Class 11 Engineering Chemistry - Redox Reactions

Assign ON to atoms of only those elements which undergo ON change in the following redox reactions and then balance the equation

$KMnO_{4}+H_{2}S+ H_{2}SO_{4}\to KHSO_{4}+MnSO_{4}+S+H_{2}O$

Solution:

Balanced chemical reaction is as follows:

$3KMnO_4+3H_2S+4H_2SO_4\rightarrow S+3KHSO_4+3MnSO_4+4H_2O$

Oxidation state of

$Mn\ +7(in\ reactant)\rightarrow +2(in\ product)$ so it is oxidizing agent

Oxidation state of

$S\ -2(in\ reactant)\rightarrow 0(in\ S)\ and\ +6\ (in\ MnSO_4)$ so it is reducing agent

Assign ON to atoms of only those elements which undergo ON change in the following redox reactions and then balance the equation: $Cl_{2}+KOH \to KOCl + KCl +H_{2}O$

$Cl_{2}+2KOH\rightarrow KOCl+KCl+H_{2}O$
1561600_1738612_ans_4230dc26504f4261ae9c75ecbe19f882.PNG

Assign ON to atoms of only those elements which undergo ON change in the following redox reactions and then balance the equation

$NO_{3}^{-} +Cl^{-}+H^{+}\to NO+Cl_{2}+H_{2}O$

1735318_1738694_ans_8c17c62dd2e34cada91bcadc6c150217.jpg

Assign ON to atoms of only those elements which undergo ON change in the following redox reactions and then balance the reaction:

$Fe_{2}(SO_{4})_{3}+Fe\to FeSO_{4}$

Balanced Chemical Reaction:

$Fe_{2}(SO_{4})_{3}+Fe\to 3FeSO_{4}$

Oxidation state of Iron in

$Fe_(SO_4)_3$ changes from (+3 to +2) and that in

$Fe$ changes from (0 to +3).

Assign ON to atoms of only those elements which undergo ON change in the following redox reactions and then balance the reaction:

$Cu(NH_{3})_{4}Cl_{2}+ KCN+H_{2}O\to K_{2}Cu(CN)_{3}+NH_{3}+KCNO+NH_{4}Cl+KCl$

Balanced Chemical Reaction:

$2Cu(NH_{3})_{4}Cl_{2}+ 7KCN+H_{2}O\to 2K_{2}Cu(CN)_{3}+6NH_{3}+KCNO+2NH_{4}Cl+2KCl$

The oxidation state of Copper changes from (+2 to +1) and that of Nitrogen in KCN changes from (-3 to -1)

Assign ON to atoms of only those elements which undergo ON change in the following redox reactions and then balance the reaction:

$Ag +KCN +H_{2}O+O_{2}\to KAg(CN)_{2}+KOH$

Balanced Chemical Reaction:

$4Ag +8KCN +2H_{2}O+O_{2}\to 4KAg(CN)_{2}+4KOH$

Oxidation state of Ag changes from (0 to +1) and that of Oxygen in

$O_2$ from (0 to -2)

Assign ON to atoms of only those elements which undergo ON change in the following redox reactions and then balance the equation:

$Cr_{2}O_{7}^{2-}+NO_{2}^{-}+H^{+}\to Cr^{3+} + NO_{3}^{-}+H_{2}O$

Balanced Reaction:

$Cr_{2}O_{7}^{2-}+3NO_{2}^{-}+8H^{+}\to 2Cr^{3+} + 3NO_{3}^{-}+4H_{2}O$

Oxidation No. of Chromium changes from (+6 to +3) and that of Nitrogen changes from (+3 to +5).

Assign ON to atoms of only those elements which undergo ON change in the following redox reactions and then balance the equation:

$S_{2}O_{3}^{2-}+ Sb_{2}O_{5}+H^{+}+H_{2}O\to SbO+H_{2}SO_{3}$

Balanced Chemical Reaction:

$3S_{2}O_{3}^{2-}+ 2Sb_{2}O_{5}+6H^{+}+3H_{2}O\to 4SbO+6H_{2}SO_{3}$

Oxidation state of Suphur changes from (+2 to +4) and that of Antimony changes from (+5 to +2)

Assign ON to atoms of only those elements which undergo ON change in the following redox reactions and then balance the reaction:

$Zn+NO_{3}^{-}+H^{+}\to Zn^{2+}+NO_{2}+H_{2}O$

Balanced Chemical Reaction:

$Zn+NO_{3}^{-}+2H^{+}\to Zn^{2+}+NO_{2}+H_{2}O$

Oxidation state of Zinc changes from (0 to +2) and that of Nitrogen from (+5 to +4)

Assign ON to atoms of only those elements which undergo ON change in the following redox reactions and then balance the reaction:

$AsO_{3}^{3-}+ IO_{3}^{-} \to AsO_{4}^{3-} +I^{-}$

Balanced Chemical Reaction:

$3AsO_{3}^{3-}+ IO_{3}^{-} \to 3AsO_{4}^{3-} +I^{-}$

The oxidation state of Arsenic changes from (+3 to +5) and that of Iodine changes from (+5 to -1)

Assign ON to atoms of only those elements which undergo ON change in the following redox reactions and then balance the reaction:

$MnO_{4}^{-}+CN^{-}+H_{2}O \to MnO_{2}+CNO^{-}+OH^{-}$

Balanced Chemical Reaction:

$2MnO_{4}^{-}+3CN^{-}+H_{2}O \to 2MnO_{2}+3CNO^{-}+2OH^{-}$

The oxidation state of Molybdenum changes from (+7 to +4) and that of Nitrogen changes from (-3 to -1).

Both ${VO}_{2}^{+}$ and ${CO}^{2+}$ are known as vanadyl ion
(a) Determine the oxidation number of vanadium in each
(b) Which one of them is oxovanadium(iv) ion and which are in dioxovanadium (v) ion

1741856_1741666_ans_1d0dff73e34a4928bc80f3856bc166c4.jpg

Write the balanced chemical equation for the following reaction and identify the type of reaction :
Ethene is burnt in presence of oxygen to form carbon dioxide , water and releases heat and light.

When ethene is burnt in presence of oxygen to form carbon dioxide, water and releases heat and light. The following reaction takes place:

$C_2H_4 + 3O_2 \rightarrow 2 CO_2 + H_2O$ + heat + light

The above reaction is a type of combustion reaction.

Combustion reaction is a reaction which takes place in the presence of oxygen to form carbon dioxide.

Calculate the oxidation number of sulphur atom in $\mathrm{Na}_{2} \mathrm{SO}_{4}$ .

Let the oxidation number of sulphur is

$4x$ $.

$N a_{2} \stackrel{x}S O_{4}$

$2 \times(+1)+x+(-2) \times 4=0$

$2+x-8=0$

$x=+6$

Calculate the oxidation number of each sulphur atom in $\mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3}$ .

Let the oxidation number of sulphur atom is

$x$ .

$N a_{2} \stackrel{x}S_{2} O_{3}$

$2 \times(+1)+2 x+(-2) \times 3=0$

$2+2 x-6=0$

$2 x-4=0$

$2 x=4$

$x=+2(\text { wrong })$

By conventional method, we can't find the oxidation number of each sulphur atom.

It will be well explained by structure. One sulphur has

$+6$ O.S, whereas another has

$-2$ .

1663143_1794818_ans_8a0f79af87cf4ae9a1a7e10d9e799ef0.png

Identify the oxidising and reducing agents in
$3 H C l(a q)+H N O_{3}(a q) \rightarrow C l_{2}(g)+N O C l(g)+2 H_{2} O(l)$

The oxidation number of

$Cl$ is increased from

$-1$ to

$0$ , it indicates that

$Cl$ undergoes oxidation and acts as a reducing agent.

The oxidation number of

$N$ is decreased from

$+5$ to

$+3$ , i.e it is reduced and acts as an oxidizing agent.

Oxidizing agent

$-\mathrm{HNO}_{3}$

Reducing agent

$- HCl$

1663350_1794851_ans_416638e038e8441abeb38c94442e2c2d.png

Calculate the oxidation number of phosphorus in $P O_{4}^{3-}$ .

Let oxidation sate of phosphorus is

$x$

$\stackrel{x}{P} O_{4}^{3-}$

$x+(-2) \times 4=-3$

$x-8=-3$

$x=-3+8$

$x=+5$

Thus, O.S. of phosphorous in this ion is +5.

Calculate the oxidation number of sulphur atom in $\mathrm{Na}_{2} \mathrm{SO}_{3}$ .

Let the oxidation number of sulphur is

$x$ .

$N a_{2} \stackrel{x}S O_{3}$

$2 \times(+1)+x+(-2) \times 3=0$

$2+x-6=0$

$x=+4$

Thus, the O.S. of sulphur is +4.

Calculate the oxidation number of each sulphur atom in $\mathrm{Na}_{2} \mathrm{S}_{4} \mathrm{O}_{6}$ .

The oxidation number of each sulphur atom in

$\mathrm{Na}_{2} \mathrm{S}_{4} \mathrm{O}_{6}$ can find from its structure.
1663168_1794819_ans_d2a6e48ee4b64f9ca0e9d5dc5a09a8a4.png

Calculate the oxidation number of phosphorus in the following species.
$HPO_{3}^{2-}$

Let the oxidation number of phosphorus is

$x$ .

$HP O_{3}^{2-}$

$+1+x+(-2) \times 3=-2$

$+1+x-6=-2$

$x-5=-2$

$x=-2+5$

$X=+3$

Thus, O.S. of phosphorous is +3

Calculate oxidation number of $S$ in $H_2S_2O_7$ .

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

$H_2S_2O_7=+6$

Calculate oxidation number of $Fe$ in $Fe(CO)_5$ .

Suppose oxidation number of

$Fe$ in

$Fe(CO)_5 =x$

$x+5\times0=0$

$x-0=0$

$x=0$

$\therefore$ Oxidation number of

$Fe\ in\ Fe(CO)_5=0$

Calculate oxidation number of $P$ in $H_3PO_3$ .

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

$H_3PO_3=+3$

Calculate oxidation number of the following compounds:
$Fe$ in $FeSO_4$

Suppose oxidation number of

$Fe$ in

$FeSO_4 =X$

$x+1(-2)=0$

$x-2=0$

$x=+2$

$\therefore$ Oxidation number of

$Fe\ FeSO_4=+2$

Calculate oxidation number of the following compounds:
$P$ in $NaH_2PO_2$

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

$NaH_2PO_2=+1$

Complete the table
Compound Oxidation state of $Fe$ Symbol of $Fe$ ions
$FeCl_2$
$FeCl_3$

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

Calculate oxidation number of the following compounds:
$C$ in $C_{12}H_{22}O_{11}$

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

$C_{12}H_{22}O_{11}=0$

Oxidation number of sulphur in ${ H }_{ 2 }{ SO }_{ 5 }$ is $+6$ . If true enter 1 else 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 $P_2O_7^{4-}$ respectively, is :

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:

Assign ON to atoms of only those elements which undergo ON change in the following redox reactions and then balance the equation:

$K_{2}Cr_{2}O_{7}+ HCl \to KCl+CrCl_{3}+H_{2}O+Cl_{2}$

Solution:

Balanced chemical reaction is as follows:

$K_2Cr_2O_7+14HCl\rightarrow 2CrCl_3+3Cl_2+2KCl+7H_2O$

O.S--

$Cr=6,\ Cl={-1}\rightarrow Cr=3,\ Cl={-1},\ Cl=0,\ Cl={-1}$

The oxidation state of chromium in the final product formed in the reaction between KI and acidified potassium dichromate solution is :

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

$+3$ .

The reaction is as follows:

$Cr_2O_7^{2-} +14H^+ + 6I^- \rightarrow 2Cr^{3+}+3I_2 +7H_2O$

$1$ mole of ${ N }_{ 2 }{ H }_{ 4 }$ loses $10$ mol of electrons to form a new compound $Y$ . Assuming that all the nitrogen appear in the new compound, what is the oxidation state of nitrogen in $Y$ ? (no change in the oxidation state $H$ ).

$\underset { \uparrow }{ { N }_{ 2 }{ H }_{ 4 } } \longrightarrow \underset { \uparrow }{ { N }_{ 2 } } (---)+10{ e }^{ - }$
In

$N_2H_4; 2N = -4$ oxidation and it is loosing 10 electron so new N have oxidation of

$=(-4+10)/2 = +3$

Consider the elements $Cs,\ Ne,\ I$ and $F$ .
Identify the element that exhibits only negative oxidation state.

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

Assign ON to atoms of only those elements which undergo ON change in the following redox reactions and then balance the equation:

$N_{2}O_{4}+BrO_{3}^{-} +H_{2}O\to NO_{3}^{-}+Br^{-}+H^{+}$

Balanced Chemical Reaction:

$3N_{2}O_{4}+BrO_{3}^{-} + 3H_{2}O\to 6NO_{3}^{-}+Br^{-}+6H^{+}$

Oxidation No. of Nitrogen changes from (+4 to +5) and Bromine from (+5 to -1)

What is the oxidation state of Fe in $Fe(CN)_6$ ?

1377300_1339037_ans_b87af1af1d0d4fb5963602498b8b5b65.jpg

What is oxidation number of carbon in $CN^-$ ?
(Give the absolute value)

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

$+2$ .

Calculate the oxidation number of $Mn$ in the product obtained on an alkaline oxidative fusion of $MnO_2$ .
(Give the absolute value of an answer)

The balanced chemical equation for the alkaline oxidative fusion of

$MnO_2$ is as given below:

$2MnO_2+ 4 KOH + O_2 \rightarrow 2K_2MnO_4 + 2H_2O$

Let

$x$ be the oxidation number of Mn in

$K_2MnO_4$ .

Hence,

$2(+1)+x+4(-2)=0$ (for a neutral compound)

or,

$x=+6$

Hence, the oxidation number of

$Mn$ in the product obtained on alkaline oxidative fusion of

$MnO_2$ is

$+6$ .

Calculate the oxidation number of $Mn$ in the product formed on strongly heating $Mn_2O_7$ .
(write absolute value of an answer)

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__________.
$NaH\underline { C } O_{ 3 }+HCl\rightarrow NaCl+H_{ 2 }O+CO_{ 2 }$

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.

The compound ${ \left( { NH }_{ 3 } \right) }_{ 3 }Cr{ O }_{ 4 }$ has peroxy linkage if oxidation number of $Cr$ is $+4$ .
If true enter 1, else enter 0.

${ \left( { NH }_{ 3 } \right) }_{ 3 }Cr{ O }_{ 4 }$
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

$({{NH }_{ 3 }})_{ 3 }Cr{( { O }_{ 2 }})_{ 2 }$ .

Calculate the oxidation number of sulphur, chromium and nitrogen in $H_2SO_5\,,Cr_2O^{2-}_7$ and $NO_3^-$ . Suggest structure of these compounds. Count for the fallacy.

(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.

Which reaction is this?
$CaO + H_{2}O \rightarrow Ca(OH)_{2}$

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.

What are redox reactions? Identify the substances that are oxidised and the substances that are reduced in the following reactions?
A. $2H_2S_{(g)} + SO_{2(g)} \rightarrow 3S_{(s)} + 2H_2O_{(1)}$
B. $CuO_{(s)} + H_{2(g)}\rightarrow Cu_{(s)} +H_2O$

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

Example:

(1) During this reaction, barium sulphate loses oxygen and is hence reduced. At the same time, Carbon gains oxygen and is therefore oxidised. This is termed as redox reaction.
2. During this reaction, hydrogen gains oxygen and is hence oxidised. At the same time, sulphur loses oxygen and is therefore reduced.
3. During this reaction,

$CuO$ loses oxygen and it is therefore reduced. At the same time, hydrogen gains oxygen and is therefore oxidised.

Find the oxidation number of $S$ in ${SO_4}^{2-}$ ion?

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

$+6$ .

What is the oxidation number of $Cr$ in $K_2Cr_2O_7$ ?

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

$+6$ .

Arrange the following in the order of increasing oxidation number of iodine: $I_2, HI, HIO_4, ICl$ .

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

Find the oxidation number of I in $KIO_3$ .

$KIO_3$

$K$ is in +1

$O$ is in -2 and the net charge on

$KIO_3$ is zero.

Let the oxidation state of I be x

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

$x= 5$

So oxidation number of I in

$KIO_3$ is

$5$ .

Find the oxidation number of Xe in $BaXeO_6$ .

Oxidation state of

$Ba$ in general

$=+2$ and of

$O=-2$

Applying formula,

$\text{Sum of total oxidation state of all atoms } = \text{ Overall charge on the compound}.$

Let oxidation state of

$Xe$ in

$BaXeO_6$ be

$x$ .

$2+x+6(-2)=0,\\x=10$

But oxidation state

$10$ is not possible for

$Xe$ . In this case the oxidation state of

$Xe$ is equal to maximum possible oxidation state for

$Xe=+8$

Calculate the oxidation state of the following.
$C$ in diamond

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

Find out oxidation states of phosporus in ${PO_4}^{3-}, P_4O_{10}$ and ${P_2O_7}^{4-}$ .

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

$P_4O_{10} \Rightarrow 4x-2 \times 10=0 \Rightarrow 4x=20\Rightarrow x=5$

$P_2O_7^{4-} \Rightarrow 2x- 14=-4 \Rightarrow 2x=10 \Rightarrow x=+5$

Oxidation state of phosphorus in all given compounds is

$+5$ .

Oxidation number of Sulphur in $S_2 O_8^{-2}$ ?

1377269_706124_ans_076a5b3b0be8442bb12cd293c399612d.jpg

Find out the oxidation state of shown carbonyl carbon.

i) Every bond

$C-C$ does not alter the oxidation state of

$C$ .

ii) when

$C$ bonded to

$O$ , oxidation state increases by

$+1$ .

$\therefore$ oxidation state of

$C=(+1)\times2=+2$

1080634_773092_ans_0951973cf16c43a5b4d1e40db1a27eb9.jpg

Assign ON to atoms of only those elements which undergo ON change in the following redox reactions and then balance the equation

$Cr(OH)_{3}+IO_{3}^-+OH^{-}\to I^{-}+CrO_{4}^{2-}+ H_{2}O$

Solution: Balanced chemical reaction is as follows:

$2Cr(OH)_3+IO_3^{-}+4OH^-\rightarrow 2CrO_4^{2-}+I^-+5H_2O$

Oxidation number of chromium in reactant side of reaction is

$3$ whereas in product side it has

$6$ oxidation state. Hence, we can conclude that

$Cr(OH)_3$ is oxidized.

$I$ has oxidation

$5$ in reactant side of reaction and

$-1$ on the product side, so it has reduced in the reaction. By this we can say that

$IO_3^-.$ The oxidation number of

$I$ changes from

$+5\rightarrow -1$ . Thus, each iodine takes

$6$ electrons and is reduced.

(i) Calculate oxidation number of $Cl$ in $Cl_{2}O_{7}$ .
(ii) $2FeCl_{3} + H_{2}S \rightarrow 2FeCl_{2} + 2HCl + S$
Write the name of oxidising agent and reducing agent in above reaction:

(i) The net charge on the molecule is 0, therefore the oxidation states of each atom combined together should give a sum equal to 0.

Since

$O$ has

$-2$ oxidation state, let that be of

$Cl$ as

$x$ .

Therefore,

$2\times x+7\times (-2)=0\\ x=+7$ .

Oxidation number of

$Cl$ in

${ Cl }_{ 2 }{ O }_{ 7 }$ is

$7$ .

(ii) Calculating the oxidation state of each specie in the given reaction we get:

$Fe\ (+3)\ \rightarrow \ Fe(+2)\qquad gain\ of\ electron\\ S\ (-2)\rightarrow \ S(0)\qquad \qquad \ \ loss\ of\ electron$

Oxidising agent tends to oxidise other specie and itself get reduced (gains electron) in the process, therefore

${ Fe }{ Cl }_{ 3 }$ is the oxidising agent and similarly

${ H }_{ 2 }S$ is the reducing agent as it losses electron and gets oxidized.

What is the oxidation number of oxygen in oxide, peroxide and superoxide?

Oxide

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

$=-2$

Peroxide

$\quad O_2^{2-}\quad$ Oxidation No.

$=\cfrac {-2}{2}=-1$

Superoxide

$\quad O_2^{-1}\quad$ Oxidation No.

$=\cfrac {-1}{2}$

The $Mn^{3+}$ ion is unstable in solution and undergoes disproportional reaction to give $Mn^{2+}, MnO_2$ and $H^+$ . Write a balanced ionic equation for the reaction.

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

Balance the following redox reaction.
$MnO^{\ominus}_4(aq)+I^{\ominus}(aq)\rightarrow MnO_2(s)+I_2(s)$ .
What is the coefficient of $I^-$ ?

$\Rightarrow MnO_{4}^{-}+I^{-}\rightarrow MnO_{2}+I_{1}$

$\left \{ 3e^{-}+4H^{+}+MnO_{4}^{-}\rightarrow MnO_{2}+2H_{2}O \right \}\times 2$

$\left \{ 2I^{-}\rightarrow I_{2}+2e^{-} \right \}\times 3$

$\boxed{8H^{+}+2MnO_{4}^{-}+6I^{-}I 2MnO_{2}+4H_{2}O+3I_{2}}$

1163153_1031318_ans_008d8e20fb744f109e7fac3cd99699ab.jpeg

What is the oxidation number of nitrogen in $NH_2OH$ ?

Let the oxidation number of

$N$ be

$x$ .

The oxidation number of H and O is +1 and -2 respectively.

$\therefore$

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

$\therefore$

$x+2-2+1=0$

$\therefore$

$x=-1$

Therefore, the oxidation number of nitrogen in

${ NH }_{ 2 }OH$ is

$-1$ .

Find oxidation number $CH_{3}CH_{2}OH$ .

Oxidation number of

${ C }_{ 1 }=-1-1-1=-3$

Oxidation number of

${ C }_{ 2 }=-1-1+1=-1$

1174145_1097887_ans_60b90584120a43f9a7e3e7f9996ebf94.jpg

What is the oxidation number of C in HNC?

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

Let, the oxidation number of

$C$ be

$x$ .

$\therefore$

$x-3+1=0$

$\therefore$

$x-2=0$

$\therefore$

$x=+2$

Therefore, the oxidation number of C in HNC is +2.

Find oxidation number of $H_{2}S_{4}O_{6}$ .

Oxidation number of

${ S }_{ 1 }=+2+1+2=+5$

Oxidation number of

${ S }_{ 2 }=0$

Oxidation number of

${ S }_{ 3 }=0$

Oxidation number of

${ S }_{ 4 }=+2+1+2=+5$

Average oxidation number of

$S=\dfrac { +5+5+0+0 }{ 4 } =+\dfrac { 5 }{ 2 }$

1052333_1097888_ans_4ae43ce902cd4c258c994fbda04d2d66.jpg

Find oxidation number of $C_{3}O_{2}$ .

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

Oxidation number of

${ C }_{ 1 }=+2$

Oxidation number of

${ C }_{ 2 }=0$

Oxidation number of

${ C }_{ 3 }=+2$

Average oxidation number of

$C=\dfrac { +2+0+2 }{ 3 } =+\dfrac { 4 }{ 3 }$

Find the oxidation numbers to the underlined species in the following compounds or ions.
$K_2Cr_2O_7$
$NaHCO_3$

1348563_1143906_ans_bb4d8925978a4b2e847f694ad4912707.jpg

Identify the substance oxidized,substance reduced ,oxidizing agent and reducing agent :
$MnO_2+4HCl\rightarrow MnCl_2+2H_2O+Cl_2$

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

Oxidation state:

$+4$

$-1$

$+2$

$0$

Increase in oxidation state

$=$ Oxidation

$\Rightarrow$

As chlorine increases from

$-1\longrightarrow 0\Rightarrow$ (Oxidation of

$HCl$ )

Decrease in oxidation state

$=$ Reduction

$\Rightarrow$ As manganese decreases from

$+4\rightarrow +2\Rightarrow$ (Reduction of

$MnO_2$ )

$\Rightarrow MnO_2$ is reduced,

$HCl$ is oxidized,

$MnO_2$ is oxidizing agent,

$HCl$ is reducing agent.

Assign ON to atoms of only those elements which undergo ON change in the following redox reactions and then balance the equation:

$P+HNO_{3}\to HPO_{3}+NO+H_{2}O$

Solution:

$3P+5HNO_3\rightarrow 3HPO_3+5NO+H_2O$

Oxidation state of

$P^0\rightarrow P^{+5}\ in\ HPO_3$

Oxidation state of

$N=(+5)\ in\ HNO_3\rightarrow (+2)\ in\ NO$

Assign oxidation number to atoms of only those elements which undergo O. No. change in the following redox reactions and then balance the equation

$Cr_{2}O_{7}^{2-}+C_{2}O_{4}^{2-}+H^{+}\to Cr^{3+}+CO_{2}+H_{2}O$

1735280_1738687_ans_590200d528f14a809baf312924b2c5bd.jpg

$12 .53 \mathrm{cm}^{3}$ of $0.051 \mathrm{M} \mathrm{SeO}_{2}$ reactsexactly with $25 \cdot 5 \mathrm{cm}^{3}$ of $0 \cdot 1 \mathrm{M}$ CrSO $_{4}$ which is oxidised to $\mathrm{Cr}_{2}\left(\mathrm{SO}_{4}\right)_{3} .$ To what oxidation state is the selenium converted during the reaction?

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,

$\dfrac{2 \cdot 55}{4-x}=0 \cdot 64 \text { or } x=0$

Match the reaction in List 1 with the colour at the final stage in List 2.

(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.

A volume of $12.5$ mL, $0.05$ M $SeO_{2}$ reacts with $25$ mL of $CrSO_{4}$ which is oxidised to $0.1$ M $Cr^{3-}$ .

The oxidation state at which selenium is converted by the reaction is_________.

$25 ml \times 0.1 M$ 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.

$1$ mol of ${ N }_{ 2 }{ H }_{ 4 }$ loses $10$ mol of electrons to form a new compound $Y$ . Assuming that all the nitrogen appear in the new compound, what is the oxidation state of nitrogen in $Y$ (no change in the oxidation state of $H$ )?

$\underset { \uparrow }{ { N }_{ 2 }{ H }_{ 4 } } \longrightarrow \underset { \uparrow }{ { N }_{ 2 } } (X)+10{ e }^{ - }$
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

$=+3$ .

Match the different compounds of $Mn$ (in List I) with oxidation number (in List II).

(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.

State True or False.
$H{NO}_{3}$ is added in $II$ group. filtrate during basic radical analysis to convert ${Fe}^{2+}$ to ${Fe}^{3+}$ as well as to oxidise ${H}_{2}S$ in solution.
Enter 1 if true else 0.

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

$HNO_3 + Fe^{+2} \rightarrow Fe^{+3} + NO$

Match the half-reactions (in List I) with change in oxidation number (in List II).

(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.

Match list 1 containing compounds with list 2 containing oxidation states of nitrogen and select answer using the codes given below the lists.
Code $\colon\,\,(a)\,\,(b)\,\,(c)\,\,(d)$
$(A)\,\,\,\,\,\,\,\,\,\,3\,\,\,\,\,\,4\,\,\,\,\,\,2\,\,\,\,\,\,1$
$(B)\,\,\,\,\,\,\,\,\,\,4\,\,\,\,\,\,3\,\,\,\,\,\,2\,\,\,\,\,\,1$
$(C)\,\,\,\,\,\,\,\,\,\,3\,\,\,\,\,\,4\,\,\,\,\,\,1\,\,\,\,\,\,2$
$(D)\,\,\,\,\,\,\,\,\,\,4\,\,\,\,\,\,3\,\,\,\,\,\,1\,\,\,\,\,\,2$

$NaN_3\longrightarrow N_3\,^-\longrightarrow N\longrightarrow -1/3$ . 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.

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

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

Write Nearst equation and explain the terms involved in it. What part of the equation represent the correction factor for nonstandard state conditions?

Let us consider a reaction,

$aA+bB\rightarrow cC+dD$

Therefore, Nernst equation is as follows :-

${ E }_{ cell }={ E }_{ cell }^{ 0 }-\dfrac { 2.303RT }{ nF } log\dfrac { { \left[ C \right] }^{ c }{ \left[ D \right] }^{ d } }{ { \left[ A \right] }^{ a }{ \left[ B \right] }^{ b } }$

where,

${ E }_{ cell }=$ potential of cell at given conditions

${ E }_{ cell }^{ 0 }=$ potential of cell at standard conditions

$n\rightarrow$ number of electrons involved in the reaction

$F=$ Faraday's constant

$=96500C$

The correction factor for non-standard state conditions is represented by

$\dfrac { -2.303RT }{ nF } log\dfrac { { \left[ C \right] }^{ c }{ \left[ D \right] }^{ d } }{ { \left[ A \right] }^{ a }{ \left[ B \right] }^{ b } }$

where,

$R$ is universal gas constant.

$T$ is temperature in kelvin.

Balance equation with oxidation number of method.
${P}_{4} + {OH}^{-}\rightarrow {PH}_{3} + {HPO}_{2}^{-}$

${ \overset { 0 }{ P } }_{ 4 }+\overset { -2 }{ O } \overset { +1 }{ { H }^{ - } } \rightleftharpoons \overset { -3 }{ P } { \overset { +1 }{ H } }_{ 3 }+{ \overset { +1 }{ H } }_{ 2 }\overset { +1 }{ P } { \overset { -2 }{ O } }_{ 2 }^{ - }$

$\Rightarrow$

${ P }_{ 4 }^{ 0 }\longrightarrow 4{ \overset { +1 }{ H } }_{ 2 }\overset { +1 }{ P } { \overset { -2 }{ O } }_{ 2 }+4{ e }^{ - }$ (oxidation)

$\Rightarrow$

${ \overset { 0 }{ P } }_{ 4 }+12{ e }^{ - }\longrightarrow 4\overset { -3 }{ P } { \overset { +1 }{ H } }_{ 3 }$ (Reduction).

Balance the oxygen atoms in oxidation reaction.

${ \overset { 0 }{ P } }_{ 4 }+8{ OH }^{ - }\longrightarrow 4{ \overset { +1 }{ H } }_{ 2 }\overset { +1 }{ P } { \overset { 2 }{ O } }_{ 2 }+4{ e }^{ - }$

$3{ \overset { 0 }{ P } }_{ 4 }+24{ OH }^{ - }\longrightarrow 12{ \overset { +1 }{ H } }_{ 2 }\overset { +1 }{ P } { \overset { +2 }{ O } }_{ 2 }+12{ e }^{ - }\quad \quad \left( O \right)$

$\underline { { \overset { 0 }{ P } }_{ 4 }+12{ e }^{ - }+12{ H }_{ 2 }O\longrightarrow 4\overset { -3 }{ P } { \overset { +1 }{ H } }_{ 3 }+12{ OH }^{ - }\quad \quad \left( R \right) }$

$4{ \overset { 0 }{ P } }_{ 4 }+24{ OH }^{ - }+12{ e }^{ - }+12{ H }_{ 2 }O\longrightarrow 12{ \overset { +1 }{ H } }_{ 2 }\overset { +1 }{ P } { \overset { -2 }{ O } }_{ 2 }+4\overset { -3 }{ P } { \overset { +1 }{ H } }_{ 3 }+12{ e }^{ - }+12{ OH }^{ - }$

Simplity the above equation.

${ \overset { 0 }{ P } }_{ 4 }+3{ OH }^{ - }+3{ H }_{ 2 }O\longrightarrow 3{ \overset { +1 }{ H } }_{ 2 }\overset { +1 }{ P } { \overset { -2 }{ O } }_{ 2 }+\overset { -3 }{ P } { \overset { +1 }{ H } }_{ 3 }$

$\boxed { { P }_{ 4 }+3{ OH }^{ - }+3{ H }_{ 2 }O\rightleftharpoons { PH }_{ 3 }+3{ H }_{ 2 }{ PO }_{ 2 } }$

In Column I, some reactions are given and in Column II, the type of some reactions on the basis of oxidation and reduction processes are given. Match the reactions with their correct type.

(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.

$A\rightarrow R$

$B\rightarrow P,T$

$C\rightarrow Q,T$

$D\rightarrow S,T$

Select the type of redox reaction from the following on the basis of type of redox changes (a) intermolecular redox; (b) intramolecular redox; (c) auto redox. If none, write none.
(a) ${C}_{6}{H}_{5}CHO\xrightarrow [ ]{ NaOH } {C}_{6}{H}_{5}{CH}_{2}OH+{C}_{6}{H}_{5}COONa$
(ii) ${Cr}_{2}{O}_{7}^{2-}+2{OH}^{-}\longrightarrow 2Cr{O}_{4}^{2-}+{H}_{2}O$
(iii) $2{Mn}_{2}{O}_{7}\longrightarrow 4Mn{O}_{2}+3{O}_{2}$
(iv) ${NO}_{3}^{-}+{H}_{2}S+{H}_{2}O+{H}^{+}\longrightarrow {NH}_{4}^{+}+H{SO}_{4}^{-}$
(v) $Fe+{N}_{2}{H}_{4}\longrightarrow {NH}_{3}+Fe{(OH)}_{2}$
(vi) $2KOH+{Br}_{2}\longrightarrow KBr+KBrO$
(vii) $2{Cu}^{+}\longrightarrow Cu+{Cu}^{2+}$
(viii) $Ag{({NH}_{3})}_{2}^{+}\xrightarrow [ ]{ 2{ H }^{ + } } {Ag}^{+}+2{NH}_{4}^{+}$
(ix) $5KI+KI{O}_{3}+6HCl\longrightarrow 3{I}_{2}+6KCl+3{H}_{2}O$

1751816_1741638_ans_b345a233d09f44228aa9aa16c2fe0b66.jpg

Assign ON to atoms of only those elements which undergo ON change in the following redox reactions and then balance the equation:

$H_{2}O_{2}+ 2I^{-}+2H^{+}\to I_{2}+2H_{2}O$

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

1694266_1738697_ans_cd790967eeef41ea8320926b4b6ca2d3.png

Match the reactions in column I with nature of reaction / type of the products in column II.

(A)

$O_2^-\rightarrow O_2+O_2^{2-}$

Here

$O_2^-$ IS oxidized to

$O_2$ and reduced to

$O_2^{2-}$ . Hence it is a redox reaction and disproportion reaction. When the same molecule is oxidised or reduced, then the reaction is called a disproportionation reaction. The reactions in which oxidation and reduction take place simultaneously are called redox reactions.

(B)

$2CrO_4^{2-}+H^+\rightarrow Cr _2O_7^{2-}$

Here

$Cr_2O_7^{2-}$ is a dimeric bridges tetrahedram metal ion.

(C)

$MnO_4^-+NO_2^-+K^+\rightarrow Mn^{2+}+NO_3^-+H_2O$

Here

$MnO_4^-$ is reduced to

$Mn^{2+}$ and

$NO_2^-$ is oxidized to

$NO_3^-$ . Also one of the products

$NO_3^-$ has a triagonal planar structure.

(D)

$NO_3^-+H_2SO_4+Fe^{2+}\rightarrow Fe^{3+}+NO+SO_4^{2-}+H_2O$

Here

$NO_3^-$ is reduced to

$NO$ and

$Fe^{2+}$ is oxidized to

$Fe^{3+}$ . Hence it is a redox reaction.

Assign ON to atoms of only those elements which undergo ON change in the following redox reactions and then balance the equation.

$C_{2}H_{5}OH + Cr_{2}O_{7}^{2-}+ H^{+} \to Cr^{3+}+C_{2}H_{4}O + H_{2}O$

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

Oxidation number of C in

$C_2H_5OH$

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

$or,2x+5-2+1=0$

$or, 2x= -4$

$or, x=-2$

Oxidation number of C in

$C_2H_4O$

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

$or, 2x+4-2=0$

$or, 2x=-2$

$x=-1$

So oxidation number of carbon is changing from -2 to -1 i.e. oxidation is occuring.

Oxidation number of

$Cr$ in

$Cr_2O_7^{2-}$

$2x + 7(-2)=2$

$or, 2x-14=-2$

$or, 2x=-2+14=12$

$x=+6$

Oxidation state of

$Cr$ in product is

$+3$ i.e.reduction is occuring.

$3C_{2}H_{5}OH+Cr_{2}O_{7}^{2-}+8H^{+}\to 2Cr^{3+}+3C_{2}H_{4}O+7H_{2}O$

Calculate the oxidation number of Mn in the product of alkaline oxidative fusion of $MnO_2$ .

1614996_1741078_ans_6b80690bf69645f2843732795d06a29e.jpg

$25 \,mL$ of $0.017M \,H_2SO_3$ in strongly acidic medium required $16.9 \,mL$ of $0.01 \,MKMnO_4$ and in neutral medium required $28.6 \,mL$ of $0.01M \,KMnO_4$ for complete conversion of $SO_3^{2-}$ to $SO_4^{2-}$ . Assign the oxidation no. of Mn in the product formed in each case.

$SO^{2-}_3 \rightarrow SO^{2-}_4$

$S^{4+} \rightarrow S^{6+} + 2e$

$\therefore$ Valence factor of

$SO_3 = 2$

In acid medium:

$Meq.$ of

$SO_3^{2-} = Meq.$ of

$MnO_4^{-}$

$25 \times 0.017 \times 2 = 16.9 \times 0.01 \times n_1$

$\therefore n_1 = 5$

$\therefore Mn^{7+} + 5e \rightarrow Mn^{2+}$

In neutral medium:

$Meq.$ of

$SO_3^{2-} = Meq.$ of

$MnO^-_4$

$25 \times 0.017 \times 2 = 28.6 \times 0.01 \times n_2$

$\therefore n_2 = 3$

$\therefore Mn^{7+} + 3e \rightarrow Mn^{4+}$

On adding dilute $HCl$ to $CuO$ powder, the solution formed is blue-green. Predict the new compound formed which imparts a blue-green color to the solution.

$\text{When}$

$HCl$

$\text{is diluted with copper oxide, the following reaction takes place:}$

$HCl + CuO \rightarrow CuCl_2 + H_2O$

$\text{The new compound formed is copper chloride}$

$(CuCl_2)$

$\text{which imparts a blue-green colour to the solution.}$

$\text{The above reaction is a type of double displacement reaction.}$

Classify the given chemical reactions in the table below.
a) $Mg + O_2 \rightarrow 2MgO$
b) $H_2 + I_2 \rightarrow 2HI$
c) $2H_2O \rightarrow 2H_2 + O_2$
d) $NaCl + AgNO_3 \rightarrow AgCl + NaNO_3$
e) $ZnSO_4 + BaCl_2 \rightarrow BaSO_4 + ZnCl_2$
f) $Zn + H_2SO_4 \rightarrow ZnSO_4 + H_2$
g) $FeSO_4 + Zn \rightarrow ZnSO_4 + Fe$
h) $CaCO_3 \rightarrow CaO + CO_2$

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

g) Displacement

h) Decomposition

What happens when:
(a) copper powder is heated in china dish?
(b) hydrogen gas is passed over hot copper (II) oxide?

$\text{(a) When the copper powder is heated in the presence of air (oxygen), copper reacts with oxygen to form copper}$

$\text{oxide. The copper oxide formed is black in color. The black color is formed due to oxidation of copper takes place}$

$\text{The chemical equation of the reaction takes place is given below:}$

$2Cu + O_2 \rightarrow 2CuO$

$Copper \ oxide$

$\text{(b) when hydrogen gas is passed over hot copper oxide, the black coating on the surface turns brown as the}$

$\text{reverse reaction takes place and copper is obtained.}$

$\text{The reaction that takes place is:}$

$CuO + H_2 \rightarrow Cu + H_2O$

Find out the oxidation number of chlorine in the following compounds and arrange
them in increasing order of oxidation number of chlorine.
$NaClO_{4}, NaClO_{3}, NaClO, KClO_{2}, Cl_{2}O, ClO_{3}, Cl_{2}O, NaCl, Cl_{2}, ClO_{2}$

Which oxidation state is not present in any of the above compounds?

$\stackrel{+1}{N a}\ \stackrel{+7}{Cl} \stackrel{-2}O_{4} \quad$ Oxidation

no. of chlorine

$=+7$

Suppose oxidation number of chlorine is

$x$ then ,

$1+x+4\times(-2)=0$

$\therefore x-7=0$

$x=+7$

Like above method,

$\stackrel{+1}{Na}\ \stackrel{+1}{Cl} \stackrel{-2}O_{3} \quad$ Oxidation no. of chlorine

$= +5$

$\stackrel{+1}{N a}\ \stackrel{+1}{Cl} \ \stackrel{-2}O \quad$ Oxidation no. of chlorine

$=+1$

$\stackrel{+1}K \ \stackrel{+3}{C l}\ \stackrel{-2}O_{2} \quad$ Oxidation no. of chlorine

$=+3$

$\stackrel{+7}{Cl}_{2} \ \stackrel {-2}O_{7}\quad$ Oxidation no. of chlorine

$=+7$

$\stackrel{+6}{C l} \ \stackrel {-2}O_3 \quad$ Oxidation

$\quad$ no. of chlorine

$=+6$

$\stackrel{+1}{Cl}_{2}\ \stackrel {-2}O \quad$ Oxidation no. of chlorine

$=+1$

$\mathrm{\stackrel {+1}{Na}Cl} \quad$ Oxidation

$\quad$ no. of chlorine

$=-1$

$C l_{2} \quad$ Oxidation

$\quad$ no. of chlorine

$=0$

$\mathrm{Cl\stackrel {-2}O}_{2} \quad$ Oxidation no. of chlorine

$=+4$

Compounds in increasing order of oxidation number of chlorine is as follows:

$NaCl, C l_{2}, C l_{2} O, N a C l O, K C l O_{2}, C l O_{2}, N a C l O_{3}, C l O_{3}, C l_{2} O_{7}, N a C l O_{4}$

Redox Reactions - Class 11 Engineering Chemistry - Extra Questions

Assign ON to atoms of only those elements which undergo ON change in the following redox reactions and then balance the equation KMnO4+H2S+H2SO4→KHSO4+MnSO4+S+H2OKMnO_{4}+H_{2}S+ H_{2}SO_{4}\to KHSO_{4}+MnSO_{4}+S+H_{2}O

Assign ON to atoms of only those elements which undergo ON change in the following redox reactions and then balance the equation: Cl2+KOH→KOCl+KCl+H2O Cl_{2}+KOH \to KOCl + KCl +H_{2}O

Assign ON to atoms of only those elements which undergo ON change in the following redox reactions and then balance the equation NO−3+Cl−+H+→NO+Cl2+H2O NO_{3}^{-} +Cl^{-}+H^{+}\to NO+Cl_{2}+H_{2}O

Assign ON to atoms of only those elements which undergo ON change in the following redox reactions and then balance the reaction:Fe2(SO4)3+Fe→FeSO4 Fe_{2}(SO_{4})_{3}+Fe\to FeSO_{4}

Assign ON to atoms of only those elements which undergo ON change in the following redox reactions and then balance the reaction:Cu(NH3)4Cl2+KCN+H2O→K2Cu(CN)3+NH3+KCNO+NH4Cl+KCl Cu(NH_{3})_{4}Cl_{2}+ KCN+H_{2}O\to K_{2}Cu(CN)_{3}+NH_{3}+KCNO+NH_{4}Cl+KCl

Assign ON to atoms of only those elements which undergo ON change in the following redox reactions and then balance the reaction:Ag+KCN+H2O+O2→KAg(CN)2+KOH Ag +KCN +H_{2}O+O_{2}\to KAg(CN)_{2}+KOH

Assign ON to atoms of only those elements which undergo ON change in the following redox reactions and then balance the equation: Cr2O2−7+NO−2+H+→Cr3++NO−3+H2OCr_{2}O_{7}^{2-}+NO_{2}^{-}+H^{+}\to Cr^{3+} + NO_{3}^{-}+H_{2}O

Assign ON to atoms of only those elements which undergo ON change in the following redox reactions and then balance the equation: S2O2−3+Sb2O5+H++H2O→SbO+H2SO3 S_{2}O_{3}^{2-}+ Sb_{2}O_{5}+H^{+}+H_{2}O\to SbO+H_{2}SO_{3}

Assign ON to atoms of only those elements which undergo ON change in the following redox reactions and then balance the reaction:Zn+NO−3+H+→Zn2++NO2+H2OZn+NO_{3}^{-}+H^{+}\to Zn^{2+}+NO_{2}+H_{2}O

Assign ON to atoms of only those elements which undergo ON change in the following redox reactions and then balance the reaction:AsO3−3+IO−3→AsO3−4+I− AsO_{3}^{3-}+ IO_{3}^{-} \to AsO_{4}^{3-} +I^{-}

Assign ON to atoms of only those elements which undergo ON change in the following redox reactions and then balance the reaction:MnO−4+CN−+H2O→MnO2+CNO−+OH− MnO_{4}^{-}+CN^{-}+H_{2}O \to MnO_{2}+CNO^{-}+OH^{-}

Both VO+2{VO}_{2}^{+} and CO2+{CO}^{2+} are known as vanadyl ion(a) Determine the oxidation number of vanadium in each(b) Which one of them is oxovanadium(iv) ion and which are in dioxovanadium (v) ion

Write the balanced chemical equation for the following reaction and identify the type of reaction :Ethene is burnt in presence of oxygen to form carbon dioxide , water and releases heat and light.

Calculate the oxidation number of sulphur atom in Na2SO4\mathrm{Na}_{2} \mathrm{SO}_{4} .

Calculate the oxidation number of each sulphur atom in Na2S2O3 \mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3}.

Identify the oxidising and reducing agents in 3HCl(aq)+HNO3(aq)→Cl2(g)+NOCl(g)+2H2O(l) 3 H C l(a q)+H N O_{3}(a q) \rightarrow C l_{2}(g)+N O C l(g)+2 H_{2} O(l)

Calculate the oxidation number of phosphorus in PO3−4 P O_{4}^{3-} .

Calculate the oxidation number of sulphur atom in Na2SO3 \mathrm{Na}_{2} \mathrm{SO}_{3} .

Calculate the oxidation number of each sulphur atom in Na2S4O6 \mathrm{Na}_{2} \mathrm{S}_{4} \mathrm{O}_{6} .

Calculate the oxidation number of phosphorus in the following species.HPO2−3HPO_{3}^{2-}

Calculate oxidation number of SS in H2S2O7H_2S_2O_7.

Calculate oxidation number of FeFe in Fe(CO)5Fe(CO)_5.

Calculate oxidation number of PP in H3PO3H_3PO_3.

Calculate oxidation number of the following compounds:FeFe in FeSO4FeSO_4

Calculate oxidation number of the following compounds:PP in NaH2PO2NaH_2PO_2

Complete the tableCompoundOxidation state of FeFeSymbol of FeFe ionsFeCl2FeCl_2FeCl3FeCl_3

Calculate oxidation number of the following compounds:CC in C12H22O11C_{12}H_{22}O_{11}

Oxidation number of sulphur in H2SO5{ H }_{ 2 }{ SO }_{ 5 } is +6+6. If true enter 1 else 0.

The oxidation number of phosphorus in PO3−4,P4O10PO^{3-}_4, P_4O_{10} and P2O4−7P_2O_7^{4-} respectively, is :

Assign ON to atoms of only those elements which undergo ON change in the following redox reactions and then balance the equation: K2Cr2O7+HCl→KCl+CrCl3+H2O+Cl2 K_{2}Cr_{2}O_{7}+ HCl \to KCl+CrCl_{3}+H_{2}O+Cl_{2}

The oxidation state of chromium in the final product formed in the reaction between KI and acidified potassium dichromate solution is :

11 mole of N2H4{ N }_{ 2 }{ H }_{ 4 } loses 1010 mol of electrons to form a new compound YY. Assuming that all the nitrogen appear in the new compound, what is the oxidation state of nitrogen in YY? (no change in the oxidation state HH).

Consider the elements Cs, Ne, ICs,\ Ne,\ I and FF.Identify the element that exhibits only negative oxidation state.

Assign ON to atoms of only those elements which undergo ON change in the following redox reactions and then balance the equation: N2O4+BrO−3+H2O→NO−3+Br−+H+ N_{2}O_{4}+BrO_{3}^{-} +H_{2}O\to NO_{3}^{-}+Br^{-}+H^{+}

What is the oxidation state of Fe in Fe(CN)6 Fe(CN)_6 ?

What is oxidation number of carbon in CN−CN^-?(Give the absolute value)

Calculate the oxidation number of MnMn in the product obtained on an alkaline oxidative fusion of MnO2MnO_2.(Give the absolute value of an answer)

Calculate the oxidation number of MnMn in the product formed on strongly heating Mn2O7Mn_2O_7.(write absolute value of an answer)

The change in oxidation number of the underlined element in the reaction given below is__________.NaHC_O3+HCl→NaCl+H2O+CO2NaH\underline { C } O_{ 3 }+HCl\rightarrow NaCl+H_{ 2 }O+CO_{ 2 }

The compound (NH3)3CrO4{ \left( { NH }_{ 3 } \right) }_{ 3 }Cr{ O }_{ 4 } has peroxy linkage if oxidation number of CrCr is +4+4.If true enter 1, else enter 0.

Calculate the oxidation number of sulphur, chromium and nitrogen in H2SO5,Cr2O2−7H_2SO_5\,,Cr_2O^{2-}_7 and NO−3NO_3^-. Suggest structure of these compounds. Count for the fallacy.

Which reaction is this?CaO+H2O→Ca(OH)2CaO + H_{2}O \rightarrow Ca(OH)_{2}

What are redox reactions? Identify the substances that are oxidised and the substances that are reduced in the following reactions?A. 2H2S(g)+SO2(g)→3S(s)+2H2O(1)2H_2S_{(g)} + SO_{2(g)} \rightarrow 3S_{(s)} + 2H_2O_{(1)} B. CuO(s)+H2(g)→Cu(s)+H2OCuO_{(s)} + H_{2(g)}\rightarrow Cu_{(s)} +H_2O

Find the oxidation number of SS in SO42−{SO_4}^{2-} ion?

What is the oxidation number of CrCr in K2Cr2O7K_2Cr_2O_7?

Arrange the following in the order of increasing oxidation number of iodine: I2,HI,HIO4,IClI_2, HI, HIO_4, ICl.

Find the oxidation number of I in KIO3KIO_3.

Find the oxidation number of Xe in BaXeO6BaXeO_6.

Calculate the oxidation state of the following.CC in diamond

Find out oxidation states of phosporus in PO43−,P4O10{PO_4}^{3-}, P_4O_{10} and P2O74−{P_2O_7}^{4-}.

Oxidation number of Sulphur in S2O−28 S_2 O_8^{-2} ?

Find out the oxidation state of shown carbonyl carbon.

Assign ON to atoms of only those elements which undergo ON change in the following redox reactions and then balance the equation Cr(OH)3+IO−3+OH−→I−+CrO2−4+H2O Cr(OH)_{3}+IO_{3}^-+OH^{-}\to I^{-}+CrO_{4}^{2-}+ H_{2}O

(i) Calculate oxidation number of ClCl in Cl2O7Cl_{2}O_{7}.(ii) 2FeCl3+H2S→2FeCl2+2HCl+S2FeCl_{3} + H_{2}S \rightarrow 2FeCl_{2} + 2HCl + SWrite the name of oxidising agent and reducing agent in above reaction:

What is the oxidation number of oxygen in oxide, peroxide and superoxide?

The Mn3+Mn^{3+} ion is unstable in solution and undergoes disproportional reaction to give Mn2+,MnO2Mn^{2+}, MnO_2 and H+H^+. Write a balanced ionic equation for the reaction.

Balance the following redox reaction.MnO⊖4(aq)+I⊖(aq)→MnO2(s)+I2(s)MnO^{\ominus}_4(aq)+I^{\ominus}(aq)\rightarrow MnO_2(s)+I_2(s).What is the coefficient of I−I^-?

What is the oxidation number of nitrogen in NH2OHNH_2OH?

Find oxidation number each carbon atom separately in CH3CH2OHCH_{3}CH_{2}OH.

What is the oxidation number of C in HNC?

Find oxidation number of Sulphur in H2S4O6H_{2}S_{4}O_{6}.

Find oxidation number of Carbon in C3O2C_{3}O_{2}.

Find the oxidation numbers to the underlined species in the following compounds or ions.K2Cr2O7K_2Cr_2O_7NaHCO3NaHCO_3

Identify the substance oxidized,substance reduced ,oxidizing agent and reducing agent :MnO2+4HCl→MnCl2+2H2O+Cl2MnO_2+4HCl\rightarrow MnCl_2+2H_2O+Cl_2

Assign ON to atoms of only those elements which undergo ON change in the following redox reactions and then balance the equation: P+HNO3→HPO3+NO+H2O P+HNO_{3}\to HPO_{3}+NO+H_{2}O

Assign oxidation number to atoms of only those elements which undergo O. No. change in the following redox reactions and then balance the equation Cr2O2−7+C2O2−4+H+→Cr3++CO2+H2O Cr_{2}O_{7}^{2-}+C_{2}O_{4}^{2-}+H^{+}\to Cr^{3+}+CO_{2}+H_{2}O

Match the reaction in List 1 with the colour at the final stage in List 2.

A volume of 12.512.5 mL,0.050.05 M SeO2SeO_{2} reacts with 2525 mL of CrSO4 CrSO_{4} which is oxidised to 0.10.1M Cr3−Cr^{3-}. The oxidation state at which selenium is converted by the reaction is_________.

11 mol of N2H4{ N }_{ 2 }{ H }_{ 4 } loses 1010 mol of electrons to form a new compound YY. Assuming that all the nitrogen appear in the new compound, what is the oxidation state of nitrogen in YY (no change in the oxidation state of HH)?

Match the different compounds of MnMn (in List I) with oxidation number (in List II).

State True or False.HNO3H{NO}_{3} is added in IIII group. filtrate during basic radical analysis to convert Fe2+{Fe}^{2+} to Fe3+{Fe}^{3+} as well as to oxidise H2S{H}_{2}S in solution.Enter 1 if true else 0.

Match the half-reactions (in List I) with change in oxidation number (in List II).

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

Write Nearst equation and explain the terms involved in it. What part of the equation represent the correction factor for nonstandard state conditions?

Balance equation with oxidation number of method.P4+OH−→PH3+HPO−2{P}_{4} + {OH}^{-}\rightarrow {PH}_{3} + {HPO}_{2}^{-}

In Column I, some reactions are given and in Column II, the type of some reactions on the basis of oxidation and reduction processes are given. Match the reactions with their correct type.

Assign ON to atoms of only those elements which undergo ON change in the following redox reactions and then balance the equation:H2O2+2I−+2H+→I2+2H2OH_{2}O_{2}+ 2I^{-}+2H^{+}\to I_{2}+2H_{2}O

Match the reactions in column I with nature of reaction / type of the products in column II.

Assign ON to atoms of only those elements which undergo ON change in the following redox reactions and then balance the equation. C2H5OH+Cr2O2−7+H+→Cr3++C2H4O+H2O C_{2}H_{5}OH + Cr_{2}O_{7}^{2-}+ H^{+} \to Cr^{3+}+C_{2}H_{4}O + H_{2}O

Assign ON to atoms of only those elements which undergo ON change in the following redox reactions and then balance the equation

$KMnO_{4}+H_{2}S+ H_{2}SO_{4}\to KHSO_{4}+MnSO_{4}+S+H_{2}O$

Assign ON to atoms of only those elements which undergo ON change in the following redox reactions and then balance the equation: $Cl_{2}+KOH \to KOCl + KCl +H_{2}O$

Assign ON to atoms of only those elements which undergo ON change in the following redox reactions and then balance the equation

$NO_{3}^{-} +Cl^{-}+H^{+}\to NO+Cl_{2}+H_{2}O$

Assign ON to atoms of only those elements which undergo ON change in the following redox reactions and then balance the reaction:

$Fe_{2}(SO_{4})_{3}+Fe\to FeSO_{4}$

Assign ON to atoms of only those elements which undergo ON change in the following redox reactions and then balance the reaction:

$Cu(NH_{3})_{4}Cl_{2}+ KCN+H_{2}O\to K_{2}Cu(CN)_{3}+NH_{3}+KCNO+NH_{4}Cl+KCl$

Assign ON to atoms of only those elements which undergo ON change in the following redox reactions and then balance the reaction:

$Ag +KCN +H_{2}O+O_{2}\to KAg(CN)_{2}+KOH$

Assign ON to atoms of only those elements which undergo ON change in the following redox reactions and then balance the equation:

$Cr_{2}O_{7}^{2-}+NO_{2}^{-}+H^{+}\to Cr^{3+} + NO_{3}^{-}+H_{2}O$

Assign ON to atoms of only those elements which undergo ON change in the following redox reactions and then balance the equation:

$S_{2}O_{3}^{2-}+ Sb_{2}O_{5}+H^{+}+H_{2}O\to SbO+H_{2}SO_{3}$

Assign ON to atoms of only those elements which undergo ON change in the following redox reactions and then balance the reaction:

$Zn+NO_{3}^{-}+H^{+}\to Zn^{2+}+NO_{2}+H_{2}O$

Assign ON to atoms of only those elements which undergo ON change in the following redox reactions and then balance the reaction:

$AsO_{3}^{3-}+ IO_{3}^{-} \to AsO_{4}^{3-} +I^{-}$

Assign ON to atoms of only those elements which undergo ON change in the following redox reactions and then balance the reaction:

$MnO_{4}^{-}+CN^{-}+H_{2}O \to MnO_{2}+CNO^{-}+OH^{-}$

Both ${VO}_{2}^{+}$ and ${CO}^{2+}$ are known as vanadyl ion
(a) Determine the oxidation number of vanadium in each
(b) Which one of them is oxovanadium(iv) ion and which are in dioxovanadium (v) ion

Write the balanced chemical equation for the following reaction and identify the type of reaction :
Ethene is burnt in presence of oxygen to form carbon dioxide , water and releases heat and light.

Calculate the oxidation number of sulphur atom in $\mathrm{Na}_{2} \mathrm{SO}_{4}$ .

Calculate the oxidation number of each sulphur atom in $\mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3}$ .

Identify the oxidising and reducing agents in
$3 H C l(a q)+H N O_{3}(a q) \rightarrow C l_{2}(g)+N O C l(g)+2 H_{2} O(l)$

Calculate the oxidation number of phosphorus in $P O_{4}^{3-}$ .

Calculate the oxidation number of sulphur atom in $\mathrm{Na}_{2} \mathrm{SO}_{3}$ .

Calculate the oxidation number of each sulphur atom in $\mathrm{Na}_{2} \mathrm{S}_{4} \mathrm{O}_{6}$ .

Calculate the oxidation number of phosphorus in the following species.
$HPO_{3}^{2-}$

Calculate oxidation number of $S$ in $H_2S_2O_7$ .

Calculate oxidation number of $Fe$ in $Fe(CO)_5$ .

Calculate oxidation number of $P$ in $H_3PO_3$ .

Calculate oxidation number of the following compounds:
$Fe$ in $FeSO_4$

Calculate oxidation number of the following compounds:
$P$ in $NaH_2PO_2$

Complete the table
Compound Oxidation state of $Fe$ Symbol of $Fe$ ions
$FeCl_2$
$FeCl_3$

Calculate oxidation number of the following compounds:
$C$ in $C_{12}H_{22}O_{11}$

Oxidation number of sulphur in ${ H }_{ 2 }{ SO }_{ 5 }$ is $+6$ . If true enter 1 else 0.

The oxidation number of phosphorus in $PO^{3-}_4, P_4O_{10}$ and $P_2O_7^{4-}$ respectively, is :

Assign ON to atoms of only those elements which undergo ON change in the following redox reactions and then balance the equation:

$K_{2}Cr_{2}O_{7}+ HCl \to KCl+CrCl_{3}+H_{2}O+Cl_{2}$

$1$ mole of ${ N }_{ 2 }{ H }_{ 4 }$ loses $10$ mol of electrons to form a new compound $Y$ . Assuming that all the nitrogen appear in the new compound, what is the oxidation state of nitrogen in $Y$ ? (no change in the oxidation state $H$ ).

Consider the elements $Cs,\ Ne,\ I$ and $F$ .
Identify the element that exhibits only negative oxidation state.

Assign ON to atoms of only those elements which undergo ON change in the following redox reactions and then balance the equation:

$N_{2}O_{4}+BrO_{3}^{-} +H_{2}O\to NO_{3}^{-}+Br^{-}+H^{+}$

What is the oxidation state of Fe in $Fe(CN)_6$ ?

What is oxidation number of carbon in $CN^-$ ?
(Give the absolute value)

Calculate the oxidation number of $Mn$ in the product obtained on an alkaline oxidative fusion of $MnO_2$ .
(Give the absolute value of an answer)

Calculate the oxidation number of $Mn$ in the product formed on strongly heating $Mn_2O_7$ .
(write absolute value of an answer)

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

The compound ${ \left( { NH }_{ 3 } \right) }_{ 3 }Cr{ O }_{ 4 }$ has peroxy linkage if oxidation number of $Cr$ is $+4$ .
If true enter 1, else enter 0.

Calculate the oxidation number of sulphur, chromium and nitrogen in $H_2SO_5\,,Cr_2O^{2-}_7$ and $NO_3^-$ . Suggest structure of these compounds. Count for the fallacy.

Which reaction is this?
$CaO + H_{2}O \rightarrow Ca(OH)_{2}$

What are redox reactions? Identify the substances that are oxidised and the substances that are reduced in the following reactions?
A. $2H_2S_{(g)} + SO_{2(g)} \rightarrow 3S_{(s)} + 2H_2O_{(1)}$
B. $CuO_{(s)} + H_{2(g)}\rightarrow Cu_{(s)} +H_2O$

Find the oxidation number of $S$ in ${SO_4}^{2-}$ ion?

What is the oxidation number of $Cr$ in $K_2Cr_2O_7$ ?

Arrange the following in the order of increasing oxidation number of iodine: $I_2, HI, HIO_4, ICl$ .

Find the oxidation number of I in $KIO_3$ .

Find the oxidation number of Xe in $BaXeO_6$ .

Calculate the oxidation state of the following.
$C$ in diamond

Find out oxidation states of phosporus in ${PO_4}^{3-}, P_4O_{10}$ and ${P_2O_7}^{4-}$ .

Oxidation number of Sulphur in $S_2 O_8^{-2}$ ?

Assign ON to atoms of only those elements which undergo ON change in the following redox reactions and then balance the equation

$Cr(OH)_{3}+IO_{3}^-+OH^{-}\to I^{-}+CrO_{4}^{2-}+ H_{2}O$

(i) Calculate oxidation number of $Cl$ in $Cl_{2}O_{7}$ .
(ii) $2FeCl_{3} + H_{2}S \rightarrow 2FeCl_{2} + 2HCl + S$
Write the name of oxidising agent and reducing agent in above reaction:

The $Mn^{3+}$ ion is unstable in solution and undergoes disproportional reaction to give $Mn^{2+}, MnO_2$ and $H^+$ . Write a balanced ionic equation for the reaction.

Balance the following redox reaction.
$MnO^{\ominus}_4(aq)+I^{\ominus}(aq)\rightarrow MnO_2(s)+I_2(s)$ .
What is the coefficient of $I^-$ ?

What is the oxidation number of nitrogen in $NH_2OH$ ?

Find oxidation number $CH_{3}CH_{2}OH$ .

Find oxidation number of $H_{2}S_{4}O_{6}$ .

Find oxidation number of $C_{3}O_{2}$ .

Find the oxidation numbers to the underlined species in the following compounds or ions.
$K_2Cr_2O_7$
$NaHCO_3$

Identify the substance oxidized,substance reduced ,oxidizing agent and reducing agent :
$MnO_2+4HCl\rightarrow MnCl_2+2H_2O+Cl_2$

Assign ON to atoms of only those elements which undergo ON change in the following redox reactions and then balance the equation:

$P+HNO_{3}\to HPO_{3}+NO+H_{2}O$

Assign oxidation number to atoms of only those elements which undergo O. No. change in the following redox reactions and then balance the equation

$Cr_{2}O_{7}^{2-}+C_{2}O_{4}^{2-}+H^{+}\to Cr^{3+}+CO_{2}+H_{2}O$

A volume of $12.5$ mL, $0.05$ M $SeO_{2}$ reacts with $25$ mL of $CrSO_{4}$ which is oxidised to $0.1$ M $Cr^{3-}$ .

The oxidation state at which selenium is converted by the reaction is_________.

$1$ mol of ${ N }_{ 2 }{ H }_{ 4 }$ loses $10$ mol of electrons to form a new compound $Y$ . Assuming that all the nitrogen appear in the new compound, what is the oxidation state of nitrogen in $Y$ (no change in the oxidation state of $H$ )?

Match the different compounds of $Mn$ (in List I) with oxidation number (in List II).

State True or False.
$H{NO}_{3}$ is added in $II$ group. filtrate during basic radical analysis to convert ${Fe}^{2+}$ to ${Fe}^{3+}$ as well as to oxidise ${H}_{2}S$ in solution.
Enter 1 if true else 0.

Balance equation with oxidation number of method.
${P}_{4} + {OH}^{-}\rightarrow {PH}_{3} + {HPO}_{2}^{-}$

Assign ON to atoms of only those elements which undergo ON change in the following redox reactions and then balance the equation:

$H_{2}O_{2}+ 2I^{-}+2H^{+}\to I_{2}+2H_{2}O$

Assign ON to atoms of only those elements which undergo ON change in the following redox reactions and then balance the equation.

$C_{2}H_{5}OH + Cr_{2}O_{7}^{2-}+ H^{+} \to Cr^{3+}+C_{2}H_{4}O + H_{2}O$

Calculate the oxidation number of Mn in the product of alkaline oxidative fusion of $MnO_2$ .

On adding dilute $HCl$ to $CuO$ powder, the solution formed is blue-green. Predict the new compound formed which imparts a blue-green color to the solution.

What happens when:
(a) copper powder is heated in china dish?
(b) hydrogen gas is passed over hot copper (II) oxide?