Redox Reactions Important Question


6532 Genius308


In this reaction the oxidation state of Chromium is decreasing from +6 to +3 while that of Iron is also decreasing from +3 to +2. This is not a redox reaction and is not written because a substance cannot get reduced until some other sentence gets oxidized.

Q2. Suggest a scheme of classification of the following redox reaction: 2NO2(g) + 2OH(aq) → NO2(aq) + NO3(aq) + H2O(liq)


This is a disproportionation reaction since here the oxidation state of nitrogen decreases from +4 in NO2 to +3 in NO2 ion, as well as increases from +4 in NO2 to +5 in NO3ion.

Q3. The oxidation number of Fe in Fe3O4 is fractional. Why?


Fe3O4 contains Fe atoms of both +2 and +3 oxidation number. It is a stoichiometric mixture of Ferrous (FeO) and Ferric (Fe2O3) oxides combined  as FeO.Fe2O3. Therefore, the oxidation number found in such cases is average oxidation number.

Q4. Give an example of disproportionation redox reaction.


Decomposition of hydrogen peroxide is a disproportionation reaction where oxygen atom undergoes disproportionation.

Q5. Find the oxidation number of Boron in BHandBF3.


Oxidation number of Boron in BH3 is -3 and in BF3    the oxidation number of Boron is +3

Q6. The standard electrode potentials of a few metals are given below: AI(-0.66V), Cu(+0.34V), Li(-3.05V), Ag(+0.80V), and Zn(-0.76V) Which of these will behave as the strongest oxidizing agent and which as the strongest reducing agent?


Li is the strongest reducing agent while Ag+ is the strongest oxidizing agent.

Q7. Define Oxidation number.


The charge which an atom appears to have when all other atoms are removed from it as ions.

Q8. Write the net ionic equation for the reaction of Potassium Dichromate (VI), K2Cr2O7 with Sodium Sulphite, Na2SO3 in acid solution to give Chromium (III) ion and Sulphate ion.


Cr2O72-(aq) + 3SO32-(aq) + 8H+(aq) →  2Cr3+(aq) + 3SO42-(aq) + 4H2O(liq)

Q9. Can a solution of 1 M CuSO4 be stored in a vessel made of Nickel-metal? Given that E0 Ni, Ni2+ = 0.25 V, E0 Cu, Cu2+ = – 0.34 V


In this problem, we want to see whether following reaction can take   place or not   Ni + CuSO4     NiSO4 + Cu    i.e.  Ni + Cu2+ Ni2+ + Cu   By convention, the cell may be represented as Ni│Ni2+ ║Cu2+│Cu   We are given the oxidation potentials as   E0Ni,Ni2+ = + 0.25 V and E0Cu,Cu2+ = – 0.34 V  

Hence, the reduction potential will be E0Ni2+│Ni = – 0.25 V and E0Cu2+│Cu = + 0.34 V  E0cell    = E0Cu2+│Cu – E0Ni2+│Ni  = +0.34 – (-0.25) = +0.59 V This EMF comes out to be positive which means CuSO4 reacts with nickel. Therefore, CuSO4 solution cannot be stored in a Nickel vessel.

Q10. In which compounds is the oxidation number of Oxygen -1 and +2?


The oxidation number in peroxides (Na2O2) is -1 and in OF2 its oxidation number is +2.

Q11. Calculate the Oxidation number of Sulphur in S2O82- ion.


The Oxidation number of Sulphur in this ion is 6

Q12. What are the steps involved in balancing a Redox equation by oxidation number method?


Step – I Write the skeletal equation of all the reactants and products of the reaction.

Step – II Indicate the Oxidation number of each element.

Step – III Calculate the increase or decrease in oxidation number per atom and identify the oxidizing and reducing agents.

Step – IV Multiply the formulae of the oxidizing and reducing agents by suitable integers so as to equalize the total increase or decrease in oxidation number as calculated in Step – IIIStep – V Balance all atoms other than Hydrogen and Oxygen.

Step – VI Finally balance Hydrogen and Oxygen atoms by adding H2O molecules using hit and trial method.

Step – VII In case of ionic reaction

a) for acidic medium, first balance oxygen atoms by adding water molecules to whatever side deficient in oxygen atoms and then balance hydrogen atoms by adding H+ ions to whatever side deficient in hydrogen atoms

b) for basic mediums, first balance oxygen atoms by adding water molecules to whatever side deficient in oxygen atoms the hydrogen atoms are then balanced by adding water molecules equal in number to the deficiency of hydrogen atoms, and an equal number of OH ions are added to the opposite side of the equation.

Q13. I2 and Br2 are added to a solution containing Br and I ions. What reaction will occur if, I2 + 2e 2I; E0 = +0.54 V and Br2 + 2e 2Br; E0 = +1.09 V?


Since E0 of Br2 is higher than that of I2, therefore, Br2 has a higher tendency to accept electrons that I2. Conversely, I ion has a higher tendency to lose electrons than Br ions. Therefore, the following reaction will occur:  2I  I2 + 2e  Br2 +2e     2Br2I + Br2     I2 +2Br–  In other words I ion will be oxidized to I2 while Br2 will be reduced to Br ion.

Q14. How many grams of Potassium Dichromate are required to oxidize 15.2 g of FeSO4 in acidic medium?


6545 pdd 65

Q15. Permanganate ion reacts with bromide ion in basic medium to give manganese dioxide and bromate ion. Write the balanced chemical equation for the reaction?


2MnO4(aq) + Br(aq) + H2O(liq)6530 image001(43) 2MnO2(s) + BrO3(aq) + 2OH(aq)


6537 Genius314


6537 Genius315

Q17. Which of the following species do not show disproportionation reaction and why? ClO, ClO2, ClO3 and ClO4


Out of the above species, ClO4 does not undergo disproportionation since in this oxoanion chlorine is already present in the highest oxidation state of +7 and hence cannot be further oxidised.

Q18. What are the steps involved in balancing a Redox equation by Ion Electron method?


Step – I Write the skeletal equation and indicate the Oxidation number of all the elements which appear in the skeletal equation above their respective symbols.

Step – II Find out the species which are oxidized and which are reduced.

Step – III Split the skeletal equation into two half reactions, i.e. oxidation half reaction and reduction half equation.

Step – IV Balance the two half equations separately by the rules described below:

  • (a) In each half-reaction, first balance the atoms of the elements which have undergone a change in oxidation number.
  • (b) Add electrons to whatever side is necessary to make up the difference in oxidation number in each half-reaction.
  • (c) Balance charge by adding H+ ions if the reaction occurs in the acidic medium and by adding OH ions if the reaction occurs in a basic medium.
  • (d) Balance oxygen atoms by adding the required number of H2O molecules to the side deficient in O atoms.
  • (e) In the acidic medium, H atoms are balanced by adding H+ ions to the side deficient in H atoms. However, in the basic medium H atoms are balanced by adding H2O molecules equal in number to the deficiency of H atoms, and equal OH ions are included in the opposite side of the equation. Remove the duplication, if any.

Step – V The two half-reactions are then multiplied by suitable integers so that the total number of electrons gained in the one-half reaction is equal to the number of electrons lost in the other half-reaction. The two half-reactions are then added up.

Q19. Justify that the reaction 2Na(s) + H2(g) → 2NaH(s) is a redox reaction.


Since NaH is an ionic compound, it may be represented as Na+ H(s) 2Na(s) + H2(g) → 2Na+ H This reaction can be split into following two half reactions: Na(s) → Na+ e (oxidation) (loss of electrons) H2(s) + 2e → 2H (reduction) (gain of electrons) Overall redox reaction, 2Na(s) + H2(g) → 2Na+ H(s)