MCQ Questions for Class 11 Medical Chemistry Redox Reactions Quiz 11

When

$MnO_{2}$ is fused with

$KOH$ , a coloured compound is formed. The product and its colour are, respectively :

0%
$K_{2}MnO_{4}$ , purple green
0%
$KMnO_{4}$ , purple
0%
$Mn_{2}O_{3}$ , brown
0%
$Mn_{3}O_{4}$ , black

Explanation

When

$MnO_2$ is fused with

$KOH$ , a purple green coloured

$K_{2}MnO_{4}$ is formed. The reaction is given below:

$2MnO_2+4KOH+O_2 \rightarrow 2K_{2}MnO_{4}+2H_2O$ .

Identity the pair of binary corresponds in which nitrogen exhibits the lowest and the highest oxidation state.

0%
$NH_{3}$ , $NO_{2}$
0%
$NH_3$ , $N_{2}O_{5}$
0%
$N_{2}$ , $HNO_{3}$
0%
$N_{2}O$ , $N_{2}O_{5}$

Number of moles of acidic

$KMnO_4$ needed for complete oxidation of

$1$ mole equimolar mixture of

$FeSO_4$ and

$FeC_2O_4$ is:

0%
$\dfrac {2}{5}$
0%
$\dfrac {1}{5}$
0%
$\dfrac {1}{10}$
0%
none of the above

Explanation

Moles of acidic

$KMnO_4$ needed for complete oxidation of

$1$ mole equimolar mixture of

$FeSO_4$ and

$FeC_2O_4$ is

$\dfrac {2}{5}$ .

$1$ mole equimolar mixture of

$FeSO_4$ and

$FeC_2O_4$ will contain

$0.5$ moles of

$FeSO_4$ and

$0.5$ moles of

$FeC_2O_4$ .

Thus, it contains

$1$ mol of

$Fe^{2+}$ ions which will be oxidized to

$Fe^{3+}$ ions and

$0.5$ moles of

$C_2O_4^{2-}$ ions which will be oxidized to

$CO_2$ .
The reactions are as follows:

$2MnO_4^- + 5C_2O_4^{2-}+16H^+ \rightarrow 2Mn^{2+} + 10CO_2 +8H_2O$

$MnO_4^- + 5Fe^{2+}+8H^+ \rightarrow 5Fe^{3+} + Mn^{2+} + 4H_2O$

Thus,

$2$ moles

$KMnO_4 =$

$5$ moles

$C_2O_4^{2-}$

$1$ mole

$KMnO_4 =$

$5$ moles

$Fe^{2+}$

Hence, moles of acidic

$KMnO_4$ needed for complete oxidation of

$1$ mole equimolar mixture of

$FeSO_4$ and

$FeC_2O_4$ is

$\dfrac {0.5}{5}+\dfrac {0.5}{5}+\dfrac {1}{5}=\dfrac {2}{5}$ .

Amongst the following, identify the species with an atom in

$+6$ oxidation state.

0%
$[MnO_4]^-$
0%
$[Cr(CN)_6]^{3-}$
0%
$Cr_2O_3$
0%
$CrO_2Cl_2$

Explanation

Correct option: $D$

Explanation:

Step: 1 calculate the oxidation state of central atom $Mn$ In $MnO_4^{-}$

Let us assume the $oxidation$ $state$ of $Mn$ be x.

Overall charge on the complex: $−1$

Thus, $Mn + 4 \times O = - 1$

$x+4(−2) = −1$ ( $O$ has $-2$ charge )

So, $x=+7$

Hence, the $oxidation$ $state$ is $+7$ .

Step: 2 calculate the oxidation state of central atom $Cr$ In $\left [ Cr\left ( CN \right )_6 \right ]^{3-}$

Let us assume the $oxidation$ $state$ of $Cr$ be x.

Overall charge on the complex: $−3$

Thus, $Cr + 6 \times CN = -1$

$x+6(−1) = −3$ (Where $CN$ has $-1$ overall charge )

So, $x=+3$

Hence, the $oxidation$ $state$ is $+3$ .

Step: 3 calculate the oxidation state of central atom $Cr$ In $Cr_2O_3$

Let us assume the $oxidation$ $state$ of $Cr$ be x.

Overall charge on the complex: $0$

Thus, $2 \times Cr + 3 \times O = 0$

$2x+3(−2) = 0$ ( $O$ has $-2$ charge )

So, $x=+3$

Hence, the $oxidation$ $state$ is $+3$ .

Step: 4 calculate the oxidation state of central atom $Cr$ In $CrO_2Cl_2$

Let us assume the $oxidation$ $state$ of $Cr$ be x.

Overall charge on the complex: $0$

Thus, $1 \times Cr + 2 \times O + 2 \times Cl$

$x+2(−2)-2 = 0$ ( $O$ has $-2$ charge while $Cl$ has $-1$ charge)

So, $x=+6$

Hence, the $oxidation$ $state$ is $+6$ .

Hence, the species having oxidation state of $+6$ is $CrO_2Cl_2$ .

The average oxidation states of sulphur in

$Na_2S_2O_3$ and

$Na_2S_4O_6$ are, respectively :

0%
$+5$ and $+2$
0%
$+2$ and $+2.5$
0%
$+5$ and $2.5$
0%
$+2$ and $+4$

Explanation

Let

$x$ be the oxidation number of sulphur in

$Na_2S_2O_3$ .
For a neutral compound, sum of oxidation states of all elements must be zero.

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

$2x=6-2$

$2x=4$

$x=+2$
Suppose

$y$ be the oxidation number of sulphur in

$Na_2S_4O_6$ .

$2+4y+3(-2)=0$

$4y=12-2$

$4y=10$

$y=+2.5$
The average oxidation states of sulphur in

$Na_2S_2O_3$ and

$Na_2S_4O_6$ are respectively

$+2$ and

$+2.5$ .

Which of the following does not exist?

0%
HS $_6$
0%
HPO $_4$
0%
FeI $_3$
0%
HClO $_4$

Explanation

The molecules

$HS_6$ and

$HPO_4$ do not exist.

Whereas,

$HClO_4$ and

$FeI_3$ exist. In

$HPO_4$ , the oxidation state of

$P$ is

$+7$ , whereas

$P$ can form a maximum of

$+5$ oxidation state. In

$HS_6$ , the oxidation state of

$S$ is

$+\frac {1}{6}$ which is not possible.

In a daniel cell, if

$A(E^o = -0.76 \ C)$ and

$B(E^o = -2.36 \ V)$ half cells are taken then:

0%
$B$ acts as anode
0%
$A$ acts as anode
0%
$B$ acts as cathode
0%
cannot be predicted

Explanation

Daniel cell converts chemical energy into electrical energy. Lower standard reduction potential is of anode and cathode has higher standard reduction potential. Emf of Daniel cell equals E

$_{red. (cathode)}$ - E

$_{ red. (anode)}$ . Emf of cell has to be greater than zero for working of cell.

How many substances have underlined atoms of different nature?

$Na_2\underline{S_4}O_6,Na_2\underline{S_2}O_3,CaO\underline{Cl_2},\underline{N_2}H_4O_3,\underline{Fe_3}O_4,K\underline{I_3},Cr\underline{O_5},K_2\underline{Cr_2}O_7$

0%
7
0%
5
0%
4
0%
8

Explanation

Seven substances have underlined atoms of different nature. These include

$Na_2\underline{S_4}O_6,Na_2\underline{S_2}O_3,CaO\underline{Cl_2},\underline{N_2}H_4O_3,\underline{Fe_3}O_4,K\underline{I_3},Cr\underline{O_5}$ . The underlined atoms in these substances have different oxidation numbers.

Oxidation state of 'S' in sodium tetrathionate is :

0%
$+5,0,0,+5$
0%
$+6,+6,0$
0%
$+4,+2,+5$
0%
$+2,0,+2, +3$

Explanation

In Sodium tetrathionate, sulphur is in

$+5, 0, 0, +5$ oxidation state.

The oxidation states of middle sulphur atoms are

$0$ each and terminal sulphur is

$+5$ each.

In an experiment,

$50$ mL of

$0.1$ M solution of a metallic salt reacted exactly with

$25$ mL of

$0.1$ M solution of sodium sulphite. In the reaction,

${SO_3}^{-}$ is oxidised to

${SO_4}^{2-}$ . If the original oxidation number of the metal in the salt was

$3$ , what would be the new oxidation number of the metal?

0%
0
0%
1
0%
2.5
0%
4

Explanation

Total number of millimoles of metallic salt present

$= 50 mL \times 0.1 M = 5$ millimoles.
Total number of millimoles of sodium sulphite

$= 25 mL \times 0.1 M = 2.5$ millimoles.
The oxidation number of S changes from +5 to +6.
Increase in the oxidation number of 1 S atom = 1.
Total number of millimoles electrons lost by sodium sulphite = 2.5
5 millimoles of metallic salt gains 2.5 millimoles of electrons.
1 molecule of metallic salt will gain 0.5 electron.
Hence, the oxidation number of metal will decrease from 3 to 2.5.

The oxidation state of the most electronegative element in the products of the reaction between

$BaO_2$ and

$H_2SO_4$ is:

0%
0 and -1
0%
-1 and -2
0%
-2 and 0
0%
-2 and +1

Explanation

Reaction between

$BaO_2$ and

$H_2SO_4$ is as follows:

$BaO_2+H_2SO_4\rightarrow H_2O_2+BaSO_4$

$-1$

$-2$
The most electronegative element is oxygen in both the products and its oxidation state is

$-1$ and

$-2$ respectively, in

$H_2O_2$ and

$BaSO_4$ .

The oxidation state of +3 for phosphorous is in :

0%
hypophosphorous acie
0%
meta-phosphoric acid
0%
ortho-phosphoric acid
0%
phosphorous acid

In which of the following, sulphur has the highest oxidation state?

0%
$SO_2$
0%
$SO_3$
0%
$H_2SO_4$
0%
$H_2S$

Explanation

Sulphur has the highest oxidation state in

$SO_3$ and

$H_2SO_4$ . The oxidation state of S in

$SO_2$ ,

$SO_3$ ,

$H_2SO_4$ and

$H_2S$ is +4, +6, +6 and -2 respectively.

Which of the following is/are correct in case of Mohr's salt?

0%
It decolourises $KMnO_4$ .
0%
It is primary standard titrant.
0%
It is a double salt.
0%
Oxidation state of Fe is +3 in the salt.

Explanation

Mohr's Salt is a double salt and it reacts with

$KMnO_4$ according to the following reaction:

$KMnO_4+ Fe^{+2} \rightarrow Fe^{+3} + Mn^{+2}$
+7 +2 +3 +2
So, it decolourises

$KMnO_4$ .

Which does not represent auto-redox process?

0%
$4CrO_5 + 6H_2SO_4 \rightarrow 2Cr_2(SO_4)_3+6H_2O+7O_2$
0%
$2C_6H_5CHO \overset{Al(OC_2H_5)_3}{\rightarrow}C_6H_5COOCH_2C_6H_5$
0%
$4KCIO_3\rightarrow 3KClO_4+KCI$
0%
$CHO-CHO \xrightarrow {NaOH} CH_2OH-CHO + COOH-CHO$

Explanation

$4CrO_5 + 6H_2SO_4 \rightarrow 2Cr_2(SO_4)_3+6H_2O+7O_2$

$+6$

$+3$
So, its an intramolecular redox reaction.

Select the incorrect statement for developing of an exposed camera film involving the reaction.

0%
Hydroquinol acts as reductant.
0%
$Ag^+$ acts as oxidant.
0%
Hydroquinol and AgBr undergoes redox change.
0%
It involves intramolecular change.

Explanation

It is an intermolecular redox change where

$AgBr$ reduces to

$Ag$ and hydroquinol oxidizes to quinone.

Which of the following statement(s) is/are true?

0%
All reactions are oxidation and reduction reactions.
0%
Oxidising agent is itself reduced.
0%
Oxidation and reduction always go side by side.
0%
Oxidation number during reduction decreases.

Choose the correct statement(s).

0%
If oxidation and reduction is carried out simultaneously in one container, the redox reactions are called direct redox changes.
0%
If oxidation and reduction is carried out simultaneously in two compartments, the redox reactions are called indirect redox changes.
0%
Direct redox changes produce electrical energy.
0%
Indirect redox changes produce electrical energy.

Which of the following is/are redox reactions?

0%
$BaO_2+H_2SO_4 \rightarrow BaSO_4+ H_2O_2$
0%
$2BaO+O_2 \rightarrow 2BaO_2$
0%
$2KClO_3 \rightarrow 2KCl+ 3O_2$
0%
$SO_2 + 2H_2S \rightarrow 2H_2O+ 3S$

Explanation

$2BaO+O_2 \rightarrow 2BaO_2$
-2 0 -1

$2KClO_3 \rightarrow 2KCl+ 3O_2$
+5 -2 -1 0

$SO_2 + 2H_2S \rightarrow 2H_2O+ 3S$
+4 -2 0

So, they all are redox reactions in which oxidation and reduction takes place simultaneously.

The oxidation number of sulphur in

$S_8, S_2F_2$ and

$H_2S$ respectively, are :

0%
$0,\: +1$ and $-2$
0%
$+2,\: +1$ and $-2$
0%
$0,\: +1$ and $+2$
0%
$-2,\: +1$ and $-2$

Explanation

The oxidation number of sulphur in

$S_8, S_2F_2$ and

$H_2S$ respectively, are

$0,\: +1$ and

$-2$
(1) Let x be the oxidation number of S in

$S_8$ .

$\displaystyle 8X = 0$

$\displaystyle X=0$
(2) Let x be the oxidation number of S in

$S_2F_2$

$\displaystyle 2X -2 =0$

$\displaystyle X = +1$
(3) Let x be the oxidation number of S in

$H_2S$

$\displaystyle X+2=0$

$\displaystyle X = -2$

Oxidation number of

$Cr$ is

$+5$ in :

0%
${ K }_{ 3 }Cr{ O }_{ 8 }$
0%
${ \left( { NH }_{ 3 } \right) }_{ 3 }Cr{ O }_{ 4 }$
0%
${ K }_{ 2 }Cr{ O }_{ 4 }$
0%
$\left[ Cr{ \left( { NH }_{ 3 } \right) }_{ 5 }\left( { H }_{ 2 }O \right) \right] { Cl }_{ 3 }$

Explanation

$\begin{array}{l}\text { There are } 4 \text { Peroxy bonds } \\\qquad \begin{aligned}3+x &+8(-1)=0 \\\Rightarrow x &=+5\end{aligned}\end{array}$

$\begin{array}{l}\left(\mathrm{NH}_{3}\right)_{3}\mathrm{CrO}_{4},\mathrm{~K}_{2}\mathrm{Cr}_{}\mathrm{O}_{4},\left[\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{5}\mathrm{(H}_{2}\mathrm{O)}\right]\mathrm{Cl}_{3} \text { contain } \\\text { Chromium oxidation states of }+4,+6,+3 \\\text { respectively }\end{array}$

1990601_180030_ans_1fe238569380452caa6f0cc38dd2a529.PNG

Consider the following unbalanced reactions:

$I: Zn+dil.\:{ H }_{ 2 }S{ O }_{ 4 }\longrightarrow ZnS{ O }_{ 4 }+{ H }_{ 2 }$

$II: Zn+conc.\:{ H }_{ 2 }S{ O }_{ 4 }\longrightarrow ZnS{ O }_{ 4 }+S{ O }_{ 2 }+{ H }_{ 2 }O\quad$

$III: Zn+HN{ O }_{ 3 }\longrightarrow Zn{ \left( { NO }_{ 3 } \right) }_{ 2 }+{ NH }_{ 4 }{ NO }_{ 3 }+{ H }_{ 2 }O$
Oxidation number of hydrogen changes in :

0%
$I,II,III$
0%
$I,II$
0%
$II,III$
0%
$I$

Explanation

Change in oxidation state of H is shown below:

$I: Zn+\underset { +1 }{ 2{ H }^{ + } } \longrightarrow \underset { 0 }{ { H }_{ 2 } }$ oxidation state changes (reduction)

$II: Zn+\underset { +1 }{ { H }^{ + } } +S{ O }_{ 4 }^{ 2- }\longrightarrow { Zn }^{ 2+ }+S{ O }_{ 2 }+\underset { +1 }{ { H }_{ 2 } } O$
no change in oxidation state of H

$III: Zn+\underset { +1 }{ { H }^{ + } } +N{ O }_{ 3 }^{ - }\longrightarrow { Zn }^{ 2+ }+\underset { \uparrow \\ +1 }{ { NH }_{ 4 } } \underset { \uparrow \\ +1 }{ { H }_{ 2 } } O$ no change
Therefore, from the above reactions, we can see oxidation number of

$H$ changes in

$I$ .

The complex

${ \left[ Fe{ \left( { H }_{ 2 }O \right) }_{ 5 }NO \right] }^{ 2+ }$ is formed in the ring-test for nitrate when freshly prepared

$FeS{ O }_{ 4 }$ solution is added to aqueous solution of

$N{ O }_{ 3 }^{ - }$ followed by addition of conc.

${ H }_{ 2 }S{ O }_{ 4 }$ . This complex is formed by charge transfer in which :

0%
${ Fe }^{ 2+ }$ changes to ${ Fe }^{ 3+ }$ and ${NO}^{+}$ changes to $NO$
0%
${ Fe }^{ 2+ }$ changes to ${ Fe }^{ 3+ }$ and $NO$ changes to ${NO}^{+}$
0%
${ Fe }^{ 2+ }$ changes to ${ Fe }^{ + }$ and $NO$ changes to ${NO}^{+}$
0%
no charge transfer takes place

Explanation

${ Fe }^{ 2+ }+{ e }^{ - }\longrightarrow { Fe }^{ + }$

$NO\longrightarrow { NO }^{ + }+{ e }^{ - }$
Based on magnetic properties of the complex, it is found that iron has three unpaired electrons. Thus, it is as

${Fe}^{+}$ formed by reduction of

${ Fe }^{ 2+ }$ by

$NO$ which is oxidised to

${NO}^{+}$ .

Which combination appears odd with respect to oxidation number per atom of the underlined?

0%
${ H }_{ 2 }\underline { S } { O }_{ 5 },{ H }_{ 2 }\underline { S } { O }_{ 8 },{ K }_{ 2 }\underline { { Cr }_{ 2 } } { O }_{ 7 }$
0%
$\underline { Cr } { O }_{ 5 },\underline { Cr } { O }_{ 4 }^{ 2- },S{ O }_{ 4 }^{ 2- }$
0%
Both (a) and (b)
0%
None of the above

Of the following acids, the one that is strongest is :

0%
$HBr{O}_{4}$
0%
$HOCl$
0%
$H{NO}_{2}$
0%
${H}_{3}{PO}_{3}$

Explanation

As oxidation state of central atom increases, acidity increases.
So, strongest acid is that which has maximum value of oxidation number (O.N) of the central atom.

$HBr{ O }_{ 4 }$ ; O.N of

$Br=+7$ (strongest acid)

$HOCl$ ; O.N of

$Cl=+1$

$H{NO}_{2}$ ; O.N of

$N=+3$

${H}_{3}{PO}_{3}$ ; O.N of

$P=+3$

For a titration of

$100\: {cm}^{3}$ of

$0.1$

$M$

${ Sn }^{ 2+ }$ to

${ Sn }^{ 4+ }$ ,

$50$

${cm}^{3}$ of

$0.40$

$M$

${ Ce }^{ 4+ }$ solution was required. The oxidation state of cerium in the reduction product is :

0%
$+1$
0%
$+2$
0%
$+3$
0%
$0$

Explanation

According to given conditions, we have

${ Sn }^{ 2+ }\longrightarrow { Sn }^{ 4+ }+2{ e }^{ - }; increase\ in\ O.N=2$

${ Ce }^{ 4+ }+n{ e }^{ - }\longrightarrow { Ce }^{ (4-n) };\ decrease\ in\ O.N=n$
On balancing, we get

$n{ Sn }^{ 2+ }+2{ Ce }^{ 4+ }\longrightarrow n{ Sn }^{ 4+ }+2{ Ce }^{ (4-n) }$
Given,
Millimoles of

${ Sn }^{ 2+ }=100\times 0.1=10$
Millimoles of

${ Ce }^{ 4+ }=50\times 0.4=20$
At equivalent point, number of equivalents of both reactant and product are same.
Therefore,

$\cfrac { n }{ 2 } =\cfrac { 10 }{ 20 }$
or,

$n=1$
Thus, oxidation state of cerium in the reduced product

$({ Ce }^{ (4-1) }={ Ce }^{ 3+ })=+3$ .

$1$ 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

$H$ )?

0%
$-1$
0%
$-3$
0%
$+3$
0%
$+5$

Explanation

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

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

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

Which statement(s) is/are not correct about the reaction?

$H_{2}O_2(aq) + 2I^- (aq) + 2H^+(aq) \rightarrow I_2(aq) + 2H_2O(l) I_2(aq) + 2Na_2S_2O_3\rightarrow Na_2S_4O_6+2NaI?$

0%
The $I_2$ formed is consumed completely in the second state is confirmed by addition of starch just before its complete consumption
0%
The reaction process is called clock reaction because the appearance of blue colour on addition of starch is like an alarm given by clock
0%
Sulphur is oxidised whereas iodine is oxidised and reduced during whole process
0%
$H_2O_2$ acts as reducing agent

Explanation

$H_{2}O_2(aq) + 2I^- (aq) + 2H^+(aq) \rightarrow I_2(aq) + 2H_2O(l)$
-1 -1 0 -2

$I_2(aq) + 2Na_2S_2O_3\rightarrow Na_2S_4O_6+2NaI$
0 +2 +2.5 -1
This is iodine clock reaction.

$Mn{O}_{4}^{2-}$ (1 mole) in neutral aqueous medium disproportionate to :

0%
$2/3$ mole ${Mn}{O}_{4}^{-}$ and $1/3$ mole $Mn{O}_{2}$
0%
$1/3$ mole ${Mn}{O}_{4}^{-}$ and $2/3$ mole $Mn{O}_{2}$
0%
$1/3$ mole of ${Mn}_{2}{O}_{7}$ and $1/3$ mole of $Mn{O}_{2}$
0%
$2/3$ mole of ${Mn}_{2}{O}_{7}$ and $1/3$ mole of $Mn{O}_{2}$

Explanation

Option (A) is correct.

$Mn{O}_{4}^{2-}$ (1 mole) in neutral aqueous medium disproportionate to

$2/3$ mole

${Mn}{O}_{4}^{-}$ and

$1/3$ mole

$Mn{O}_{2}$ .

$3K_2MnO_4 + 2H_2O \longrightarrow 2KMnO_4 + MnO_2 +4KOH$

Compound(s) of

$Cu$ and

$Ag$ showing

$+3$ oxidation state is\are :

0%
$K_3 [CuF_6]$
0%
$Na[AgF_4]]$
0%
both a and b
0%
none of these

Explanation

Compound of

$Cu$ and

$Ag$ showing

$+3$ oxidation state are

$K_3 [CuF_6]$ and

$Na[AgF_4]$ .
Let

$x$ be the oxidation number of

$Cu$ in

$K_3 [CuF_6]$ .
Since the overall charge on the compound is

$0$ , the sum of oxidation states of all elements in it should be equal to

$0$ .
Therefore,

$3(+1)+x+6(-1)=0$
or,

$x=-3$
Let

$y$ be the oxidation number of Ag in

$Na[AgF_4]$ .
Since the overall charge on the compound is

$0$ , the sum of oxidation states of all elements in it should be equal to

$0$ .
Therefore,

$+1+y+4(-1)=0$
or,

$y=+3$

Quinhydrone half cell is not reversible to:

0%
H $^+$
0%
quinone
0%
quinol
0%
OH $^-$

Explanation

Quinhydrogene half cell is only reversible to

${ H }^{ + }$ and not to

${ OH }^{ - }$ and any other species like quinone, quinol.

Find the oxidation number of elements in each case :

$S$ in

$Na_2S_2O_3$ ,

$S_4$ ,

$S_8$ and

$Na_2S_2O_7$ .

0%
$+2,\: 0,\: 0,\: +6$
0%
$+1,\: 0,\: +1,\: +6$
0%
$+1,\: 0,\: 0,\: +4$
0%
None of the above

Explanation

The oxidation number of

$S$ in

$Na_2S_2O_3$ ,

$S_4$ ,

$S_8$ and

$Na_2S_2O_7$ are

$+2,\: 0,\: 0,\: +6$ respectively.
Let X be the oxidation number of

$S$ in

$Na_2S_2O_3$ ,

$\displaystyle 2+2X+3(-2)=0$

$\displaystyle X = +2$
Let X be the oxidation number of

$S$ in

$S_4$ ,

$\displaystyle 4X=0$

$\displaystyle X=0$
Let X be the oxidation number of

$S$ in

$S_8$

$\displaystyle 8X = 0$

$\displaystyle X = 0$
Let X be the oxidation number of

$S$ in

$Na_2S_2O_7$

$\displaystyle 2+2x+7(-2) = 0$

$\displaystyle X=+6$ .

The coefficients of

$I^{-}, IO_{3}^{-}$ and

$H^{+}$ in the redox reaction,

$I^{-}+ IO_{3}^{-}+H^{+}\longrightarrow I_{2}+H_{2}O$

in the balanced form respectively are

0%
$5,1,6$
0%
$1,5,6$
0%
$6,1,5$
0%
$5,6,1$

Out of

$\underset { (I) }{ { H }_{ 2 }S{ O }_{ 4 } } ,\ \underset { (II) }{ { H }_{ 2 }{ S }_{ 2 }{ O }_{ 7 } } ,\ \underset { (III) }{ { H }_{ 2 }S{ O }_{ 5 } }$ and

$\underset { (IV) }{ { H }_{ 2 }{ S }_{ 2 }{ O }_{ 3 } }$ :

0%
oxidation number of sulphur is $+6$ in all except $IV$
0%
peroxy linkage is in $I,II$ and $III$
0%
peroxy linkage is in $II$ and $III$
0%
oxidation number of sulphur is $-2$ in $I$ and $IV$

In which of the following pairs is there the greatest difference in the oxidation number of the underlined elements?

0%
$\underline { N } { O }_{ 2 }$ and $\underline { { N }_{ 2 } } { O }_{ 4 }$
0%
$\underline { S } { O }_{ 3 }^{ 2- }$ and $\underline { S } { O }_{ 4 }^{ 2- }$
0%
$\underline { { S }^{ 2- } }$ and $\underline { S } { O }_{ 3 }^{ 2- }$
0%
$\underline { { S }^{ 2- } }$ and $\underline { S } { O }_{ 4 }^{ 2- }$

Explanation

$\underline { { S }^{ 2- } }$ and

$\underline { S } { O }_{ 4 }^{ 2- }$ the difference in the oxidation number of the underlined elements is greatest.
(A)

$\underline { N } { O }_{ 2 }$ and

$\underline { { N }_{ 2 } } { O }_{ 4 }$
The oxidation numbers of the underlined elements are +4 and +4 respectively. The difference in the oxidation number is 0.
(B)

$\underline { S } { O }_{ 3 }^{ 2- }$ and

$\underline { S } { O }_{ 4 }^{ 2- }$
The oxidation numbers of the underlined elements are +4 and +6 respectively. The difference in the oxidation number is 2.
(C)

$\underline { { S }^{ 2- } }$ and

$\underline { S } { O }_{ 3 }^{ 2- }$
The oxidation numbers of the underlined elements are -2 and +4 respectively. The difference in the oxidation number is 6.
(D)

$\underline { { S }^{ 2- } }$ and

$\underline { S } { O }_{ 4 }^{ 2- }$
The oxidation numbers of the underlined elements are -2 and +6 respectively. The difference in the oxidation number is 8.

What is the oxidation number of the underlined elements?
a.

${ H }_{ 2 }\underline { S }$ b.

${ H }_{ 2 }\underline { S } { O }_{ 4 }$ c.

${ Na }_{ 2 }\underline { { S }_{ 2 } } O_3$ d.

${ Na }_{ 2 }\underline { { S }_{ 4 } } { O }_{ 6 }$

0%
a. -2, b. +6, c. +2, d. +2.5
0%
a. -2, b. +4, c. +2, d. +2.5
0%
a. -2, b. +6, c. +3, d. +2.5
0%
None of these

Which of the following represent redox reactions?
I.

${ Cr }_{ 2 }{ O }_{ 7 }^{ 2- }+2\overset { \ominus }{ O } H\longrightarrow 2Cr{ O }_{ 4 }^{ 2- }+{ H }_{ 2 }O$
II.

$Zn+CuS{ O }_{ 4 }\longrightarrow ZnS{ O }_{ 4 }+Cu$
III.

$2Mn{ O }_{ 4 }^{ \ominus }+3{ Mn }^{ 2+ }+4\overset { \ominus }{ O } H\longrightarrow 5Mn{ O }_{ 2 }+2{ H }_{ 2 }O$
IV.

$2{ Cu }^{ \oplus }\longrightarrow Cu+{ Cu }^{ 2+ }$

0%
I, II
0%
I, III
0%
III, IV
0%
II, III, IV

Explanation

The redox reactions are given below:
I.

${ Cr }_{ 2 }{ O }_{ 7 }^{ 2- }+2\overset { \ominus }{ O } H\longrightarrow 2Cr{ O }_{ 4 }^{ 2- }+{ H }_{ 2 }O$
+6 +6
II.

$Zn+CuS{ O }_{ 4 }\longrightarrow ZnS{ O }_{ 4 }+Cu$
0 +2 +2 0
III.

$2Mn{ O }_{ 4 }^{ \ominus }+3{ Mn }^{ 2+ }+4\overset { \ominus }{ O } H\longrightarrow 5Mn{ O }_{ 2 }+2{ H }_{ 2 }O$
+7 +2 +4
IV.

$2{ Cu }^{ \oplus }\longrightarrow Cu+{ Cu }^{ 2+ }$
+1 0 +2
Hence, only II, III and IV are redox reactions.

In the compound

$Y{ Ba }_{ 2 }{ Cu }_{ 3 }{ O }_{ 7 }$ which shows superconductivity, what is the oxidation state of

$Cu$ ?
Assume that the rare earth element yttrium is in its usual +3 oxidation state.

0%
$+\displaystyle\frac{7}{3}$
0%
$-\displaystyle\frac{7}{3}$
0%
$\displaystyle\frac{5}{3}$
0%
$-\displaystyle\frac{5}{3}$

Explanation

Let oxidation state of

$Cu =x$

So, in

$\left( Y{ Ba }_{ 2 }{ Cu }_{ 3 }{ O }_{ 7 } \right)$ ,

$3+\left( 2\times 2 \right) +3x-14=0$

$\therefore x=+\displaystyle\frac { 7 }{ 3 }$

In which of the following cases is the oxidation state of

$N$ atom wrongly calculated?

0%
$\begin{matrix} Compound & Oxidation\quad state \\ N{ H }_{ 4 }Cl & -3 \end{matrix}$
0%
$\begin{matrix} Compound & Oxidation\quad state \\ { \left( { N }_{ 2 }{ H }_{ 5 } \right) }_{ 2 }S{ O }_{ 4 } & +2 \end{matrix}$
0%
$\begin{matrix} Compound & Oxidation\quad state \\ { Mg }_{ 3 }{ N }_{ 2 } & -3 \end{matrix}$
0%
$\begin{matrix} Compound & Oxidation\quad state \\ NH_{ 2 }OH & -1 \end{matrix}$

Explanation

$\begin{matrix} Compound & Oxidation\quad state \\ { \left( { N }_{ 2 }{ H }_{ 5 } \right) }_{ 2 }S{ O }_{ 4 } & +2 \end{matrix}$ the oxidation state of

$N$ atom is wrongly calculated.
Let X be the oxidation state of N in

$(N_2H_5)_2SO_4$

$2X+5=+1$

$2X=-4$

$X=-2$

The oxidation number of

$Cr$ is

$+6$ in :

0%
$Fe{ Cr }_{ 2 }{ O }_{ 4 }$
0%
$KCr{ O }_{ 3 }Cl$
0%
$Cr{ O }_{ 5 }$
0%
${ \left[ Cr{ \left( OH \right) }_{ 4 } \right] }^{ \ominus }$

Explanation

${ Fe }^{ +2 }{ \left( { Cr }_{ 2 }{ O }_{ 4 } \right) }^{ 2- }:$ Oxidation state of

$Cr = +3$
b.

${ K }^{ \oplus }{ \left[ Cr{ O }_{ 3 }Cl \right] }^{ \ominus }:$ Oxidation state of

$Cr = +6$
c.

$CrO_5$ : Oxidation state of

$Cr = +6$ (Two peroxo linkage)
d.

${ \left[ Cr{ \left( OH \right) }_{ 4 } \right] }^{ \ominus }:$ Oxidation state of

$Cr = +3$
Hence, options B and C are correct.

Which of the following has/have been arranged in order of decreasing oxidation number of sulphur?

0%
${ H }_{ 2 }{ S }_{ 2 }{ O }_{ 7 }>{ Na }_{ 2 }{ S }_{ 4 }{ O }_{ 6 }>{ Na }_{ 2 }{ S }_{ 2 }{ O }_{ 3 }>{ S }_{ 8 }$
0%
$S{ O }^{ 2+ }>S{ O }_{ 4 }^{ 2- }>S{ O }_{ 3 }^{ 2- }>HS{ O }_{ 4 }^{ \ominus }$
0%
${ H }_{ 2 }S{ O }_{ 5 }>{ H }_{ 2 }S{ O }_{ 3 }>S{ Cl }_{ 2 }>{ H }_{ 2 }S$
0%
${ H }_{ 2 }S{ O }_{ 4 }>S{ O }_{ 2 }>{ H }_{ 2 }S>{ H }_{ 2 }{ S }_{ 2 }{ O }_{ 8 }$

Explanation

Let

$x$ be the oxidation state of S.
a. Oxidation state of

$S$ in

${ H }_{ 2 }{ S }_{ 2 }{ O }_{ 7 } : 2 + 2x - 14 = 0 \Rightarrow x = 6$

${ Na }_{ 2 }{ S }_{ 4 }{ O }_{ 6 } : 2 + 4x - 12 = 0 \Rightarrow x = 2.5$

${ Na }_{ 2 }{ S }_{ 2 }{ O }_{ 3 } : 2 + 2x - 6 = 0 \Rightarrow x = 2$

${ S }_{ 8 } : 8x = 0 \Rightarrow x = 0$
Therefore, the decreasing order is

${ H }_{ 2 }{ S }_{ 2 }{ O }_{ 7 }>{ Na }_{ 2 }{ S }_{ 4 }{ O }_{ 6 }>{ Na }_{ 2 }{ S }_{ 2 }{ O }_{ 3 }>{ S }_{ 8 }$ .
c.

${ H }_{ 2 }S{ O }_{ 5 }$ (Peroxo linkage)
Oxidation state of

$S = +6$

${ H }_{ 2 }S{ O }_{ 3 } : 2 + x - 6 = 0$
Oxidation state of

$S = +4$

$S{ Cl }_{ 2 } : x - 2 = 0$
Oxidation state of

$S = +2$

${ H }_{ 2 }S : 2 + x = 0$
Oxidation state of

$S = -2$
Therefore, the decreasing order is

${ H }_{ 2 }S{ O }_{ 5 }>{ H }_{ 2 }S{ O }_{ 3 }>S{ Cl }_{ 2 }>{ H }_{ 2 }S$ .
Similarly, we can check for options b and d, we will find them to be incorrect.

What is the oxidation number of tungsten in the ion

$W_6O_6Cl^{2-}_{12}$ ?

0%
2.7
0%
3.3
0%
3.7
0%
4.3

Explanation

$W_{6}O_{6}Cl_{12}^{2-}$

$6\times (x)+(-2)\times 6+12(-1)=-2$

$6x-12-12=-2$

$6x-24=-2$

$6x=24-2$

$6x=22$

$x=\dfrac{22}{6}$ =

$3.666$

$x=3.67$

$x=3.7$

It required

$40.05$ mL of

$\displaystyle 1 M \:Ce^{4+}$ to titrate

$20$ mL of

$\displaystyle 1 M \:Sn^{2+}$ to

$\displaystyle Sn^{4+}$ . What is the oxidation state of the cerium in the product?

0%
$+1$
0%
$+2$
0%
$0$
0%
$+3$

Explanation

$\displaystyle Ce^{+4}\:\:+Sn^{+2}\rightarrow Sn^{+4}\:\:+Ce^{+n}$

$40.05$

$20$ Volume required

$1 M$

$1 M$ Molarity

Meq. of

$\displaystyle Ce^{+4} =$ Meq. of

$Sn^{+2}$

$\displaystyle 40.05\times 1\times \left ( 4-n \right )=20\times 1\times 2$

$\displaystyle \left ( 4-n \right )=\frac{20\times 2}{40.05}\simeq 1$

$\therefore\displaystyle n=3$

Therefore, the oxidation state of

$Ce$ in the product is

$+3$ .

Which one of the following statements is/are not correct?

0%
Oxidation number of $S$ in $(NH_4)_2S_2O_8$ is $+7$ .
0%
Oxidation number of $Os$ in $OsO_4$ is $+8$ .
0%
Oxidation number of $S$ in $H_2SO_5$ is $+8$ .
0%
Oxidation number of $O$ in $KO_2$ is $+\frac{1}{2}$ .

Explanation

$H_2SO_5=2+x-10=0$

$\Rightarrow x=+8$

This is an impossible oxidation state of sulphur as

$S:3s^23p^4$ Maximum oxidation state=

$+6$

Therefore,

$H_2SO_3. O_2$

$\downarrow$

Oxide

$(-2)$ Peroxide

$(-1)$

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

$\Rightarrow 2+x-6-2=0$

$\Rightarrow x=6$ .

$KO_2$ is a superoxide. It is made up of

$K^=$ and

$O_2^-$ .

Therefore, single oxygen atom has

$-0.5$ oxidation state. So,

$KO_2=x-(0.5\times 2)=0$

$\Rightarrow x-1=0$

$\Rightarrow x=+1$ .

Oxidation state of phosphorus in the

$PH_4^+ ,PO_2^{3-}, PO_4^{3-}$ and

$PO_3^{3-}$ respectively, are :

0%
$-4, \:+1,\: +3,\: +5$
0%
$-3,\: +3,\: +5,\: +1$
0%
$+3,\: -3,\: +5,\: +1$
0%
$-3,\: +1,\: +5,\: +3$

Explanation

Let

$x$ be the oxidation number of

$P$ in

$PH_4^+$ .
Since, the overall charge on the complex is

$0$ , the sum of oxidation states of all elements in it should be equal to

$0$ .
Therefore,

$x+4(+1)=+1$
or,

$x=-3$
Similarly,
Let

$x$ be the oxidation number of P in

$PO_2^{3-}$ .
Therefore,

$x+2(-2)=-3$
or,

$x=+1$
Let

$x$ be the oxidation number of P in

$PO_4^{3-}$ .
Therefore,

$x+4(-2)=-3$
or,

$x=+5$
Let

$x$ be the oxidation number of P in

$PO_3^{3-}$ .
Therefore,

$x+3(-2)=-3$
or,

$x=+3$
The oxidation state of phosphorus in the

$PH_4^+ ,PO_2^{3-}, PO_4^{3-}$ and

$PO_3^{3-}$ are

$-3, +1, +5$ and

$+3$ respectively.

The oxidation state of oxygen of

$H_2O_2$ in the final products when it reacts with

$As_2O_3$ is :

0%
$0$
0%
$1$
0%
$-1$
0%
$-2$

Explanation

As the reaction between

$As_2O_3$ and

$H_2O_2$ is as follows:

$As_2O_3 + 3 H_2O_2 \rightarrow 3 H_2O + 2 As$

So here, the final product is

$H_2O$ , where O has Oxidation state to be -2.

Which of the following does not represent redox reaction?

0%
$Cr_2O^{2-}_7+2\overset{\circleddash}{O}H\rightarrow CrO^{2-}_4+H_2O$
0%
$SO^{2-}_5+2I^{\circleddash}+2H^{\bigoplus}\rightarrow I_2+SO^{2-}_4$
0%
$2Ca(OH)_2+2Cl_2\rightarrow Ca(ClO)_2+CaCl_2+2H_2O$
0%
$PCl_5\rightarrow PCl_3+Cl_2$

Explanation

$PCl_5\rightarrow PCl_3+Cl_2$
+5 +3 redox reaction

$2Ca(OH)_2+2Cl_2\rightarrow Ca(ClO)_2+CaCl_2+2H_2O$
0 +1 -1 redox reaction

$SO^{2-}_5+2I^{\circleddash}+2H^{\bigoplus}\rightarrow I_2+SO^{2-}_4$
-1 0 redox reaction

$Cr_2O^{2-}_7+2\overset{\circleddash}{O}H\rightarrow CrO^{2-}_4+H_2O$
+6 +6
Since, there is no change in oxidation state in this reaction, it is not a redox reaction.

The oxidation state of oxygen of

$H_2O_2$ in the final products when it reacts with

$ClO^{\circleddash}_3$ is :

0%
$0$
0%
$1$
0%
$-1$
0%
$-2$

Explanation

$H_2O_2 + 2ClO_2 \rightarrow 2ClO_2^- + O_2 + 2H^+$

This is now balanced. The charges are the same on both sides and the number of atoms of each element is the same on both sides.
As here, we have

$O_2$ as final product in this reaction, which is an element so this has oxidation state to be 0.

In the reaction

$SO_2+\frac{1}{2}O_2\xrightarrow [ ]{V_2O_5}SO_3$ , the vanadium changes from

$V^{5+}$ to:

0%
$V^{4+}$
0%
$V^{6+}$
0%
$V^{3+}$
0%
$V^{2+}$

Explanation

The catalyst only serves to increase the rate of reaction as it does not change the position of the thermodynamic equilibrium. The mechanism for the action of the catalyst comprises two steps:

Oxidation of

$SO_2$ into

$SO_3$ by

$V^{5+}$ :

$2SO_2 + 4V^{5+} + 2O^{2−} \rightarrow 2SO_3 + 4V^{4+}$

So, there is a change in oxidation state from the +5 to +4, means

$V^{5+}$ to

$V^{4+}$

Hence, option

$A$ is correct.

Consider a galvanic cell using solid

$Cu$ and

$Fe$ metals with their corresponding solutions.
What is the

$E^{\circ}_{cell}$ ?

Standard Potential (V)	Reduction Half-Reaction
$2.87$	$F_{2}(g) + 2e^{-} \rightarrow 2F^{-}(aq)$
$1.51$	$MnO_{4}^{-}(aq) + 8H^{+}(aq) + 5e^{-}\rightarrow Mn^{2+}(aq) + 4H_{2}O(l)$
$1.36$	$Cl_{2}(aq) + 3e^{-} \rightarrow 2Cl^{-}(aq)$
$1.33$	$Cr_{2}O_{7}^{2-} (aq) + 14H^{+}(aq) + 6e^{-} \rightarrow 2Cr^{3+}(aq) + 7H_{2}O(l)$
$1.23$	$O_{2}(g) + 4H^{+}(aq) + 4e^{-}\rightarrow 2H_{2}O(l)$
$1.06$	$Br_{2}(l) + 2e^{-} \rightarrow 2Br^{-}(aq)$
$0.96$	$NO_{3}^{-}(aq) + 4H^{+}(aq) + 3e^{-}\rightarrow NO(g) + H_{2}O(l)$
$0.80$	$Ag^{+}(aq) + e^{-} \rightarrow Ag(s)$ $
$0.77$	$Fe^{3+} (aq) + e^{-} \rightarrow Fe^{2+}(aq)$
$0.68$	$O_{2}(g) + 2H^{+}(aq) + 2e^{-}\rightarrow H_{2}O_{2}(aq)$
$0.59$	$MnO_{4}^{-}(aq) + 2H_{2}O(l) + 3e^{-}\rightarrow MnO_{2}(s) + 4OH^{-}(aq)$
$0.54$	$I_{2}(s) + 2e^{-}\rightarrow 2I^{-}(aq)$
$0.40$	$O_{2}(g) + 2H_{2}O(l) + 4e^{-} \rightarrow 4OH^{-}(aq)$
$0.34$	$Cu^{2+}(aq) + 2e^{-} \rightarrow Cu(s)$
$0$	$2H^{+}(aq) + 2e^{-}\rightarrow H_{2}(g)$
$-0.28$	$Ni^{2+}(aq) + 2e^{-}\rightarrow Ni(s)$
$-0.44$	$Fe^{2+}(aq) + 2e^{-}\rightarrow Fe(s)$
$-0.76$	$Zn^{2+}(aq) + 2e^{-}\rightarrow Zn(s)$
$-0.83$	$2H_{2}O(l) + 2e^{-}\rightarrow H_{2}(g) + 2OH^{-}(aq)$
$1.66$	$Al^{3+}(aq) + 3e^{-}\rightarrow Al(s)$
$-2.71$	$Na^{+}(aq) + e^{-} \rightarrow Na(s)$
$-3.05$	$Li^{+}(aq) + e^{-}\rightarrow Li(s)$

0%
$-0.78 V$
0%
$0.10 V$
0%
$0.78 V$
0%
$-0.10 V$

Explanation

$E^0$ for Cu = 0.34 V

$E^0$ for Fe = -0.44V

so,

$E^0_{cell} = 0.34 - (-0.44) = 0.78 V$

CBSE Questions for Class 11 Medical Chemistry Redox Reactions Quiz 11 - MCQExams.com

0:0:1

Practice Class 11 Medical Chemistry Quiz Questions and Answers

CBSE Questions for Class 11 Medical Chemistry Redox Reactions Quiz 11 - MCQExams.com

0:0:1

Practice Class 11 Medical Chemistry Quiz Questions and Answers

Support mcqexams.com by disabling your adblocker.