MCQ Questions for Class 11 Medical Chemistry Redox Reactions Quiz 13

The oxidation number of carbon in

$CH_2F_2$ is:

0%
$0$
0%
$1$
0%
$4$
0%
$6$

The sum of oxidation states of all atoms in

$ClO_4^-$ is:

0%
0
0%
-4
0%
-1
0%
+2

In which following compounds, phosphorous exhibits its highest oxidation state?

0%
$H_3PO_2$
0%
$H_4P_2O_7$
0%
$H_4P_2O_6$
0%
$H_4P_2O_5$

Oxidation number of chlorine atoms in

$CaOCl_2$ are

0%
$0,0$
0%
$-1, -1$
0%
$-1, +1$
0%
$-2,+7$

Find the potential of a half-cell having reaction,

$Ag_{2}S+2e\rightarrow2Ag+S^{2-}$ in a solution buffered

$pH=3$ and which is also saturated with

$0.1\ M\ H_{2}S$ . For

$H_{2}S:\ K_{1}=10^{-8}$ and

$K_{2}=2\times 10^{-13},\ K_{sp}(Ag_{2}S)=2\times 10^{-48},\ E_{Ag,ag}^{o}=0.8$

0%
$0.432\ V$
0%
$1.658\ V$
0%
$-0.245\ V$
0%
$-0.144\ V$

The number of mole of ferrous oxalate oxidized by one mole of

$KMnO_{4}$ in acidic medium is:

0%
1.66
0%
5
0%
0.6
0%
2

The oxidation state of iron in chalco pyrite

$(CuFeS_2)$ is

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

How many moles of

$KMnO_{4}$ are required to oxidise one mole of

$SnCl_{2}$ in acidic medium ?

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

Explanation

Equate equivalents.

Equivalents = $n-factor\times moles$

For $KMnO_4$ n-factor is 5. Let moles be 'm'.

So equivalents of $KMnO_4 = 5\times m= 5m$ .

n-factor for $SnCl_2$ is 2 as it goes from $+2-+4$ oxidation state. It's moles is 1.

So equivalents is 2.

$5m = 2$

$m=\cfrac{2}{5}$

$Zn|Zn^{+2}|| Ag^+|Ag$ , how will cell potential be affected if KI is added to

$Ag^{+}$ half cell?

0%
$E_{cell}$ will increase
0%
$E_{cell}$ will decrease
0%
$E_{cell}$ will remain unaffected
0%
$E_{cell}$ is first increase then decrease

Which is a redox reaction ?

0%
${ 2CuI }_{ 2 }\rightarrow CuI+{ I }_{ 2 }$
0%
$NaCl+AgNO_{ 3 }\rightarrow AgCl+NaNO_{ 3 }$
0%
${ NH }_{ 4 }Cl+NaOH\rightarrow { NH }_{ 3 }+NaCl+H_{ 2 }O$
0%
${ Cr }_{ 2 }{ \left( { SO }_{ 4 } \right) }_{ 3 }+6KOH\rightarrow 2Cr{ \left( OH \right) }_{ 3 }+{ 3K }_{ 2 }{ SO }_{ 4 }$

When ethane is burnt in excess of oxygen, the oxidation number of carbon changes by

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

What is the sum of oxidation number of all carbon in

$C_6H_5CHO$ ?

0%
-4
0%
5
0%
6
0%
7

Which of the following is correct about oxidation number?

0%
The O.N. of all atoms in elementary states is 0
0%
Alkali and alkaline earth metals have +1 and +2 oxidation states respectively
0%
The O.N. of a monoatomic ion is the same as its charge
0%
All of the above

Explanation

The oxidation number of a monatomic ion is the same as its charge. E.g.

$Na^+$ ,

$Al^{+3}$ ,

$S^{-2}$ .

Option D is correct.
2025375_1359729_ans_480b8248f4ab4a6cb55f31b197018a8c.jpg

Average oxidation state of

${\rm{Fe}}$ in

${\rm{F}}{{\rm{e}}_{\rm{3}}}{{\rm{O}}_{\rm{4}}}$ is

0%
$\frac{2}{3}$
0%
$\frac{4}{5}$
0%
$\frac{5}{4}$
0%
$\frac{8}{3}$

${\text{2Fe}}{\left( {{\text{CO}}} \right)_{\text{5}}}\xrightarrow{{{\text{Photolysis}}}}{\text{Fe}}{\left( {{\text{CO}}} \right)_{\text{9}}}{\text{ + CO}}$ Oxidation of

${\text{Fe}}$ in

${\text{F}}{{\text{e}}_{\text{2}}}{\left( {{\text{CO}}} \right)_{\text{9}}}$

0%
$4.5$
0%
$2$
0%
$3$
0%
$0$

Find out the values of a,b,c and d in the following equation:

${IO_{3}}^{-}+aI^{-}+bH^{+}$

$\rightarrow$

$cH_{2}O+dI_{2}$

0%
a=3, b=2, c=1, d=1
0%
a=5, b=6, c=3, d=3
0%
a=5, b=6, c=3, d=6
0%
a=10, b=12, c=6, d=6

Following reaction describes the rusting of iron

$4{\text{Fe}} + 3{{\text{O}}_2} \to 4{\text{F}}{{\text{e}}^{3 + }} + {60^{2 - }}$ , which one of the following statement is incorrect?

0%
This is an example of a redox reaction
0%
Metallic iron is reduced to ${\text{F}}{{\text{e}}^{{\text{3 + }}}}$
0%
${\text{F}}{{\text{e}}^{{\text{3 + }}}}$ is an oxidizing agent
0%
Metallic iron is a reducing agent

Of the following transition metals, the maximum number of oxidation states is shown by

0%
$Cr$
0%
$Mn$
0%
$Fe$
0%
$Ti$

Which of the following is a redox:

0%
$2NaAg{(CN)}_{2}+Zn\longrightarrow {Na}_{2}Zn{(CN)}_{4}+2Ag$
0%
$Ba{O}_{2}+{H}_{2}{SO}_{4}\longrightarrow Ba{SO}_{4}+{H}_{2}O$
0%
${N}_{2}{O}_{5}+{H}_{2}O\longrightarrow 2H{NO}_{3}$
0%
$Ag{NO}_{3}+KI\longrightarrow AGI+K{NO}_{3}$

Consider the half-cell reaction(s)

${Cu}^{2+}+{e}^{-}\rightarrow {Cu}^{+};{E}^{o}=0.15V$

${Cu}^{2+}+2{e}^{-}\rightarrow {Cu};{E}^{o}=0.33V$

${E}^{o}$ for the half-cell reaction:

${Cu}^{+}+{e}^{-}\rightarrow Cu$ will be:

0%
$0.48V$
0%
$0.18V$
0%
$0.31V$
0%
$0.51V$

What is the oxidation state of iodine when it is oxidized in the form

${{\text{I}}^{\text{ - }}}$ in neutral medium by permanganate ion

0%
$+ 5$
0%
$+ 7$
0%
$+ 3$
0%
$+ 1$

A compound contains atoms

$\mathrm { A } , \mathrm { B }$ and

$\mathrm { C }$ . The oxidation number of

$\mathrm { A }$ is

$+ 2 ,$ of

$\mathrm { B }$ is

$+ 5$ and of

$\mathrm { C }$ is

$- 2 .$ The possible formula of the compound is:

0%
$A B C _ { 2 }$
0%
$\mathrm { A } _ { 2 } \left( \mathrm { BC } _ { 3 } \right) _ { 2 }$
0%
$A_3(BC_4)_2$
0%
$\mathrm { A } _ { 3 } \left( \mathrm { B } _ { 4 } \mathrm { C } \right) _ { 2 }$

An element A in a compound ABD has oxidation number -n. It is oxidized by

${\text{C}}{{\text{r}}_{\text{2}}}{{\text{O}}_{\text{7}}}^{{\text{2 - }}}$ in acid medium. In the experiment,

$1.68 \times {10^{ - 3}}$ moles of

$\;{{\text{K}}_{\text{2}}}{\text{C}}{{\text{r}}_{\text{2}}}{{\text{O}}_{\text{7}}}$ were used for

$1.68 \times {10^{ - 3}}$ mole ABD. The new oxidation number of A after oxidation is :

0%
3
0%
3 - n
0%
n - 3
0%
+ n

In the reaction

$Cu + {H_2}S{O_4} \to CuS{O_4} + {H_2}O + S{O_2}$

0%
${{\text{H}}^{\text{ + }}}$ is the oxidising agent
0%
${\text{S}}{{\text{O}}_{{\text{2 }}}}$ is the oxidising agent
0%
Both $\left( {\text{A}} \right)$ and $\left( {\text{B}} \right)$ are correct
0%
None of the above is correct

What is the oxidation number of

$N$ in

$NH_2OH$ ?

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

Which of the following arrangement represents increasing O.N. of central atom.

0%
${ CrO }_{ 2 }^{ - },{ ClO }_{ 3 }^{ - },{ CrO }_{ 4 }^{ 2- },{ MnO }_{ 4 }^{ - }$
0%
${ ClO }_{ 3 }^{ - },{ CrO }_{ 4 }^{ 2- },{ MnO }_{ 4 }^{ - },{ CrO }_{ 2 }^{ - }$
0%
${ CrO }_{ 2 }^{ - },{ ClO }_{ 3 }^{ - },{ MnO }_{ 4 }^{ - },{ CrO }_{ 4 }^{ 2- }$
0%
${ CrO }_{ 4 }^{ 2- },{ MnO }_{ 4 }^{ - },{ CrO }_{ 2 }^{ - },{ ClO }_{ 3 }^{ - }$

The oxidation states of the most electronegative element in the product of the reaction of

$Ba{ O }_{ 2 }$ with dilute

${ H }_{ 2 }{ SO }_{ 4 }$ are:

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

The oxidation number of

$P$ in

$Mg{ NH }_{ 4 }{ PO }_{ 4 }$ is

0%
+3
0%
+5
0%
+2
0%
+7

Match List I and List-II and select the correct answer using the codes given below the lists:

	List-I (compound)		List-II ( oxidation states of N )
(A)	$NO_2$	(1)	+5
(B)	$HNO_2$	(2)	-3
(C)	$NH_3$	(3)	+4
(D)	$N_2O_5$	(4)	+3

The oxidation number of ''V'' in the complex

${\text{R}}{{\text{b}}_{\text{4}}}{\text{Na}}\left[ {{\text{H}}{{\text{V}}_{{\text{10}}}}{{\text{O}}_{{\text{28}}}}} \right]$ is

0%
+3
0%
+5
0%
+7
0%
+6

Oxidation numbers of sulphur in

$H_2S_3$ molecule are

0%
-1, -1 and -1
0%
-1, +1 and +1
0%
-1, 0 and -1
0%
+1, 0 and -1

Oxidation number of oxygen in

$NaO_{ 3 }, { KO }_{ 2 }$ respectively are:

0%
3, 2
0%
1, 0
0%
0, 1
0%
-0.33, -0.5

The oxidation number of nitrogen in

$NC{l_3}$ is

0%
+3
0%
-3
0%
zero
0%
-1/3

When

$Cl_2$ gas reacts with hot and concentrated sodium hydroxide solution, the oxidation number of chlorine changes from____

0%
zero to +1 and zero to -5
0%
zero to -1 and zero to +5
0%
zero to -1 and zero to +3
0%
zero to +1 and zero to +3

Which of the following is a non - redox process ?

0%
$SO_{4}^{2-}\rightarrow SO_{3}$
0%
$Cr_{2}O_{7}^{2-}\rightarrow CrO_{4}^{2-}$
0%
$PO_{4}^{3-}\rightarrow P_{2}O_{7}^{4-}$
0%
$C_{2}O_{4}^{-2}\rightarrow CO_{2}$

Explanation

In Redox reaction oxidation state changes

(a) In

$SO_{4}^{-2},$ Its

$S^{+6}$
In

$SO_{3},$ Its

$S^{+6}$

$\therefore$ In

$SO_{4}^{-2}\rightarrow SO_{3},$ there is no change
in oxidation state of

$O^{+5}$

$\therefore$ Non- redox reaction

(b) In

$Cr_{2}O_{7}^{-2}$
Lets oxidation state of Cr be x

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

$\therefore 2x+(-14) = -2$

$\therefore 2x = 12$

$\therefore x = 16$

$CrO_{4}^{-2}$

Let oxidation Sate of Cr be y

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

$y-8 = -2$

$y = +6$

$\therefore$ oxidation of Cr remain same

$\therefore$ Non- redox Readtion

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

$P^{+}$
and In

$P_{2}O_{7}^{-4}$
Let oxidation sate of P be x

$2x+7(-2) = -4$

$2x-14 = -4$

$\therefore 2x = 10$

$\therefore$ Non - redox Reaction

$\therefore x = 5$

(d) In

$C_{2}O_{4}^{-2},$ Its

$C^{+3}$
In

$Co_{2},$ Its

$C^{+4}$

$\therefore$ Redox reaction

Hence, the correct options are

$\text{A}$ ,

$\text{B}$ and

$\text {C}$

Balance the following redou reaction

$x M n O _ { 4 } ^ { - } + y F e ^ { + 2 } + z H ^ { + } \rightarrow a M n ^ { + 2 } + b F e ^ { + 3 } + c H _ { 2 } O$
What is the ratio of

$( x + y + z ) / ( a + b + c ) ?$

0%
$5 : 7$
0%
$7 : 5$
0%
$4 : 7$
0%
$7 : 4$

Which of the following reaction is non-redox?

0%
$2NaNO_{3}\rightarrow 2NaNO_{2}+O_{2}$
0%
$CaO+SiO_{2}\rightarrow CaSiO_{3}$
0%
$Fe+H_{2}SO_{4}\rightarrow FeSO_{4}+H_{2}$
0%
$4Ag+8CN^{-}+O_{2}+2H_{2}O\rightarrow 4[Ag(CN)_{2}]^{-}+4OH^{-}$

$HOCl(aq)$ is the molecule that kills bacteria wen chlorine is added to water.
The following reaction produces this molecule

${Cl}_{2}(g)+{H}_{2}O(l)\rightleftharpoons HOCl(aq)+{H}^{+}(aq)+{Cl}^{-}(aq)$
Which statement about this reaction is correct?

0%
Chlorine is both oxidised and reduced
0%
Chlorine is oxidised but not reduced
0%
Hydrogen is both oxidised and reduced
0%
Hydrogen is oxidised but not reduced

Oxidation number of oxygen in ozone

$\left( \mathrm { O } _ { 3 } \right)$ is:

0%
$+ 3$
0%
$- 3$
0%
$- 2$
0%
0

The oxidation state of A, B, and C in a compound are +2, +5, and -2, respectively. The correct formula for the compound is:

0%
${ A }_{ 2 }\left( BC \right) _{ 2 }$
0%
${ A }_{ 2 }\left( BC \right) _{ 3 }$
0%
${ A }_{ 3 }\left( BC_{ 4 } \right) _{ 2 }$
0%
${ A }_{ 2 }\left( BC_{ 4 } \right) _{ 3 }$

Which of the following carbonyl compounds undergoes self redox reaction in presence of concentrated base?

0%
3-Methylpentanal
0%
2-Chlorobutanal
0%
2,2-Dimethylpropanal
0%
tert-butyl methyl ketone.

Which of the following statement is correct?

0%
$MnO^-_4$ is very good oxidant in acidic medium and liberates iodine when added to $KI$ solution by forming $MnO_2$ .
0%
$MnO^-_4$ is a very good oxidant in basic medium and liberates iodine when added to $KI$ solution by forming $MnO_2$ .
0%
$MnO^-_4$ is very good oxidant in acidic medium and liberates iodine when added to $KI$ solution by forming $Mn^{2+}$ .
0%
$MnO^-_4$ is very good oxidant in basic medium and liberates iodine when added to $KI$ solution by forming $Mn^{2+}$ .

What is the molar concentration of the triiodide solution,

$KI_3$ (aq), if 30.00 mL of the solution is required to completely react with 25.00 mL of a

$0.200\ mol/L$ , thiosulphate solution,

$K_2S_2O_3$ (aq) ?
The chemical equation for the reaction is:

$2S_2O_3^{2-}(aq) + I_3^-(aq)\rightarrow S_4O_6^{2-} (aq)+3 I^-(aq)$

0%
0.0333 M
0%
0.12 M
0%
0.0833 M
0%
0.00012 M

Balance the following redox reaction

$xMnO_{4}^{-} + yFe^{+2} + zH^{+} \rightarrow aMn^{+2} + bFe^{+3} + cH_2O$
What is the ratio of

$(x + y + z)/ (a + b + c)$ ?

0%
$5 : 7$
0%
$7 : 5$
0%
$4 : 7$
0%
$7 : 4$

The dissociation constant for

$CH_{3}COOH$ is

$1.8\times 10^{-5}$ at

$298\ K$ . The electrode potential for the half-cell:

$Pt|H_{2}\ (1\ bar)\|0.5\ M\ CH_{3}COOH$ , at

$298\ K$ is:

$(log\ 2 = 0.3\ ;\ log\ 3 = 0.48\ ;\ 2.303\ RT/F = 0.06)$

0%
$-0.3024\ V$
0%
$-0.1512\ V$
0%
$+0.3024\ V$
0%
$+0.1512\ V$

From the following

$E^{\circ}$ values for the half-cells:

$(i)\ D\rightarrow D^{2+}+2e^{-};E^{\circ}=-1.5V$

$(ii)\ B^{+}+e^{-}\rightarrow B;E^{\circ}=-0.5V$

$(iii)\ A^{3-}\rightarrow A^{2-}+e^{-};E^{\circ}=1.5V$

$(iv)\ C^{2+}+e^{-}\rightarrow C^{+};E^{\circ}= +0.5V$

Which combination of two half-cells would result in a cell with the largest potential?

0%
$(i)$ and $(iii)$
0%
$(i)$ and $(iv)$
0%
$(iii)$ and $(iv)$
0%
$(ii)$ and $(iv)$

The value of

$n$ in the reaction is:

$Cr_{2}O_{7}^{2-}+14H^{+}+nFe^{2+}\rightarrow 2Cr^{3+}+nFe^{3+}+7H_{2}O$

0%
$2$
0%
$3$
0%
$6$
0%
$7$

The atomic number of an element isThe highest oxidation state exhibited by it in its compounds is

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

Explanation

Electronic configuration for

$Z=22$ is

$[Ar]3d^24s^2.$

Highest oxidation state

$=$ no. of

$ns$ electron

$+$ no. of

$(n-1)d$ electrons.

$=\,\,\,2+\,\,\,2=4$

For the feasibility of a redox reaction in a cell, the emf should be

0%
Positive
0%
Fixed
0%
Zero
0%
Negative

Explanation

Answer A

Solution :

Any redox reaction would occur spontaneously if the free energy charnge

$(ΔG)$ is negative.

$ΔG^∘=nFE^2$

Where n is the number of electrons involved, F is the value of faraday and

$E^∘$ is the cell emf.

$ΔG^2$ can be negative if

$2$ is

$\textbf{positive.}$

When

$KMnO_4$ is reduced with oxalic acid in acidic solution, the oxidation number of Mn changes from

0%
7 to 4
0%
6 to 4
0%
7 to 2
0%
4 to 2

Explanation

When

$KMnO_4$ is reduced with oxalic acid in acidic solution, the oxidation number of Mn changes from +7 to +2.

$2MnO_4^- +6H^++ 5C_2H_2O_4 \rightarrow 2Mn^{2+} +8H_2 O + 10 CO_2$

CBSE Questions for Class 11 Medical Chemistry Redox Reactions Quiz 13 - 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 13 - MCQExams.com

0:0:1

Practice Class 11 Medical Chemistry Quiz Questions and Answers

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