Consider the table of standard reduction potentials shown below. Half-reaction $$E^{\circ}$$ $$Cl_{2} + 2e^{-}\rightarrow 2Cl^{-}$$ $$1.36\ V$$ $$O_{2} + 4H^{+} + 4e^{-} \rightarrow 2H_{2}O$$ $$1.23\ V$$ $$2H_{2}O + 2e^{-} \rightarrow H_{2} + 2OH^{-}$$ $$-0.83\ V$$ $$Rb^{+} + e^{-} \rightarrow Rb$$ $$-2.93\ V$$ Use the information from the table and your knowledge of electrochemistry to predict the CORRECT net ionic equation for the reaction that will occur when an aqueous solution of rubidium chloride undergoes electrolysis.

$$2H_{2}O \rightarrow 2H_{2} + O_{2}$$

$$H_{2} + 2OH^{-} + 2Rb^{+} \rightarrow 2Rb + 2H_{2}O$$

$$4Cl^{-} + O_{2} + 4H^{+} \rightarrow 2H_{2}O + 2Cl_{2}$$

Fill up the table from the given choice. Element Oxidation number Oxygen -2 in most compounds ____(i) in $$H_2O_2$$ and _____(ii) in $$ OF_2$$ Halogen -1 for_____(iii) in all its compounds Hydrogen _____(iv) in most of its compounds ______(v)in binary metallic hydrides Sulphur ______(vi) in all sulphides

$$(i) -1 (ii) +2 (iii) F (iv) +1 (v) -1 (vi) -2 $$

$$(i) +1 (ii) +1 (iii) Cl (iv) +1 (v) -1 (vi) +2 $$

$$(i) -1 (ii) +1 (iii) F (iv) +1 (v) +2 (vi) +2$$

$$(i) +1 (ii) +2 (iii) Cl (iv) +1 (v) +1 (vi) +6$$

Match the column I with column II with the type of reaction and mark the appropriate choice. Column I Column II $$(A)$$ $$3Mg_(s) + N_{2(g)}\xrightarrow{\Delta } Mg_3N_{2(s)}$$ $$(i)$$ Displacement $$(B)$$ $$NaH_(s) + H_2O_(l)\rightarrow NaOH_{(aq)} + H_{2(g)}$$ $$(ii)$$ Decomposition $$(C)$$ $$3Cl{ O }_{( aq) }^{ - }\rightarrow 2C{ l}_{( aq) }^{ - } + Cl{ O_3 }_{ (aq) }^{ - }$$ $$(iii)$$ Combination $$(D)$$ $$2KClO_{3(s)}\rightarrow 2KCl_(s) + 3O_{2(g)}$$ $$(iv)$$ Disproportionation

$$(A)\rightarrow (iii) , (B)\rightarrow (i) , (C)\rightarrow (iv) ,D\rightarrow (ii)$$

$$(A)\rightarrow (i) , (B)\rightarrow (iii) , (C)\rightarrow (ii) ,D\rightarrow (iv)$$

$$(A)\rightarrow (iv) , (B)\rightarrow (iii) , (C)\rightarrow (ii) ,D\rightarrow (i)$$

$$(A)\rightarrow (ii) , (B)\rightarrow (i) , (C)\rightarrow (iii) ,D\rightarrow (iv)$$

MCQ Questions for Class 11 Medical Chemistry Redox Reactions Quiz 12

Statement 1 : The oxidation state of

$\displaystyle Cr$ in

$\displaystyle { { Al }_{ 2 }\left( { Cr }_{ 2 }{ O }_{ 7 } \right) }_{ 3 }$ is +3.
Statement 2 : As a neutral compound, the sum of oxidation numbers of all the atoms must equal to zero.

0%

Both Statement 1 and Statement 2 are correct and Statement 2 is the correct explanation of Statement 1.
0%

Both Statement 1 and Statement 2 are correct and Statement 2 is not the correct explanation of Statement 1.
0%

Statement 1 is correct but Statement 2 is not correct.
0%

Statement 1 is not correct but Statement 2 is correct.
0%

Both the Statement 1 and Statement 2 are not correct.

Which one has the highest oxidation number ?

0%
$NO$
0%
$N_2O$
0%
$NO_2$
0%
$N_2O_4$
0%
$NO_{3}{^-}$

Explanation

Nitrate ion

$\displaystyle NO_3^-$ has the highest oxidation number.
The oxidation number of N in nitrate ion is +5.

Oxidation number of iodine in

${ IO }_{ 3 }^{ - }, { IO }_{ 4 }^{ - }, KI$ and

$I_2$ respectively is _____

0%
-2, -5, -1 ,0.
0%
+5, +7, -1 , 0.
0%
+2, +5, +1 , 0.
0%
-1, +1 , 0, +1.

Explanation

Cosider the oxidation number of iodine as

$x$ . We know the charges on oxygen and potassium atoms are

$-2$ and

$+1$ respectively. The oxidation states of the compounds of iodine are calculated as follows:

$IO_{3}^- \rightarrow x + 3(-2) = -1 \Rightarrow x = +5$

$IO_{4}^- \rightarrow x + 4(-2) = -1 \Rightarrow x = +7$

$KI \rightarrow +1 + x = 0 \Rightarrow x = -1$

$I_2 \rightarrow 0\Rightarrow x = 0$

$\displaystyle 2Al\left( s \right) +6HCl\left( aq \right) \rightarrow 2{ Al }^{ 3+ }\left( aq \right) +6{ Cl }^{ - }\left( aq \right) +3{ H }_{ 2 }\left( g \right)$
Which of the following is correct about the above reaction?

0%
Precipitation reaction
0%
Acid/base reaction
0%
Redox reaction
0%
Combustion reaction
0%
Decomposition reaction

Explanation

$2Al + 6HCl \rightarrow 2Al^3+ + 6Cl^- + 3H_2$
here,

$Al$ is oxidised to

$Al^{3+}$ and

$H^+$ ( in

$HCl$ ) is reduced to

$H^0$ ( in

$H_2$ )
Redox is a contraction of the name for a chemical reduction–oxidation reaction. Any such reaction involves both a reduction process and a complementary oxidation process. Redox reactions include all chemical reactions in which atoms have their oxidation state changed; in general, redox reactions involve the transfer of electrons between chemical species. The chemical species from which the electron is stripped is said to have been oxidized, while the chemical species to which the electron is added is said to have been reduced. Oxygen is not necessarily included in such reactions as other chemical species can serve the same function.

The term "redox" comes from two concepts involved with electron transfer: reduction and oxidation.[1] It can be explained in simple terms:
Oxidation is the loss of electrons or an increase in oxidation state by a molecule, atom, or ion.
The reduction is the gain of electrons or a decrease in oxidation state by a molecule, atom, or ion.

Consider the table of standard reduction potentials shown below.

Half-reaction	$E^{\circ}$
$Cl_{2} + 2e^{-}\rightarrow 2Cl^{-}$	$1.36\ V$
$O_{2} + 4H^{+} + 4e^{-} \rightarrow 2H_{2}O$	$1.23\ V$
$2H_{2}O + 2e^{-} \rightarrow H_{2} + 2OH^{-}$	$-0.83\ V$
$Rb^{+} + e^{-} \rightarrow Rb$	$-2.93\ V$

Use the information from the table and your knowledge of electrochemistry to predict the CORRECT net ionic equation for the reaction that will occur when an aqueous solution of rubidium chloride undergoes electrolysis.

0%
$2Rb^{+} + 2Cl^{-} \rightarrow 2Rb + Cl_{2}$
0%
$2H_{2}O \rightarrow 2H_{2} + O_{2}$
0%
$H_{2} + 2OH^{-} + 2Rb^{+} \rightarrow 2Rb + 2H_{2}O$
0%
$4Cl^{-} + O_{2} + 4H^{+} \rightarrow 2H_{2}O + 2Cl_{2}$

Explanation

From the given table, we have;

Rb as the lowest reduction potential therefore it undergo oxidation

and

$Cl_2$ have the highest oxidation potential therefore it undergo reduction.

so, the net reaction becomes

$2Rb^+ + 2Cl^- \rightarrow 2Rb + Cl_2$

A steady current of

$10.0$ Amps is passed through a nickel production cell of

$15$ minutes.
Which of the following is the correct expression for calculating the number of grams of nickel produced?
Note:

$\bullet 1\ faraday = 96,500\ Coulombs$

$\bullet$ The electroytic cell involves the following half-reaction:

$Ni_{(aq)}^{2+} + 2e^{-} \rightarrow Ni_{(s)}$

0%
$\dfrac {(10.0)(15)(96,500)}{(59)(60)} g$
0%
$\dfrac {(10.0)(15)(59)}{(60)(96,500)} g$
0%
$\dfrac {(10.0)(15)(60)(59)}{(96,500)(2)} g$
0%
$\dfrac {(96,500)(59)}{(10)(15)(60)(2)} g$

Explanation

According to Faraday's Law

$W= \cfrac {Z \times I \times t}{F}$

$W$ = mass of compound

$Z$ = Equivalent mass

$F$ = 96500

$I$ = Current in Ampere

$t$ = time in seconds

$\Rightarrow W= \left(\cfrac {59}{2}\right)\cfrac {\times 15\times 60 sec\times 10A}{96500}$

$\Rightarrow W= \cfrac {(10.0)(15)(60)(59)}{(96500)(2)}g$

Oxidation number of the nitrogen atom in

$HNO_3$ is :

0%
-1
0%
+1
0%
0
0%
+4
0%
+5

Explanation

Let X be the oxidation number of the nitrogen atom in

$\displaystyle HNO_3$ .

$\displaystyle +1 + X+ 3(-2) = 0 \\ X - 5 = 0 \\ X = +5$
Hence, the oxidation number of the nitrogen atom in

$\displaystyle HNO_3$ is +5.

Match the List - I with List - II and select the correct matching from the codes given below:

Column - I	Column - II
A. $NaN_3$	+5
B. $N_2H_4$	+1
C. $NH_2OH$	-1/3
D. $N_2O_5$	-2

0%
A-3 B-4 C-2 D-1
0%
A-4 B-3 C-1 D-2
0%
A-3 B-4 C-1 D-2
0%
A-4 B-3 C-2 D-1

For the redox process,

$Zn(s) + Cu^{2+} \rightleftharpoons Zn^{2} + Cu(s)\ E_{cell}^{\circ} = + 1.10\ V$
which graph correctly represents

$E_{cell}$ (Y-axis) as a function of

$\log \dfrac {[Zn^{2+}]}{[Cu^{2+}]}$ (X-axis)?

Explanation

At the initial stages voltage is greater than

$e^o$ because the reaction quotient is less than

$1(Q<1)$ .

As the reaction proceeds

the ratio Q=

$\cfrac {[Zn^{+2}]}{[Cu^{+2}]}$ steadily increases and cell voltage steadily decreases then graph looks like

(Refer to Image)

1122426_645737_ans_9aa5a944e53e42dbb59a16640f816eb0.PNG

The Metal with the highest oxidation state present in

$K_2CrO_4$ ,

$NbCl_5$ and

$MnO_2$ is:

0%
$Nb$
0%
$Mn$
0%
$K$
0%
$Cr$

Explanation

Oxidation state of

$Cr$ in

$K_2Cr_2O_7=+6$

Oxidation state of

$Nb$ in

$NbCl_5=+5$

Oxidation state of

$Mn$ in

$MnO_2=+4$

Hence option D is correct.

Oxidation number of N in

$NH_4NO_3$ is:

0%
$-3$
0%
$+5$
0%
$-3$ and $+5$
0%
$+3$ and $-5$

Explanation

$NH_4NO_3$ exists as

$NH_4^{+}$ and

$NO_3^{-}$ .Oxidation state of

$N$ in

$NH_4^{+}$ is -3 and in

$NO_3^{-}$ is +5.

Note: Oxidation state of oxygen is -2 and of hydrogen is +1.

A mixture of potassium chlorate, oxalic acid and sulphuric acid is heated. During the reaction which element undergoes maximum change in the oxidation number?

0%
Cl
0%
C
0%
S
0%
H

Among the following, identify the species with an atom in

$+6$ oxidation state.

0%
$MnO^-_4$
0%
$Cr(CN)^{3-}_6$
0%
$NiF^{2-}_6$
0%
$CrO_2Cl_2$

Explanation

Hint:

The total number of electrons an atom can lose or gain to form a bond with other elements is called the oxidation state of that atom.

Explanation:

In $MnO_{4}^{-}$ , let the oxidation state of $Mn$ be $x$

$x\ +\ (4\left ( -2 \right ))=-1$ $\Rightarrow x=+7$ .

In $Cr\left ( CN \right )_{6}^{3-}$ , let the oxidation state of $Cr$ be $y$

$y\ +\ (6\left ( -1 \right ))=-3$ $\Rightarrow y=+3$ .

In $NiF_{6}^{2-}$ , let the oxidation state of $Ni$ be $z$

$z\ +\ (6\left ( -1 \right ))=-2$ $\Rightarrow z=+4$ .

In $CrO_2Cl_2$ , let the oxidation state of $Cr$ be $q$

$q\ +\ (2\left ( -2 \right ))\ +\ (2\left ( -1 \right ))=0$ $\Rightarrow q=+6$ .

Hence, $CrO_2Cl_2$ has $+6$ oxidation state.

Therefore, the correct answer is option $D.$

The oxidation numbers of the sulphur atoms in peroxomonosulphuric acid

$(H_{2}SO_{5})$ and peroxodisulphuric acid

$(H_{2}S_{2}O_{8})$ are respectively

0%
$+8$ and $+7$
0%
$+3$ and $+3$
0%
$+6$ and $+6$
0%
$+4$ and $+6$

Explanation

Both the peroxomonosulphuric acid

$(H_{2}SO_{5})$ and peroxodisulphuric acid

$(H_{2}S_{2}O_{8})$ have peroxide linkage and have the following structures.

Thus, the oxidation number of

$S$ in

$H_{2}SO_{5}$ is calculated as:

$\underset {(for\ H)}{2\times (+1)} + \underset {(for\ S)}{X} + \underset {for\ O}{3\times (-2)} + \underset {(for\ peroxide)}{2\times (-1)}=0$

$2 + x - 6 - 2 = 0$

$\therefore x = +6$

Thus, oxidation number of

$S$ in

$H_{2}S_{2}O_{8}$ is calculated as

$\underset {(for\ H)}{2\times (+1)} + \underset {(for\ S)}{2\times (x)} + \underset {(for\ O)}{6\times (-2)} + \underset {(for\ peroxide)}{2\times (-1)} = 0$

$2 + 2x - 12 + 2 = 0$

$\therefore x = + 6$ .

$CH_3COOH$ is neutralized by

$NaOH$ . Conductometric titration curve will be of the type:

Explanation

When

$CH_3COOH$ is neutralized with

$NaOH$ . The conductometric curve is as follows-

This is because, weak acid dissouates partiaaly into

$H^+$ &

$CH_3COO^-$ ions & show no, of ions few in the solution. Hence the solution has low conductance value. When, few drops of strong base which dissociates as

$(Na^+ + OH^-)$ completely acid-base reaction occurs, the

$OH^-$ ion consumes

$H^+$ to form

$H_2O$ . So, by adding few drops of base at first conductance decrease slightly, ith further addition of more drops of base all

$H^+$ are consumed & more drops of base, all

$H^+$ are consumed & dissociation of weak acid increases which is reflected by gradual increase in conductance.

1117501_667497_ans_431df48ab4d442b9b4bbb43d5a1950aa.png

Fill up the table from the given choice.

Element	Oxidation number
Oxygen	-2 in most compounds ____(i) in $H_2O_2$ and _____(ii) in $OF_2$
Halogen	-1 for_____(iii) in all its compounds
Hydrogen	_____(iv) in most of its compounds ______(v)in binary metallic hydrides
Sulphur	______(vi) in all sulphides

0%
$(i) +1 (ii) +1 (iii) Cl (iv) +1 (v) -1 (vi) +2$
0%
$(i) -1 (ii) +2 (iii) F (iv) +1 (v) -1 (vi) -2$
0%
$(i) -1 (ii) +1 (iii) F (iv) +1 (v) +2 (vi) +2$
0%
$(i) +1 (ii) +2 (iii) Cl (iv) +1 (v) +1 (vi) +6$

Explanation

Oxygen: -2 in most compounds (

$2 \times 1+2x=0 \Rightarrow x=-1)\ \underline { -1 }$ (i) in

$H_2O_2$ and (

$x-2 \Rightarrow x=+2)\ \underline { +2 }$ (ii) in

$OF_2$

Halogen: -1 for

$\underline { Fluorine(F) }$ (iii) in all its compounds

Hydrogen:

$\underline { +1 }$ (iv) in most of its compounds

$\underline { -1 }$ (v) in binary metallic hydrides

Sulphur:

$\underline { -2 }$ (vi) in all sulphides.

In the reaction,

$4Fe+{ 3O }_{ 2 }\rightarrow { 4Fe }^{ 3+ }+{ 6O }^{ 2- }$ which of the following statement is incorrect?

0%
It is redox reaction
0%
Metallic iron is a reducing agent
0%
${ Fe }^{ 3+ }$ is an oxidising agent
0%
Metallic iron is reduced to ${ Fe }^{ 3+ }$

Explanation

$\begin{array}{l} \text { The given reaction is redox reaction } \\ \text { as Fe oxidises to Fe }^{3+} \text { and } \mathrm{O}_{2} \\ \text { reduces to } O^{2-} \text { . } \end{array}$

$\begin{array}{l}\text{In this reaction metallic iron }\\ \text { is acting like a reducing agent. }\end{array}$

$\begin{array}{l} \text { If, we reverse the given reaction, } \\\text{Fe}^{3+}\text{ will behave as an oxidizing }\\ \text { agent. } \end{array}$

${ H }_{ 2 }{ O }_{ 2 }(aq)+{ Cr }_{ 2 }{ O }_{ 7 }^{ 2- }\longrightarrow { O }_{ 2 }(g)+{ Cr }^{ 3+ }(aq)$

This reaction occurs in acidic medium.

0%
True
0%
False

Explanation

$H_{2}O_{2}(aq)+Cr_{2}O_{7}^{2-}\rightarrow O_{2}(g)+Cr^{3+}(aq)$

The above reaction occurs in acidic medium.

Arrange the following in increasing order of oxidation state of

$Ni$ .

$K_2[Ni(CN)_4],K_2[NiF_6],Ni(CO)_4$

0%
$Ni(CO)_4,K_2[Ni(CN)_4],K_2[NiF_6]$
0%
$K_2[Ni(CN)_4],Ni(CO)_4,K_2[NiF_6]$
0%
$Ni(CO)_4,K_2[NiF_6],K_2[Ni(CN)_4]$
0%
$K_2[NiF_6],K_2[Ni(CN)_4],K_2[Ni(CN)_4]"$

Explanation

Let the oxidation number of

$Ni$ be

$x$ .

Oxidation number of

$K=+1$ ,

$CN=F=-1$ and of

$CO=0$ .

(A)

$K_2[Ni(CN)_4]$ :

$2 \times 1+x+4 \times (−1)=0$

$2+x-4=0$ or

$x=+2$

(B)

$K_2[NiF_6]$ :

$2 \times 1+x+6 \times (−1)=0$

$2+x-6=0$ or

$x=+4$

(C)

$Ni(CO)_4]$ :

$x+4 \times 0=0$

$x+0=0$ or

$x=0$

thus the increasing order of oxidation state of

$Ni$ is

$0<+2<+4$ in:

$Ni(CO)_4<K_2[Ni(CN)_4] < K_2[NiF_6]$

When a piece of sodium metal is dropped in water, hydrogen gas evolved because:

0%
sodium is reduced and acts as an oxidising agent
0%
water is oxidised and acts as a reducing agent
0%
sodium loses electrons and is oxidised while water in reduced
0%
water loses electrons and is oxidised to hydrogen

Explanation

Reaction of sodium metal with water results in the evolution of hydrogen gas and formation of sodium hydroxide as:

$2\overset { 0 }{ Na } +2\overset { +1 }{ { H }_{ 2 } } O\quad \longrightarrow 2\overset { +1 }{ Na } OH+\overset { 0 }{ { H }_{ 2 } }$

Thus, the metal (Na) gets oxidized by lossing one electron and water gets reduced to hydrogen gas by gaining electron.

For the redox reaction,

$MnO_{4}^{-} + C_{2}O_{4}^{2-} + H^{+}\rightarrow Mn^{2+} + CO_{2} + H_{2}O$ , the correct coefficients of the reactants for the balanced equation are _____________.

0%
$MnO_{4}^{-} = 2, C_{2}O_{4}^{2-} = 16, H^{+} = 5$
0%
$MnO_{4}^{-} = 16, C_{2}O_{4}^{2-} = 5, H^{+} = 2$
0%
$MnO_{4}^{-} = 5, C_{2}O_{4}^{2-} = 16, H^{+} = 2$
0%
$MnO_{4}^{-} = 2, C_{2}O_{4}^{2-} = 5, H^{+} = 16$

Explanation

The unbalanced redox reaction is

$\displaystyle MnO_{4}^{-} + C_{2}O_{4}^{2-} + H^{+}\rightarrow Mn^{2+} + CO_{2} + H_{2}O$
Balance C atoms.

$\displaystyle MnO_{4}^{-} + C_{2}O_{4}^{2-} + H^{+}\rightarrow Mn^{2+} + 2CO_{2} + H_{2}O$

The oxidation number of Mn decreases from

$\displaystyle +7$ to

$\displaystyle +2$ .
The decrease in the oxidation number is

$\displaystyle 7-2=5$
The oxidation number of C increases from

$\displaystyle +3$ to

$\displaystyle +4$ .
The increase in the oxidation number for 1 C atom is

$\displaystyle 4-3=1$
The increase in the oxidation number for 2 C atom is

$\displaystyle 2 \times 1=2$

To balance the increase in the oxidation number with decrease in the oxidation number, multiply Mn containing species with 2 and C containing species with 5.

$\displaystyle 2MnO_{4}^{-} + 5C_{2}O_{4}^{2-} \rightarrow 2Mn^{2+} + 10CO_{2}$

Balance O atoms by adding 8 water molecules to the products side.

$\displaystyle 2MnO_{4}^{-} + 5C_{2}O_{4}^{2-} \rightarrow 2Mn^{2+} + 10CO_{2} + 8H_{2}O$

Balance H atoms by adding 16

$\displaystyle H^+$ ions to the reactants side.

$\displaystyle 2MnO_{4}^{-} + 5C_{2}O_{4}^{2-} + 16H^{+}\rightarrow 2Mn^{2+} + 10CO_{2} + 8H_{2}O$

This is the balanced equation.

In the chemical reaction

${ K }_{ 2 }{ Cr }_{ 2 }{ O }_{ 7 }+x{ H }_{ 2 }{ SO }_{ 4 }+y{ SO }_{ 4 }\longrightarrow { K }_{ 2 }{ SO }_{ 4 }+{ Cr }_{ 2 }{ \left( { SO }_{ 4 } \right) }_{ 3 }+z{ H }_{ 2 }O$
Values of x, y and z are:

0%
1, 3, 1
0%
4, 1, 4
0%
3, 2, 3
0%
2, 1, 2

A metal, M forms chlorides in its +2 and +4 oxidation states. Which of the following statements about these chlorides is correct?

0%
$MCl_2$ is less ionic than $MCl_4$
0%
$MCl_2$ is less easily hydrolysed than $MCl_4$
0%
$MCl_2$ is more volatile than $MCl_4$
0%
$MCl_2$ is more soluble in anhydrous ethanol than $MCl_4$

Which of the following reaction(s) is/are example(s) of intra-redox reaction?

0%
$2NaAg{\left( {CN} \right)_2} + Zn \to N{a_2}Zn{\left( {CN} \right)_4} + 2Ag$
0%
$Ba{O_2} + {H_2}S{O_4} \to BaS{O_4} + {H_2}O$
0%
${N_2}{O_5} + {H_2}O \to 2HN{O_3}$
0%
$AgN{O_3} + KI \to AgI + KN{O_3}$

For the redox reaction:

$Zn+N{ O }^-_3\rightarrow Zn^{2+}+NH^+_4$

in a basic medium, coefficients of

$Zn, \ NO^-_3$ and

$OH^-$ in the balanced equation respectively are:

0%
$4, 1, 7$
0%
$7, 4, 1$
0%
$4, 1, 10$
0%
$1, 4, 10$

Explanation

$4 Zn+1 \mathrm{NO}_{3}+10 \mathrm{H}^{+} \longrightarrow 4Zn^{+2}+\mathrm{NH}_{4}^{+}+3 \mathrm{H}_{2} \mathrm{O}$

$\mathrm{Zn}, \mathrm{NO}_{3}^- \& \mathrm{^-OH}=4,1,10$

2004611_1025188_ans_9d5e8131141f43a5839a1a6581483a45.jpeg

Half cell reactions for some electrodes are given below:

$A+{ e }^{ - }\longrightarrow { A }^{ - }\ ;\quad \quad \ { E }^{ 0 }=0.96V$

II.

${ B }^{ - }+{ e }^{ - }\longrightarrow { B }^{ 2- }\ ;\quad { E }^{ o }=-0.12V$
III.

${ C }^{ + }+{ e }^{ - }\longrightarrow C\ ;\quad \ { E }^{ o }=+0.18V$

IV.

${ D }^{ 2+ }+{ 2e }^{ - }\longrightarrow D\ ;\quad { E }^{ o }=-1.12V$

The largest potential will be generated in which cell?

0%
${ A }^{ - }|A\parallel { B }^{ - }|{ B }^{ 2 }$
0%
$D|{ D }^{ 2+ }\parallel A|{ A }^{ - }$
0%
${ B }^{ 2 }|{ B }^{ - }\parallel { C }^{ + }|C$
0%
$D|{ D }^{ 2+ }\parallel { C }^{ + }|C$

Match the column I with column II with the type of reaction and mark the appropriate choice.

	Column I		Column II
$(A)$	$3Mg_(s) + N_{2(g)}\xrightarrow{\Delta } Mg_3N_{2(s)}$	$(i)$	Displacement
$(B)$	$NaH_(s) + H_2O_(l)\rightarrow NaOH_{(aq)} + H_{2(g)}$	$(ii)$	Decomposition
$(C)$	$3Cl{ O }_{( aq) }^{ - }\rightarrow 2C{ l}_{( aq) }^{ - } + Cl{ O_3 }_{ (aq) }^{ - }$	$(iii)$	Combination
$(D)$	$2KClO_{3(s)}\rightarrow 2KCl_(s) + 3O_{2(g)}$	$(iv)$	Disproportionation

0%
$(A)\rightarrow (i) , (B)\rightarrow (iii) , (C)\rightarrow (ii) ,D\rightarrow (iv)$
0%
$(A)\rightarrow (iv) , (B)\rightarrow (iii) , (C)\rightarrow (ii) ,D\rightarrow (i)$
0%
$(A)\rightarrow (ii) , (B)\rightarrow (i) , (C)\rightarrow (iii) ,D\rightarrow (iv)$
0%
$(A)\rightarrow (iii) , (B)\rightarrow (i) , (C)\rightarrow (iv) ,D\rightarrow (ii)$

Explanation

	Column I	Column II	Reason
$A$	$3Mg(s)+N_2(g) \rightarrow Mg_3N_2(s)$	(iii) Combination	A combination reaction is defined as a chemical reaction in which two or more substances combine to form a single substance under suitable conditions.
$B$	$NaH(s)+H_2O(l)\rightarrow NaOH(aq)+H_2(g)$	(i) Displacement	Displacement reaction is a chemical reaction in which a more reactive element (NaH) displaces a less reactive element ( $H_2O \rightarrow H_2$ ) from its compound. Both metals and non-metals take part in displacement reactions.
$C$	$3ClO^−(aq) \rightarrow 2Cl^−(aq)+ClO_3^−(aq)$	(iv) Disproportionation	Disproportionation reaction is a redox reaction in which a compound of intermediate oxidation state converts to two different compounds, one of higher and one of lower oxidation states as: $3\overset {+1}{Cl}O^−(aq) \rightarrow 2\overset {-1}{Cl^−}(aq)+\overset {+5}{Cl}O_3^−(aq)$
$D$	$2KClO_3(s) \rightarrow 2KCl(s)+3O_2(g)$	(ii) Decomposition	A decomposition reaction is a type of chemical reaction in which a single compound breaks down into two or more elements or new compounds.

$(A)-(iii),\ (B)-(i),\ (C)-(iv),\ D-(ii)$

$KMnO_4$ acts as an oxidizing agent in acidic medium. The number of moles of

$KMnO_4$ that will be needed to react with one mole of sulphide ions in acidic solution is:

0%
$\dfrac{2}{5}$
0%
$\dfrac{3}{5}$
0%
$\dfrac{4}{5}$
0%
$\dfrac{1}{5}$

Explanation

$5\overline { e } +Mn{ O }_{ 4 }^{ -2 }\longrightarrow M{ n }^{ +2 }\times 2\\ +\quad \quad \quad { 5 }^{ -2 }+2\overline { e } \longrightarrow S\times 2\\ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \\ 2Mn{ O }_{ 4 }^{ - }+5{ S }^{ -2 }\longrightarrow 2Mn^{ +2 }+5S$

Divide reaction by

$5.$

$\cfrac { 2 }{ 5 } Mn{ O }_{ 4 }^{ - }+{ 5 }^{ -2 }\longrightarrow 2Mn^{ +2 }+5S$

Hence

$1$ mole of sulphide ions will require

$\cfrac { 2 }{ 5 }$ moles of

$Mn{ O }_{ 4 }^{ - }$ ions.

In the electrolysis of aqueous sodium chloride solution which of the half-cell reaction will occur at the anode?

0%
$Na^+_{(aq)}+e^-\rightarrow Na_{(s)}; E^o_{cell}=-2.71$ V
0%
$2H_2O_{(l)}\rightarrow O_{2(g)}+4H^+_{(aq)}+4e^-; E^o_{cell}=1.23$ V
0%
$H^+_{aq}+e^-\rightarrow \dfrac{1}{2}H_{2(g)}; E^o_{cell}=0.00$ V
0%
$Cl^-_{(aq)}\rightarrow \dfrac{1}{2}Cl_{2(g)}+e^-; E^o_{cell}=1.36$ V

Explanation

At anode, oxidation takes place.

According to preferential discharge theory, more is the

$\varepsilon_{cell}$ of anode potential more is the tendency for reaction to take place.

hence,

$Cl_2$ gas is evolved.

$Cl^{\circleddash}\longrightarrow \cfrac{1}{2}Cl_2+e^-\quad\quad \varepsilon^o_{cell}=1.36$

$V$

Which of the following is

$\textbf{not}$ an example of redox reaction?

0%
$CuO + H_{2} \rightarrow Cu + H_{2}O$
0%
$Fe_{2}O_{3} + 3CO \rightarrow 2Fe + 3CO_{2}$
0%
$2K + F_{2} \rightarrow 2KF$
0%
$BaCl_{2} + H_{2}SO_{4} \rightarrow BaSO_{4} + 2HCl$

Explanation

Redox reaction involves simultaneous oxidation and reduction of reacting species. Thus change in oxidation state will decide whether a reaction is redox or not. Thus assigning the oxidation states as:

$\overset {+2}{Cu}O + \overset {0}{H_2} \rightarrow \overset {0}{Cu} + \overset {+1}{H_2}O$ : oxidation of H and reduction of Cu

$\overset {+3}{Fe_2}O_3 + 3\overset {+2}{C}O \rightarrow 2\overset {0}{Fe} + 3\overset {+4}{C}O_2$ : oxidation of C and reduction of Fe

$2\overset {0}{K} + \overset {0}{F_2} \rightarrow 2\overset {+1}{K}\overset {-1}{F}$ : oxidation of K and reduction of F

$\overset {+2}{Ba}Cl_2 + H_2SO_4 \rightarrow \overset {+2}{Ba}SO_4 + 2HCl$ : not a redox reaction.

Based on the cell notation for spontaneous reaction, at the anode:

$Ag(s)|AgCl(s)|{ Cl }^{ - }(aq)\parallel { Br }^{ - }(aq)|{ Br }_{ 2 }(I)|C(s)$

0%
$AgCl$ gets reduced
0%
$Ag$ gets oxidized
0%
${ Br }^{ - }$ gets oxidized
0%
${ Br }_{ 2 }$ gets reduced

Explanation

$Ag\left( s \right) \left| AgCl\left( s \right) \right| { Cl }^{ \left( - \right) }\left| { Br }^{ - } \right| \left| { Br }_{ 2 } \right| C\left( s \right)$

Anode :

$Ag+{ Cl }^{ \left( - \right) }\rightarrow AgCl+{ e }^{ - }$

Cathode :

$\dfrac { 1 }{ 2 } { Br }_{ 2 }+{ e }^{ \left( - \right) }\rightarrow { Br }^{ \left( - \right) }$

$\therefore$ at anode, oxidation takes place.

Hence,

$Ag$ gets oxidized.

The oxidation number and coordination number of

$M$ in the compound,

$[M(SO_4)(NH_3)_5]$ will be:

0%
10 and 3
0%
1 and 6
0%
6 and 4
0%
2 and 6

Explanation

Since

$M$ has 6 ligands attached to it, the co-ordination no. of the compound is 6. Here, the anionic species inside the co-ordination sphere is

$(SO_{4})^{2-}$ . Since the compound as a whole is neutral, the oxidation no. of

$M$ should be 2.

Hence, the correct option is

$D$

The magnetic moment of

$M^{x+}$ (atomic number

$= 25$ ) is

$\sqrt{15}$ . Then, the oxidation number

$x$ of

$M$ is:

0%
4
0%
3
0%
2
0%
none of these

Explanation

$\sqrt { n\left( n+2 \right) } =\sqrt { 15 }$

Squaring both sides

$\Rightarrow n\left( n+2 \right) =15$

$\Rightarrow { n }^{ v }+2n-15=0$

$\Rightarrow { n }^{ v }+5n-3n-15=0$

$\Rightarrow n\left( n+5 \right) -3\left( n+5 \right) =0$

$\Rightarrow \left( n+5 \right) \left( n-3 \right) =0$

Solving we get,

$n=3$

Thus there are

$3$ unpaired electrons in the element.

The configuration of an element with atomic number

$2S$ is

$\rightarrow$

${ 1S }^{ 2 }{ 2S }^{ 2 }{ 2P }^{ 6 }{ 3S }^{ 2 }{ 3P }^{ 6 }{ 4S }^{ 2 }{ 3d }^{ 5 }$ . This configuration has electrons in the valence shell.

We have been given that the symbol of the element is

${ M }^{ x+ }$ .

So, oxidation number of

$M=5-3$

$=2$

Oxidation number of

$'Co'$ in

$Hg[Co(SCN)_{4}]$

0%
$+2$
0%
$+1$
0%
$+3$
0%
$+5$

In acid medium, the standard reduction potential of

$NO$ converted to

${ N }_{ 2 }O$ is

$1.59 V$ . Its standard potential in alkaline medium would be:

0%
$-1.59 V$
0%
$-0.764 V$
0%
$0.764 V$
0%
$0.062 V$

Explanation

$NO \longrightarrow N_2O$

$E^o= 1.59V$

The reaction in acidic medium,

$2NO+2H^+ \longrightarrow N_2O+H_2O$ ;

$E^o=1.59$

In basic medium,

$2NO+H_2O \longrightarrow N_2O+2OH^-$ ; Let

$E^o=x$

Now, in alkaline medium,

$H_2O \rightleftharpoons H^++OH^-$

$\Rightarrow K=10^{-14}$

or,

$2H_2O \rightleftharpoons 2H^++2OH^-$ ,

$K= 10^{-28}$

We know,

$E^o= \cfrac {0.059}{n}\log K$

$=\cfrac {0.059}{2}\log 10^{-28}= 0.826$

For standard potential,

$E^o_{acidic}+E^o_{basic}=E^o_{cell}$

$\Rightarrow 1.59+x= 0.826$

$\Rightarrow x= 0.764$

Write balanced equations for the half reactions and calculate the reduction potentials at

$25^0C$ for the following half cells :

$Cl^-(1.2M) | Cl_2(g, 3.6 atm) | Pt E^o = 1.36 V$

0%
$1.372 V$
0%
$1.363 V$
0%
$1.358 V$
0%
$1.342 V$

The pair of compounds in which both the metals are in the highest possible oxidation state is

0%
$FeSO_{4},CuS_{2}$
0%
$CrO_{2}Cl_{2}, MnO_{4}^{-}$
0%
$TiO_{2},MnO_{2}$
0%
$[Co(CN_{3})]^{3-}, MnO_{3}$

Oxidation number of Cl in

$KClO_3$ :

0%
-1
0%
-5
0%
+1
0%
+5

In the given reaction, the increase in the oxidation number of sulphur is:

${H_2}{O_2} + PbS \to {H_2}O + PbS{O_4}$

0%
$2$ units
0%
$4$ units
0%
$6$ units
0%
$8$ units

The oxidation state of A,B and C in compaund are -2, +5, and -2, respectively. the compound is ?

0%
$A_3( BC)_{2}$
0%
$A_2( BC )_{3}$
0%
$A_2 ( BC )_{4}$
0%
$A_2 ( BC )_{5}$

Justify that the following reactions are redox reactions.

0%
$CuO\left( s \right) +{ H }_{ 2 }\left( g \right) \rightarrow Cu\left( s \right) +{ H }_{ 2 }O\left( g \right)$
0%
${ Fe }_{ 2 }{ O }_{ 3 }\left( s \right) +3CO\left( g \right) \rightarrow 2Fe\left( s \right) +3{ CO }_{ 2 }\left( g \right)$
0%
$4{BCl }_{ 3 }\left( g \right) +3Li{AlH }_{ 4 }\left( s \right) \rightarrow 2{ B }_{ 2 }{ H }_{ 6 }\left( s \right) +3LiCl\left( s \right)+3{AlCl }_{ 3 }\left( s \right)$
0%
$2K\left( s \right) +{ F }_{ 2 }\left( g \right) \rightarrow { 2K }^{ + }{ F }^{ - }\left( s \right)$

Explanation

$(A)$

$\overset { +2 }{ Cu } O\left( s \right) +{ \overset { 0 }{ H } }_{ 2 }\left( g \right) \longrightarrow \overset { 0 }{ Cu } \left( s \right) +{ \overset { +1 }{ H } }_{ 2 }O\left( g \right)$

Here, reduction of

$\overset { +2 }{ Cu } O$ to

$\overset { 0 }{ Cu }$ and oxidation of

${ \overset { 0 }{ H } }_{ 2 }$ to

${ \overset { +1 }{ H } }_{ 2 }O$ takes place. Therefore, it is a redox reaction.

$(B)$

${ \overset { +3 }{ Fe } }_{ 2 }{ O }_{ 3 }\left( s \right) +3\overset { +2 }{ CO } \left( g \right) \longrightarrow 2\overset { 0 }{ Fe } \left( s \right) +3{ \overset { +4 }{ CO } }_{ 2 }\left( g \right)$

Here, reduction of

${ \overset { +3 }{ Fe } }_{ 2 }{ O }_{ 3 }$ to

$\overset { 0 }{ Fe }$ and oxidation of

$\overset { +2 }{ CO }$ takes place. Therefore, it is a redox reaction.

$(C)$

$4\overset { +3 }{ B } { Cl }_{ 3 }\left( g \right) +3Li\overset { +3 }{ AI } { H }_{ 4 }\left( s \right) \longrightarrow 3{ \overset { +3 }{ B } }_{ 2 }{ H }_{ 6 }\left( s \right) +3LiCl\left( s \right) +3\overset { +3 }{ AI } { Cl }_{ 3 }\left( s \right)$

Here, no change of oxidation states take place. Therefore, it is not a redox reaction.

$(D)$

$2\overset { 0 }{ K } \left( s \right) +{ \overset { 0 }{ F } }_{ 2 }\left( g \right) \longrightarrow 2{ K }^{ + }{ F }^{ - }\left( s \right)$

Here,

$\overset { 0 }{ K }$ is oxidized to

${ K }^{ +1 }$ and

${ \overset { 0 }{ F } }_{ 2 }$ is reduced to

${ F }^{ -1 }$ . So, it is a redox reaction.

Which of the following is a redox

0%
$2NaAg(CN{ ) }_{ 2 }+Zn\quad \longrightarrow N{ a }_{ 2 }Zn(CN{ ) }_{ 4 }+2\quad Ag$
0%
$Ba{ O }_{ 2 }+{ H }_{ 2 }S{ O }_{ 4 }\quad \longrightarrow BaS{ O }_{ 4 }+{ H }_{ 2 }{ O }_{ 2 }$
0%
${ N }_{ 2 }{ O }_{ 5 }+{ H }_{ 2 }O\longrightarrow 2HN{ O }_{ 3 }$
0%
$AgN{ O }_{ 3 }+KI\longrightarrow AgI+KN{ O }_{ 3 }\\$

In the balanced equation

${MnO_{4}}^{-}+H^{+}+{C_{2}O_{4}}^{2-}\rightarrow Mn^{2+}+CO_{2}+H_{2}O$ , the moles of

$CO_{2}$ formed are :-

0%
2
0%
4
0%
5
0%
10

Oxidation numbers of

$P$ in

$PO_{4}^{3-}$ ,

$S$ in

$SO_{4}^{2-}$ and that of

$Cr$ in

$Cr_{2}O_{7}^{2-}$ are respectively::

0%
+3, +6 and +5
0%
+5, +3 and +6
0%
-3, +6 and +6
0%
+5, +6 and +6

Which of the following is a strong oxidising agent?

0%
$AlCl_{3}$
0%
$TlCl_{3}$
0%
$NF_{3}$
0%
$PCl_{3}$

What is the oxidation number of xenon(Xe) in the

$Ba_2XeO_6$ ?

0%
ZERO
0%
+8
0%
+6
0%
None of these

A compound contains three elements A,B and C, if the oxidation number of

$A=+2, B=+5$ and

$C= -2,$ the possible formula of the compound is :

0%
$A_3(B_4C)_2$
0%
$A_3(BC_4)_2$
0%
$A_2(BC_3)_2$
0%
$ABC_2$

The formal charge on carbon atom in carbonate ion is

0%
$+1$
0%
$-1$
0%
$+4$
0%
$Zero$

Explanation

Carbonate ion

$CO_{3}^{2-}$

Formal charge

$F=V-(L+\cfrac{1}{2}B)$

where

$V=$ Number of valence electrons .

$L=$ Number of lone pair electrons.

$B=$ Number of bonding electrons on that particular atom.

$\therefore$ Formal charge for

$C$ atom is

$F=4-(0+\cfrac{1}{2}\times 8)$

$F=0$

The compound

$YBa_2Cu_3O_7$ , which shows super conductivity, has copper in the oxidation state:

0%
$+\dfrac{7}{3}$
0%
$+\dfrac{3}{7}$
0%
$3$
0%
$7$

The value of n in the molecular formula

$Be_n Al_{2}Si_{6}O_{18}$ is

0%
1
0%
2
0%
3
0%
4

CBSE Questions for Class 11 Medical Chemistry Redox Reactions Quiz 12 - MCQExams.com

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Practice Class 11 Medical Chemistry Quiz Questions and Answers

CBSE Questions for Class 11 Medical Chemistry Redox Reactions Quiz 12 - MCQExams.com

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Practice Class 11 Medical Chemistry Quiz Questions and Answers

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