NCERT Exemplar: Redox Reactions Notes | EduRev

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JEE : NCERT Exemplar: Redox Reactions Notes | EduRev

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MULTIPLE CHOICE QUESTIONS - I

Q.1. Which of the following is not an example of redox reaction?
(i) CuO + H2 → Cu + H2O
(ii) Fe2O3 + 3CO → 2Fe  + 3CO2
(iii) 2K + F2 → 2KF
(iv) BaCl2 + H2SO4 → BaSO4 + 2HCl
Ans.
(d)
Solution.
BaCl2 + H2SO4 → BaSO4 + 2HCl is not a redox reaction. It is an example of double displacement reactions.

Q.2. The more positive the value of Eθ, the greater is the tendency of the species to get reduced. Using the standard electrode potential of redox couples given below find out which of the following is the strongest oxidising agent.
Eθ values: Fe3+/Fe2+ = + 0.77; I2(s)/I = + 0.54;
Cu2+/Cu = + 0.34; Ag+/Ag = + 0.80V
(1) Fe3+
(2) I2(s)
(3) Cu2+
(4) Ag+

Ans.(4)
Solution.
Since Ag+/Ag has highest positive value of E°, therefore, Ag+ is the strongest oxidizing agent with highest tendency to get reduced.

Q.3. Eθ values of some redox couples are given below. On the basis of these values choose the correct option.
Eθ values : Br2/Br = + 1.90;  Ag+ /Ag(s) = + 0.80
Cu2+/Cu(s) = + 0.34;  I2(s)/I = + 0.54
(1) Cu will reduce Br
(2) Cu will reduce Ag
(3) Cu will reduce I
(4) Cu will reduce Br2
Ans.
(4)
Solution.

Copper will reduce Br2, if the E° of the redox reaction, 2Cu + Br2
CuBr2 is +ve.
Now,
NCERT Exemplar: Redox Reactions Notes | EduRev
Since E° of this reaction is +ve, therefore, Cu can reduce Br2 and hence option (d) is correct.

Q.4. Using the standard electrode potential, find out the pair between which redox reaction is not feasible.
Eθ values : Fe3+/Fe2+ = + 0.77; I2/I– = + 0.54;
Cu2+/Cu = + 0.34; Ag+/Ag = + 0.80 V
(1) Fe3+ and I
(2) Ag+ and Cu
(3) Fe3+ and Cu
(4) Ag and Fe3+
Ans. 
(4)
Solution.

NCERT Exemplar: Redox Reactions Notes | EduRev
NCERT Exemplar: Redox Reactions Notes | EduRev
NCERT Exemplar: Redox Reactions Notes | EduRev
NCERT Exemplar: Redox Reactions Notes | EduRev

Q.5. Thiosulphate reacts differently with iodine and bromine in the reactions given below:
NCERT Exemplar: Redox Reactions Notes | EduRev
Which of the following statements justifies the above dual behavior of thiosulphate?
(1) Bromine is a stronger oxidant than iodine.
(2) Bromine is a weaker oxidant than iodine.
(3) Thiosulphate undergoes oxidation by bromine and reduction by iodine in these reactions.
(4) Bromine undergoes oxidation and iodine undergoes reduction in these reactions.
Ans.
(1)
Solution.

NCERT Exemplar: Redox Reactions Notes | EduRev
Bromine being stronger oxidizing agent than I2, it oxidises S of S2O2-3 to SO2-4 whereas I2 oxidises it only into S4O62- ion.

Q.6. The oxidation number of an element in a compound is evaluated on the basis of certain rules. Which of the following rules is not correct in this respect?
(1) The oxidation number of hydrogen is always +1.
(2) The algebraic sum of all the oxidation numbers in a compound is zero.
(3) An element in the free or the uncombined state bears oxidation number zero.
(4) In all its compounds, the oxidation number of fluorine is – 1.
Ans. 
(1)
Solution.

Oxidation number of hydrogen is -1 in metal hydrides like NaH.

Q.7. In which of the following compounds, an element exhibits two different oxidation states.
(1) NH2OH
(2) NH4NO3
(3) N2H4
(4) N3H
Ans.(2)
Solution.

NH4NO3 is an ionic compound consisting of NH4+ and NO3- ion.
The oxidation number of N in two species is different as shown below:
In NH,
Let the oxidation state of N in NH4+ be x.
x + 4x(+1) = +1
x = -3
In NO3-
Let the oxidation state of N in NO3-  be y,
y + 3 x (-2) = -1
y = +5

Q.8. Which of the following arrangements represent increasing oxidation number of the central atom?
(1) CrO2  , ClO3    , CrO42– , MnO4–
(2) ClO3 , CrO42– , MnO4 , CrO2
(3) CrO2 , ClO3 , MnO4 , CrO42–
(4) CrO42–, MnO4 , CrO2 , ClO3
Ans. 
(1)
Solution.
Writing the O.N. of Cr, Cl and Mn on each species in the four set of ions, we have,
NCERT Exemplar: Redox Reactions Notes | EduRev

Q.9. The largest oxidation number exhibited by an element depends on its outer electronic configuration. With which of the following outer electronic configurations the element will exhibit largest oxidation number?
(1) 3d14s2
(2) 3d34s2
(3) 3d54s1
(4) 3d54s2
Ans.(4)
Solution.

Highest O.N. of any transition element = (n – 1)d electrons +ns electrons. Therefore, larger the number of electrons in the 3d orbitals, higher is the maximum O.N.
(a) 3d14s= 3;
(b) 3d24s2 = 3 + 2 = 5;
(c) 3d54s= 5 + 1=6
(d) 3d54s= 5 + 2 = 7

Q.10. Identify disproportionation reaction
(1) CH4 + 2O2 → CO2 + 2H2O
(2) CH4 + 4Cl2 → CCl4 + 4HCl
(3) 2F2 + 2OH → 2F + OF2 + H2O
(4) 2NO2 + 2OH– → NO2  + NO3  + H2O
Ans.
(4)
Solution.

Reactions in which the same substance is oxidized as well as reduced are called disproportionation reactions. Writing the O.N. of each element above its symbol in the given reactions,
NCERT Exemplar: Redox Reactions Notes | EduRev
Thus, in reaction (d), N is both oxidized as well as reduced since the O.N. of N increases from +4 in NO2 to +5 in NO3 and decreases from +4 in N02 to +3 in N02.

Q.11. Which of the following elements does not show disproportionation tendency?
(1) Cl
(2) Br
(3) F
(4) I
Ans. 
(3)
Solution.

Being the most electronegative element, F can only be reduced and hence it always shows an oxidation number of-1. Further, due to the absence of d-orbitals, it cannot be oxidized and hence it does not show +ve oxidation numbers. In other words, F cannot be simultaneously oxidized as well as reduced and hence does not show disproportionation reactions. Thus, option (3) is correct.


MULTIPLE CHOICE QUESTIONS - II

In the following questions two or more options may be correct.
Q.12. Which of the following statement(s) is/are not true about the following decomposition reaction. 

2KClO3 → 2KCl + 3O2
(1) Potassium is undergoing oxidation
(2) Chlorine is undergoing oxidation
(3) Oxygen is reduced
(4) None of the species are undergoing oxidation or reduction
Ans.(1, 2, 3, 4)
Solution.
Writing the oxidation number of each element above its symbol,
NCERT Exemplar: Redox Reactions Notes | EduRev
(a) The O.N. of K does not change, K. undergoes neither reduction nor oxidation. Thus, option (a) is not correct.
(b) The O.N. of chlorine decreases from +5 in KClO3 to -l in KCl, hence, Cl undergoes reduction.
(c) Since, O.N. of oxygen increases from -2 in KClO3 to O in O2, oxygen is oxidized.
(d) This statement is not correct because Cl is undergoing reduction and O is undergoing oxidation.

Q.13. Identify the correct statement (s) in relation to the following reaction:
Zn + 2HCl  → ZnCl2 + H2
(1) Zinc is acting as an oxidant
(2) Chlorine is acting as a reductant
(3) Hydrogen ion is acting as an oxidant
(4) Zinc is acting as a reductant
Ans.
(3, 4)
Solution.

NCERT Exemplar: Redox Reactions Notes | EduRev
(a) The O.N. of Zn increases from 0 to +2 (in ZnCl2) and therefore, Zn acts as a reductant and not as an oxidant. Hence, option (a) is not correct.
(b) The O.N. of Cl does not change and therefore, it neither acts as a reductant nor an oxidant. Hence, option (b) is not correct.
(c) The O.N. of H decreases from +1 in H+ to O in H2. Therefore, H+ acts an oxidant. This option is correct.
(d) Zinc acts as reductant because its O.N. changes from 0 to +2. This option is correct.

Q.14. The exhibition of various oxidation states by an element is also related to the outer orbital electronic configuration of its atom. Atom(s) having which of the following outermost electronic configurations will exhibit more than one oxidation state in its compounds.
(1) 3s1
(2) 3d14s2
(3) 3d24s2
(4) 3s23p3
Ans. 
(2, 3, 4)
Solution.

Elements which have only s-electron in the valence shell do not show more than one oxidation state. Thus, element with 3s1 as outer electronic configuration shows only one oxidation state of +1.
Transition elements, i.e., elements (2, 3) having incompletely filled orbitals in the penultimate shell show variable oxidation states. Thus, element with outer electronic configuration as 3d1 4s2 shows variable oxidation states of +2 and +3 and the element with outer electronic configuration as 3d24s2 shows variable oxidation states of +2, +3 and +4.
p-Block elements also show variable oxidation states due to a number of reasons such as involvement of J-orbitals and inert pair effect. For example, element (4) with 3s2 3p3 as (i.e., P) as the outer electronic configuration shows variable oxidation states of +3 and +5 due to involvement of d-orbitals.

Q.15. Identify the correct statements with  reference to the given reaction
NCERT Exemplar: Redox Reactions Notes | EduRev

(1) Phosphorus is undergoing reduction only.
(2) Phosphorus is undergoing oxidation only.
(3) Phosphorus is undergoing oxidation as well as reduction.
(4) Hydrogen is undergoing neither oxidation nor reduction.
Ans. 
(3, 4)
Solution.

NCERT Exemplar: Redox Reactions Notes | EduRev
Because O.N. of P increases from 0 (P4) to +1 (H2P02) and decreases from 0 (P4) to -3 (PH3), therefore, P has undergone both oxidation as well as reduction. So, option (c) is correct. Option (d) is also correct because O.N. of H remains +1 in all the compounds and hence hydrogen is undergoing neither oxidation nor reduction.

Q.16. Which of the following electrodes will act as anodes, when connected to Standard Hydrogen Electrode?

 Al/Al3+ Eθ = –1.66
 Fe/Fe2+ Eθ = – 0.44
 Cu/Cu2+ Eθ = + 0.34
 F2 (g)/2F (aq) 
 Eθ = + 2.87

Ans.(1, 2)
Solution.

The electrodes having negative electrode potentials are stronger reducing agents than H2 gas and therefore, will act as anodes.


SHORT ANSWER TYPE QUESTIONS

Q.17. The reactionNCERT Exemplar: Redox Reactions Notes | EduRevrepresents the process of bleaching. Identify and name the species that bleaches the substances due to its oxidising action.
Ans.

NCERT Exemplar: Redox Reactions Notes | EduRev
In this reaction, O.N. of Cl increases from 0 (in Cl2) to +1 (in ClO) and decreases to -1 (in Cl). Therefore, Cl2 is both oxidized to ClO and reduced to Cl. Since Cl ion cannot act as an oxidizing agent (because it cannot decrease its O.N. lower than -1), therefore, Cl2 bleaches substances due to oxidizing action of hypochlorite, ClO ion.

Q.18. Mn042- undergoes disproportionation reaction in acidic medium but MnO4 does not. Give reason.
Ans.
In Mn04, Mn is in the highest oxidation state of+7 (i.e.. cannot be oxidized further) and hence it cannot undergo disproportionation.
In contrast, the O.N. of Mn in Mn042- is +6. Therefore, it can increase its O.N. to+7 or decrease its O.N. to some lower value. Thus, Mn042- undergoes disproportionation according to the following reaction.
NCERT Exemplar: Redox Reactions Notes | EduRev
Here, the O.N. of Mn increases from +6 in Mn042- to +7 in Mn04 and decreases to +4 in MnO2. Thus, Mn042- undergoes disproportionation in acidic medium.

Q.19. PbO and PbO2 react with HCl according to following chemical equations :
NCERT Exemplar: Redox Reactions Notes | EduRev
Why do these compounds differ in their reactivity?
Ans.
(i)
NCERT Exemplar: Redox Reactions Notes | EduRev
In reaction (i), O.N. of none of the atoms undergo a change. Therefore, it is not a redox reaction. It is an acid-base reaction, because PbO is a basic oxide which reacts with HCl acid. The reaction (ii) is a redox reaction in which PbO2   gets reduced and acts as an oxidizing agent.

Q.20. Nitric acid is an oxidising agent and reacts with PbO but it does not react with PbO2. Explain why?
Ans.
Nitric acid is an oxidizing agent and reacts with PbO to give a simple acid- base reaction without any change in oxidation state. In PbO2, Pb is in +4 oxidation state and cannot be oxidized further, hence no reaction takes place between PbO2 and HNO3.
NCERT Exemplar: Redox Reactions Notes | EduRev

Q.21. Write balanced chemical equation for the following reactions:
(1) Permanganate ion (MnO4) reacts with sulphur dioxide gas in acidic medium to produce Mn2+ and hydrogensulphate ion. (Balance by ion electron method)
(2) Reaction of liquid hydrazine (N2H4) with chlorate ion (ClO3) in basic medium produces nitric oxide gas and chloride ion in gaseous state. (Balance by oxidation number method)
(3) Dichlorine heptaoxide (Cl2O7) in gaseous state combines with an aqueous solution of hydrogen peroxide in acidic medium to give chlorite ion (ClO2) and oxygen gas. (Balance by ion electron method)
Ans.
(1) 2MnO4 + 5SO2 + 2H2O + H+→5HS04 + 2Mn2+
Balancing by ion-electron method:
NCERT Exemplar: Redox Reactions Notes | EduRev
Oxidation half: SO2 → HSO-4
Reduction half: MnO4- →Mn2+
Oxidation half: SO→ HSO-4 + 2e-
NCERT Exemplar: Redox Reactions Notes | EduRev...........(i)
(Add 2H2O molecules to balance O atoms)
Reduction half:
NCERT Exemplar: Redox Reactions Notes | EduRev
(Add 4H2O molecules to balance O atoms and H atoms)
Add oxidation and reduction half
NCERT Exemplar: Redox Reactions Notes | EduRev
(2)NCERT Exemplar: Redox Reactions Notes | EduRev
Balancing by oxidation number method:
NCERT Exemplar: Redox Reactions Notes | EduRev
NCERT Exemplar: Redox Reactions Notes | EduRev
(3) NCERT Exemplar: Redox Reactions Notes | EduRev
Balancing by ion-electron method:
Oxidation half:
H2O→ O2 + 2e-
Adding 2H+ on right side to balance H atoms and charge.
NCERT Exemplar: Redox Reactions Notes | EduRev
Reduction half:
NCERT Exemplar: Redox Reactions Notes | EduRev
Adding H2O and H+ to balance H and O atoms
NCERT Exemplar: Redox Reactions Notes | EduRev
Adding oxidation and reduction half
NCERT Exemplar: Redox Reactions Notes | EduRev

Q.22. Calculate the oxidation number of phosphorus in the following species.
(1) HPO2-3 and
(2) PO3-4
Ans.
(1) Let the oxidation number of P in HPO2-3 be x. So, + 1 + x + (-6) = -2 x = +3
(2) Let the oxidation number of P in P03-4 be x. So, x + (-8) = -3
x = +5

Q.23. Calculate the oxidation number of each sulphur atom in the following compounds:
(1) Na2S2O3
(2) Na2S4O6
(3) Na2SO3
(4) Na2SO
4
Ans. (1)
Solution. (a)
Structure of Na2S2O3
NCERT Exemplar: Redox Reactions Notes | EduRev
The oxidation state of Sis -2
Let oxidation state of Sbe x
NCERT Exemplar: Redox Reactions Notes | EduRev
Thus, the oxidation states of two S atoms in Na2S2O3 are -2 and +6.
(b) Na2S4O6
NCERT Exemplar: Redox Reactions Notes | EduRev
From the left, S1 = (-2) + (-2) + (-1) = +5
S2 = 0
S3 = 0
S4 = +5
(c) Na2SO3
2 x (+1) + x + 3 x (-2) = 0
x = +4
(d) Na2SO4
+2 + x + (-8) = 0
x = +6

Q.24. Balance the following equations by the oxidation number method.
(i)NCERT Exemplar: Redox Reactions Notes | EduRev
(ii)NCERT Exemplar: Redox Reactions Notes | EduRev
(iii)NCERT Exemplar: Redox Reactions Notes | EduRev
(iv)NCERT Exemplar: Redox Reactions Notes | EduRev
Ans.
NCERT Exemplar: Redox Reactions Notes | EduRev
(a) Balance the increase and decrease in O.N.
NCERT Exemplar: Redox Reactions Notes | EduRev
(b) Balancing H and O atoms by adding H+ and H2O molecules
NCERT Exemplar: Redox Reactions Notes | EduRev
(ii)
NCERT Exemplar: Redox Reactions Notes | EduRev
Total increase in O.N. = 5 × 2 = 10
Total decrease in O.N. = 1
To equalize O.N. multiply NO3-, by 10
I2 + 10no3→ 10no2 + IO3-
Balancing atoms other than O and H
I2 + 10no3-→ 10NO2 + 2 IO3-
Balancing O and H
I2 + lO no3- + 8H+→ 10NO2 + 2 IO3- + 4H20
(iii) NCERT Exemplar: Redox Reactions Notes | EduRev
Total increase in O.N. = 0.5 x 4 = 2
Total decrease in O.N. = 1x2 = 2
To equalize O.N. multiply S2O2-3 and I- by 2.
NCERT Exemplar: Redox Reactions Notes | EduRev
(iv) NCERT Exemplar: Redox Reactions Notes | EduRev
Total increase in O.N. = 1 x 2 = 2
Total decrease in O.N. = 2
To equalize O.N. multiply COby 2.
MnO2 + C2O2-4 → Mn2+ + 2CO2
Balance H and O by adding 2H2O on right side, and 4H+ on left side of equation.
MnO2 + C2O2-4 + 4H+ → Mn2+ + 2CO2 + 2H2O


Q.25. Identify the redox reactions out of the following reactions and identify the oxidising and reducing agents in them.
NCERT Exemplar: Redox Reactions Notes | EduRev

NCERT Exemplar: Redox Reactions Notes | EduRev
Ans.(i)
NCERT Exemplar: Redox Reactions Notes | EduRev
Oxidising agent - HNO3
Reducing agent - HCl
(ii) This is not a redox reaction.
(iii)NCERT Exemplar: Redox Reactions Notes | EduRev
Fe - Oxidising agent
CO - Reducing agent
(iv)NCERT Exemplar: Redox Reactions Notes | EduRev
Above reaction is not a redox reaction.
NCERT Exemplar: Redox Reactions Notes | EduRev
Reducing agent — NH3
Oxidising agent — O2

Q.26. Balance the following ionic equations
NCERT Exemplar: Redox Reactions Notes | EduRev
NCERT Exemplar: Redox Reactions Notes | EduRev
Ans.
(i)NCERT Exemplar: Redox Reactions Notes | EduRev
NCERT Exemplar: Redox Reactions Notes | EduRev
Dividing eq. by 2
NCERT Exemplar: Redox Reactions Notes | EduRev
NCERT Exemplar: Redox Reactions Notes | EduRev
NCERT Exemplar: Redox Reactions Notes | EduRev
On adding (i) and (ii)
NCERT Exemplar: Redox Reactions Notes | EduRev
(iii) NCERT Exemplar: Redox Reactions Notes | EduRev
Reduction half:
NCERT Exemplar: Redox Reactions Notes | EduRev
Oxidation half:
NCERT Exemplar: Redox Reactions Notes | EduRev
NCERT Exemplar: Redox Reactions Notes | EduRev
(iv) NCERT Exemplar: Redox Reactions Notes | EduRev
Reduction half:
NCERT Exemplar: Redox Reactions Notes | EduRev
NCERT Exemplar: Redox Reactions Notes | EduRev


MATCHING TYPE QUESTIONS

Q.27. Match Column I with Column II for the oxidation states of the central atoms.

 Column I Column II
 (i) Cr2O72- (a) +3
 (ii) MnO4- (b) +4
 (iii)  VO3- (c) +5 
 (iv) FeF63- (d) +6

 (e) +7

Ans. (i) → (d); (ii) → (e); (iii) → (c); (iv) → (a)

Q.28. Match the items in Column I with relevant items in Column II.

 Column I Column II
 (i) Ions having positive charge (a) +7
 (ii) The sum of oxidation number of all atoms in a neutral molecule (b) –1 
(iii) Oxidation number of hydrogen ion (H+) (c) +1
(iv)  Oxidation number of fluorine in NaF (d) 0 
 (v) Ions having negative charge (f) Cation

 (g) Anion

Ans. (i) → (e); (ii) → (d); (iii) → (c); (iv) → (b); (v) → (f)


ASSERTION AND REASON TYPE QUESTIONS

In the following questions a statement of assertion (A) followed by a statement of reason (R) is given. Choose the correct option out of the choices given below each question.
Q.29. Assertion (A) : Among halogens fluorine is the best oxidant.
Reason (R) : Fluorine is the most electronegative atom.
(1) Both A and R are true and R is the correct explanation of A.
(2) Both A and R are true but R is not the correct explanation of A.
(3) A is true but R is false.
(4) Both A and R are false.

Ans.(1) Both A and R are true and R is the correct explanation of A.
Explanation.
Fluorine is most electro-negative element that is why it is best oxidant among halogens.

Q.30. Assertion (A): In the reaction between potassium permanganate and potassium iodide, permanganate ions act as oxidising agent.
Reason (R) : Oxidation state of manganese changes from +2 to +7 during the reaction.
(1) Both A and R are true and R is the correct explanation of A.
(2) Both A and R are true but R is not the correct explanation of A.
(3) A is true but R is false.
(4) Both A and R are false.
Ans.
(3) A is true but R is false.
Explanation.
As permanganate ion changes to MnO2.

Q.31. Assertion (A) : The decomposition of hydrogen peroxide to form water and oxygen is an example of disproportionation reaction.
Reason (R) : The oxygen of peroxide is in -1 oxidation state and it is converted to zero oxidation state in O2 and -2 oxidation state in H2O.
(1) Both A and R are true and R is the correct explanation of A.
(2) Both A and R are true but R is not the correct explanation of A.
(3) A is true but R is false.
(4) Both A and R are false.
Ans. 
(1) Both A and R are true and R is the correct explanation of A.
Explanation.
NCERT Exemplar: Redox Reactions Notes | EduRev
Here the oxygen of peroxide, which is present in -1 state, is converted to zero oxidation state in O2 undergoing oxidation and decreases to -2 oxidation state in H2O undergoing reduction.

Q.32. Assertion (A) : Redox couple is the combination of oxidised and reduced form of a substance involved in an oxidation or reduction half cell.
Reason (R) : In the representation NCERT Exemplar: Redox Reactions Notes | EduRevand Cu2+/Cu are redox couples.
(1) Both A and R are true and R is the correct explanation of A.
(2) Both A and R are true but R is not the correct explanation of A.
(3) A is true but R is false.
(4) Both A and R are false.
Ans. 
(2) Both A and R are true but R is not the correct explanation of A.
Explanation.

A redox couple is defined as pair of compounds or elements having together the oxidised and reduced forms of it and taking part in an oxidation or reduction half reaction.


LONG ANSWER TYPE QUESTIONS

Q.33. Explain redox reactions on the basis of electron transfer. Give suitable examples.
Ans. NCERT Exemplar: Redox Reactions Notes | EduRevis a redox change.
NCERT Exemplar: Redox Reactions Notes | EduRevand the other half reaction is:
NCERT Exemplar: Redox Reactions Notes | EduRev
This splitting of the reaction into two half reactions automatically reveals here that sodium is oxidised and hydrogen is reduced. Any sustance which loses electron is oxidised and gains electron is reduced hence is the case of sodium and hydrogen atoms respectively. Therefore, the complete reaction is a redox change.

Q.34. On the basis of standard electrode potential values, suggest which of the following reactions would take place? (Consult the book for Eθ value).
(1) Cu + Zn2+ → Cu2+ + Zn
(2) Mg + Fe2+ → Mg2+ + Fe
(3) Br2 + 2Cl– → Cl2 + 2Br
(4) Fe + Cd2+ → Cd + Fe2+
Ans.
On the basis of standard reduction potential suggested in the reactivity series (ii) reaction can take place as Mg has more negative value of E° cell. Thus, Mg will be oxidized by losing electron and iron will be reduced by gaining electron.

Q.35. Why does fluorine not show disporportionation reaction?
Ans.
Flourine shows disproportionation reaction because fluorine is most electronegative element of its period as well as group. It shows only -1 oxidation state and is smallest in size of all the halogens.

Q.36. Write redox couples involved in the reactions (i) to (iv) given in question 34.
Ans.
NCERT Exemplar: Redox Reactions Notes | EduRev

Q.37. Find out the oxidation number of chlorine in the following compounds and arrange them in increasing order of oxidation number of chlorine.
NaClO4, NaClO3, NaClO, KClO2, Cl2O7, ClO3, Cl2O, NaCl, Cl2, ClO2.
Which oxidation state is not present in any of the above compounds?
Ans.

NCERT Exemplar: Redox Reactions Notes | EduRevOxidation no. of chlorine = +7
NCERT Exemplar: Redox Reactions Notes | EduRevOxidation no. of chlorine = +5
NCERT Exemplar: Redox Reactions Notes | EduRev Oxidation no. of chlorine= +1
NCERT Exemplar: Redox Reactions Notes | EduRevOxidation no. of chlorine =+3
NCERT Exemplar: Redox Reactions Notes | EduRev Oxidation no. of chlorine =+7
NCERT Exemplar: Redox Reactions Notes | EduRevOxidation no. of chlorine =+6
NCERT Exemplar: Redox Reactions Notes | EduRevOxidation no. of chlorine = + l
NaCl Oxidation no. of chlorine = - l
Cl2 Oxidation no. of chlorine =0
NCERT Exemplar: Redox Reactions Notes | EduRevOxidation no. of chlorine = +4
NaCl, Cl2, Cl2O, NaClO, KClO2, ClO2, NaClO3, ClO3, NaClO4

Q.38. Which method can be used to find out strength of reductant/oxidant in a solution? Explain with an example.
Ans. In redox systems, the titration method can be adopted to determine the strength of a reductant/oxidant using a redox sensitive indicator. The usage of indicators in redox titration is illustrated below: (i) In one situation, the reagent itself is intensely coloured, e.g.. permanganate ion, MnO-4. Here, MnO4- acts as the self indicator. The visible end point in this case is achieved after the last of the reductant (Fe2+ or C2O2-4) is oxidised and the first lasting tinge of pink colour appears at MnO4- concentration as low as 10-6 mol (10-6 mol L-1). This ensures a minimal ‘overshoot’ in colour beyond the equivalence point, the point where the reductant and the oxidant are equal in terms of their mole stoichiometry. 

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,

Extra Questions

,

Free

,

Previous Year Questions with Solutions

,

Sample Paper

,

pdf

,

Objective type Questions

,

MCQs

,

past year papers

,

Summary

,

shortcuts and tricks

,

mock tests for examination

,

NCERT Exemplar: Redox Reactions Notes | EduRev

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