Redox Reactions - Class 11 Engineering Chemistry - Extra Questions

Solution:

$$ Cl_{2}+2KOH\rightarrow KOCl+KCl+H_{2}O $$

Balanced Chemical Reaction:

Balanced Chemical Reaction:

Balanced Chemical Reaction:

Balanced Reaction:

Balanced Chemical Reaction:

Balanced Chemical Reaction:

Balanced Chemical Reaction:

Balanced Chemical Reaction:

The oxidation number of each sulphur atom in $$ \mathrm{Na}_{2} \mathrm{S}_{4} \mathrm{O}_{6} $$ can find from its structure.

Suppose oxidation number of $$S$$ in $$H_2S_2O_7=x$$
$$2(+1)+2x+7(-2)=0$$
$$2+2x-14=0$$
$$2x-12=0$$
$$2x=12$$
$$x=\dfrac {12}{2}=6$$
$$\therefore $$ Oxidation number of $$S$$ in $$H_2S_2O_7=+6$$

Suppose oxidation number of $$Fe$$ in $$Fe(CO)_5 =x$$
$$x+5\times0=0$$
$$x-0=0$$
$$x=0$$
$$\therefore $$ Oxidation number of $$Fe\ in\  Fe(CO)_5=0$$

Suppose oxidation number of $$P$$ in $$H_3PO_3=x$$
$$3(+1)+x+3(-2)=0$$
$$3+x-6=0$$
$$x-3=0$$
$$x=+3$$
$$\therefore $$ Oxidation number of $$P$$ in $$H_3PO_3=+3$$

Suppose oxidation number of $$Fe$$ in $$FeSO_4 =X$$
$$x+1(-2)=0$$
$$x-2=0$$
$$x=+2$$
$$\therefore $$ Oxidation number of $$Fe\ FeSO_4=+2$$

Suppose oxidation number of $$P$$ in $$NaH_2PO_2=x$$
$$1(+1)+2(+1)+x+2(-2)=0$$
$$1+2+x-4=0$$
$$x-1=0$$
$$x=+1$$
$$\therefore $$ Oxidation number of $$P$$ in $$NaH_2PO_2=+1$$

Compound	Oxidation state of $$Fe$$	Symbol of $$Fe$$ ions
$$FeCl_2$$	$$+2$$	$$Fe^{2+}$$
$$FeCl_3$$	$$+3$$	$$Fe^{3+}$$

Suppose oxidation number of $$C$$ in $$12x+22(+1)+11(-2)=0$$
$$12x+22-22=0$$
$$12x+0=0$$
$$12x=0$$
$$x=0$$
$$\therefore $$ Oxidation number of $$C$$ in $$C_{12}H_{22}O_{11}=0$$

$${ H }_{ 2 }{ SO }_{ 5 }+{ H }_{ 2 }O\longrightarrow {2 H }_{ 2 }{ SO }_{ 4 }+{ H }_{ 2 }{ O }_{ 2 }$$
$$H_2SO_5$$ consists of peroxide linkage. Each oxygen atom in peroxide has oxidation number of $$-1$$.
Hence, the oxidation state of $$H_2SO_5$$ is $$+6$$.

The oxidation number of phosphorus in $$PO^{3-}_4, P_4O_{10}$$ and $$P_2O_7^{4-}$$ is the same i.e. $$5$$.
The structures are shown below:

Solution:

The oxidation state of chromium in the final product formed in the reaction between KI and acidified potassium dichromate solution is $$+3$$. 

$$\underset { \uparrow  }{ { N }_{ 2 }{ H }_{ 4 } } \longrightarrow \underset { \uparrow  }{ { N }_{ 2 } } (---)+10{ e }^{ - }$$
In $$N_2H_4;  2N = -4$$ oxidation and it is loosing 10 electron so new N have oxidation of $$=(-4+10)/2 = +3$$

Fluorine is the most electronegative element in the periodic table. It exhibits only negative oxidation state of -1.

Balanced Chemical Reaction:

Let $$x$$ be the oxidation number of $$C$$ in $$CN^-$$ ion. The oxidation number of N is $$-3$$.
The sum of the oxidation numbers is equal to the charge on the ion.
Hence, $$x+(-3)=-1$$
or, $$x=+2$$.
Thus, the oxidation number of $$C$$ in cyanide ion is $$+2$$.

The balanced chemical equation for the alkaline oxidative fusion of $$MnO_2$$ is as given below:

When $$2Mn_2O_7$$ is strongly heated, $$MnO_2 $$ is formed. The reaction is given below:
$$2Mn_2O_7\overset{\Delta }{\rightarrow}4 MnO_2 + 3O_2$$.
Let $$x$$ be the oxidation number of $$Mn$$ in $$MnO_2$$. 
Hence, $$x+2(-2)=0$$   (equated to zero for a neutral compound)
or, $$x=4$$.
Hence, the oxidation number of $$Mn$$ in $$MnO_2 $$ is $$+4$$.

The change in oxidation number of the underlined element in the reaction given below is 0.
$$NaH\underline { C } O_{ 3 }+HCl\rightarrow NaCl+H_{ 2 }O+CO_{ 2 }$$ The oxidation number of C in both $$NaH  C  O_{ 3 }$$ and $$CO_2$$ is +4.

$${ \left( { NH }_{ 3 } \right)  }_{ 3 }Cr{ O }_{ 4 }$$
Let us take oxidation number of $$O=x$$
Then, $$0+4-4x$$ ; $$x=-1$$
Thus, oxygen atoms are in the form of peroxy linkage.
The compound can be written as $$({{NH }_{ 3 }})_{ 3 }Cr{( { O }_{ 2 }})_{ 2 }$$.

(a) $$\displaystyle  H_2SO_5$$ by conventional method.
Let x be the oxidation number of S
$$\displaystyle 2(+1) +x+5(-2)=0 $$
$$\displaystyle x=+8 $$
+8 Oxidation state of S is not possible as S cannot have oxidation number more than 6. The fallacy is overcomed if we calculate the oxidation number from its structure $$\displaystyle HO-S(O_2)-O-O-H $$.
$$\displaystyle -1+X+2(-2)+2(-1)+1=0 $$
$$\displaystyle x=+6 $$

(b) Dichromate ion
Let x be the oxidation number of Cr in dichromate ion
$$\displaystyle 2x+7(-2) =-2 $$
$$\displaystyle  x = +6$$
Hence the oxidation number of Cr in dichromate ion is +6. This is correct and there is no fallacy.

(c) Nitrate ion, by conventional method
Let x be the oxidation number of N in nitrate ion.
$$\displaystyle  x+3(-2) =-1$$
From the structure $$\displaystyle  O^--N^+(O)-O^-$$
$$\displaystyle x+1(-1)+1(-2)+1(-2)=0 $$
$$\displaystyle  x = +5$$
Thus there is no fallacy.

A combination reaction is a reaction where two or more elements or compounds combine to form a single compound (product). Such reactions may be represented by equations of the following form: X + Y → XY.   $$CaO$$ combines with water in to form copper hydroxide hence it is a combination reaction.

Redox Reactions: Redox reactions are a chemical reaction is which oxidation and reduction take place simultaneously.

Let the oxidation number of $$S$$ be $$x$$.
We know that, Ox. no. of $$O$$$$=-2$$
So, Ox. no. S $$+$$ $$4$$(Ox. no. $$O$$)$$=-2$$
or $$x + 4(-2)=-2$$
or $$x - 8 =-2$$
or $$x=+8-2=+6$$
The oxidation number of $$S$$ in $$SO^{2-}_4$$ ion is $$+6$$.

Let the Ox. no. of $$Cr$$ in $$K_2Cr_2O_7$$ be x.
We know that, Ox. no. of $$K=+1$$
Ox. no. of O$$=-2$$
So, $$2(Ox. no. K)+2(Ox. no. Cr)+7(Ox. no. O)=0$$
$$2(+1) + 2(x) + 7(-2)=0$$
or $$+2 + 2x - 14=0$$
or $$2x=+14-2=+12$$
or $$x=+\displaystyle\frac{12}{2}=+6$$
Hence, oxidation number of $$Cr$$ in $$K_2Cr_2O_7$$ is $$+6$$.

$$HI(-1), I_2(0), ICl(+1), HIO_4(+7)$$.

In $$KIO_3$$ $$K$$ is in +1 $$O$$ is in -2 and the net charge on $$KIO_3$$ is zero.

Oxidation states of carbon atoms in pure ekement is zero. Hence, oxidation number of carbon in diamond is zero.

$$PO_4^{-3} \Rightarrow x-2 \times 4= -3 \Rightarrow x=+5$$

i)  Every bond $$C-C$$ does not alter the oxidation state of $$C$$.

Oxide           $$\quad O^{-2}\quad $$ Oxidation No. $$=-2$$

$$2Mn^{+3}(aq.)+2H_2O(l)\longrightarrow MnO_2(s)+Mn^{+2}(aq.)+4H^+(aq)$$

Let the oxidation number of $$N$$ be $$x$$.

$$\overset { +1 }{ H } -\overset { -3 }{ N } =\ddot { C } :$$

$${ C }_{ 3 }{ O }_{ 2 }\Rightarrow \overset { -2 }{ O } =\overset { +2 }{ \underset { \left( 1 \right)  }{ C }  } =\overset { 0 }{ \underset { \left( 2 \right)  }{ C }  } =\overset { +2 }{ \underset { \left( 3 \right)  }{ C }  } =\overset { -2 }{ O } $$

                             $$MnO_2+4HCl\longrightarrow MnCl_2+2H_2O+Cl_2$$

Solution:

Let O.N. of Se in the new compound $$ =x $$
$$\begin{array}{l}\text { Now } 12.53 \mathrm{cm}^{3} \text { of } 0.051 \mathrm{M} \mathrm{SeO}_{2} \\=12.53 \times 0.051 \\=0.64 \mathrm{millimoles} \text { of } \mathrm{SeO}_{2} \\\text { and } 25 \cdot 5 \mathrm{cm}^{3} \text { of } 0 \cdot 1 \mathrm{M} \mathrm{CrSO}_{4}=25 \cdot 5 \times 0 \cdot 1 \\=2.55 \text { millimoles of } \mathrm{CrSO}_{4}\end{array}$$
But according to balanced redox equation, $$ (4-x) $$ moles of $$ \mathrm{CrSO}_{4} $$ reduce 1 mole of $$ \mathrm{SeO}_{2} $$
$$ \therefore \quad 2 \cdot 55 $$ millimoles of $$ \mathrm{CrSO}_{4} $$ will reduce $$ \mathrm{SeO}_{2} $$
$$=\dfrac{2 \cdot 55}{(4-x)} \text { millimoles }$$
But $$ \mathrm{SeO}_{2} $$ actually reduced $$ =0 \cdot 64 $$ millimoles
Equating these two values, we have,
$$\dfrac{2 \cdot 55}{4-x}=0 \cdot 64 \text { or } x=0$$

(A) During the titration $${MnO_4}^-/H^+/Fe^{2+}$$, the solution at the final stage is colorless.
(B) During the titration $${MnO_4}^-/ conc. OH^-/Mn^{2+}$$, the solution at the final stage is green.
(C) During the titration $${MnO_4}^-/H_2O/ dil. OH^-$$, the solution at the final stage is blackish brown.
(D) During the titration $$CuSO_4/KI$$, the solution at the final stage is deep reddish brown.

$$25 ml \times 0.1 M$$ of $$Cr = 4 \times 12.5 mL \times 0.05 M$$ of $$Se$$
The oxidation state of Cr increases from +2 to +3.
Hence, the oxidation state of Se will decrease from  +4 to 0.

$$\underset { \uparrow  }{ { N }_{ 2 }{ H }_{ 4 } } \longrightarrow \underset { \uparrow  }{ { N }_{ 2 } } (X)+10{ e }^{ - }$$
O.N of two $$N$$ atoms of $${ { N }_{ 2 }{ H }_{ 4 } } = - 4$$
O.N of two $$N$$ atoms in oxidised species $$=-4+10=+6$$ as 10 electrons are lost.
Thus, oxidation state of $$N$$ in new compound $$=+3$$.

(A) The oxidation state of Mn in $$Mn{O}_{3}HS{ O }_{ 4 }$$ is +7.
(B) The oxidation state of Mn in $${ K }_{ 2 }Mn{ O }_{ 4 }$$ is +6.
(C) The oxidation state of Mn in  $$Mn{ O }_{ 2 }$$ is +4.
(D) The oxidation state of Mn in $${Mn}_{2}{O}_{3}$$ is +3.
(E) The oxidation state of Mn in $$Mn{ \left( HC{ O }_{ 3 } \right)  }_{ 2 }$$ is +2.

Nitric acid is added to oxidise ferrous ion into ferric ion as

(A) The change in the oxidation state of Cl in $${ Cl }^{ - }\longrightarrow Cl{ O }_{ 4 }^{ - }\ $$ is +8.
(B) The change in the oxidation state of Cr in $${ Cr }^{ 3+ }\longrightarrow Cr{ O }_{ 5 }$$ is +3.
(C) The change in the oxidation state of O in $${ H }_{ 2 }{ O }_{ 2 }\longrightarrow { O }_{ 2 }$$ is +2
(D) The change in the oxidation state of Cr in $$Cr{ O }_{ 2 }^{ 2+ }\longrightarrow Cr{ O }_{ 4 }^{ 2- }$$ is +3.

$$NaN_3\longrightarrow N_3\,^-\longrightarrow N\longrightarrow -1/3$$. Thus, the oxidation stat of N is $$\-dfrac{1}{3}$$.
$$N_2H_2\longrightarrow N\longrightarrow -1$$. Thus, the oxidation staet of N is $$-1$$.
$$N\,O\longrightarrow N\longrightarrow +2$$. Thus, the oxidation state of N is $$+2$$.
$$N_2O_5\longrightarrow N\longrightarrow +5$$. Thus, the oxidation state of N is $$+5$$.
Hence, the option A represents the correct answer.

The substances alongwith oxidation states of C are shown below.

Substance	Oxidation number of C
$$\displaystyle CH_2Cl_2 $$	0
$$\displaystyle  FC \equiv CF$$	+1
$$\displaystyle HC \equiv CH $$	-1
$$\displaystyle CHCl_3, CO $$	+2
$$\displaystyle CH_3Cl $$	-2
$$\displaystyle  Cl_3C-CCl_3$$	+3
$$\displaystyle  H_3C-CH_3$$	-3
$$\displaystyle  CCl_4, CO_2$$	+4
$$\displaystyle  CH_4$$	-4

The substances alongwith oxidation states of N are shown below.

Substance	Oxidation number of N
$$\displaystyle  N_2$$	0
$$\displaystyle  N_2O$$	+1
$$\displaystyle  N_2H_2$$	-1
$$\displaystyle  NO$$	+2
$$\displaystyle  N_2H_4$$	-2
$$\displaystyle  N_2O_3$$	+3
$$\displaystyle  NH_3$$	-3
$$\displaystyle  NO_2$$	+4
$$\displaystyle  N_2O_5$$	+5

Let us consider a reaction, $$aA+bB\rightarrow cC+dD$$

(A) $$3Ca^{+2}O^{-2}+2P_{2}^{+5}O_{5}^{-2}\rightarrow Ca_{3}^{+2}(PO^{+5}u^{-2})_{2}$$
No change in oxidation state of element
$$\therefore $$ Non-redox
(B) $$2Cu^{+}\rightarrow e^{-}\rightarrow Cu$$
$$Cu^{+}+e^{-}\rightarrow Cu$$
$$Cu^{+}\rightarrow Cu^{2+}+e^{-}$$
Same element goes in reduction and oxidation and form two products.
So it is redox and disproportionation reaction.
(C) $$NH_{4}NO_{2}\rightarrow N_{2}+2H_{2}O$$
$$NH_{4}^{+1}\rightarrow N^{-3}$$
$$NO_{2}^{-}\rightarrow N^{+3}$$
$$N^{0}$$ same element from different oxidation state goes to same oxidation state, by gaining or losing $$e^{-}$$.
$$\therefore $$ It is redox and comproportionation reaction.
(D) $$CH_{4}^{-1}+2O_{2}^{0}\rightarrow CO_{2}+2H_{2}O$$
$$H^{-1}\rightarrow H^{+}+2e$$
$$O^{0}+2e\rightarrow O^{-2}$$ combustion + Redox.
$$A\rightarrow R$$
$$B\rightarrow P,T$$
$$C\rightarrow Q,T$$
$$D\rightarrow S,T$$

Oxidation No. changes for Oxygen from (-1 to -2) and Iodine from (-1 to 0) 

1. $$C_2H_5OH + Cr_2O_7^{2-} + H^{+} \rightarrow Cr^{3+} +C_2H_4O +H_2O$$

a) Combination
b) Combination
c) Decomposition
d) Double displacement
e) Double displacement
f) Displacement