Explore topic-wise InterviewSolutions in Current Affairs.

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1.

E ^(@) of the cell, zn| Zn^(2+)(aq) ||Cu^(2+) (aq) |Cu is 1.10Vat 25^(@)C. The equilibrium constant for the cell reaction Zn + Cu^(2+) (aq) hArr Cu + Zn ^(2+) (aq) is of the order of

Answer»

`10^(-37)`
`10^(37)`
`10^(-17)`
`10^(17)`

Solution :`E_(cell)^(@) =(0.059)/(2) log K_(C)or (1.10xx2)/(0.059) =log K_(c)`
`therefore K_(C) =1.9 xx10^(37)`
Discussion
2.

E^(@) of some oxidants are given as : I_(2)+2e rarr 2I^(-) " "E^(@) = +0.54V MnO_(4)^(-)+8H^(+)+5e rarr Mn^(2+)+4H_(2)O "" E^(@) = +1.52V Fe^(3+)+e rarr Fe^(2+) " " E^(@) = +0.77V Sn^(4+)+2e rarr Sn^(2+) ""E^(@) = +0.1V (a) Select the strongest reductant and oxidant in these. (b) Select the weakest reductant and oxidant in these. (c) Select the spontaneous reaction from the changes given below : (i) Sn^(4+)+2Fe^(2+) rarr Sn^(2+)+2Fe^(3+) (ii) 2Fe^(2+)+I_(2) rarr 2Fe^(3+)+2I^(-) (iii) Sn^(4+)+2I^(-) rarr Sn^(2+)+I_(2) (iv) Sn^(2+)+I_(2) rarr Sn^(4+)+2I^(-)

Answer»

Solution :(a) More or `+ve` the `E_(OP)^(@)`, more is the tendency form oxidation or stronger is reductance.
Therefore, since maximum `E_(OP)^(@)`, stands for :
`Sn^(2+) rarr Sn^(4+)+2e, E_(OP)^(@) = -0.1V`
`:.` STRONGEST reducant :` Sn^(2+)`
and weakest OXIDANT : `Sn^(4+)`
(b) More or `+ve` is `E_(RP)^(@)`, more is the tendency form reduction or stronger is oxidant. Therefore, since maximum `E_(RP)^(@)` stands for : `MnO_(4)^(-)+8H^(+)+5e rarr Mn^(2+)+4H_(2)O, E_(RP)^(@) = +1.52V`
`:.` Strongest oxidant : `MnO_(4)^(-)`
and weakest REDUCTANT : `Mn^(2+)`
Note: Stronger is oxident, weaker is its conjugate reductant and vice-versa.
(c) For (i)
`E_(cell)^(@) = E_(OP_(Fe^(2+)//Fe^(3+)))^(@)+E_(RP_(Sn^(2+)//Sn^(3+)))^(@) = -0.77+0.1`
`becauseFe^(2+)` oxidizes and `Sn^(4+)` reduces in change.
`E_(cell)^(@) = -0.67V`
`E_(Cell)^(@)` isnegative.
`:.` (i) Is non-spontaneous change.
For (ii), `E_(Cell)^(@) =E_(OP_(Fe^(2+)//Fe^(3+)))^(@)+E_(RP_(I_(2//I^(-))))`
`= -0.770+0.54 = -0.23V`
(ii) Is non-spontaneouos change.
(iii) `E_(Cell)^(@) = E_(OP_(I^(-)//I_(2)))^(@)+E_(RP_(Sn^(4+)//Sn^(2+)))^(@)`
`= -0.54 + 0.1 = -0.44 V`
`:.` (iii) Is non-spontaneous change.
For (IV), `E_(Cell)^(@) = E_(RP_(Sn^(2+)//Sn^(4+)))^(@)+E_(RP_(I_(2)//I^(-)))`
`= -0.1+0.54 = +0.44V`
Discussion
3.

E^@ of zinc electrode is -0.762V. Calculate the single electrode potential of Zn electrode in decimolar ZnSO_4 solution.

Answer»

Solution :SIMPLIFIED Nernst equation for single electrode POTENTIAL,
`E=E^(@)+(0.059)/(2) log[ZN^(2+)]`
SUBSTITUTING the values in the expression for the single electrode potential of Zn.
`E = -0.762+0.0295log 10^(-1)`
`= - 0.7915V`.
Discussion
4.

E^(@) of three mctals A, B, C are +0.44 V, +1.37 V and -1.35 V. Arrange the metals in increasing order of their reducing power.

Answer»

SOLUTION :As `E^(@)` value decreases REDUCING power increases
`THEREFORE B
Discussion
5.

E^(@) values of Mg^(2+) //Mg is -2.37V, of Zn^(2+)//Zn is -0.76V and Fe^(2+) //Fe is -0.44V. Which of the following statement is correct ?

Answer»

Mg oxidises FE
`ZN oxidises Fe`
`Zn REDUCES Mg^(2+)`
`Zn reduces Fe^(2+)`

Answer :D
Discussion
6.

E^@ of Fe|Fe^(2+) is +0.44V , E^@ of Cu|Cu^(2+) is -0.32V. Then in the cell

Answer»

Cu OXIDISES `Fe^(2+)`ION
`Cu^(2+)` oxidises iron
Cu reduces `Fe^(2+)` ion
`Cu^(2+)` ion reduces Fe

Answer :B
Discussion
7.

E^(@) of Cu is +0.34V while that of Zn is -0.76V. Explain

Answer»

Solution :In CASE of copper, the hydration enthalpy of `CU^(2+)` does not balance the summation of sublimation enthalpy and ionization ENTHALPIES and so it has positive `E^(@)` VALUES. However in case of ZN the second ionization enthalpy is low because electrons is to be removed from 4s-orbital to attain stable `3d^(10)` configuration. Hence Zn has negative `E^(@)` value
Discussion
8.

E^(@) of Cu is + 0.34 V while that of Zn is – 0.76 V. Explain.

Answer»

Solution :High ionisation enthalpy to change Cu (s) to `Cu^(2+)` is not balanced by hydration enthalpy. Therefore, it EXHIBITS a positive `E^(@)` value. However, Zn exhibits a lower value of ionization enthalpy because a stable `3d^(10)` CONFIGURATION is attained after LOSING two electrons. The hydration energy for `Zn^(2+)` is COMPARABLE to that of `Cu^(2+)`. Therefore, `Eo^(@)` for Zn is negative.
Discussion
9.

E^(@) of a cell aA=bBtocC+dD is

Answer»

`-(RT)/(nF)"LOG"([C]^(C)[D]^(d))/([A]^(a)[B]^(b))`
`-Rt"log"([a]^(A)[b]^(B))/([a]^(C)[d]^(D))`
`-(RT)/(nF)"log"([C]^(c)[d]^(D))/([A]^(a)[B]^(b))`
`-(RT)/(nF)"log"([C]^(c)[d]^(D))/([a]^(A)[B]^(b))`

ANSWER :A
Discussion
10.

E^(0) for the reaction Fe +Zn^(2+) to Zn + Fe^(2+) is -0.35 V . The given cell reaction is

Answer»

FEASIBLE
not feasible
EXPLOSIVE
slow

Answer :B
Discussion
11.

E^(@) for two reactions are given below : Cr^(+3) + 3e^(-) to Cr , E^(@) = - 0.74 V Ocl^(-) + H_2O + 2e^(-) to Cl^(-) + 2OH^(-) , 2OH , E^(@) = 0.94 V What will be the E^(@) for 3OCl^(-) + 2Cr + 3H_2O to 2Cr^(+3) + 3Cl^(-) + 6OH^(-)

Answer»

`-1.68 V `
`1.68 V `
`-0.20 V `
`+ 0.20 V `

SOLUTION :` 2 XX 3 xx E^(0) = -(-0.74 xx 3 xx 2) + (0.94 xx 2 xx 3) , E^(0) = 0.74 + 0.94 = 1.68V`
Discussion
12.

E^(0) for the half cell Zn^(2+) // Znis - 0.76 V . Emfof the cell Zn// Zn^(2+)(1M) // //H^(+)(1M) // H_2 at 1 atm is

Answer»

`-0.76`V
`+0.76` V
`- 0.38V`
`+0.38V`

ANSWER :B
Discussion
13.

E^@ for the half cell reactions are as,Zn→Zn^(2+) +2e,E^@=+0.76VFe→Fe^(2+)+2e,E^@=+0.41VThe E^@ for the cell reaction.Fe^(2+) +ZnrarrZn^(2+)+Fe:

Answer»

`-035V`
`+0.35V`
`+1.17V`
`-0.17V`

ANSWER :B
Discussion
14.

E^@ for Mg to 2e^(-) + Mg^(2+)is + 2.37 V and for Cu to 2e^(-) + Cu is -0.34V. What will be the standard potential of cell constructed with these electrodes ? Which electrode will be positive terminal to draw the current?

Answer»


ANSWER :2.71 V, CU ELECTRODE
Discussion
15.

E^(@) for Fe^(2+)+2e^(-) too Feis -0.4 volts and E^(@) for Zn^(2+)+2e^(-) to Znis -0.76 volts . Its means that

Answer»

FE is more electropositive
ZN is more electropositive
Zn is more electronegative
Fe is more electronegative

Solution :`E_("RED")^(0)Fe gt E_("red")^(0)Zn`
Discussion
16.

E^(@) forFe^(2+) + 2e to Feis -0.44 V , E^(@) for Zn^(2+) + 2e to Znis - 0.76 V . Then

Answer»

ZN is more ELECTROPOSITIVE than FE
Fe is more ELECTRONEGATIVE than Zn
Zn is more electronegative
NONE of the above
Discussion
17.

E^@ for Fe^(2+) +2e rarrFe is -0.44 volt and E^@ for Zn^(2+) +2e rarrZn is -0.76 volt thus:

Answer»

ZN is more ELECTROPOSITIVE than Fe
Fe is more electropositive than Zn
Zn is more electronegative
None

Answer :A
Discussion
18.

E^@ for F_2 +2erarr 2F^- is 2.8V,E^@ for 1//2F_2+erarr F^- is :

Answer»

`2.8V`
`1.4V`
`-2.8V`
`-1.4 V`

ANSWER :A
Discussion
19.

E^(0) for F_2 + 2e^(-) to 2F^(-) is 2.8 V , E^(0) for1//2 F_2 + e^(-) to F^(-) is

Answer»

2.8V
1.4V
`-2.8V`
`-1.4V`

ANSWER :A
Discussion
20.

(E) dissolves in aqueous solution of ammonia giving a colourless solution of (F). Identify (A) to (F) and give chemical equations for reactions involved insteps (i) to (v).

Answer»

Solution :`(A) -Ag_(2)S` (B) `-NaAg(CN)_(2)`
(C) `-Na_(2)SO_(4)` (D) `AG`
(E) `AgCl` (F) `-Ag(NH_(3))_(2)Cl`
`underset((E))(AgCl)+underset((F))(2NH_(3) rarr)Ag(NH_(3))_(2)Cl`
`4AgCl+2Na_(2)CO_(3) rarr 4Ag+4 NaCl+2 CO_(2)+O_(2)`
Discussion
21.

E^(@) for a cell having, Fe to Fe^(2+) + 2e, E^(@) = 0.40 V Zn to Zn^(2+) +2e, E^(@) = 0.76 V

Answer»

0.36 V
`-0. 36 V`
`-1. 16 V`
`1.16 V`

Answer :A
Discussion
22.

d_(z^(2))-orbital has:

Answer»

TWO LOBES ALONG z-axis
Ring along yz-plane
Ring along xy-plane
Ring along X axis

Answer :A
Discussion
23.

Dyeting of fibre involves the process of

Answer»

adsorption
absorption
sorption
all of these

Answer :D
Discussion
24.

Dyes are generally from the substances obtained from

Answer»

PETROLEUM products
Gasoline
Coal-tar
Water gas

Answer :A
Discussion
25.

Dyeing of fibre involves the process of

Answer»

ADSORPTION
absorption
SORPTION
all of these

Solution :It involves sorption. Both PROCESSES of adsorption and absorption take PLACE simultaneously.
Discussion
26.

Dyeing of fibre involves the process of :

Answer»

Adsorption
Absorption
Sorption
All

Answer :D
Discussion
27.

Dye test is used to distinguish

Answer»

ETHANAMINE and METHANAMINE
Methanamine and propanamine -I
Benzenamine and ethanamine
Urea and ACETANILIDE

Answer :C
Discussion
28.

Dye test can be used to distinguish

Answer»

ETHYLAMINE and acetamide
Ethylamine and aniline
Urea and acetamide
Methylamine and ethylamine

SOLUTION :Dye test is given by aromatic PRIMARY amines and not by aliphatic amines.
Aniline gives diazonium chloride with `HNO_(2)(NaNO_(2)//HCl)` at `0^(@)C` which form coloured dye with PHENOLS. This is dye-tesopt. Aliphatic amines form alcohol with evolution of `N_(2)` on treating with `HNO_(2)`.
Discussion
29.

d_(xy) orbital has lobes between x- and y- axes. The wave function of two lobes are positive and those of other two are negative. The positive wave function signifies that :

Answer»

both `X` and `y` are positive
both `x` and `y` are negative
either `x` IR `y` US negative
None of these

Answer :A, B
Discussion
30.

Dye test can be used to distinguish between:

Answer»

ETHYLAMINE and acetamide
Ethylamine and aniline
Urea and acetamide
Methylamine and ethylamine

Answer :B
Discussion
31.

Dyamite is a mixture of :

Answer»

Nitroglycerine `+` Saw DUST.
Nitroglycerine `+` HCl.
Hydrogen bomb `+ H_(2)SO_(4)`
GLYCERINE `+ H_(2)SO_(4)`.

SOLUTION :Dynamite is a mixture of nitroglycerine and saw dust.
Discussion
32.

Dust is an example of-

Answer»

SOLID SOLUTION
liquid solution
GAS solution
none

Solution :Dust is the PARTICLES in AIR
Discussion
33.

Draw the structures of all eight structural isomers that have the molecular formula C_(5)H_(11)Br. Name each isomer according to IUPAC system and classify them as primary, secondary or tertiary bromide.

Answer»

SOLUTION :EIGHT ISOMERS are possible. Their structures and IUPAC NAMES are given below:
Discussion
34.

During World War I ........... gas mask was employed

Answer»

SOLUTION :CHARCOAL
Discussion
35.

Draw the structures of (a) cis-dichloridotetracyanochromate (III) (b) Mer-triamminetrichloridocobalt (III) (c ) Fac-triaquatrinitro-N-cobalt (III)

Answer»

SOLUTION :
Discussion
36.

During winters , moisture condenses in the form of dew and can be seen on plant leaves and grass, The entropy of the system in such cases decreases as liquids process lesser disorder as compared to gases. With reference to the second law, which statement is correct , for the above process?

Answer»

The randomeness of the universe decreases
The randomness of the surrounding decreases
INCREASE is randomness of surrounding equals to the decrease in randomness of SYSTEM
The increase in randomness of the surroundings in greater as COMPARED to the decrease in randomness of the system.

Solution :As dew formation is spontaneous PROCESS, therefore, entrophy or randomness of the universe will increase. As randomeness of the system has decreased but randomness of the surrounding will increase larger so that change is positive.
Discussion
37.

Draw the structures, give names according to IUPAC and indicate primary, secondary and tertiary amines : (a) eight isomeric amines of formula C_(4)H_(11)N. (b) five isomeric amines of formula C_(7)H_(9)N that contain a benzene ring.

Answer»

SOLUTION :(a) Eight isomeric amines of formula `C_(4)H_(11)N`
(i) `underset("BUTANAMINE"(1^(@)))(CH_(3)CH_(2)CH_(2)CH_(2)NH_(2))` (ii) `underset("Butan-2-amine(1^(@)))(CH_(3)CH_(2)underset(NH_(2))underset(|)(CHCH_(3)))` (iii) `underset("2-Methylpropanamine"(1^(@)))(CH_(3)underset(CH_(3))underset(|)(CHCH_(2))NH_(2))`
(iv) `underset("2-Methylpropan-2-amine"(1^(@)))(CH_(3)-underset(CH_(3))underset(|)overset(CH_(3))overset(|)(C)-NH_(2))` (v) `underset("N-Methylpropanamine"(2^(@)))(CH_(3)CH_(2)CHNHCH_(2))` (vi) `underset("N-Methylpropan-2-amine"(2^(@)))(CH_(3)overset(CH_(3))overset(|)CHNHCH_(3))`
(vii) `underset("N-Ethylethanamine"(2^(@)))(CH_(3)CH_(2)NHC_(2)H_(5))` (viii) `underset("N,N-Dimethylethanamine"(3^(@)))(CH_(3)CH_(2)N(CH_(3))_(2))`
(b) Five isomeric amines of formula `C_(7)H_(9)N` containing a benzene ring

Discussion
38.

During war, arsenic containing poisonous gas called Lewisite is sometimes used. The soldiers may get affected due to poisoning by arsenic or it may affect the people living in the nearby areas. Similarly, we use a number of articles containing lead. If somehow it enters into our body, e.g., along with water if lead pipes are used for transport of water, there may be lead poisoning. The person inflicted with arsenic or lead poisoning has to be given an antidote. Now, answer the following questions : (i) What values are expressed in the above paragraph? (ii) Whihc antidote is given for arsenic poisoning and which one for lead poisoning?

Answer»

SOLUTION :(i) (a) Lead pipes should not be used for transport of drinking water.
(b) People LIVING in the areas close to war field should take the antidote if inflicted.
(ii) BRITISH Anti-Lewisite (BAL) for ARSENIC poisoning and EDTA for lead poisoning.
Discussion
39.

Draw the structures and write the lUPAC name of the following compounds. () Benzyl alcohol (i) Allyl alcohol (i) Cyclohexyl alcohol

Answer»

SOLUTION :
Discussion
40.

During which of the following extraction of metal poling process is used :-

Answer»

Zn
Cu
Mg
Both (2) and (3)

ANSWER :D
Discussion
41.

During war , accidents or street quarrels , sometime people get deep injuries which require stitching, of wounds. Earlier these wounds used to be stitched by nylon thread which was non - biodegradable. This thread used to be pulled out after healing of wounds. This process caused pain to the patients. But these days biodegradable polymer is used for stitching of wounds which gets degraded by itself , within a weak or so. Now answer the following questions : Write the name of the biodegradable polymer used for stitching of wounds after operation. (ii) What are the monomer units of this polymer ? (iii) What are the degradation products of this polymer and what happens to them in the body ?

Answer»

Solution :(i) Name of the biodegradable polymer in nylon -2-nylon -6.
(ii) The monomer units of this polymer are : `underset("GLYCINE")(H_(2)N-CH_(2)-COOH)`
`underset("6 -Aminohexanoic acid")(H_(2)N-(CH_2)_(5)-COOH`
(iii) The degradation products of this polymer are the monomer molecules themselves. These are either oxidation in the BLOOD to `CO_2` and `H_2O` or glycine MAY be used in the synthesis of PROTEINS needed by the body.
Discussion
42.

During thhe preparation of H_(2)S_(2)O_(8) (per disulphuric acid) O_(2) gas also releases at anode as byproduct. When 9.72 of H_(2) releases at cathode and 2.35L O_(2) at anode at STP, the weight of H_(2)S_(2)O_(8) produced in gram is

Answer»

87.12
43.56
83.42
51.74

Solution :`(9.72)/(22.4)xx2=(2.35)/(22.4)xx4+(W)/(194)xx2` or `W=43.47g`
Discussion
43.

Draw the structure of xenon oxyfluoride which is isoelectronic with IF_5.

Answer»

SOLUTION :Xe ATOM in `XeOF_4` is `sp^3 d^2`hybridized and has square PYRAMIDAL STRUCTURE shown below:
Discussion
44.

During the working of the Daniel cell , which of the following happen ?

Answer»

the size of the Zn rod as well as the INTENSITY of the COLOUR of `CuSO_(4)` solution remain unchanged
the size of the Zn rod is REDUCED andthe blue colour of `CuSO_(4)` solution become faint
the size of the Zn rod remains same but the blue colour of `CuSO_(4)` solution become faint
the size of the Zn rod is reduced but there is no change in the intensity of colour of the `CuSO_(4)` solution

Solution :Zn dissolves to give `Zn^(2+)` ION at anode and `Cu^(2+)` deposited at cathode.
Discussion
45.

During the vulcanization of rubber, sulphur cross linking occurs at a) Double bonds b) Allylic - CH_(2) - groups c)Methyl groups The correct answer is

Answer»

only a
only B
only a and b
all

ANSWER :C
Discussion
46.

Draw the structure of xenon fluoride molecule which is isoelectronic with IF_5.

Answer»

SOLUTION :` XeOF_4` is ISOELECTRONIC with `IF_5`. It has `sp^3d^(2)` hybridisation and exists in SQUARE PYRAMIDAL structure.
Discussion
47.

During the transformation of .^(b)X_(a) rarr .^(d)Y_(c) the number of beta-particles emitted is

Answer»

`((b -d))/(4)`
`(c -a) + (1)/(2) (b -d)`
`(a -c) - (1)/(2) ( b-d)`
`(b -d) + 2 (c -a)`

Solution :`._(a)^(b) X rarr ._(c)^(d)Y + x_(1)^(4) alpha + y_(-1)^(0)beta`
`b = d + 4x + 0y " " a = c + 2X - y`
or, `b =d+ 4x " " or, a = c + 2.(b-d)/(4) -y`
or, `x = (b-d)/(4) "" or, a = c + (b -d)/(2) - y`
or, `y = (c -a) + ((b -d))/(2)`
Discussion
48.

Draw the structure of XeF_4 molecule.

Answer»

SOLUTION :
Discussion
49.

During the transformation of ""_(a)^(c)X to ""_(d)^(b)Y the number of beta-particles emitted are :

Answer»

`(a-b)/(4)`
`d+(a-b)/(2)+C`
`d+((a-b)/(2))-c`
`2c-d+a-b`

ANSWER :C
Discussion
50.

During the titration of sodium thiosulphate and iodien, the following reaction takesplace : 2Na_2 S_2O_3 + I_2 to Na_2 S_4 O_6 + 2NaI Which of the following statement is correct ?

Answer»

SODIUM thiosuphate gets oxidised because OXIDATION number of S in it INCREASES from+ 2 to + 4.
Oxidation number of S increases by 0.5 while that iodine decreases by -1.
Iodine behaves as reducing agent because its oxidation number inreases.
The product `Na_2S_4O_6` is sodium hexathionate and oxidation number of S in it is 2.5.

Answer :B
Discussion