Chemistry
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Easy
Question 1:
Define oxidation and reduction in terms of oxidation number.
Answer 1:
Increase in Oxidation Number is Oxidation and decrease in Oxidation Number is
called reduction.
Question 2:
What is meant by disproportionation? Give one example.
Answer 2:
In a disproportionation reaction an element simultaneously oxidized and reduced.
P4 + 3OH– + 3H2O → PH3 + 3H2PO2–
Question 3:
What is Oxidation Number of sulphur in H2SO4?
Answer 3:
+6.
Question 4:
Identify the central atom in the following and predict their O.S. HNO3.
Answer 4:
Central atom N and O.S. +5.
Question 5:
Out of Zn and Cu which is more reactive?
Answer 5:
Zn.
Question 6:
What is galvanization?
Answer 6:
Coating of a less reactive metal with a more reactive metal e.g. coating of iron
surface with Zn to prevent rusting of iron.
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Question 7:
How is standard cell potential calculated using standard electrode potential?
Answer 7:
0
E cell = E0cathode − E0anode
Question 8:
What is oxidation state of oxygen in H2O2?
Answer 8:
– 1.
Question 9:
The formation of sodium chloride from gaseous sodium and gaseous chloride is
a redox process. Justify.
Answer 9:
Na atom get oxidize and Cl is reduced.
Average
Question 1:
Write the balanced redox reaction:
a) MnO4– (aq) + Fe2+ (aq) → Mn2+ (aq) + Fe3+ (aq) [acidic medium]
b) Cr2O72– + Fe2+ → Cr3+ + Fe3+ [Acidic medium]
Answer 1:
a) MnO4– (aq) + 5Fe2+ (aq) + 8H+ (aq) → Mn2+ (aq) + 5Fe3+ (aq) + 4H2O(l)
b) Cr2O72– + 6Fe2+ + 14H+ → 2Cr3+ + 6Fe3+ + 7H2O.
Question 2:
Identify the strongest & weakest reducing agent from the following metals:
Zn, Cu, Na, Ag, Sn.
Answer 2:
Strongest reducing agent : Na
Weakest reducing agent : Ag
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Question 3:
Determine the oxidation number of all the atoms in the following oxidants:
KMnO4, K2Cr2O7 and KClO4.
Answer 3:
In KMnO4 K = +1,
Mn = +7, O = –2
In K2Cr2O7 K = +1,
Cr = +6, O = –2
In KClO4 K = +1,
Cl = +7, O = –2.
Question 4:
Determine the oxidation number of all the atoms in the following species:
Na2O2 and OF2.
Answer 4:
In Na2O2, Na = +1,
O = –1
In OF2,
F = –1,
O = +2.
Question 5:
Is it possible to store?
(i)
H2SO4 in Al container?
(ii) CuSO4 solution in Zn vessel?
Answer 5:
(i)
Yes
(ii) No
Question 6:
Identify the oxidizing and reducing agents in the following equations:
a) MnO4– (aq) + 5Fe2+ (aq) + 8H+ (aq) → Mn2+ (aq) + 5Fe3+ (aq) + 4H2O (l)
b) Cr2O72– + 6Fe2+ + 14H+ → 2Cr3+ + 6Fe3+ + 7H2O.
Answer 6:
Oxidizing agent = MnO4– and Reducing agent = Fe2+
Oxidizing agent = Cr2O72– and Reducing agent = Fe2+.
Question 7:
Predict all the possible oxidation states of Cl in its compounds.
Answer 7:
0, –1, +1, +3, +5, +7.
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Question 8:
Formulate possible compounds of Cl when its oxidation state is 0, -1, +1, +3, +5
and +7.
Answer 8:
Cl2, HCl, HOCl, HOClO, HOClO2, HOClO3 respectively
Question 9:
List three measures used to prevent rusting of iron.
Answer 9:
 Galvanization (coating iron by a more reactive metal)
 Greasing/oiling
 Painting.
Difficult
Question 1:
Write short notes on:
a) Electrochemical series
b) Redox reactions
c) Oxidizing agents
Answer 1:
a) Electrochemical series: Arrangement of metals (non-metals also) in
increasing order of their reducing power or vice versa.
b) Redox reaction: Reactions in which both Oxidation and reduction take
place simultaneously are REDOX reactions.
c) Oxidizing agents: Chemical specie which can oxidize the other one or can
reduce itself.
Question 2:
Calculate oxidation states of sulphur in the following oxoacids of S:
H2SO4, H2SO3, H2S2O8 and H2S2O7.
Answer 2:
H2SO4
+6
H2SO3
+4
H2S2O8
+6
H2S2O7
+6
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Question 3:
Explain role of salt bridge in Daniel cell.
Answer 3:
 It completes the electric circuit in the cell.
 It maintains the electric neutrality in the cell.
Question 4:
Account for the followings:
a) Sulphur exhibits variable oxidation states.
b) Fluorine exhibits only –1 oxidation state.
c) Oxygen can’t extend its valency from 2.
Answer 4:
a) Due to the presence of vacant d orbitals in sulphur.
b) It is most electronegative element.
c) Small size/unavailability of vacant d orbitals in O.
Question 5:
Complete and balance the following equations:
a) H+ + Cr2O72– + Br– → 2Cr3+ + ? + ?
b) H2O2 + Cl– → OH– + ?
c) Zn + Cu2+ → Zn2+ + ?
Answer 5:
a) 14H+ + Cr2O72– + 6 Br– → 2Cr3+ + 3Br2 + 7H2O
b) H2O2 + 2Cl– → 2OH– + Cl2
c) Zn + Cu2+ → Zn2+ + Cu.
Question 6:
Identify the oxidizing and reducing agents in the following equations:
Fe + H2SO4
→
FeSO4 + H2
H2 + Cl2
→
2HCl
MnO2 + 4HCl
→
MnCl2 + 2H2O + Cl2
Answer 6:
Oxidizing agents = H2SO4 and Reducing agents = Fe
Oxidizing agents = Cl2 and Reducing agents = H2
Oxidizing agents = MnO2 and Reducing agents = HCl.
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Question 7:
Arrange the following in increasing order of their reducing power:
Cu, Ag, Au, Zn, Fe, Al, Na, Mg, Pt(SHE), Hg, Ca, K.
Answer 7:
Au, Hg, Ag, Cu, Pt(SHE), Fe, Zn, Al, Mg, Na, Ca, K.
Question 8:
What is SHE? What is its use?
Answer 8:
Standard Hydrogen Electrode (SHE) has been selected to have zero standard
potential at all temperatures. It consists of a platinum foil coated with platinum
black (finely divided platinum) dipping partially into an aqueous solution in
which the activity (approximate concentration 1M) of hydrogen ion is unity and
hydrogen gas is bubbled through the solution at 1 bar pressure. The potential of
the other half cell is measured by constructing a cell in which reference electrode
is standard hydrogen electrode. The potential of the other half cell is equal to the
potential of the cell.
Question 9:
We spend crore of Rupees and even thousands of lives every year due to
corrosion. How can it be prevented? Explain.
Answer 9:
 By Galvanization: Coating of a less reactive metal with a more reactive
metal e.g. coating of iron surface with Zn to prevent rusting of iron.
 By greasing/oiling (to keep away the object from the contact of air &
moisture.)
 By painting (to keep away the object from the contact of air &moisture.)
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