Home NCERT Solutions NCERT Exemplar CBSE Sample Papers NCERT Books Class 12 Class 10

Chapter 8 Redox Reactions

At Saralstudy, we are providing you with the solution of Class 11 Chemistry Redox Reactions according to the latest NCERT (CBSE) Book guidelines prepared by expert teachers. Here we are trying to give you a detailed answer to the questions of the entire topic of this chapter so that you can get more marks in your examinations by preparing the answers based on this lesson. We are trying our best to give you detailed answers to all the questions of all the topics of Class 11th chemistry Redox Reactions so that you can prepare for the exam according to your own pace and your speed.

Download pdf of NCERT Solutions for Class Chemistry Chapter 8 Redox Reactions

Download pdf of NCERT Examplar with Solutions for Class Chemistry Chapter 8 Redox Reactions

Exercise 1

  • Q1

    Assign oxidation number to the underlined elements in each of the following species:

    (a) NaH2PO4

    (b) NaHSO4

    (c) H4P2O7

    (d) K2MnO4

    (e) CaO2

    (f) NaBH4

    (g) H2S2O7

    (h) KAl(SO4)2.12 H2O

    Ans:

    (a) NaH2PO4

    Let assume oxidation number of P is x.

    We know that,

    Oxidation number of Na = +1

    Oxidation number of H = +1

    Oxidation number of O = -2

    Then we have

    1(+1) + 2(+1) + 1 (x) + 4(-2) = 0

    ⇒ 1 + 2 + x - 8 = 0

    ⇒ x - 5 = 0

    ⇒ x = + 5

    Hence the oxidation number of P is +5

     

    (b)  NaHSO4

    Let assume oxidation number of S is x.

    Oxidation number of Na = +1

    Oxidation number of H = +1

    Oxidation number of O = -2

    Then we have:

    1(+1) + 1(+1) + 1 (x) + 4(-2) = 0

    ⇒ 1 + 1 + x - 8 = 0

    ⇒ x-6 = 0

    ⇒ x = +6

    Hence the oxidation number of S is +6

     

    (c)  H4P2O7

    Let assume oxidation number of P is x.

    Oxidation number of H = +1

    Oxidation number of O = -2

    Then we have:

    4(+1) + 2(x) + 7 (-2) = 0

    ⇒ 4 + 2x - 14 = 0

    ⇒ 2x - 10 = 0

    ⇒ 2x  = +10

    ⇒ x  = +5

    Hence, Oxidation number of P is +5

     

    (d) K2MnO4

    Let assume oxidation number of Mn is x.

    Oxidation number of K = +1

    Oxidation number of O = -2

    Then we have:

    2(+1) + 1(x) + 4 (-2) = 0

    ⇒ 2 + x - 8 = 0

    ⇒ x  - 6 = 0

    ⇒ x  = +6

    Hence, Oxidation number of Mn is +6

     

    (e) CaO2

    Let assume oxidation number of O is x.

    Oxidation number of Ca = +2

    Then we have:

    1(+2) + 2(x) = 0

    ⇒ 2 + 2x  = 0

    ⇒ 2x = -2

    ⇒ x  = -1

    Hence, Oxidation number of O is -1

     

    (f) NaBH4

    Let assume oxidation number of B is x.

    Oxidation number of Na = +1

    Oxidation number of H = -1

    Then we have:

    1(+1) + 1(x) + 4(-1)  = 0

    ⇒ 1 + x -4 = 0

    ⇒ x - 3 = 0

    ⇒ x  = +3

    Hence, Oxidation number of B is +3.

     

    (g) H2S2O7

    Let assume oxidation number of S is x.

    Oxidation number of O = -2

    Oxidation number of H = +1

    Then we have:

    2(+1) + 2(x) + 7(-2)  = 0

    ⇒ 2 + 2x - 14 = 0

    ⇒ 2x - 12 = 0

    ⇒ x  = +6

    Hence, Oxidation number of S is +6.

     

    (h) KAl(SO4)2.12 H2O

    Let assume oxidation number of S is x.

    Oxidation number of K = +1

    Oxidation number of Al  = +3

    Oxidation number of O  = -2

    Oxidation number of H  = +1

    Then we have:

    1(+1) + 1 (+3) + 2(x) + 8(-2) + 24(+1) + 12 (-2)  = 0

    ⇒ 1 + 3 + 2x -16 +24 -24 = 0

    ⇒ 2x - 12 = 0

    ⇒ 2x  = +12

    ⇒ x  = +6

    Hence, Oxidation number of S is +6.

     


    Q2

    What are the oxidation number of the underlined elements in each of the following and how do you rationalise your results ?

    (a) KI3

    (b) H2S4O6

    (c) Fe3O4

    (d) CH3CH2OH

    (e) CH3COOH

    Ans:

    (a) KI3

    Let assume oxidation number of l is x.

    In KI3, the oxidation number (O.N.) of K is +1.

    1(+1) + 3(x) = 0

    ⇒ +1 +3x = 0

    ⇒ 3x = -1

    ⇒ x = -1/3

    Hence, the average oxidation number of I is - 1/3

    However, O.N. cannot be fractional. Therefore, we will have to consider the structure of KI3 to find the oxidation states. In a KI3 molecule, an atom of iodine forms a coordinate covalent bond with an iodine molecule.

    Hence, in a KI3 molecule, the O.N. of the two I atoms forming the I2 molecule is 0, whereas the O.N. of the I atom forming the coordinate bond is –1.

     

    (b) H2S4O6

     

    Let assume oxidation number of S is x.

    The oxidation number (O.N.) of H is +1.

    The oxidation number (O.N.) of O is -2.

    2(+1) + 4(x) + 6(-2) = 0

    ⇒ 2 + 4x - 12 = 0

    ⇒ 4x -10 = 0

    ⇒ 4x  =  +10

    ⇒ x  = +10/4

    However, O.N. cannot be fractional. Hence, S must be present in different oxidation states in the molecule. 

     

    The O.N. of two of the four S atoms is +5 and the O.N. of the other two S atoms is 0. 

     

    (c) Fe3O

     

    Let assume oxidation number of Fe is x.

    The oxidation number (O.N.) of O is -2.

    3(x) + 4(-2) = 0

    ⇒ 3x  - 8 = 0

    ⇒ 3x  = 8

    ⇒ x  = 8/3

    However, O.N. cannot be fractional. 

    Here, one of the three Fe atoms exhibits the O.N. of +2 and the other two Fe atoms exhibit the O.N. of +3. 

     

    (d) CH3CH2OH

    Let assume oxidation number of C is x.

    The oxidation number (O.N.) of O is -2.

    The oxidation number (O.N.) of H is +1.

    x + 3(+1) + x + 2(+1) + 1(-2) + 1(+1) = 0

    ⇒ x +3 + x +2 - 2  + 1 = 0

    ⇒ 2x  + 4 = 0

    ⇒ 2x  = -4

    ⇒ x  = -2

    Hence, the oxidation number of C is -2.

     

    (e) CH3COOH

    Let assume oxidation number of C is x.

    The oxidation number (O.N.) of O is -2.

    The oxidation number (O.N.) of H is +1.

    x + 3(+1) + x + (-2) + (-2) + 1(+1) = 0

    ⇒ 2x + 3 - 2 -  2  +  1 = 0

    ⇒ 2x + 0 = 0

    ⇒ x = 0

    However, 0 is average O.N. of C.

    The two carbon atoms present in this molecule are present in different environments. Hence, they cannot have the same oxidation number. Thus, C exhibits the oxidation states of +2 and –2 in CH3COOH. 

     


    Q3

    Justify that the following reactions are redox reactions:

    (a) CuO(s) + H2(g) → Cu(s) + H2O(g)

    (b) Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g)

    (c) 4BCl3(g) + 3LiAlH4(s) → 2B2H6(g) + 3LiCl(s) + 3 AlCl3 (s)

    (d) 2K(s) + F2(g) → 2K+F(s)

    (e) 4 NH3(g) + 5 O2(g) → 4NO(g) + 6H2O(g)

    Ans:

    (a) CuO(s) + H2(g) → Cu(s) + H2O(g)

    Let us write the oxidation number of each element involved in the given reaction as: 

    +2  -2        0             0             +1    -2

    Cu O(s) +  H2(g)  →  Cu(s)  +  H2  O(g)

    Here, the oxidation number of Cu decreases from +2 in CuO to 0 in Cu i.e., CuO is reduced to Cu. Also, the oxidation number of H increases from 0 in H2 to +1 in H2O i.e., H2 is oxidized to H2O. Hence, this reaction is a redox reaction.

     

    (b) Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g)

    Let us write the oxidation number of each element involved in the given reaction as:

    +3   -2          +2  -2            0          +4  -2

    Fe2 O3(s) + 3C  O(g) → 2Fe(s) + 3C O2(g)

    Here, the oxidation number of Fe decreases from +3 in Fe2O3 to 0 in Fe i.e., Fe2O3 is reduced to Fe. On the other hand, the oxidation number of C increases from +2 in CO to +4 in CO2 i.e., CO is oxidized to CO2. Hence, the given reaction is a redox reaction.

     

    (c) 4BCl3(g) + 3LiAlH4(s) → 2B2H6(g) + 3LiCl(s) + 3 AlCl3 (s)

    Let us write the oxidation number of each element involved in the given reaction as:

     +3  -1            +1   +3   -1            -3   +1         +1  -1             +3   -1

    4B Cl3(g) + 3  Li   Al   H4(s) → 2B2 H6(g) + 3Li   Cl(s)   + 3 Al  Cl3 (s)

    In this reaction, the oxidation number of B decreases from +3 in BCl3 to –3 in B2H6. i.e., BCl3 is reduced to B2H6. Also, the oxidation number of H increases from –1 in LiAlH4 to +1 in B2H6 i.e., LiAlH4 is oxidized to B2H6. Hence, the given reaction is a redox reaction.

     

    (d) 2K(s) + F2(g) → 2K+F(s)

    Let us write the oxidation number of each element involved in the given reaction as:

    0               0                    +1   -1

    2K(s)   +   F2(g)    →    2KF(s)

    In this reaction, the oxidation number of K increases from 0 in K to +1 in KF i.e., K is oxidized to KF. On the other hand, the oxidation number of F decreases from 0 in F2 to – 1 in KF i.e., F2 is reduced to KF.

    Hence, the above reaction is a redox reaction.

     

    (e) 4 NH3(g) + 5 O2(g) → 4NO(g) + 6H2O(g)

    Let us write the oxidation number of each element involved in the given reaction as:

      -3   +1            0              +2  -2         +1    -2

    4 N  H3(g) + 5 O2(g) → 4N  O(g) + 6HO(g)

    Here, the oxidation number of N increases from –3 in NH3 to +2 in NO. On the other hand, the oxidation number of O2 decreases from 0 in O2 to –2 in NO and H2O i.e., O2 is reduced. Hence, the given reaction is a redox reaction.


    Q4

    Fluorine reacts with ice and results in the change:

    H2O(s) + F2(g) → HF(g) + HOF(g)

    Justify that this reaction is a redox reaction.

    Ans:

    Let us write the oxidation number of each atom involved in the given reaction above its symbol as:

    Here, we have observed that the oxidation number of F increases from 0 in F2 to +1 in HOF. Also, the oxidation number decreases from 0 in F2 to -1 in HF. Thus, in the above reaction, F is both oxidized and reduced. Hence, the given reaction is a redox reaction.


    Q5

    Calculate the oxidation number of sulphur, chromium and nitrogen in H2SO5, Cr2O2- 7 and NO 3. Suggest structure of these compounds. Count for the fallacy.

    Ans:

    (a) H2SO5

    Let assume oxidation number of S is x.

    We know that,

    Oxidation number of H = +1

    Oxidation number of O = -2

    Then we have

    2(+1) + (x) + 5 (-2) = 0

    ⇒ 2 + x - 10 = 0

    ⇒ x  =  +8

    However, the O.N. of S cannot be +8. S has six valence electrons. Therefore, the O.N. of S cannot be more than +6. The structure of H2SO5 is shown as follows:

    Now , 2(+1) + 1(x) + 3(-2)  + 2 (-1) = 0

    ⇒ 2 + x - 6 -2 = 0

    ⇒ x  =  +6

    Therefore, the O.N. of S is +6.

     

    (b) Cr2O2-7

    Let assume oxidation number of Cr is x.

    We know that,

    Oxidation number of O = -2

    Then we have

     2(x) + 7 (-2) =  -2

    ⇒ 2x  -14  = -2

    ⇒ 2x   =  +12

    x  = +6

    Here, there is no fallacy about the O.N. of Cr in Cr2O72-

    The structure of Cr2O72- is shown as follows: 

    Here, each of the two Cr atoms exhibits the O.N. of +6. 

     

    (c) NO3-

    Let assume oxidation number of N is x.

    We know that,

    Oxidation number of O = -2

    Then we have

    1(x) + 3 (-2) =  -1

    ⇒ x  - 6 = -1

    ⇒ x   =  +5

    Here, there is no fallacy about the O.N. of N in NO3-

    The structure of  NO3is shown as follows:

    The N atom exhibits the O.N. of +5. 

     


    Q6

    Write the formulae for the following compounds:
    (a) Mercury(II) chloride
    (b) Nickel(II) sulphate
    (c) Tin(IV) oxide
    (d) Thallium(I) sulphate
    (e) Iron(III) sulphate
    (f) Chromium(III) oxide

    Ans:

    The representation of a molecule in terms of the symbols of the constituting atoms is called its chemical formula.

    Rules for building the formula for a compound:

    1) Write side by side the symbols of the radicals involved, with their valencies at the top, the positive radical to the left & the negative radical to the right.

    2) Remove the common factor, if any, to get a simple ratio of the valencies of combining radicals.

    3) Criss cross the valencies i.e, shift the valencies cross wise to the lower right of radical.

    (a) Mercury (II) chloride:

    HgCl2

    (b) Nickel (II) sulphate:

    NiSO4

    (c) Tin (IV) oxide:

    SnO2

    (d) Thallium (I) sulphate:

    Tl2SO4

    (e) Iron (III) sulphate:

    Fe2(SO4)3

    (f) Chromium (III) oxide:

    Cr2O3


    Q7

    Suggest a list of the substances where carbon can exhibit oxidation states from -4 to +4 and nitrogen from -3 to +5.

    Ans:

    The substances where carbon can exhibit oxidation states from -4 to +4 are listed in the following table.

     Substance  Formula  Oxidation State of (C)
     Methan   CH4   -4
     Ethane   C2H6   -3
     Ethene   C2H4   -2
     Ethyne   C2H2   -1
     Dichloromethane   CH2Cl2    0
     Chloroform   ChCl3   +2
     Oxalic Acid   (COOH)2   +3
     Carbon TetraChloride   CCl4   +4
     Carbon dioxide   CO2   +4

     

    The substances where nitrogen can exhibit oxidation states from -3 to +5 are listed in the following table.

     Substance  Formula  Oxidation State of Nitrogen
     Ammonia  NH3   -3
     Hydrazine  N2H4   -2
     Hydride  N2H2   -1
     Dinitrogen Gas  N2    0
     Nitrous Oxide  N2O   +1
     Nitric Oxide  NO   +2
     Dinitrogen Trioxide  N2O3   +3
     Dinitrogen Dioxide  NO2   +4
     Nitrogen Pentroxide  N2O5   +5

     


    Q8

    While sulphur dioxide and hydrogen peroxide can act as oxidising as well as reducing agents in their reactions, ozone and nitric acid act only as oxidants. Why?

    Ans:

    In sulphur dioxide (SO2), the oxidation number (O.N.) of S is +4 and the range of the O.N. that S can have is from +6 to -2.

    Therefore, SO2 can act as an oxidising as well as a reducing agent.

    In hydrogen peroxide (H2O2), the O.N. of O is -1 and the range of the O.N. that O can have is from 0 to -2. O can sometimes also attain the oxidation numbers +1 and +2. Hence, H2O2 can act as an oxidising as well as a reducing agent.

    In ozone (O3), the O.N. of O is zero and the range of the O.N. that O can have is from 0 to -2. Therefore, the O.N. of O can only decrease in this case. Hence, O3 acts only as an oxidant.

    In nitric acid (HNO3), the O.N. of N is +5 and the range of the O.N. that N can have is from +5 to -3. Therefore, the O.N. of N can only decrease in this case. Hence, HNO3 acts only as an oxidant.


    Q9

    Consider the reactions:

    (a) 6CO2(g) + 6H2O(l) → C6H12O6(aq) + 6O2(g)

    (b) O3(g) + H2O2(l) → H2O(l) + 2O2(g)

    Why it is more appropriate to write these reactions as:

    (a) 6CO2(g) + 12H2O(l) → C6H12O6(aq) + 6H2O(l) + 6O2(g)

    (b) O3(g) + H2O2(l) → H2O(l) + O2(g) + O2(g)

    Also suggest a technique to investigate the path of the above (a) and (b) redox reactions.

    Ans:

    (a) The process of photosynthesis involves two steps.

    Step 1:

    H2O decomposes to give H2 and O2.

    2H2O(l)  →  2H2(g)  + O2(g)

     

    Step 2:

    The H2 produced in step 1 reduces CO2, thereby producing glucose (C6H12O6) and H2O.

    6CO2(g)  +  12H2(g)  →  C6H12O6(s)  +  6H2O(l)

    Now, the net reaction of the process is given as:

    It is more appropriate to write the reaction as given above because water molecules are also produced in the process of photosynthesis.

    The path of this reaction can be investigated by using radioactive H2O18 in place of H2O.

     

    (b) O2 is produced from each of the two reactants O3 and H2O2. For this reason, O2 is written twice.

    The given reaction involves two steps. First, O3 decomposes to form O2 and O. In the second step, H2O2 reacts with the O produced in the first step, thereby producing H2O and O2.

    The path of this reaction can be investigated by using H2O18 or O183.


    Q10

    The compound AgF2 is an unstable compound. However, if formed, the compound acts as a very strong oxidizing agent. Why?

    Ans:

    The oxidation state of Ag in AgF2 is +2. But, +2 is an unstable oxidation state of Ag. Therefore, whenever AgF2 is formed, silver readily accepts an electron to form Ag+. This helps to bring the oxidation state of Ag down from +2 to a more stable state of +1. As a result, AgF2 acts as a very strong oxidizing agent.


    Q11

    Whenever a reaction between an oxidising agent and a reducing agent is carried out, a compound of lower oxidation state is formed if the reducing agent is in excess and a compound of higher oxidation state is formed if the oxidising agent is in excess. Justify this statement giving three illustrations.

    Ans:

    Whenever a reaction between an oxidising agent and a reducing agent is carried out, a compound of lower oxidation state is formed if the reducing agent is in excess and a compound of higher oxidation state is formed if the oxidising agent is in excess. This can be illustrated as follows:

     

    (i) P4 and F2 are reducing and oxidising agents respectively.

    If an excess of P4 is treated with F2, then PF3 will be produced, wherein the oxidation number (O.N.) of P is +3.

    P4 (excess) + F2 → P+3F3

    However, if P4 is treated with an excess of F2, then PF5 will be produced, wherein the O.N. of P is +5.

    P4  + F2 (excess) → P+5F5

     

    (ii) K acts as a reducing agent, whereas O2 is an oxidising agent. If an excess of K reacts with O2, then K2O will be formed, wherein the O.N. of O is -2.

    4K (excess) + O2 →  2K2O-2

    However, if K reacts with an excess of O2, then K2O2 will be formed, wherein the O.N. of O is -1.

    2K + O2 (excess)  →  K2O-12

     

    (iii) C is a reducing agent, while O2 acts as an oxidising agent.

    If an excess of C is burnt in the presence of insufficient amount of O2, then CO will be produced, wherein the O.N. of C is +2.

    C (excess) + O2 →  C+2O

    On the other hand, if C is burnt in an excess of O2, then CO2 will be produced, wherein the O.N. of C is +4.

    C + O2 (excess)  →  C+4O2


    Q12

    How do you count for the following observations?

    (a) Though alkaline potassium permanganate and acidic potassium permanganate both are used as oxidants, yet in the manufacture of benzoic acid from toluene we use alcoholic potassium permanganate as an oxidant. Why? Write a balanced redox equation for the reaction.

    (b) When concentrated sulphuric acid is added to an inorganic mixture containing chloride, we get colourless pungent smelling gas HCl, but if the mixture contains bromide then we get red vapour of bromine. Why?

    Ans:

    (a) In the manufacture of benzoic acid from toluene, alcoholic potassium permanganate is used as an oxidant because of the following reasons.

    (i) In a neutral medium, OH- ions are produced in the reaction itself. As a result, the cost of adding an acid or a base can be reduced.

    (ii) KMnO4 and alcohol are homogeneous to each other since both are polar. Toluene and alcohol are also homogeneous to each other because both are organic compounds. Reactions can proceed at a faster rate in a homogeneous medium than in a heterogeneous medium. Hence, in alcohol, KMnO4 and toluene can react at a faster rate.

    The balanced redox equation for the reaction in a neutral medium is give as below:

     

    (b) When conc. H2SO4 is added to an inorganic mixture containing bromide, initially HBr is produced. HBr, being a strong reducing agent reduces H2SO4 to SO2 with the evolution of red vapour of bromine.

    2NaBr +  2H2SO  →    2NaHSO4  +  2HBr

    2HBr + H2SO4   →   Br + SO2  +  2H2O

                                 (Red Vapour)

    But, when conc. H2SO4 is added to an inorganic mixture containing chloride, a pungent smelling gas (HCl) is evolved. HCl, being a weak reducing agent, cannot reduce H2SO4 to SO2.

    2NaCl  +   2H2SO4    →   2NaHSO4  +   2HCl


    Q13

    Identify the substance oxidised, reduced, oxidising agent and reducing agent for each of the following reactions:

    (a) 2AgBr (s) + C6H6O2(aq) → 2Ag(s) + 2HBr (aq) + C6H4O2(aq)

     

    (b) HCHO(l) + 2[Ag (NH3)2]+(aq) + 3OH-(aq) → 2Ag(s) + HCOO-(aq) + 4NH3(aq) + 2H2O(l)

     

    (c) HCHO (l) + 2Cu2+(aq) + 5 OH-(aq) → Cu2O(s) + HCOO-(aq) + 3H2O(l)

     

    (d) N2H4(l) + 2H2O2(l) → N2(g) + 4H2O(l)

     

    (e) Pb(s) + PbO2(s) + 2H2SO4(aq) → 2PbSO4(s) + 2H2O(l)

    Ans:

    a) AgBr is reduced to Ag

    C6H6O2 is oxidised to C6H4O2

    AgBr is an oxidising agent

    C6H6O2 is a reducing agent.

     

    b) [Ag(NH3)+2 reduced to Ag+

    HCHO is oxidised to HCOO-

    [Ag(NH3)+2 is an oxidising agent.

     

    c) HCHO is oxidised to HCOO-

    Cu2+ is reduced to Cu(I) state.

     

    d) N2H4 is reduced to H2O

    H2O2 is an oxidising agent

    N2H4 is reducing agent.

     

    e) Pb has been oxidised to PbSO4

    PbO2 is reduced to PbSO4

    PbO2 is an oxidising agent

    Pb is a reducing agent.


    Q14

    Consider the reactions :

    2 S2O2– 3 (aq) + I2(s) → S4 O2– 6(aq) + 2I (aq)

    S2O2– 3(aq) + 2Br2(l) + 5 H2O(l) → 2SO2–4(aq) + 4Br(aq) + 10H+(aq)

    Why does the same reductant, thiosulphate react differently with iodine and bromine ?

    Ans:

    The average oxidation number (O.N.) of S in  S2O2– 3 is +2. Being a stronger oxidising agent than I2, Br2 oxidises S2O2– 3 to SO2-4 , in which the O.N. of S is +6. However, I2 is a weak oxidising agent.

    Therefore, it oxidises S2O2– 3 to S4O2– 6 , in which the average O.N. of S is only +2.5. As a result, S2O2– 3 reacts differently with iodine and bromine.


    Q15

    Justify giving reactions that among halogens, fluorine is the best oxidant and among hydrohalic compounds, hydroiodic acid is the best reductant.

    Ans:

    F2 can oxidize Cl- to Cl2, Br-  to Br2, and I- to I2 as:

    F2(aq) + S2cl-(s)  →  2F-(aq)  +  Cl (g)

    F2(aq) + 2Br-(aq)  →  2F-(aq)  + Br 2(l)

    F2(aq) + 2l-(aq)  →  2F-(aq)  + 12(s)

     

    On the other hand, Cl2, Br2, and I2 cannot oxidize F- to F2. The oxidizing power of halogens increases in the order of I2 < Br2 < Cl2 < F2. Hence, fluorine is the best oxidant among halogens.

    HI and HBr can reduce H2SO4 to SO2, but HCl and HF cannot. Therefore, HI and HBr are stronger reductants than HCl and HF.

    2HI + H2SO4  →  l2  +  SO2  +  2H2O

    2HBr + H2SO4  →  Br2 + SO2  +  2H2O

    Again, I- can reduce Cu2+ to Cu+, but Br- cannot.

    4l-(aq)  +   2Cu2+(aq)  →  Cu2l2(s)  + l2(aq)

    Hence, hydroiodic acid is the best reductant among hydrohalic compounds.

    Thus, the reducing power of hydrohalic acids increases in the order of HF < HCl < HBr < HI.


    Q16

    Why does the following reaction occur ?

    XeO4– 6(aq)  +  2F (aq) + 6H + (aq)  →   XeO3(g)  +  F2(g) + 3H2O(l)

    What conclusion about the compound Na4XeO6 (of which XeO4– 6 is a part) can be drawn from the reaction.

    Ans:

    The given reaction occurs because XeO4– 6 oxidises F- and F- reduces XeO4-6.

    In this reaction, the oxidation number (O.N.) of Xe decreases from +8 in XeO4– 6 to +6 in XeO3 and the O.N. of F increases from -1 in F- to O in F2.

    Hence, we can conclude that is a stronger oxidising agent than F-.


    Q17

    Consider the reactions:

    (a) H3PO2(aq) + 4 AgNO3(aq) + 2 H2O(l) → H3PO4(aq) + 4Ag(s) + 4HNO3(aq)

     

    (b) H3PO2(aq) + 2CuSO4(aq) + 2 H2O(l) → H3PO4(aq) + 2Cu(s) + H2SO4(aq)

     

    (c) C6H5CHO(l) + 2 [Ag (NH3)2]+(aq) + 3OH-(aq) → C6H5COO-(aq) + 2Ag(s) + 4NH3(aq) + 2 H2O(l)

     

    (d) C6H5CHO(l) + 2Cu2+(aq) + 5OH-(aq) → No change observed.

     

    What inference do you draw about the behaviour of Ag+and Cu2+from these reactions?

    Ans:

    Ag+ and Cu2+ act as oxidising agents in reactions (a) and (b) respectively.

    In reaction (c), Ag+ oxidises C6H5CHO to C6H5COO-, but in reaction (d), Cu2+ cannot oxidise C6H5CHO.

    Hence, we can say that Ag+ is a stronger oxidising agent than Cu2+.


    Q18

    Balance the following redox reactions by ion – electron method :

    (a) MnO4 (aq) + I(aq) → MnO2 (s) + I2(s) (in basic medium)

    (b) MnO4 (aq) + SO2 (g) → Mn2+ (aq) + HSO4 (aq) (in acidic solution)

    (c) H2O2 (aq) + Fe 2+ (aq) → Fe3+ (aq) + H2O (l) (in acidic solution)

    (d) Cr2O7 2– + SO2(g) → Cr3+ (aq) + SO42– (aq) (in acidic solution)

    Ans:

    Step 1:

    The two half reactions involved in the given reaction are:

                                               -1              0

    Oxidation half reaction:  l (aq)  →  l2(s)

                                                      +7                         +4

    Reduction half reaction: Mn O-4(aq)   →  MnO2(aq)

     

    Step 2:

    Balancing I in the oxidation half reaction, we have:

    2l-(aq)  →  l2(s)

    Now, to balance the charge, we add 2 e- to the RHS of the reaction.

    2l-(aq) →  l2(s) + 2e-

     

    Step  3 :

    In the reduction half reaction, the oxidation state of Mn has reduced from +7 to +4. Thus, 3 electrons are added to the LHS of the reaction.

    MnO-4(aq) + 3e-   →MnO2(aq)

    Now, to balance the charge, we add 4 OH- ions to the RHS of the reaction as the reaction is taking place in a basic medium.

    MnO-4(aq) + 3e-   →MnO2(aq)  +  4OH-

     

    Step  4:

    In this equation, there are 6 O atoms on the RHS and 4 O atoms on the LHS. Therefore, two water molecules are added to the LHS.

    MnO-4(aq) + 2H2O + 3e-   →MnO2(aq)  +  4OH-

    Step 5:

    Equalising the number of electrons by multiplying the oxidation half reaction by 3 and the reduction half reaction by 2, we have:

    6l-(aq) →  3l2(s) + 2e-

    2MnO-4(aq) + 4H2O + 6e-   →  2MnO2(s)  +  8OH-(aq)

    Step 6:

    Adding the two half reactions, we have the net balanced redox reaction as:

    6l-(aq)  +   2MnO-4(aq) + 4H2O(l)  →   3l2(s) + 2MnO2(s)  +  8OH-(aq)

     

    (b) Following the steps as in part (a), we have the oxidation half reaction as:

    SO2(g)  +  2H2O(l)     →  HSO-4(aq) + 3H+(aq)  + 2e-(aq)

    And the reduction half reaction as:

    MnO-4(aq)  +  8H+(aq)  +  5e-   →  Mn2+(aq)  +  4H2O(l)

    Multiplying the oxidation half reaction by 5 and the reduction half reaction by 2, and then by adding them, we have the net balanced redox reaction as:

    2MnO-4(aq)  + 5SO2(g) + 2H2O(l) +  H+(aq)  → Mn2+(aq)  +  HSO-4(aq)

     

    (c) Following the steps as in part (a), we have the oxidation half reaction as:

    Fe2+(aq)  →   Fe3+(aq)  +  e-

    And the reduction half reaction as:

    H2O2(aq)  +  2H+(aq) +  2e-  →  2H2O(l)

    Multiplying the oxidation half reaction by 2 and then adding it to the reduction half reaction, we have the net balanced redox reaction as:

    H2O2(aq)  +  2Fe2+(aq)  +  2H+(aq)    →   2Fe3+(aq) +  2H2O(l)

     

    (d) Following the steps as in part (a), we have the oxidation half reaction as:

    SO2(g)  +  2H2O(l)    →  SO2-4(aq)  +   4H+ (aq)   +  2e-

    And the reduction half reaction as:

    Cr2O2-7(aq) + 14H+(aq)  +  6e-   →  2Cr3+(aq)  + 3SO2-4(aq)  +  H2O(l)

    Multiplying the oxidation half reaction by 3 and then adding it to the reduction half reaction, we have the net balanced redox reaction as:

    Cr2O2-7(aq) +  3SO2(g)  +  2H+(aq)   →  2Cr3+(aq)  + 3SO2-4(aq) +  H2O(l)

     


    Q19

    Balance the following equations in basic medium by ion-electron method and oxidation number methods and identify the oxidising agent and the reducing agent.

    (a) P4(s) + OH (aq)  →  PH3(g) + HPO2 – (aq)

    (b) N2H4(l) + ClO3 – (aq)  →  NO(g)  +  Cl(g)

    (c) Cl2O7 (g)  +  H2O2(aq)  →  ClO2(aq)   +  O2(g)  +  H + (aq)

    Ans:

    (a) The O.N. (oxidation number) of P decreases from 0 in P4 to -3 in PH3 and increases from 0 in P4 to + 2 in HPO-2. Hence, P4 acts both as an oxidizing agent and a reducing agent in this reaction.

    Ion-electron method:

    The oxidation half equation is:

    P4(s) →  H2PO-(aq)

    The P atom is balanced as:

    P0 4(s)  →  4H2P+1O-(aq)

    The O.N. is balanced by adding 4 electrons as:

    P4(s)    →   4H2PO-(aq)    +   4e-

    The charge is balanced by adding 8OH- as:

    P4(s)    +  8OH - (aq)  →   4H2PO-2(aq)

    The O and H atoms are already balanced. The reduction half equation is:

    P4(s)    →    PH3(g)

    The P atom is balanced as

    P04(s)    →   4 P-3H3(g)

    The O.N. is balanced by adding 12 electrons as:

    P4(s)  +  12e-   →    4 PH3(g)

    The charge is balanced by adding 12OH- as:

    P4(s)  +  12e-   →    4 PH3(g)   +  12OH-(aq)  .....(i)

    The O and H atoms are balanced by adding 12H2O as:

    P4(s) + 12H2O(l)  +  12e-   →    4 PH3(g)   +  12OH-(aq)  -- (ii)

    By multiplying equation (i) with 3 and (ii) with 2 and then adding them, the balanced chemical equation can be obtained as:

    P4(s) + 3OH-(aq) + 3H2O  →  PH3   +   3H2PO-2(aq)

     

    (b)

    The oxidation number of N increases from -2 in N2H4 to +2 in NO and the oxidation number of Cl decreases from + 5 in CIO-3 to -1 in Cl-. Hence, in this reaction, N2H4 is the reducing agent and CIO-3 is the oxidizing agent. Ion-electron method:

    The oxidation half equation is:

    N-22 H4(l)  →  N+2 O(g)

    The N atoms are balanced as:

    N2H4(l)   →   2NO(g)

    The oxidation number is balanced by adding 8 electrons as:

    N2H4(l)   →    2NO(g)  +  8e-

    The charge is balanced by adding 8 OH-ions as:

    N2H4(l)  + 8OH-(aq)  →  2NO(g)  +  8e-

    The O atoms are balanced by adding 6H2O as:

    N2H4(l)  + 8OH-(aq)  → 2NO(g)  + 6H2O(l) +  8e- .... (i)

    The reduction half equation is:

    C+5IO-3(aq)  →  C-1l-(aq)

    The oxidation number is balanced by adding 6 electrons as:

    CIO-3(aq)  +  6e → Cl-(aq)

    The charge is balanced by adding 6OH- ions as:

    CIO-3(aq)  +  6e-  →    Cl-(aq) + 6OH-(aq) 

    The O atoms are balanced by adding 3H2O as:

    CIO-3(aq)  + 3H2O(l)  + 6e-  →  Cl-(aq) + 6OH-(aq) .... (ii)

    The balanced equation can be obtained by multiplying equation (i) with 3 and equation (ii) with 4 and then adding them as:

    3N2H4(l)  +  4CIO-3(aq)   →  6NO(g)  +  4Cl-(aq) + 6H2O(l) 

     

    Oxidation number method:

    Total decrease in oxidation number of N = 2 × 4 = 8

    Total increase in oxidation number of Cl = 1 × 6 = 6

    On multiplying N2H4 with 3 and CIO-3 with 4 to balance the increase and decrease in O.N., we get:

    3N2H4(l)  + 4CIO-3(aq)  → NO(g)  + Cl-(aq)

    The N and Cl atoms are balanced as:

    3N2H4(l)  + 4CIO-3(aq)  → 6NO(g)  + 4Cl-(aq)

    The O atoms are balanced by adding 6H2O as:

    3N2H4(l)  + 4CIO-3(aq)  → 6NO(g)  + 4Cl-(aq) + 6H2O(l)

    This is the required balanced equation.

     

    (c)

    The oxidation number of Cl decreases from + 7 in Cl2O7 to + 3 in CIO-2and the oxidation number of O increases from -1 in H2O2 to zero in O2. Hence, in this reaction, Cl2O7 is the oxidizing agent and H2O2 is the reducing agent.

    Ion-electron method:

    The oxidation half equation is:

    H2O-12(aq)  →  O02(g)

    The oxidation number is balanced by adding 2 electrons as:

    H2O2(aq)  →  O2(g)  + 2e-

    The charge is balanced by adding 2OH-ions as:

    H2O2(aq)  +  2OH-(aq)  →   O2(g)  + 2e-

    The oxygen atoms are balanced by adding 2H2O as:

    H2O2(aq)  +  2OH-(aq)  →   O2(g)  + 2H2O(l) + 2e- ... (i)

    The reduction half equation is:

    C+7l2O7(g)   →  C+3lO-2(g)

    The Cl atoms are balanced as:

    Cl2O7(g)   →   2ClO-2(g)

    The oxidation number is balanced by adding 8 electrons as:

    Cl2O7(g)    +   8e- →   2ClO-2(g)

    The charge is balanced by adding 6OH- as:

    Cl2O7(g)    +   8e- →   2ClO-2(g)  +  6OH- (aq)

    The oxygen atoms are balanced by adding 3H2O as:

    Cl2O7(g)   +    3H2O(l) +  8e- →   2ClO-2(g)  +  6OH- (aq)  .... (ii)

    The balanced equation can be obtained by multiplying equation (i) with 4 and adding equation (ii) to it as:

    Cl2O7(g)   +    4H2O2(aq) + 2OH- (aq)    →  2ClO-2(aq)  +  4O2(g)  + 5H2O(l)

     

    Oxidation number method:

    Total decrease in oxidation number of Cl2O7 = 4 × 2 = 8

    Total increase in oxidation number of H2O2 = 2 × 1 = 2

    By multiplying H2O2 and O2 with 4 to balance the increase and decrease in the oxidation number, we get:

    Cl2O7(g)   +    4H2O2(aq)  →   CIO-2(aq)  +  4O2(g)

    The Cl atoms are balanced as:

    Cl2O7(g)   +    4H2O2(aq)  →   2CIO-2(aq)  +  4O2(g)

    The O atoms are balanced by adding 3H2O as:

    Cl2O7(g)   +    4H2O2(aq)  →   2CIO-2(aq)  +  4O2(g)  +  3H2O(l)

    The H atoms are balanced by adding 2OH- and 2H2O as:

    Cl2O7(g)   +   4H2O2(aq)   +   2OH-(aq) →   2CIO-2(aq)  +  4O2(g)  +  5H2O(l)

    This is the required balanced equation.


    Q20

    What sorts of informations can you draw from the following reaction ?

    (CN)2(g)  +  2OH-(aq)  →  CN-(aq) + CNO-(aq) + H2O(l)

    Ans:

    The oxidation numbers of carbon in (CN)2, CN- and CNO- are +3, +2 and +4 respectively.

    These are obtained as shown below:

    Let the oxidation number of C be x.

    (CN)2

    2(x - 3) = 0

    ∴x = 3

    CN-

    x - 3 = -1

    ∴x = 2

    CNO-

    x -3 - 2 =-1

    ∴x = 4

    The oxidation number of carbon in the various species is:

    (C+3N)2(g)  +  2OH-(aq)  →  C+2N-(aq) + C+4NO-(aq) + H2O(l)

    It can be easily observed that the same compound is being reduced and oxidised simultaneously in the given equation. Reactions in which the same compound is reduced and oxidised is known as disproportionation reactions. Thus, it can be said that the alkaline decomposition of cyanogen is an example of disproportionation reaction.


    Q21

    The Mn3+ ion is unstable in solution and undergoes disproportionation to give Mn2+, MnO2, and H+ ion. Write a balanced ionic equation for the reaction.

    Ans:

    The given reaction can be represented as:

    Mn3+(aq)  →  Mn2+(aq)  + MnO2(s)  +  H+(aq)

    The oxidation half equation is:

    M+3 n3+(aq)  →  M+4 nO2(s)

    The oxidation number is balanced by adding one electron as:

    Mn3+(aq)     →  M nO2(s) + e-

    The charge is balanced by adding 4H+ ions as:

    Mn3+(aq)     →  M nO2(s)  +  4H+(aq)   +    e-

    The O atoms and H+ ions are balanced by adding 2H2O molecules as:

    Mn3+(aq)   +  2H2O(l)   →    MnO2(s)  + 4H+(aq)   +    e..... (i)

    The reduction half equation is:

    Mn3+(aq)     →   Mn2+(aq)  

    The oxidation number is balanced by adding one electron as:

    Mn3+(aq)   +    e-    →   Mn2+(aq)   ....... (ii)

    The balanced chemical equation can be obtained by adding equation (i) and (ii) as:

    2Mn3+(aq)   +  2H2O(l)   →   MnO2(s)  + 2Mn2+(aq) + 4H+(aq)


    Q22

    Consider the elements:

    Cs, Ne, I and F

    (a) Identify the element that exhibits only negative oxidation state.

    (b) Identify the element that exhibits only postive oxidation state.

    (c) Identify the element that exhibits both positive and negative oxidation states.

    (d) Identify the element which exhibits neither the negative nor does the positive oxidation state.

    Ans:

    (a) F exhibits only negative oxidation state of -1.

    (b) Cs exhibits positive oxidation state of +1.

    (c) I exhibits both positive and negative oxidation states. It exhibits oxidation states of - 1, + 1, + 3, + 5, and + 7.

    (d) The oxidation state of Ne is zero. It exhibits neither negative nor positive oxidation states.


    Q23

    Chlorine is used to purify drinking water. Excess of chlorine is harmful. The excess of chlorine is removed by treating with sulphur dioxide. Present a balanced equation for this redox change taking place in water.

    Ans:

    The given redox reaction can be represented as:

    Cl2(s)  + SO2(aq)  +  H2O(l)   →  Cl-(aq)  +   SO2-4(aq)

    The oxidation half reaction is:

    S+4 O2(aq)  →   S+6O2-4(aq)

    The oxidation number is balanced by adding two electrons as:

    SO2(aq)  →   SO2-4(aq)  + 2e-

    The charge is balanced by adding 4H+ ions as:

    SO2(aq)   →   SO2-4(aq)  +   4H+(aq) + 2e-

    The O atoms and H+ ions are balanced by adding 2H2O molecules as:

    SO2(aq)  + 2 H2O(l)   → SO2-4(aq)  + 4H+(aq) + 2e-      ..........(i)

    The reduction half reaction is: The chlorine atoms are balanced as:

    Cl2(s)    →  Cl-(aq) 

    The oxidation number is balanced by adding electrons

    Cl2(s)   + 2e-    →   2 Cl-(aq) ............(ii)

    The balanced chemical equation can be obtained by adding equation (i) and (ii) as:

    Cl2(s)   + SO2(aq)  + 2 H2O(l)   →   2 Cl-(aq)  + SO2-4(aq)  +  4H+(aq)


    Q24

    Refer to the periodic table given in your book and now answer the following questions:

    (a) Select the possible non metals that can show disproportionation reaction.

    (b) Select three metals that can show disproportionation reaction.

    Ans:

    In disproportionation reactions, one of the reacting substances always contains an element that can exist in at least three oxidation states.

    (a) P, Cl, and S can show disproportionation reactions as these elements can exist in three or more oxidation states.

    (b) Mn, Cu, and Ga can show disproportionation reactions as these elements can exist in three or more oxidation states.


    Q25

    In Ostwald's process for the manufacture of nitric acid, the first step involves the oxidation of ammonia gas by oxygen gas to give nitric oxide gas and steam. What is the maximum weight of nitric oxide that can be obtained starting only with 10.00 g. of ammonia and 20.00 g of oxygen?

    Ans:

    The balanced chemical equation for the given reaction is given as:

    4NH3(g)   + 5O2(g)  →  4NO(g) + 6 H2O(g)

    4x17g     2x32g              4x30g        6x18g

    = 68g        =160g             =120g       108g

    Thus, 68 g of NH3 reacts with 160 g of O2.

    Therefore, 10g of NH3 reacts with 160x10 / 68g of O2, or 23.53 g of O2.

    But the available amount of O2 is 20 g.

    Therefore, O2 is the limiting reagent (we have considered the amount of O2 to calculate the weight of nitric oxide obtained in the reaction).

    Now, 160 g of O2 gives 120g of NO.

    Therefore, 20 g of O2 gives120x20 / 160 g of N, or 15 g of NO.

    Hence, a maximum of 15 g of nitric oxide can be obtained.


    Q26

    Using the standard electrode potentials given in the Table 8.1, predict if the reaction between the following is feasible:

    (a) Fe3+(aq) and I-(aq)

    (b) Ag+(aq) and Cu(s)

    (c) Fe3+ (aq) and Cu(s)

    (d) Ag(s) and Fe3+(aq)

    (e) Br2(aq) and Fe2+(aq)

    Ans:

    (a) The possible reaction between is given by, Fe3+(aq) +  I-(aq)

    E° for the overall reaction is positive. Thus, the reaction between Fe3+(aq) and  I-(aq) is feasible.

     

    (b) The possible reaction between is given by, Ag+(aq) + Cu(s)

    E° positive for the overall reaction is positive. Hence, the reaction between Ag+(aq) and Cu(s) is feasible.

     

    (c) The possible reaction between and is given by, Fe3+ (aq) +  Cu(s)

    E° positive for the overall reaction is positive. Hence, the reaction between Fe3+ (aq)  and Cu(s) is feasible.

     

    (d) The possible reaction betweenand is given by, Ag(s) + Fe3+(aq)

    Here,E° for the overall reaction is negative. Hence, the reaction between Ag(s) and Fe3+(aq) is not feasible.

     

    (e) The possible reaction betweenand is given by, Br2(aq)  +  Fe2+(aq)

    Here, E° for the overall reaction is positive. Hence, the reaction between Br2(aq) and Fe2+(aq) is feasible.


    Q27

    Predict the products of electrolysis in each of the following:

    (i) An aqueous solution of AgNO3 with silver electrodes

    (ii) An aqueous solution AgNO3 with platinum electrodes

    (iii) A dilute solution of H2SO4 with platinum electrodes

    (iv) An aqueous solution of CuCl2 with platinum electrodes.

    Ans:

    (i) AgNO3 ionizes in aqueous solutions to form Ag+ and NO-3 ions.

    On electrolysis, either Ag+ ions or H2O molecules can be reduced at the cathode. But the reduction potential of Ag+ ions is higher than that of H2O.

    Ag+(aq)  +  e-  →  Ag(s)  ;  E° = +0.80V

    2H2O(l)  +  2e- →  H2(g) + 2OH-(aq)  ; E° = -0.83V

    Hence, Ag+ ions are reduced at the cathode. Similarly, Ag metal or H2O molecules can be oxidized at the anode. But the oxidation potential of Ag is higher than that of H2O molecules.

    Ag(s) →  Ag+(aq)  +  e-  ;   E° = -0.80V

    2H2O(l)  →  O2(g) + 4 H+(aq) + 4e-   ; E° = -1.23V

    Therefore, Ag metal gets oxidized at the anode.

     

    (ii) Pt cannot be oxidized easily. Hence, at the anode, oxidation of water occurs to liberate O2. At the cathode, Ag+ ions are reduced and get deposited.

     

    (iii) H2SO4 ionizes in aqueous solutions to give H+ and SO2-4 ions.

    H2SO4(aq)  →  2H+(aq) + SO2-4(aq)

    On electrolysis, either of H+ ions or H2O molecules can get reduced at the cathode. But the reduction potential of H+ ions is higher than that of H2O molecules.

    2H+(aq) + 2e- →  H2(g) ; E° = 0.0V

    2H2O(aq) + 2e- →  H2g + 2OH-(aq)  ; E° = -0.83V

    Hence, at the cathode, H+ ions are reduced to liberate H2 gas.

    On the other hand, at the anode, either of SO2-4 ions or H2O molecules can get oxidized. But the oxidation of SO2-4 involves breaking of more bonds than that of H2O molecules.Hence, ions have a lower oxidation potential than H2O. Thus, H2O is oxidized at the anode to liberate O2 molecules.

     

    (iv) In aqueous solutions, CuCl2 ionizes to give Cu2+ and Cl- ions as:

    CuCl2(aq)  →  Cu2+(aq) + 2Cl-(aq)

    On electrolysis, either of Cu2+ ions or H2O molecules can get reduced at the cathode. But the reduction potential of Cu2+ is more than that of H2O molecules.

    Cu2+(aq) + 2e-  →  Cu(aq)  ;  E° = +0.34V

    H2O(l)  + 2e-  →  H2(g) + 2OH-  ;  E° = -0.83V

    Hence, Cu2+ ions are reduced at the cathode and get deposited.

    Similarly, at the anode, either of Cl- or H2O is oxidized. The oxidation potential of H2O is higher than that of Cl-.

    2Cl-(aq) →  Cl2(g)  + 2e-  ;  E° = -1.36V

    2H2O(l)  →  O2(g) + 4H+(aq) + 4e-   ;  E° =  -1.23V

    But oxidation of H2O molecules occurs at a lower electrode potential than that of Cl- ions because of over-voltage (extra voltage required to liberate gas). As a result, Cl- ions are oxidized at the anode to liberate Cl2 gas.


    Q28

    Arrange the following metals in the order in which they displace each other from the solution of their salts.

    Al, Cu, Fe, Mg and Zn.

    Ans:

    A metal of stronger reducing power displaces another metal of weaker reducing power from its solution of salt.

    The order of the increasing reducing power of the given metals is Cu < Fe < Zn < Al < Mg.

    Hence, we can say that Mg can displace Al from its salt solution, but Al cannot displace Mg.

    Thus, the order in which the given metals displace each other from the solution of their salts is given below:

    Mg>Al> Zn> Fe>Cu


    Q29

    Given the standard electrode potentials,

    K+/K = -2.93V, Ag+/Ag = 0.80V,

    Hg2+/ Hg = 0.79V

    Mg2+/ Mg = -2.37V. Cr3+/ Cr = -0.74V

    Arrange these metals in their increasing order of reducing power.

    Ans:

    The lower the electrode potential, the stronger is the reducing agent. Therefore, the increasing order of the reducing power of the given metals is Ag < Hg < Cr < Mg < K.


    Q30

    Depict the galvanic cell in which the reaction Zn(s) + 2Ag+(aq) → Zn2+(aq) + 2Ag(s) takes place, further show:

    (i) which of the electrode is negatively charged,

    (ii) the carriers of the current in the cell, and

    (iii) individual reaction at each electrode.

    Ans:

    The galvanic cell corresponding to the given redox reaction can be represented as:

    (i) Zn electrode is negatively charged because at this electrode, Zn oxidizes to Zn2+ and the leaving electrons accumulate on this electrode.

    (ii) Ions are the carriers of current in the cell.

    (iii) The reaction taking place at Zn electrode can be represented as:

    Zn(s)  →  Zn2+(aq)  +  2e-

    And the reaction taking place at Ag electrode can be represented as:

    Ag+(aq)  +  e-  →  Ag(s)


Popular Questions of Class 11 Chemistry

Recently Viewed Questions of Class 11 Chemistry