MnO42- undergoes disproportionation reaction in acidic medium but MnO4- does not?
Due to the highest oxidation state of Mn in MnO42-
Due to the highest oxidation state of Mn in MnO4-
Due to the endothermic nature of disproportionation reaction
Due to the exothermic nature of disproportionation reaction
Nitric acid reacts with PbO but does not react with PbO2 , because -
PbO is a base while PbO2 is a strong oxidizing reagent
PbO is a base while PbO2 is a weak oxidizing reagent
PbO is neutral while PbO2 is a strong oxidizing reagent
PbO is acid while PbO2 is a strong oxidizing reagent
The oxidation state of P in HPO32- is-
+3
+4
+2
+5
the oxidation state of P in PO43-?
The oxidation state of two S-atoms in Na2S2O3 is -
+2 and +4
+3 and -2
+4 and -2
+6 and -2
(a) EK+/K o= -2.93 V ; EAg+/Ago= 0.80 V
(b) EHg+2/Hgo= 0.79V; EMg+2/Mgo= -2.37V
(c) ECr+3/Cro=-0.74 V
Based on standard electrode potentials given above, what is the correct arrangement for increasing order of reducing power of elements?
Ag < Hg < Cr < Mg < K
Ag > Cr > Mg > Hg > K
K > Mg < Cr < Hg > Ag
K< Mg < Cr < Hg < Ag
The compound AgF2 (unstable) acts as a/ an:
Oxidising agent.
Reducing agent.
Both oxidising and reducing agent.
Neither oxidising and reducing agent.
Among the following hydrohalic compounds, the best reductant is -
1. HCl2. HBr3. HI4. HF
The incorrect oxidation number of the underlined atom in the following species
Identify the incorrect option.
Cu2O is -1
ClO3- is +5
K2Cr2O7 is +6
HAuCl4 is +3
Identify the reaction from having the top position in the EMF series (standard reduction potential) according to their electrode potential at 298 K.
Mg2++ 2e–→Mg(s)
Standard reduction potentials of the half reactions are given below :
F2(g) + 2e-→2F-(aq) ; E° =+2.85 V
Cl2(g) + 2e-→2Cl-(aq) ; E° =+1.36 V
Br2(g) + 2e-→2Br-(aq) ; E° =+1.06 V
I2(g) + e-→2I-(aq) ; E° =+0.53 V
The strongest oxidizing and reducing agents, respectively, are:
Br2 and Cl-
Cl2 and Br-
Cl2 and I2
F2 and l-
A solution contains Fe2+, Fe3+ and I– ions. This solution was treated with iodine at 35°C. E° for Fe3+/Fe2+ is +0.77 V and E° for I2/2I– = 0.536 V.The favourable redox reaction is-
Fe2+ will be oxidized to Fe3+
I2 will be the reduced to I–
There will be no redox reaction
I– will be oxidized to I2
The compound that contains zero oxidation state of Fe :
[Fe (CN )6]-4
[Fe (CN )6]-3
Fe (CO )5
All of the above .
a Zn + b NO3- + cH+ → d NH4+ + e H2O +f Zn+2 ; a , b, c, d, e and f respectively are :
Which is the best description of the behavior of bromine in the reaction given below:-
H2O+Br2→HOBr+HBr
Both oxidized and reduced
Oxidized only
Reduced only
Proton acceptor only
Oxidation numbers of A, B and C are + 2, +5 and –2 respectively possible formula of compound is :
A2(BC2)2
A3(BC4)2
A2(BC3)2
A3(B2C)2
A non feasible reaction among the following is : -
2 KI+Br2→2KBr+I2
2 KBr+I2→2KI+Br2
2 KBr+Cl2→2KCl+Br2
2H2O+2F2→4HF+O2
Standard electrode potentials are
Fe+2/Fe E° = -0.44
Fe+3/Fe+2 E° = 0.77
If Fe+2 , Fe+3 and Fe block are kept together, then :-
Fe+3 increases
Fe+3 decreases
Fe+2Fe+3 remains unchanged
Fe+2 decreases
The oxidation states(O.S.) of sulphur in the anions SO32-, S2O42- and S2O62- follow the order -
S2O42-<SO32-<S2O62-
SO32-<S2O42-<S2O62-
S2O42-<S2O62-<SO32-
S2O62-<S2O42-<SO32-
Fluorine reacts with ice as per the following reaction
H2O(s) + F2(g) → HF(g) + HOF(g)
This reaction is a redox reaction because-
F2 is getting oxidized.
F2 is getting reduced.
Both (1) and (2)
None of the above.
The oxidation number of sulphur and nitrogen in H2SO5 and NO3- are respectively-
-6,-6
+8, +6
+6, +5
-8, -6
What will be the formulas for the following compounds
(a) Mercury(II) chloride and (b) Thallium(I) sulphate
HgCl2, Tl2SO4
Hg2Cl2, Tl2SO4
HgCl2, TlSO4
HgCl2, Tl3SO4
Consider the following reactions:
(a) 6 CO2(g) + 6H2O(l) → C6H12O6(aq) + 6O2(g)
(b) O3(g) + H2O2(l) → H2O(l) + 2O2(g)
It is more appropriate to write these reactions as following because:
(a) 6CO2(g) + 12H2O(l) → C6H12O6(aq) + 6H2O(l) + 6O2(g)
(b) O3(g) + H2O2(l) → H2O(l) + O2(g) + O2(g)
Water molecules are also the product of photosynthesis
Water molecules are not a product of photosynthesis.
Oxygen molecules are also the product of photosynthesis.
Oxygen molecules are not a product of photosynthesis.
In the manufacture of benzoic acid from toluene, we use alcoholic potassium permanganate as an oxidant because -
The cost of adding an acid or a base can be reduced.
The reactions can proceed at a faster rate.
Both '1' and '2'
Neither '1' nor '2'
Consider the following reactions :
2 S2O32–(aq) + I2(s) → S4O62–(aq) + 2I–(aq)
S2O32–(aq) + 2Br2(l) + 5H2O(l) → 2SO42–(aq) + 4Br–(aq) + 10H+(aq)
The same reductant, thiosulphate react differently with iodine and bromine because-
Br2 is a stronger oxidizing agent than I2
I2 is a stronger oxidizing agent than Br2
The correct statement(s) about the given reaction is -
XeO6aq4-+2F-1aq+6H+aq→XeO3(g)+F2(g)+3H2O(l)
XeO64- oxidises F-
The oxidation number of F increases from -1 to zero
XeO64- is a stronger oxidizing agent that F-
All of the above.
(a) H3PO2(aq) + 4AgNO3(aq) + 2H2O(l) → H3PO4(aq) + 4Ag(s) + 4HNO3(aq)
(b) H3PO2(aq) + 2CuSO4(aq) + 2H2O(l) → H3PO4(aq) + 2Cu(s) + H2SO4(aq)
c C6H5CHO(l) + 2AgNH32+(aq) + 3OH-(aq)→C6H5COO-(aq) + Ag(s) + 4NH3(aq) +H2O(l)(d) C6H5CHO(l) + 2Cu2+(aq) + 5OH-(aq)→No change observed
Draw the inference about the behaviour of Ag+ and Cu2+ ions from these reactions
Ag+ is a stronger oxidizing agent than Cu2+
Cu2+ is a stronger oxidizing agent than Ag+
Ag+ and Cu2+ are reducing agents
Ag+ is a reducing agent and Cu2+ is an oxidising agent
The oxidising agent and reducing agent in the given reaction are
5P4s+12H2Ol+12HO-aq→8PH3g+12HPO2-aq
Oxidising agent = P4; Reducing agent = P4
Oxidising agent = P4; Reducing agent = H2O
Oxidising agent = H2O; Reducing agent = P4
None of the above
The correct statement about the given reaction is-
(CN)2(g) + 2OH-(aq) →CN-(aq) + CNO-(aq) + H2O(l)
The reaction is an example of a disproportionation reaction
Hydrogen atom gets oxidized
Reaction occurs in acidic medium
The Mn3+ ion is unstable in solution and undergoes disproportionation reaction to give Mn2+, MnO2 and H+ ion. The balanced ionic equation for the reaction is-
2Mn3+(aq) + 2H2O(l) → MnO2(s) + Mn2+(aq) + 4H+(aq)
Mn3+(aq) + H2O(l) → MnO2(s) + 2Mn2+(aq) + 4H+(aq)
5Mn3+(aq) + 2H2O(l) → MnO2(s) + 3Mn2+(aq) + 4H+(aq)
2Mn3+(aq) + 2H2O(l) → 2MnO2(s) + 2Mn2+(aq) + 4H+(aq)
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