Oxidation Number Method
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Q. what volume of solution of HCL containing 73 gram per litre would sufficient gor the exact neutralization NaOH obtained by allowing 0.46g of metallic sodium to react with water.
Q. a solution of Naoh is 4gl^{-1} what volume of hcl gas at STP will neutralize 50ml of the alkali solutio
Q. 16. So2cl2 reacts with H2o gives H2so4 and Hcl determine volume of 0.2M Ba(oH)2 required to neutralised 25ml of 0.1 M so2cl2
Q. Which of the following doesnt reduce tollens solution?
Catechol
Phenol
Resorcinol
Quinol
Q. Match the reactions given in List I with the number of electrons lost or gained in List II
Column−IColumn−IIReactionNumber of electrons lost or gained(P)Mn(OH)2+H2O2→MnO2+2H2O(1)8(Q)AlCl3+3K→Al+3KCl(2)2(R)3Fe+4H2O→Fe3O4+4H2(3)3(S)3H2S+2HNO3→3S+2NO+4H2O(4)6
Column−IColumn−IIReactionNumber of electrons lost or gained(P)Mn(OH)2+H2O2→MnO2+2H2O(1)8(Q)AlCl3+3K→Al+3KCl(2)2(R)3Fe+4H2O→Fe3O4+4H2(3)3(S)3H2S+2HNO3→3S+2NO+4H2O(4)6
- PQRS2314
- PQRS3214
- PQRS2134
- PQRS1432
Q. Balance the following chemical equation:Cr2O−27(aq)+SO−23(aq)→Cr+3(aq)+SO−24(aq) (Acidic medium)
- Cr2O−27(aq)+3SO−23(aq)+H+(aq)→2Cr+3(aq)+3SO−24(aq)+4H2O(aq)
- Cr2O−27(aq)+SO−23(aq)+8H+(aq)→2Cr+3(aq)+3SO−24(aq)+4H2O(aq)
- Cr2O−27(aq)+3SO−23(aq)+8H+(aq)→2Cr+3(aq)+3SO−24(aq)+4H2O(aq)
- Cr2O−27(aq)+3SO−23(aq)+8H+(aq)→Cr+3(aq)+3SO−24(aq)+4H2O(aq)
Q.
In which of the following has the oxidation number of oxygen been arranged in increasing order?
BaO2< KO2< O3< OF2
OF2< KO2< BaO2< O3
BaO2< O3< OF2< KO2
KO2< OF2< O3< BaO2
Q. Balance the following redox reaction in basic medium:
ClO−+CrO−2→Cl−+CrO2−4
ClO−+CrO−2→Cl−+CrO2−4
- 2ClO−+3CrO−2+2OH−→3Cl−+CrO2−4+2H2O
- 3ClO−+2CrO−2+2OH−→3Cl−+2CrO2−4+H2O
- ClO−+2CrO−2+3OH−→2Cl−+3CrO2−4+2H2O
- 3ClO−+2CrO−2+OH−→3Cl−+3CrO2−4+2H2O
Q. The number of moles of KMnO4 that will be needed to react completely with one mole of ferrous oxalate in an acidic solution is:
- 35
- 25
- 45
- 15
Q. The reaction of the permanganate ion with the oxalate ion in acidic solution forms manganese(II) ion and carbon dioxide. What is the stoichiometric coefficient in front of manganese(II) in the balanced redox equation?
MnO−4+C2O2−4→Mn2++CO2
MnO−4+C2O2−4→Mn2++CO2
- 2
- 1
- 3
- 5
Q. The given unbalanced redox reaction
S2O2−3+Sb2O5+H+→SbO+H2SO3
is balanced by oxidation number method. The stoichiometric coefficient of Sb2O5 and SbO respectively, will be:
S2O2−3+Sb2O5+H+→SbO+H2SO3
is balanced by oxidation number method. The stoichiometric coefficient of Sb2O5 and SbO respectively, will be:
- 2, 4
- 4, 8
- 3, 6
- 1, 2
Q.
N factor for the reaction (BrO3-) + (Br-)= Br2
Q. What would be the value of stoichiometric coefficients of Mg and VO3−4 respectively, in the balanced net ionic equation for the given redox reaction in acidic medium:
Mg(s)+VO3−4(aq)→Mg2+(aq)+V2+(aq)
Mg(s)+VO3−4(aq)→Mg2+(aq)+V2+(aq)
- 3, 2
- 8, 16
- 1, 1
- 5, 10
Q. What will be the correct balanced redox reaction for the given equation? (Use oxidation number method)
Tl2O3(s)+NH2OH(aq)→TlOH(s)+N2(g)
Tl2O3(s)+NH2OH(aq)→TlOH(s)+N2(g)
- Tl2O3(s)+2NH2OH(aq)→2TlOH(s)+N2(g)+5H2O
- 2Tl2O3(s)+4NH2OH(aq)→4TlOH(s)+2N2(g)+5H2O
- Tl2O3(s)+4NH2OH(aq)→2TlOH(s)+2N2(g)+5H2O
- Tl2O3(s)+2NH2OH(aq)→2TlOH(s)+N2(g)+5H2O
Q. In the following reaction :
H2O2(aq)+Cl2O7(aq)→ClO−2(aq)+O2(g)
calculate the moles of OH− and H2O respectively present in a balanced equation in a basic medium :
H2O2(aq)+Cl2O7(aq)→ClO−2(aq)+O2(g)
calculate the moles of OH− and H2O respectively present in a balanced equation in a basic medium :
- 3, 6
- 4, 6
- 2, 5
- 4, 4
Q. Consider the following redox reaction:
Mn(OH)2(s)+MnO−4(aq)→MnO2(s)
If x, y, z are the stoichiometric coefficients of Mn(OH)2 , MnO−4 and MnO2 respectively when the above reaction is balanced in basic medium.
What will be the value of x+y+z ?
Mn(OH)2(s)+MnO−4(aq)→MnO2(s)
If x, y, z are the stoichiometric coefficients of Mn(OH)2 , MnO−4 and MnO2 respectively when the above reaction is balanced in basic medium.
What will be the value of x+y+z ?
- 5
- 8
- 10
- 12
Q. The oxidation state of Fe in Fe3O8 is :
- 45
- 54
- 32
- 163
Q. How do you form an equation for such problems?
Q.54. When potassium permanganate is titrated against ferrous ammonium sulphate, the equivalent weight of ferrous ammonium sulphate is:
(a) M
(b) M / 2
(c) M / 5
(d) M / 3
Q.54. When potassium permanganate is titrated against ferrous ammonium sulphate, the equivalent weight of ferrous ammonium sulphate is:
(a) M
(b) M / 2
(c) M / 5
(d) M / 3
Q. Consider a redox reaction :
x Na2HAsO3+y NaBrO3+z HCl→NaBr+H3AsO4+NaCl
where x, y, z are the stoichiometric coefficients of Na2HAsO3, NaBrO3, HCl respectively.
The value of x+y+z in the above balanced redox reaction is :
x Na2HAsO3+y NaBrO3+z HCl→NaBr+H3AsO4+NaCl
where x, y, z are the stoichiometric coefficients of Na2HAsO3, NaBrO3, HCl respectively.
The value of x+y+z in the above balanced redox reaction is :
- 3
- 9
- 10
- 12
Q.
How many H2O should be added to the RHS of the following equation to balance number of oxygen atoms on the left ?
6Fe2++Cr2O2−7+14H+⟶6Fe3++2Cr3++ ___________
2
4
5
7
Q. The number of moles of KMnO4 that will be needed to react completely with 1 mole of ferrous oxalate in acidic medium is
Q. Find the equivalent weight of Cr2O2−7 in the given reaction, if the molar mass of Cr2O2−7 is 'M' ?
Cr2O2−7H+→Cr3+
Cr2O2−7H+→Cr3+
- M3
- M
- M6
- 2M
Q. Which of the following equations is a balanced one?
- 2BiO−3+4H++Mn2+→2Bi3++2H2O+MnO−4
- 5BiO−3+14H++2Mn2+→5Bi3++7H2O+2MnO−4
- 5BiO−3+22H++Mn2+→5Bi3++7H2O+MnO−4
- 6BiO−3+12H++3Mn2+→6Bi3++6H2O+3MnO−4