Redox Reactions Assignment – 1 with solution

 
1. Define oxidation and reduction in terms of electrons.
2. What is a redox reaction ? Give one example.
3. Show that the formation of sodium chloride from gaseous sodium and gaseous chlorine is a  redox reaction.
4. Define oxidizing and reducing agents in terms of electron.
5. What is the oxidation number of
=> C in CH₂O
=> Pt in [Pt(C₂H₂)Cl₃]^-
 
6. What is oxidation state of
=> Ni in Ni(CO)₄?
 
7. What is oxidation number of N in HNO₄?
 
8. Define oxidation and reduction in terms of oxidation number.
 
9. When magnesium ribbon is burnt in air, two products are formed, magnesium oxide and magnesium nitride. Point out the oxidizing and reducing agents.
 
10. In the reaction, MnO₂ + 4 HCl    =>  MnCl₂ + Cl₂ + 2 H₂O which species is oxidized?
 
11. Define disproportionation reaction? Give One Example.

Answers Of Redox Reactions Assignment – 1 :
 
1. Oxidation involves loss and reduction involves gain of electrons.
 
2. Oxidation-Reduction reaction are redox reaction.     Mg   +   F₂     =>     Mg²⁺ (F^-)₂.
 
3. Na(g)    =>    Na⁺(g)   +   e^- ]    ×    2                      ….Oxidation
   Cl₂   +   2e^-  =>    2Cl^-(g)                                      ….Reduction
   2Na    +    Cl2(g)      =>     2Na   +   Cl^-(s)               ….overall redox reaction

 
4. Oxidizing agents are elctron acceptors while reducing agents are elctron donors. For example,
Oxidizing agent:    MnO^4- + 8H⁺ + 5e^- => Mn²⁺ + 4H₂O
Permanganate ion
Reducing agent:    Fe²⁺     =>     Fe³⁺    +     e^-

 
5. (i)     x + 2(+1) + 1(-2) = 0,   
             x=0
   (ii)      x + 0 + 3(-1) = -1,
              x = +2
Oxidation Number of C₂H₄ (neutral) is zero. 
 
6. Zero. Oxidation state of metal in metal carbonyls is zero.
 
7. By Convention method, O.N. of N is :
 +1 + 4(-2) + x = 0
or      x=0     or      x= +7
This is impossible because N cannot have an O.N. of more than +5 since it has only 5 electrons in the valence shell. Therefore, O.N. of N in HNO₄ (peroxy-nitric acid) must be calculated by chemical bonding method. The structure of HNO₄ is O
                            O
                             ||
            H--O--O--N -->O
Since it contains a peroxide linkage, therefore, each of the two oxygen atoms of the peroxide bond has an O.N. of -1.
Further, since O is more electronegative than N, therefore, each of the reamaining two oxygen atoms has an O.N. of -2. Thus, if x is the O.N. of N, then, +1-1-1+x-2-2=0 or x=+5.
 
8. Oxidation involves increase in O.N. while reduction involves decrease in O.N.
  Sn²⁺     +      2 Hg²⁺    =>     Sn⁴⁺     +    (Hg₂⁾²⁺
  Here, Sn²⁺ gets oxidized to Sn⁴⁺ because O.N. increases from +2 to +4 while Hg²⁺ gets reduced to (Hg₂)²⁺  because its O.N. decreases from +2 to +1.
 
9.       Reducing Agent                 Oxidizing Agents

              0                                                                                                       +2        -2           
         2 Mg(s)                     +              O2(g)                ====>                      2 Mg     O  (s)
                                                                                                                    Mag. oxide

             0                                                                                                        +2         -3  
        3Mg(s)                       +               N₂(g)                ====>                        Mg₃     N₂ (s)
                                                                                                                     Mag. nitride

10. HCl is oxidized to Cl₂ since the O.N. increases from –1 in HCl to 0 in Cl₂.

11. A reaction in which the same species is both oxidized as well as reduced is called a “disproportionation    reaction". e.g.
 P is reduced from 0 to -3
0                                                              -3                                            +1               
P₄    +    NaOH   +    3H₂O    ====>    PH₃     +      3 NaH₂    PO₂
        
P is oxidized from 0 to +1.