Organization Panel

Add Question Remove Question View Key View Solution View Paper 2 Column

Paper Set 1 Paper Set 2 Paper Set 3 Paper Set 4

	QPID #4593 (SET 1)		NEET chemistry for C-section Subjective
Class/Batch: C-Section	Duration: 1 hour	Date: 23 Nov,2022	Total Mark:180

Instructions: NEET 2 chapter test

CHEMISTRY: Redox Reactions, Electrochemistry,

Section 1 (CHEMISTRY) 50X4=200

#161768

Mark error

Quantitatively O₃ is estimated by using :

1)
standard $KMn{O_4} + $ standard Hypo
2)
standard${K_2}C{r_2}{O_7} + $standard kI
3)
Excess KI+ standard Hypo
4)
Excess pbs+standard ${H_2}S{O_4}$

Solution:
$\begin{gathered} {O_3} + KI \to {I_2} \hfill \\ \,\,\,{I_2} + 2N{a_2}{S_2}{O_3} \to N{a_2}{S_4}{O_6} + 2NaI \hfill \\ \end{gathered} $

#2578239

Mark error

Which reaction involves neither oxidation nor reduction?

1)
$C r O_{4}^{2 -} \to C r_{2} O_{7}^{2 -}$
2)
$C r \to C r C l_{3}$
3)
$N a \to N a^{+}$
4)
$2 S_{2} O_{3}^{2 -} \to S_{4} O_{6}^{2 -}$

Solution:
Ox. no. of Cr on both side is + 6.

#2578262

Mark error

‘Oxidation number of H in NaH, CaH₂ and LiH, respectively is

1)
+1, +1, –1
2)
–1, +1, + 1
3)
+1, + 1, + 1
4)
–1,–1, –1

Solution:
Oxidation number of hydrogen when it is bonded to metals in binary compounds is –1

#2578270

Mark error

The oxidation number of sulphur in S₈, S₂F₂, H₂S respectively, are

1)
0, +1 and –2
2)
`+2, +1 and –2
3)
0, +1 and +2
4)
–2, +1 and –2

Solution:
(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;

#2578277

Mark error

Which of the following transition metal has zero oxidation state ?

1)
[Fe(CO)₅]
2)
NH₂.NH₂
3)
NOClO₄
4)
CrO₅

Solution:
Fe(CO)₅ is metal carbonyl, hence O.N. of Fe is zero.

#2578282

Mark error

In which of the following compounds, the oxidation number of iodine is fractional ?

1)
IF₇
2)
$I \bar{3}$
3)
IF₅
4)
IF₃

Solution:
O.N. of iodine in $I \bar{3}$ is –1/3

#2578325

Mark error

The following species will not exhibit disproportionation reaction

1)
ClO^–
2)
ClO $\bar{2}$
3)
ClO $\bar{3}$
4)
ClO $\bar{4}$

Solution:
In disproportionation reaction, one element of a compound will simultaneously get reduced and oxidised. In ClO $\bar{4}$ , oxidation number of Cl is + 7 and it can not increase it further. So, ClO $\bar{4}$ will not get oxidised and so will not undergo disporportionation reaction.

#2578329

Mark error

Which of the following elements does not show disproportionation tendency?

1)
Cl
2)
Br
3)
F
4)
I

Solution:
F

#2578332

Mark error

How will you balance the total ionic charge of reactant and products if reaction is carried out in acidic solution?

1)
By using H⁺ ions
2)
By using OH^– ions
3)
Adding H₂O molecules to the reactant or product
4)
Multiplying by suitable coefficients

Solution:
H⁺ ions are added to the expression on the appropriate side so that the total ionic charges of reactants and products become equal.

10)

#2578341

Mark error

2MnO $\bar{4}$ + 5H₂O₂+ 6H⁺ $\to$ 2 Z + 5O₂ + 8H₂O. In this reaction Z is

1)
Mn⁺²
2)
Mn⁺⁴
3)
MnO₂
4)
Mn

Solution:
2MnO $\bar{4}$ + 5H₂O₂ +6H⁺ $\to$ 2Mn ²⁺+ 5O₂ 8H₂O.

11)

#2578348

Mark error

The values of x and y in the following redox reaction
x Cl₂ + 6OH^- $\to$ ClO $\bar{3}$ + yCl^- + 3H₂ O are

1)
x = 5, y = 3
2)
x = 2, y = 4
3)
x = 3, y = 5
4)
x = 4, y = 2

Solution:
on balancing the eq we get
3Cl₂+ 6OH^- $\to$ ClO $\bar{3}$ +5Cl ^-+3H₂O

12)

#2578360

Mark error

Electrode potential data are given below :

$F e_{(a q)}^{+ 3} + e^{-} \to F e_{(a q)}^{+ 2}; E ° = + 0.77 V$

$A 1_{(a q)}^{3 +} + 3 e^{-} \to A 1_{(s)}; E ° = - 1.66 V$

$B r_{2 (a q)} + 2 e^{-} \to 2 B r_{(a q)}^{-}; E ° = + 1.08 V$

Based on the data, the reducing power of Fe²⁺, Al and Br^– will increase in the order

1)
Br^– < Fe²⁺ < Al
2)
Fe²⁺ < Al < Br^–
3)
Al < Br^– < Fe²⁺
4)
Al < Fe²⁺ < Br^–

Solution:
Fe    Al            Br
0.77    –1.66     1.08 E°Red
–0.77           1.66 –1.08 E°Oxi
Hence, reducing power Al > Fe²⁺ > Br^–

13)

#2578368

Mark error

E^$\emptyset$ Values of some redox couples are given below. On the basis of these values choose the correct option.

E values : Br₂/Br ^– = + 1.90; Ag⁺ /Ag(s) = + 0.80 Cu²⁺ /Cu(s) = + 0.34; I₂(s) /I– = 0.54

1)
Cu will reduce Br ^–
2)
Cu will reduce Ag
3)
Cu will reduce I ^–
4)
Cu will reduce Br₂

Solution:
Cu will reduce Br₂

14)

#2578374

Mark error

The standard electrode potentials of four elements A, B, C and D are –3.05, –1.66, –0.40 and +0.80. The highest chemical reactivity will be exhibited by

1)
A
2)
B
3)
C
4)
D

Solution:
Standard electrode potential i.e. reduction potential of A is minimum (–3.05V) i.e. its oxidation potential is maximum which implies ‘A’ is most reactive chemically.

15)

#2578416

Mark error

Among NH₃, HNO₃, NaN₃ and Mg₃N2 the number of molecules having nitrogen in negative oxidation state is

1)
1
2)
2
3)
3
4)
4

Solution:
Calculating the oxidation state of nitrogen in given molecules;
Oxidation state of N in NH₃ is x + 3 × (+ 1) = 0 or x = – 3
Oxidation state on N in NaNO₃ is 1 + x + 3 × (– 2) = 0 or x = + 5

Oxidation state of N in NaN₃ is

+ 1 + 3x = 0 or x = – $\frac{1}{3}$

Oxidation state of N in Mg₃N₂ is 3 × 2 + 2x = 0 or x = –3
Thus 3 molecules (i.e. NH₃, NaN₃ and Mg₃N₂ have nitrogen in negative oxidation state.

16)

#2578419

Mark error

Fill up the table from the given choice.

Element Oxidation number

Oxygen –2 in most compounds (i) in H2O2 and (ii) in OF2
Halogen                      –1 for (iii) in all its compounds
Hydrogen    (iv) in most of its compounds (v) in binary metallic hydrides
Sulphur                         (vi) in all sulphides

	(i)	(ii)	(iii)	(iv)	(v)	(vi)
(a)	+1	+1	Cl	+1	–1	+2
(b)	–1	+2	F	+1	–1	–2
(c)	–1	+1	F	+1	+2	+2
(d)	+1	+2	Cl	+1	+1	+6

1)
A
2)
B
3)
C
4)
D

Solution:
B

17)

#2578426

Mark error

When Cl₂ gas reacts with hot and concentrated sodium hydroxide solution, the oxidation number of chlorine changes from

1)
zero to +1 and zero to –5
2)
zero to –1 and zero to +5
3)
zero to –1 and zero to +3
4)
zero to +1 and zero to –3

Solution:
On reaction with hot and concentrated alkali a mixture of chloride and chlorate is formed
3Cl2 + 3 NaOH_(excess) $\overset{H o t}{\to}$

$5 N a \overset{- 1}{C l} + N a \overset{+ 5}{C l} O_{3} + 3 H_{2} O$

18)

#2578437

Mark error

Which of the following act as reducing agents ?

(i) PO $\overset{3 -}{4}$ (ii) SO₃
(iii) PO $\overset{2 -}{3}$ (iv) NH₃

1)
(i), (ii) and (iii)
2)
Only (iii)
3)
(i), (iii) and (iv)
4)
(iii) and (iv)

Solution:
In (i) and (ii) both P and S are in highest oxidation state. In (iii) and (iv) ; P has oxidation state of +4 which can be oxidized to +5 state, while in case of NH₃ nitrogen has oxidation state of –3 which can be oxidised.

19)

#2578440

Mark error

What products are expected from the disproportionation reaction of hypochlorous acid?

1)
HCl and Cl₂O
2)
HCl and HClO₃
3)
HClO₃ and Cl₂O
4)
HClO₂ and HClO₄

Solution:
During disproportionation same compound undergo simultaneous oxidation reduction.

20)

#2578458

Mark error

In the balanced chemical reaction

$I O_{3}^{-} + a l^{-} + b H^{-} \to c H_{2} O + d I_{2}$

a, b, c and d respectively corresponds to

1)
5, 6, 3, 3
2)
5, 3, 6, 3
3)
3, 5, 3, 6
4)
5, 6, 5, 5

Solution:
Given reaction is

$I O_{3}^{-} + a I^{-} + b H^{+} \to c H_{2} O + d l_{2}$    Ist half reaction

$I^{-} \to I_{2}$                                                                                ......(i)

-1           0    (oxidation)

IInd half reaction

$I O_{3}^{-} \to I_{2}$                                                                   ........(ii)

+5                     0    (reduction)

On balancing equation (ii) we have

$10 e^{-} + 2 l o_{3}^{-} + 12 H^{+} \to I_{2} + 6 H_{2} O$           ..........(iii)

Now, balance equation (i)

$2 I^{-} \to I_{2} + 2 e^{-}$                                                           .....(iv)

Multiply eqn (iv) by 5 and add it to eqn (iii), we get

$2 I O_{3}^{-} + 10 I^{-} + 6 I_{2} + 6 H_{2} O o r, l O_{3}^{-} + 5 l^{-} + 6 H^{+} \to 3 l_{2} + 3 H_{2} O$

Hence a = 5, b = 6, c = 3, d = 3

21)

#2578460

Mark error

If equal volume of reactants are used, than no. moles of KMnO₄ (moles per liter) used in acidic medium required to completely oxidises the 0.5 M FeSO₃?

1)
0.3
2)
0.1
3)
0.2
4)
0.4

Solution:
Both Fe(ii) and S(iv) in SO $\overset{2 -}{3}$ can be oxidised to Fe(iii) and (SO₄)^2– respectively hence (3/5)×0.5 =0.3 moles / litre.

$[|M n O_{4}^{-} + 5 e^{-} + 8 H^{+} \to M n^{2 +} 4 H_{2} O|] \times \frac{3}{5}$

Fe²⁺ $\to$ Fe³⁺+le^-

SO $\overset{2 -}{3}$ $\to$ SO $\overset{2 -}{4}$ +2e^-

$\frac{8}{3}$ MnO $\bar{4}$ + $\frac{24}{5}$ H⁺+ Fe²⁺+SO $\overset{2 -}{3}$ $\to$ 3Mn²⁺ +4H₂O +Fe³⁺+SO $\overset{2 -}{4}$

22)

#2578462

Mark error

Acidic medium used in KMnO₄ can be made from which of the following acids?

1)
HCl
2)
H₂SO₄
3)
HI
4)
HBr

Solution:
If one uses HCl, HBr or HI, to make acidic medium for KMnO₄ than all the halide ion can be oxidized as the reduction potential of KMnO₄ is very high in acidic medium, while in case of H₂SO₄, sulphur is already in its highest oxidation state cannot be further oxidized.

23)

#2578471

Mark error

If rod of a metal (x) is put in a metal ion solution which is blue in colour , solution turn colorless. The metal rod and solution respectively are?

1)
Zinc and Cu(II)
2)
Zinc and Ni(II)
3)
Aluminium and Cu(II)
4)
Both (a) and (c)

Solution:
Reduction potential of Cu(II) is greater than that of Zn(II) and Al(III) thus can be easily replaced by these ions. Moreover solution of copper is blue in color.

24)

#2578473

Mark error

What could be the X^–in the system, Where X signifies halogen ; formation of shown below X₂ takes place, when F₂is purge into aqueous solution of X^–?

1)
Br ^–
2)
Cl ^–
3)
I ^–
4)
All of these

Solution:
F₂ is strongest oxididing agent among halogens thus X^–can be possibly Br^–, Cl^– or I^–.

25)

#2652452

Mark error

Standard reduction potential values are given ${E^{0}}_{c u^{+ 2} / c u} = 0.34 V a n d {E^{0}}_{c u^{+} / c u} = 0.52 V, {E^{0}}_{H^{+} / \frac{1}{2} H_{2}} = 0.0 V$

I) $C u^{+ 1} (a q)$ is stronger oxidising agent than $C u^{+ 2} (a q)$

II) $C u^{+ 1} (a q)$ is less stable than $C u^{+ 2} (a q)$

III) $C u^{+ 1} (a q)$ readily reduced than $C u^{+ 2} t o C u (s)$

IV) $E^{0} C u / C u^{+ 2} = - 0.34 V,$ is –ve quantity, Cu can not evolve hydrogen gas from $1 N H_{2} S O_{4}$

The correct statements are

1)
All are correct
2)
I, II and III are only correct
3)
IV is only correct
4)
II, III and IV are only correct.

Solution:
All are correct

26)

#2652459

Mark error

Ostwald dilution law of weak electrolyte H can be given as : ‘A’

1)
$K_{a} = \frac{C \cdot α^{2}}{(1 - α)}$
2)
$K_{a} = C \cdot α^{2} i f α < < < 1$
3)
$K_{a} = \frac{C Λ^{2}}{Λ_{0} (Λ_{0} - Λ)}$
4)
All of these

Solution:
All of these

27)

#2652644

Mark error

Which of the following statements is incorrect

1)
Adsorption always leads to a decrease in enthalpy and entropy of the system.
2)
Adsorption arises due to unsaturation of valence forces of atoms or molecules on the surface.
3)
Adsorption increases with rise in temperature.
4)
Adsorption decreases surface energy.

Solution:
Adsorption increases with rise in temperature.

28)

#2652651

Mark error

$A g | \underset{I L}{A g^{+}} (1 M) | | \underset{I L}{A g^{+}} (2 M) | A g,$

0.5 F of electricity in anode and 1F electricity in the cathode is first passed making them independent electrolytic cell at 298K, EMF of cell after electrolysis will be

1)
increased
2)
decreased
3)
no change
4)
time is also required to find it

Solution:
no change

29)

#2652655

Mark error

For the net cell reaction of the cell Zn(s) | Zn²⁺ || Cd²⁺ | Cd(s) $∆$ G⁰ in kJ at 25°C is ( $E_{c e l l}^{0} = 0.360 V)$ :

1)
112.5
2)
69.47
3)
– 34.73
4)
–69.48

Solution:
–69.48

30)

#2652657

Mark error

Equivalent conductance of saturated BaSO₄ is 400 ohm^–1cm² equiv^–1 and specific conductance is 8 x 10^–5 ohm^–1 cm^–1. Hence K_sp of BaSO₄ is:

1)
4 x 10^–8 M²
2)
1 x 10^–8M²
3)
2 x 10^–4 M²
4)
1 x 10^–4 M²

Solution:
1 x 10^–8M²

31)

#2652662

Mark error

Which of the following is/are incorrect regarding dry cell

1)
Mn is reduced from +4 to +3 state
2)
NH₃ gas is evolved
3)
Zn is used as anode
4)
A paste of NH₄Cl and ZnCl₂ is used.

Solution:
NH₃ gas is evolved

32)

#2652666

Mark error

The electrolysis of aqueous solution of CuBr₂ using Pt electrodes would lead to evolving of

1)
Br₂ gas at anode
2)
Br₂ gas at anode and O₂ gas at cathode
3)
Copper at anode
4)
H₂O gas at cathode

Solution:
Br₂ gas at anode

33)

#2652669

Mark error

The molar conductance of Acetic acid at infinite dilution if $\land^{0}$ for CH₃COONa, NaCl and HCl are 91.0, 126.5 and 426.2 S.cm²mol^-1 respectively.

1)
517.2
2)
552.7
3)
390.7
4)
217.5

Solution:
390.7

34)

#2652677

Mark error

Which of the following is correct?

1)
Specific conductivity of a solution decreases with dilution, whereas molar conductivity increases with dilution
2)
Specific conductivity of a solution increases with dilution, whereas molar conductivity decreases with dilution
3)
Both specific conductivity and molar conductivity decreases with dilution
4)
Both specific conductivity and molar conductivity increases with dilution

Solution:
Specific conductivity of a solution decreases with dilution, whereas molar conductivity increases with dilution

35)

#2652700

Mark error

$A l^{3 +} (a q .) + 3 e^{-} \to A l (s); E ° = - 1.66 V$

$C u^{2 +} (a q .) + 2 e^{-} \to C u (s); E ° = + 0.34 V$

What voltage is produced under standard conditions by combining the half reactions with these standard electrode potentials.

1)
1.32 V
2)
2.00 V
3)
2.30 V
4)
4.34 V

Solution:
2.00 V

36)

#2652701

Mark error

What is the approximate value of the equilibrium constant, Keq at 25⁰C for the reaction; $[\frac{2.303 R T}{F} = 0.06]$ $3 {Ag}^{+} (aq) + Cr (s) \to {Cr}^{3 +} (aq) + 3 Ag (s)$

$A g^{+} (a q) + e^{-} \to A g (s); E^{0} = + 0.80$

${Cr}^{3 +} (aq) + 3 e^{-} \to Cr (s); E^{0} = - 0.74$

1)
10²²
2)
10²⁶
3)
10³³
4)
10⁷⁷

Solution:
10⁷⁷

37)

#2652704

Mark error

Equal volumes of 0.015 M CH₃COOH and 0.015 M NaOH are mixed together. What would be molar conductivity of mixture if conductivity of CH₃COONa is $6.3 \times 10^{- 4} S c m^{- 1}$

1)
$8.4 S c m^{2} m o l^{- 1}$
2)
$84 S c m^{2} m o l^{- 1}$
3)
$4.2 S c m^{2} m o l^{- 1}$
4)
$42 S c m^{2} m o l^{- 1}$

Solution:
$84 S c m^{2} m o l^{- 1}$

38)

#2652707

Mark error

In the electrochemical cell $Pt H_{2} (g), 1 atm | H^{+} (1 M) | | {Cu}^{2 +} (1 M) | Cu (s),$

Which one of the following statements is true?

1)
H₂ is cathode, Cu is anode
2)
Oxidation occurs at Cu electrode
3)
Reduction occurs at H₂ electrode
4)
Pt is anode, Cu is cathode

Solution:
Pt is anode, Cu is cathode

39)

#2652709

Mark error

Which products are formed first in the electrolysis of an aqueous solution of AlCl₃?

(I) Al(s) (II) Cl₂(g) (III) H₂ (g) (IV) O₂ (g)

1)
I and III only
2)
I and IV only
3)
II and III only
4)
II and IV only

Solution:
II and III only

40)

#2652712

Mark error

A current of 2.0 A is used to plate Ni(s) from 500 mL of a 1.0 M Ni²⁺ (aq) solution. What is the [Ni²⁺] (aq) after 3.0 h?

1)
0.39 M
2)
0.46 M
3)
0.78 M
4)
0.89 M

Solution:
0.78 M

41)

#2652715

Mark error

$2 A g^{+} (a q) + C u (s) \to C u^{2 +} (a q) + 2 A g (s)$ The standard potential for this reaction is 0.46 V. Which change will increase the potential the most?

1)
Doubling the $[A g^{+}]$
2)
Halving the $[C u^{2 +}]$
3)
Doubling the size of the Cu(s) electrode
4)
Decreasing the size of the Ag electrode by one half

Solution:
Doubling the $[A g^{+}]$

42)

#2652719

Mark error

How many moles of electrons must be transvered from each mole of toluene, C₆H₅CH₃, when it is oxidized to benzoic acid. C₆H₅COOH?

1)
1
2)
2
3)
4
4)
6

Solution:
6

43)

#2652724

Mark error

How many faradays are required to reduce all the chromium in 0.150 L of 0.115 M of $C r_{2} O_{7}^{2 -} t o C r^{2 +}$ ?

1)
0.920 F
2)
0.690 F
3)
0.138 F
4)
0.069 F

Solution:
0.138 F

44)

#2652728

Mark error

Which of the following statement is NOT correct about an inert electrode in a cell ?

1)
It is a connector of internal and external parts of cell
2)
It does not participate in cell reaction
3)
It provides surface for redox reaction
4)
It provides surface for conduction of electrons

Solution:
It provides surface for redox reaction

45)

#2652798

Mark error

The standard reduction potential of $C u^{2 +} / C u$ and $C u^{2 +} / C u^{+}$ are $0.337 V$ and $0.153 V$ respectively. The standard reduction potential of $C u^{1 +} / C u$ half cell is

1)
$0.184 V$
2)
$0.827 V$
3)
$0.521 V$
4)
$0.490 V$

Solution:
$0.521 V$

46)

#2652809

Mark error

The charge required for the reduction of 1 mole of $I_{2} t o I^{-}$ is

1)
$96500 C$
2)
$2 \times 96500 C$
3)
$\frac{1}{2} \times 96500 C$
4)
$4 \times 96500 C$

Solution:
$2 \times 96500 C$

47)

#2652813

Mark error

Electrolysis of an acetate solution produces ethane according to the Kolbe reaction :

$2 C H_{3} C O O^{-} \to C_{2} H_{6} + 2 C O_{2} + 2 e^{-}$

What volume of ethane is produced at 270C and 740 mm Hg, if a current of 0.500 A were passed through the solution for 7 h and the electrode reaction is 82% efficient?

1)
2.70 L
2)
5.4 L
3)
1.35L
4)
0.65 L

Solution:
1.35L

48)

#2652815

Mark error

Electrolysis of dilute aqeous NaCl solution was carried out bypassing 10 milli ampere current. The time required to liberate 0.01mole of H₂ gas at the cathode is (1 Faraday = 96500 C)

1)
9.65 x 10⁴ sec
2)
19.3 x 10⁴ sec
3)
28.95 x 10⁴ sec
4)
38.6 x 10⁴ sec

Solution:
19.3 x 10⁴ sec

49)

#2652818

Mark error

A fuel cell develops an electrical potential from the combustion of butane at 1 bar and 298 K $C_{4} H_{10} (g) + 6.5 O_{2} (g) \to 4 C O_{2} (g) + 5 H_{2} O (l); Δ_{r} G^{0} = - 2746 k J / m o l$ what is $E^{0}$ of a cell?

1)
4.74 V
2)
0.547 V
3)
4.37 V
4)
1.09 V

Solution:
1.09 V

50)

#2652820

Mark error

The reduction potential of a $H g_{2}^{2 +} / H g$ electrode, if a 0.04 M KCl solution is present alongwith $H g_{2} C l_{2} (s)$ , is :

$(G i v e n E_{H g / H g_{2}^{2 +}}^{0} = - 0.79 V, K_{s p} (H g_{2} C l_{2}) = 1.6 \times 10^{- 18}, \frac{2.303 R T}{F} = 0.06)$

1)
0.34V
2)
0.11V
3)
1.01V
4)
1.24V

Solution:
0.34V

QPID #4593 (SET 1)

Solution:

Solution:

Solution:

Solution:

Solution:

Solution:

Solution:

Solution:

Solution:

Solution:

Solution:

Solution:

Solution:

Solution:

Solution:

Solution:

Solution:

Solution:

Solution:

Solution:

Solution:

Solution:

Solution:

Solution:

Solution:

Solution:

Solution:

Solution:

Solution:

Solution:

Solution:

Solution:

Solution:

Solution:

Solution:

Solution:

Solution:

Solution:

Solution:

Solution:

Solution:

Solution:

Solution:

Solution:

Solution:

Solution:

Solution:

Solution:

Solution:

Solution: