Very Short Answer Questions
1.
battery, and thus how much energy can still be delivered?
[HOTS]
Which instrument is used by automechanic to determine
how much sulphuric acid is present, in the lead storage
Short Answer Questions
1.
Manu went on to buy a new watch for himself along with his
father. He chose a digital watch and enquired about the cells
with the salesman. The salesman told him that these were
mercury cells, commonly called button cells and asked him
not to throw away the cells after use, but to give them for
reprocessing.
4.
5.
Now, answer the following question:
(a) What type of a cell is mercury cell?
(b) Write the reaction involved.
2.
3.
(c) Why the used cell should not be thrown away? What
values were shown by the salesman? [Value Based]
After the end of summers Sameer was painting his desert
cooler before keeping it in godown. Sameer suggested his
neighbour Shyam to do the same to prevent his cooler
from rusting. Shyam argues that it is waste of time & the
body of the cooler is quite strong.
After reading the above passage, answer the following
questions.
(a) Whose opinion is correct according to you?
(b) Are there any other ways to prevent corrosion other
than painting?
(c) Mention the values shown by the Sameer.
[Value Based]
Two iron plates, with dents were allotted to two students
for plating and study of prevention of rusting. The first
student selected tin for coating his plate, while the second
choose zinc for the same purpose.
(a) Who do you think made the right choice, and why?
(b) Mention values associated with student making
correct choice.
[Value Based]
6.
Consider the following diagram in which an electrochemical
cell is coupled to an electrolytic cell. What will be the
polarity of electrodes ‘A’ and ‘B’ in the electrolytic cell?
[Exemplar]
Solutions of two electrolytes ‘A’ and ‘B’ are diluted. The
L m of ‘B’ increases 1.5 times while that of A increases 25
times. Which of the two is a strong electrolyte? Justify
your answer.
[Exemplar]
When acidulated water (dil.H2SO4 solution) is electrolysed,
will the pH of the solution be affected? Justify your
answer.
[Exemplar]
Why on dilution the L m of CH3COOH increases
drastically, while that of CH3COONa increases gradually?
[Exemplar]
8.
A solution containing one mole per litre of each Cu(NO3)2;
AgNO3; Hg2(NO3)2; is being electrolysed by using inert
electrodes. The values of standard electrode potentials in
volts (reduction potentials) are :
Ag/Ag+ = +0.80, 2Hg/Hg2++ = +0.79
Cu/Cu++ = +0.34, Mg/Mg++ = –2.37
With increasing voltage, what will be the sequence of
deposition of metals on the cathode.
[HOTS]
9.
The 'Tins' used for containing food stuffs are actually tin
coated steel, known as 'tin plate'. Explain why tin coatings
are preferred to zinc in the environment inside the tin of a
canned fruit.
[HOTS]
10. A mercury cell is widely used where high reliability and long
life are required. Some of the devices in which such a cell is
used are digital watches, pace makers etc.
Answer the following questions for such a cell.
(a) Name the electrolyte used in this cell.
(b) What are anode and cathode made of?
(c) Write down the over all reaction of the cell. [HOTS]
7.
2
Long
LongAnswer
AnswerQuestions
Questions
1.
2.
3.
A group of students were asked to prepare a presentation on
the disadvantage of thermal power plants and its alternative
which would not cause much pollution. For their presentation
they searched a lot and visited to the site of thermal power
plant. They asked the people who were living around the
area of thermal power plant about their problems related to
pollution. The students gave the presentation, which was
followed by question round as follows:
[Value Based]
(a) Why are thermal power plants harmful for the
environment?
(b) What is the alternative of these plants? What fuel will
be used in these?
(c) What are the advantages of this alternative cell over
thermal power plants? How are they benefinal for the
environment?
(d) List the values shown by the students.
Akshay wanted to fix a photoframe on the wall. He searched
for a box of iron nails. On locating the box, he found that most
of the nails were rusted. He remembered that his teacher had
tought him about rusting of iron and its disadvantages.
Keeping that in mind he went to market and bought some
new iron nails.
[Value Based]
(a) What is rust chemically? Is it an electrochemical process?
(b) Give some other examples of corrosion.
(c) How can we prevent rusting of iron to reduce national
wastage of money?
(d) What are the values shown by Akshay?
Rahul carried out an experiment in the laboratory with the set
up given below.
[Value Based]
4.
Answer the following questions.
(a) How long did the current flow? What mass of Cu was
deposited?
(b) Which law of electrolysis is applicable in the above
experiment?
(c) How is electroysis important for industries?
(d) What are the values shown by Rahul?
Consider the given figure and answer the following questions.
(i)
5.
Cell 'A' has ECell = 2V and Cell 'B' has ECell = 1.1 V
which of the two cells 'A' or 'B' will act as an
electrolytic cell. Which electrode reactions will
occur in this cell?
(ii) If cell 'A' has ECell = 0.5V and cell 'B' has ECell = 1.1
V then what will be the reactions at anode and
cathode?
[Exemplar]
An excess of liquid mercury is added to an acidified solution
of 1.0 × 10–3 M Fe3+. It is found that 5% of Fe3+ remains at
o
equilibrium at 25°C. Calculate E Hg 2 +
reaction that occurs is
2Hg(l) + 2Fe3+ ¾¾
® Hg22+ + 2Fe2+
Hg
, assuming the only
æ
ö
o
çè Given: E Fe3+ Fe2 + = 0.77V÷ø
6.
[HOTS]
Calculate the solubility product of AgBr in water at 25°C
from the cell :
Ag, Ag + Br - ( std. soln ) | AgBr ( s ) , Ag.
The standard potentials are
E oAgBr, Ag = 0.071 V; E o
Ag + , Ag
He first cleaned the electrodes and connected two cells A
and B containing AgNO3 and CuSO4 in series. He again
checked the connection and passed a steady current of 1.50A
through them until 1.45g of Ag got deposited on the cathode
of cell A.
[HOTS]
= 0.799 V
3
SOLUTIONS
PART C : HOTS/ Exemplar/ Value Based Questions
Very Short Answer Questions
1.
The automechanic measures the density of the battery fluid
(i.e., the sulphuric acid solution) with a hydrometer to
determine how much sulphuric acid is still present in it.
Short Answer Questions
1.
(a) Mercury cell is an example of primary cell, i.e. the redox
reaction occurs only once. After some time, the cell
becomes dead and cannot be used again
(b) At anode :
Zn ( s ) + 2OH- ® ZnO ( s ) + H 2 O ( l ) + 2e At Cathode :
HgO ( s ) + H2O ( l ) + 2e- ® Hg ( l ) + 2OH -
2.
3.
4.
5.
Overall reaction : Zn ( s ) + HgO ( s ) ® ZnO (s ) + Hg ( l )
(c) The used cells should be reprocessed for mercury
recovery or treated to prevent mercury and mercury
compounds from entering atmosphere and causing
pollution. Concern about the environment and society,
awareness etc. are the values shown by salesman.
(a) Sameer's opinion is correct because painting will
prevent his coolar to get corroded.
(b) Galvanizing, anodizing etc.
(c) Carful and sensitive towards his belonging, feels free
to give suggestion and opinion, sensitive towards
others, presence of mind etc. are some values shown
by sameer.
(a) Student selected Zn for coating his plate made
correct choice. As Zn is more electropositive than Sn.
(b) Decision making, knowledgable, intelligence, presence
of mind etc. are some values displayed by student
made correct choice.
‘A’ will have negative polarity
‘B’ will have positive polarity
Electrolyte ‘B’ is strong as on dilution the number of ions
remains the same, only interionic attraction decreases
® CH3COO– + H3O+
CH3COOH + H2O ¾¾
In the case of strong electrolyte the number of ions
remains the same but the interionic attraction decreases.
8.
The reduction potentials (as given) of the ions are in the
order : Ag+ > Hg22+ > Cu2+ > Mg2+
Mg2+ (aq.) will not be reduced as its reduction potential is
much lower than water (–0.83 V).
Hence the sequence of deposition of the metals will be Ag,
Hg, Cu.
9.
The aqueous environment inside the tins might be quite
acidic (in fruits), Tin is much less reactive towards acids
than zinc and due to this tin coating is preferred over zinc
coating inside the tin of a canned fruit.
10. (a) In such a cell KOH is used as an electrolyte.
[Note: It is an alkaline cell and sealed to prevent the
loss of KOH electrolyte]
(b) Anode is amalgamated zinc mixed with KOH.
Cathode is mercury (II) oxide mixed with graphite
[Note: Both are compressed powders]
(c) The overall reaction is
Zn(s) + HgO(s) + H2O ¾¾
® Zn(OH)2 + Hg(l)
Long Answer Questions
1.
therefore increase in L m is small.
6.
pH of the solution will not be affected as [H+] remains
constant.
® O2 + 4H+ + 4e–
At anode : 2H2O ¾¾
4H+
7.
4e–
® 2H2
¾¾
At cathode
+
In the case of CH3COOH, which is a weak electrolyte, the
number of ions increase on dilution due to an increase in
degree of dissociation.
2.
(a) Thermal power plants use the chemical energy (i.e heat
of combustion) of fossil fuels like coal gas or oil to
generate electricity, thus causing pollution.
(b) An alternative is fuel cells in which energy produced
from the combustion of fuels like H2, CO, CH4, etc. is
directly converted into electrical energy. An example is
H2-O2 fuel cell.
(c) Advantages of fuel cells:
1. Due to continuous supply, such cells never become
dead.
2. They are more efficient (60-70%) than thermal power
plants (40%).
3. They do not cause any pollution problem unlike
thermal plants which burn fossil fuels. Hence, fuel
cells do not create any pollution threat for the
environment, they are eco-friendly.
(d) Dedication towards their work, co-operation, hard work
etc. are some values shown by the students.
(a) Chemically, rust is hydrated ferric oxide, Fe2O3. xH2O.
Yes, it is an electrochemical process.
(b) Some other examples include tarnishing of silver,
development of green coating on copper and bronze
etc.
4
(c) Ways to prevent corrosion are (i) painting, oiling,
greasing i.e. barrier protection, and (ii) sacrificial
protection like galvanization.
(d) Knowledgable, awareness and good memorising ability
etc. are some values shown by Akshay.
3.
Equili
o
E cell = Ecell
Charge (Q) = 1F = 96500C for 108g of Ag
Fe3 + Fe 2 +
Initial
(t = 0)
–
0
M
2
- Eo
Hg 22 + Hg
= 0.77 - E o
Hg 22 + Hg
At equilibrium Ecell = 0
\ t = Q / I = 1295.6/1.5 = 863.7s
(b) Faradays 2nd law of electrolysis
(c) (i) Production of hydrogen by electrolysis of water.
(ii) Electroplating and electrolytic refining of metals.
(iii) Electrolytic extraction of metals like Na, K, Mg,
Cu, Al, etc.
(d) Following the correct procedure, perfection and
dedication etc. are some values shown by Rahul.
(i) Cell 'B' will act as electrolytic cell as it has lower emf
The electrode reactions will be :
Zn2+ + 2e– ® Zn at cathode
Cu ® Cu2+ + 2e– at anode
(ii) Now cell 'B' acts as galvanic cell as it has higher emf
and will push electrons into cell 'A'.
The electrode reaction will be :
At anode : Zn ® Zn2+ + 2e–
At cathode : Cu2+ + 2e– ® Cu
The reaction that takes place between Fe3+ and Hg is 2Fe3
+ Hg ¾¾
® 2Fe2+ + Hg22+
For this reaction the initial concentration and the equilibrium
concentrations can be represented as follows
2Fe3 + 2Hg ¾¾
+ Hg22+
® 2Fe2+
1×10–3M
-3
[Fe 2+ ][Hg 22+ ]
0.0591
log10
n
[Fe3+ ]2
= Eo
96500
´ 1.45 = 1295.6 C
Q=
108
5.
0.95 ´ 10
o
o
o
E cell
= E cathode
- E anode
\ For 1.45g Ag,
4.
0.95×10–3M
Making use of Nernst's equation, we have
(a) Ag+ + e- ® Ag
\
0.05×10–3M
0
æ
\ 0 = ç 0.77 - E 0 2 +
è
Hg 2
ö 0.0591
log10
÷2
Hg ø
æ 0.95 ´ 10-3 ö
(0.95 ´ 10-3 )2 ´ ç
÷
2
è
ø
(0.05 ´ 10-3 )2
Solving the above equation, we get E o
Hg 22 + Hg
6.
= 0.792 V
ˆˆ† Ag ( s) + Br - ( aq) ; Eo = 0.071 V
R.H.E. : AgBr ( s) + e- ‡ˆˆ
R
ˆˆ† Ag + ( aq ) + e - ; EoL = 0.799 V
L.H.E. : Ag(s) ‡ˆˆ
Hence, for the overall reaction
ˆˆ† Ag + ( aq ) + Br - ( aq ) ,
AgBr ( s) ‡ˆˆ
o
- E oL = - 0.728 V
Eo = ER
=
\
{
0.0591
log é Ag + ù éBr - ù
ë
ûë
û
1
log K sp =
}
eqbm
0.728
Eo
=
0.0591 0.0591
where Ksp = 4.81 ´ 10-11
= 0.0591 log K sp
Q1. Complete the following :
(A) Hydrocarbons and halogen derivatives.
(i)CH3CH2CH = CH2 + HBr
(ii) CH3CH2CH = CH2 + HBr
(AI 2009, Foreign 2011)
Peroxide
(Delhi 2008)
CCl4
(Delhi 2008, AI 2008)
(iii) CH2 = CH2 + Br2
(iv) C6H5N2Cl + KI
H
H
C=C
+ HBr
H
(v)
(Foreign 2011)
Heat
(vi) (CH3)3CBr + H2O
Heat
(vii) CH3CH2Cl + SbF3
(viii) CH3CH2C ≡ CNa + CH2 = CHCH2Br
CH3
CH
CH3
(ix)
+ Br2
Liq. NH3
heat
OH
red P
I2
(x)
(xi) CH3CH2Cl + KNO2
Br C2H5ONa
(xii)
CH C2H5OH
Two products
3
(xiii)
+ Br2
heat light
(Delhi 2012C)
1
2
OH
(xiv)
+ HCl
CH2OH
CH3
(xv)
(Delhi 2012C, AI 2009, Foreign 2011)
+ HI
(xvi) CH3CH = CHCH2Cl + CN–
(A) + (B)
(2012C)
(B) Alcohols, phenols and ethers.
(i)
OH + SOCl2
(ii)
CH2OH
(Delhi 2009)
(Delhi 2009)
+ HCl
HO
(iii)
OC2H5 + HBr
(Foreign 2012)
(iv) (CH3)3C – O – C2H5 + HI
(v) CH3CH2CH2OCH3 + HBr
(vi)
(Foreign 2012)
(Foreign 2012)
+ HI
(C) Aldehydes, ketones and carboxylic acids.
2+
(i)CH3C ≡ CH
(ii) .....
(iii)
Hg
H2SO4
O3
H 2O
(Delhi 2012)
O
2
(Delhi 2008, AI 2011)
CHO (Delhi 2008, AI 2011)
CH2
CH3
(i) CrO2Cl2
(iv)
+
(ii) H3O
NO2
(Delhi 2012)
3
COCH3
anhy. AlCl3
+ .....
(v)
(Delhi 2009)
O
C
(vi)
+ ..... .....
(vii)
+ C2H5COCl
(Delhi 2009)
anhy. AlCl3
CS2
(Delhi 2011C, 2012, AI 2012C)
O
CH3
+ CH3CH2NH2
(viii)
H2NCONHNH2
(ix) C6H5CHO
(x)
C
Conc. KOH
H
(xi) C6H5CHO + C6H5COCH3
C
Zn Hg
CH3
Conc. HCl
+
(AI 2011C)
(AI 2011)
(Delhi 2013)
O
(xii)
H
OH
–
293 K
(Delhi 2011C)
(AI 2013)
O
(xiii) C6H5–COCl
H2
Pd BaSO4
(Delhi 2009A, 2011C, AI 2013)
O
CH3
(xiv)
H2CrO4
(AI 2011C)
CHO
(xv)
HNO3/H2SO4
CH2CH3
(xvi)
(Delhi 2013)
273-283 K
KMnO4
......
KOH, heat
(Delhi 2008, 2011, Foreign 2011)
4
KMnO4
(xvii)
H2SO4, heat
(AI 2011C)
COOH
SOCl2
(xviii)
(Delhi 2011, AI 2011, Foreign, 2011)
heat
COOH
(xix) CH3COOH
Br2/P
(Delhi 2013)
COOH
Br2/FeBr3
(xx)
(AI 2013)
O
(xxi) R
NH2
(xxii) C6H5CONH2
LiAlH4
H 3O
(AI 2009)
H 2O
+
heat
(AI 2009)
O
(xxiii) CH3
NH2
Br2 + NaOH
(Delhi 2011, Foreign 2011)
O
(xxiv)
O
CH2
+ NaBH4
OCH3
(D) Amines.
(i)C2H5NH2 + C6H5SO2Cl
(ii) C6H5NH2 + CH3COCl
(AI 2009)
(iv) RNH2 + CHCl3 + KOH
(v) C6H5NH2 + Br2 (aq.)
(vi) C6H5NH2 + HCl (aq.)
(vii) C6H5N2Cl + KI
(viii) Diazonium group +
(AI 2009)
(iii) C2H5NH2 + HNO2
(AI 2009)
(Delhi 2010, AI 2013)
(AI 2009, 2012, 2013, Foreign 2012)
(AI 2013)
........
(Delhi 2008)
–I
(Delhi 2008, AI 2008)
(ix) C6H5N2Cl + H3PO2 + H2O
(AI 2009, 2012, Foreign 2012, AI 2013)
(x) C6H5N2Cl + CH3CH2OH
(Delhi 2010)
5
OH
(xi) C6H5N2Cl + C6H5NH2
(xii) C6H5N2+ Cl–
H 2O
room temp.
(Delhi 2010)
(AI 2013)
Q2. Identify the missing compounds (reagents) in the following:
(i)C2H5Cl
NaCN
KCN
(ii) CH3CH2Br
NaNO2
(iii) C6H5NH2
HCl
(iv) C6H5N2+Cl–
(vi) CH3COOH
CH3CHO
B
LiAlH4
B
A
CuCN
NH3
∆
Reduction
Ni/H2
A
A
Sn + HCl
(v) C6H5NO2
(vii)
A
A
A
C6H5NH2
HNO2
(AI 2013)
C
0°C
(AI 2010)
B OH
H2O/H
(AI 2010)
+
B
NH3
heat
NaNO2 + HCl
NaOH + Br2
H2O/H
B
273 K
(Delhi 2013)
C
+
C
(Delhi 2013)
heat
CHCl3
B
C (AI 2013)
alc. KOH
(i) C2H5MgCl
A
Conc. H2SO4
B
HBr
peroxide
(ii) H2O
C
(Delhi 2011C)
(viii)
(AI 2008C)
(ix)
(AI 2009C)
COOH
(x)
+ NH3
COOH
A
Heat
strong
B heating C
(AI 2011C)
6
COOH
Conc. HNO3
(xi)
Conc. H2SO4
A
SOCl2
B
(i) NaBH4
+
C
(ii) H3O
SOCl2
H2, Pd, BaSO4
D
S or quinoline
E
CH3
273-278 K
(xii)
A
+ CrO3 + (CH3CO)2O
+
H3O , heat
B
KMnO4, KOH,
heat
Conc. NaOH
D
COONa
+
H3O
C +
E
(AI 2010C)
7
SOLUTION
(A) Hydrocarbons and halogen derivatives. 1.(i) CH3CH2CH = CH2 + HBr
Mark. addition
CH3CH2CHCH3
|
Br
Peroxide
anti-markow.
addition
CH3CH2CH2CH2Br
(ii) CH3CH2CH = CH2 + HBr
(iii) H2C = CH2 + Br2
(iv) C6H5N2Cl + KI
H
H
C
H
(v)
Br H
CCl4
BrCH2 – CH2Br
C6H5I + N2 + KCl
(vi) (CH3)3COH
Heat
(vii) 3CH3CH2Cl + SbF3
3CH3CH2F + SbCl3
(viii) CH3CH2C ≡ CCH2CH = CH2
(ix)
(
)
OH
(x)
I
red P
I2
O
(xi) CH3CH2Cl + KNO2
N
Nitroethane
CH3
Ethyl nitrite
CH3
Br
(xii)
+ CH3CH2 O N=O
O
C2H5ONa/C2H5OH
( HBr)
1-Methylcyclohexene
Br
(xiii)
3-Bromocyclohexene
8
In presence of heat and light, allylic bromination takes place, while
in dark bromine adds on double bond to form dibromoderivative.
OH
HCl reacts with alcohols (–CH2OH) and not
with phenols.
(xiv)
CH2Cl
CH3
I
HI adds according to Markownikof’s rule.
(xv)
( xvi) The reactant is allylic compound, so it can form two products.
CH3–CH = CH–CH2Cl
CH3CH = CH–CH2 + Cl–
+
+
CH3CH = CH CH2
CN
CH3CH CH = CH2
–
CN
CH3CH = CH CH2CN
–
CH3CH CH = CH2
CN
(B) Alcohols, phenols and ethers.
(i)
(ii)
Cl + SO2 + HCl
CH2OH
CH2Cl
+ HCl
+ H2O
HO
HO
(Phenolic –OH is not easily replaced, hence does not react with
HCl)
(iii)
OC2H5
OH
+ HBr
+ C2H5Br
C6H5–O bond has some double bond character (see below), thus it
is shorter and stronger, and hence difficult to be cleaved.
+
C2H5
C2H5
+
C2H5
9
Moreover,
C2H5+
H
C6H5OC2H5
is more stable than
C6H5+.
+
C6H5O+ C2H5
C6H5O+C2H5
C6H5+ + HOC2H5
(less stable)
+
C6H5OH +
C2H5
(more stable)
Br
C2H5Br
(iv) (CH3)3C – O – C2H5 + HBr
(CH3)3CBr + C2H5OH
(CH3)3C+ (a 3° carbocation) is more stable than the C2H5+, hence
it is easily formed.
(v) CH3CH2CH2OCH3 + HBr
CH3CH2CH2Br + CH3OH
+
CH3CH2CH2 is more stable than CH3+ due to inductive effect
(CH3CH2 →– CH2+ ) as well as hyperconjugation.
(vi)
CH2I
+ HI
+ HO
+
CH2
(benzyl carbocation) is more stable due to resonance than
+
(C) Aldehydes, ketones and carboxylic acids.
2+
(i) CH3C ≡ CH
Hg , H2SO4
(hydration)
CH3
= CH2
CH3
OH
O3
(ii)
O
O
Zn-H2O
In ozonolysis,
(iii) 3
2
CH3
C=C is always cleaved to C=O + O=C .
CH2
B2H6/THF
CH
B
3
H2O2
NaOH
10
3
PCC
CH2OH
3
CHO
Cyclohexane carbaldehyde
CH3
CHO
(i) CrO2Cl2
(iv)
+
(ii) H3O
(Etard oxidation)
NO2
(v)
NO2
+ CH3COCl
COCH3
anhy. AlCl3
(Friedel-Craft reaction)
Acetophenone
O
COCl
(vi)
C
anhy. AlCl3
+
(Friedel-Craft
reaction)
Benzolphenone
COC2H5
(vii)
+ C2H5COCl
anhy. AlCl3, CS2
(Friedel-Craft
reaction)
Propiophenone
CH3
(viii)
O
+ H2NCH2CH3
H
CH3
+
NCH2CH3
Schiff’s base
+
(ix) C6H5CHO + H2NNHCONH2 H C6H5CH = NNHCONH2 + H2O
Benzaldehyde semicarbazone
Note that the –NH2 end of –CONH2 does not react because its
O
H
N C N H2
O
H
+
N C = N H2
nucleophilic character is decreased due to resonance.
(x) 2H
H
Conc. KOH
(Cannizzaro reaction)
O
CH3OH + HCOONa
Methanol
Sod. methanoate
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