Conceptual Questions on Redox Reactions

1. What is the name of the reaction,2CH₃CH₂CH₂SH         ===>   CH₃CH₂CH₂—S---S---CH₃CH₂CH₃ (Weather condensation, oxidation, reduction or polymerization).

Ans.1. This is an Example of oxidation reaction since two H-atoms have been removed.

2.Write the following redox reactions using half equations.

(i)     Zn (s)    +   PbCl₂ (aq)     ===>    Pb (s)   +    ZaCl₂ (aq)

(ii)  2 Fe³⁺ (aq)  +     2I^-  (aq)       ===>     I₂  (s)    +    2Fe²⁺ (aq)

(iii) 2 Na (s)       +   Cl₂   (g)         ===>      2 NaCl (s)
(iv)   Mg (s)       +   Cl₂  (g)          ===>       MgCl₂ (s)

(v)    Zn  (s)       +    2 H+ (aq)     ====>  Zn²⁺ (aq)     +   H₂(g)
In each of the reaction given above, mention(

a) Which reactant is oxidized? To what?             

(b) Which reactant is the oxidizer?

(c) Which reactant is reduced? To What?        

 (d) Which reactant is the reducer?

Ans. 2.     (i)     Zn (s)    ====>    Zn²⁺    + 2e^-  (Oxidation)     and  

Pb²⁺ (aq)   +   2e-     ===>    Pb(s)    (Reduction)

Zn is oxidised to Zn²⁺, Pb²⁺ is reduced to Pb. and
Pb²⁺  is Oxidizer and Zn s reducer  

(ii)    2Fe³⁺    +    3e^-    ====>    3Fe²⁺    (Reduction)

2I-    ====>2    +    2e-    (Oxidation)

Fe³⁺ is reduced to Fe²⁺, I- is oxidized to I2. and I^- is the reducer and Fe³⁺ is the oxidizer.

(iii)   2Na (s)    ====>     2Na+    + 2e-    (Oxidation)

Cl₂    +     2e-      =====>    2Cl-    (Reduction)

Na is oxidized to Na⁺ and Cl₂ is reduced to Cl^-.
Na is the reducer and Cl₂ is oxidizer.

(iv) Mg    ====>     Mg²⁺    +    2e-    (Oxidation)Cl₂    +    2e-    =====>   2Cl-Mg is oxidized to Mg²⁺ and Cl₂ is reduced to cl-.
Mg is reducer and Cl₂ is oxidizer.

(v) Zn    ===>    Zn     +     2e-     (Oxidation)

2H+    +     2e-     ====> H₂    (Reduction)

Zn is oxidized to Zn²⁺ while H⁺ is reduced to H₂.

Zn is reducer and H⁺ is oxidizer.

3. What is the maximum and minimum oxidation states for Na, Mg, Al and Mn?

Ans. 3. The minimum oxidation state for all metals is 0.
The maximum oxidation state for s- and p- block element is equal to the number of electrons present in the valence shell.
Thus, maximum oxidation state of Na = +1, that of Mg is +2, that of Al is +3 and that of Sn is +4.

4. What are the maximum and minimum oxidation number of N, S and Cl?

Ans. 4.  (i) The highest oxidation number (O.N.) of N is +5 since it has five electrons in the valence shell (2s² 2p³) and its minimum O.N. is –3 since it can accept three more electrons to acquire the nearest inert gas (Ne) configuration.

(ii) Similarly, the highest O.N. of S is +6 since it has six electrons in the valence shell (3s² 3p⁴) and its minimum O.N. is –2 since it needs two more electrons to acquire the nearest inert gas (Ar) configuration.

(iii) Likewise the maximum O.N. is +7 since it has seven electron in the valence shell (3s² 3p⁵) and its minimum O.N. is –1 since it needs only on more electrons to acquire the nearest (Ar) gas configuration. 

5. Nitric acid acts only as an oxidizing agent while nitrous acid acts both as an oxidizing as well as  reducing agent. Why?

Ans. 5.   (i) HNO₃   :   Oxidation number of N in HNO₃ = +5Maximum oxidation number of N = +5Minimum oxidation number of N = –3Since the oxidation number of N in HNO₃ is maximum(+5), therefore, it can only decrease by accepting electrons. Hence, HNO₃ acts only as an oxidizing agent.

(ii)   HNO₂  :   Oxidation number of N in HNO₂= +3Maximum oxidation number of N = +5Minimum oxidation number of N = –3Thus, the oxidation number of N can either increase by losing electrons or can decrease its O.N. by accepting electrons. Therefore, HNO₂ acts both as an oxidizing as well as reducing agent.

6. How does Cu₂O act as both oxidant and reductant ? Explain with proper reactions showing the change of oxidation numbers in each example.

Ans. 6. Cu⁺ undergoes disproportionation to form Cu²⁺ and Cu          

 2Cu⁺ (aq)    ====>    Cu²⁺ (aq)    +    Cu(s)

Thus, Cu+ or Cu₂O acts both as an oxidant as well reductant.

(i)    When heated in air, Cu₂O is oxidized to CuO.

         +1  -2                 0                      +2  -2
         Cu₂O    +    1/2 O₂    ===>      2Cu O       

i.e., Cu₂O acts an a reductant and reduces O₂ to O^2-.(ii) When heated with Cu₂S, it oxidizes S^2- to SO₂ and hence Cu₂O acts as an oxidant

            +1     -2            +1    -2                      0           +4
        2  Cu₂   O     +    Cu₂   S    ====>    6Cu    +    S O₂

7. Can the reaction, (Cr₂O₇)^2-     +   H₂O    ⇌     2 (CrO₄)^2-    +    2 H⁺ be regarded as a redox reaction?

Ans. 7.  Oxidation number of Cr in (Cr₂O₇)^2-  =  +6 Oxidation number of Cr in in (CrO₄)^2-  =  +6Since during this reaction, the oxidation number of Cr has neither decreased nor increased, therefore, the above reaction cannot be regarded as a redox reaction.

8. Find out the oxidation numbers of 

(i) S atoms in Na₂S₂O₃ and 

(ii) Cl atoms in bleaching powder, CaOCl₂.

Ans. 8. Oxidation numbers of S atoms in Na₂S₂O₃                                                                                                                                  x        2       -2
(a) By Convention method : Na₂     S₂     O₃     or 

 2 × (+1) + 2x + 3 ×(-2) = 0 or x = +2 (wrong)

But this is wrong because both the Sulphur atoms cannot be in the same oxidation state as is evident from the fact that when 

Na₂S₂O₃ is treated with dil. H₂SO4, one S atom gets precipitated while the other gets converted into SO₂. The Oxidation numbers of these two S atoms can, however, be determined by the chemical bonding method.                                                                                                                                                                                                                                                                                                                                                                                                                                               S                                                                                                                                                                  ↑ 

(b) By chemical bonding method, The structure of Na₂S₂O₃ is  Na^{+  -}O—S—O^- Na⁺                                                                                                         ||                                                                                                                                                                                                                                                                O

Since there is a coordinate bond between the two S atoms, therefore, the acceptor S atom has an O.N. of –2.

 The O.N. of the other S atom can be calculated as follows:

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

or   +2 – 6 + x –2 = 0 or x = +6

Thus, the two S atoms in Na₂S₂O₃ have oxidation numbers of –2 and +6.

(ii)  Oxidation number of chlorine in bleaching powder, CaOCl₂
Average O.N. of Cl in CaOCl₂ is     Ca    O    Cl₂    or   

2x + 2 –2 = 0 or x = 0(a) by stoichiometry. The composition of bleaching powder is Ca²⁺ (OCl^-)Cl^-.

Here, O.N. of Cl in OCl^- is +1 while that in Cl^- is –1. and the average of two oxidation numbers

= 1 × (+1) + 1 × (-1) = 0.

9. Find out the Oxidation states of two types of Fe atoms in Fe₄[Fe(CN)₆]³ and rewrite the formula in stock notation form. 

Ans. 9. From our knowledge of coordinate compounds, we know that the species Fe₄ which lies outside the complex ion, i.e., [Fe(CN)₆]³ is the +ve part while the complex ion itself is the –ve part. 

In other words, the +ve charge on 4 Fe atoms outside the coordination sphere is balanced by the –ve charge on the complex ion. Since Fe has two oxidation states, i.e., +2 and +3 and oxidation number of CN- = 1, 

therefore, Total –ve charge on the complex ion, 

[Fe(CN)₆]³ = 3[(+2 × 1) + 6 (-1) ] = –12

Total –ve charge on 4 Fe atoms = 4 × (+3) = + 12

Fe in the complex ion has an O.N. of +2 while the Fe atoms outside the coordination sphere have an O.N. of +3.Thus, the stock notation for Fe₄[Fe(CN)₆]³ is Fe₄^III [Fe^II(CN)₆]³.