → 4 , B → 3 , C → 1, 5 , D → 2`

Solution:

The process of zinc plating on iron sheet is known

as galvanisation.
Correct Answer is `=>` (D) `A → 4 , B → 3 , C → 1, 5 , D → 2`
 
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