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Classical idea of redox reactions - class-XII

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Minerals associated with  redox reactions are

  1. Na, Cu

  2. N, Cu

  3. Fe, Cu

  4. Ca, Fe


Correct Option: C

Which of the following is a redox reaction?

  1. $C{ a }CO _{ 3 }\quad \rightarrow \quad { { C }{ a }O+{ CO } _{ 2 } }$

  2. $H _{ 2 }+Cl _{ 2 }\quad \rightarrow \quad 2HCl$

  3. ${ C }{ a }O+2HCl\quad \rightarrow \quad { C }{ a }{ Cl } _{ 2 }+{ H } _{ 2 }O$

  4. ${ N }{ a }OH+HCl\quad \rightarrow \quad { N }{ a }{ Cl }+{ H } _{ 2 }O$


Correct Option: A

Which reaction is not a redox reaction?

  1. $Mg+2HNO _3\rightarrow Mg(NO _3) _2+H _2$

  2. $2Mg(NO _3) _2\rightarrow 2MgO+4NO _2+O _2$

  3. $SO _2+NO _2\rightarrow SO _3+NO$

  4. $SO _3+H _2O\rightarrow H _2SO _4$


Correct Option: C

How many moles of electrons are involved in the reduction of one mole of $MnO^- _4$ ion in alkaline medium to $MnO^- _3$?

  1. $2$

  2. $1$

  3. $3$

  4. $4$


Correct Option: A
Explanation:
Reaction takes place as follows:
$Mn{ O } _{ 4 }^{ - }+{ H } _{ 2 }O+{ 2e }^{ - }\rightarrow Mn{ O } _{ 3 }^{ - }+2O{ H }^{ - }$
So, 2 electrons are involved in reduction of 1 mole of ${MnO} _4^{-}$ ion in alkaline medium to ${MnO} _3^{-}$.

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:

  1. $-1.59 V$

  2. $-0.764 V$

  3. $0.764 V$

  4. $0.062 V$


Correct Option: C
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,
$E^o _{acidic}+E^o _{basic}=E^o _{cell}$
$\Rightarrow 1.59+x= 0.826$
$\Rightarrow x= 0.764$

A solution contains ${Fe}^{+2},{Fe}^{+3}$ and ${I}^{-}$ ions. This solution was treated with iodine at ${35}^{o}C$. Then the favourable redox reaction is: 

(Given that ${ E } _{ { Fe }^{ +3 }/{ Fe }^{ +2 } }^{ o }=+0.77V;\quad { E } _{ { I } _{ 2 }/{ I }^{ - } }^{ o }=0.536V$)

  1. ${I} _{2}$ will be reduced to ${I}^{-}$

  2. there will be no redox reaction

  3. ${I}^{-}$ will be oxidised to ${I} _{2}$

  4. ${Fe}^{+}$ will be oxidised to ${Fe}^{+3}$


Correct Option: A
Explanation:

${ 2I }^{ - }\rightarrow { I } _{ 2 }+{ 2e }^{ - }(oxidation\quad half-reaction)$

${ E } _{ Oxidation }^{ 0 }=-0.536V$

${ Fe }^{ 3+ }+e^{ - }\rightarrow { Fe }^{ 2+ }(reduction\quad half-reaction)$

$E^{ 0 } _{ Reduction }=-0.77V$
----------------------------------------------------------------------------------------------------------------------
${ 2Fe }^{ 3+ }+{ 2I }^{ - }\rightarrow { 2Fe }^{ 2+ }+{ I } _{ 2 }$

$E^{ 0 }={ E } _{ oxidation }^{ 0 }+E _{ Reduction }^{ 0 },+ve$

Hence the reaction will take place.

${ 2I }^{ - }\rightarrow { I } _{ 2 }+{ 2e }^{ - }(oxidation\quad half-reaction)$

${ E }^{ 0 } _{ Oxidation }=-0.536V$

${ Fe }^{ 3+ }+{ e }^{ - }\rightarrow { Fe }^{ 2+ }(reduction\quad half-reaction)$

$E _{ Reduction }^{ 0 }=-0.77V$
------------------------------------------------------------------------------------------------------------------------
$2Fe^{ 3+ }+{ 2I }^{ - }\rightarrow { 2Fe }^{ 2+ }+{ I } _{ 2 };$

${ E }^{ 0 }={ E } _{ oxidation }^{ 0 }+{ E } _{ reduction }^{ 0 };+ve$

$HOCl(aq)$ 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?

  1. Chlorine is both oxidised and reduced

  2. Chlorine is oxidised but not reduced

  3. Hydrogen is both oxidised and reduced

  4. Hydrogen is oxidised but not reduced


Correct Option: A
Explanation:

Oxidation and reduction are two types of chemical reactions that often work together. Oxidation and reduction reactions involve an exchange of electrons between reactants.

  • Reduction and oxidation occur simultaneously in a type of chemical reaction called a reduction-oxidation or redox reaction.
  • Oxidation Involves Loss of electrons
  • Reduction Involves Gain of electrons.
As we can see the above case with the chlorine .
hence chlorine is both oxidised and reduced

Which of the following reaction is non-redox?

  1. $2NaNO _{3}\rightarrow 2NaNO _{2}+O _{2}$

  2. $CaO+SiO _{2}\rightarrow CaSiO _{3}$

  3. $Fe+H _{2}SO _{4}\rightarrow FeSO _{4}+H _{2}$

  4. $4Ag+8CN^{-}+O _{2}+2H _{2}O\rightarrow 4[Ag(CN) _{2}]^{-}+4OH^{-}$


Correct Option: B

A redox reaction is

  1. exothermic

  2. endothermic

  3. neither exothermic nor endothermic

  4. either exothermic or endothermic


Correct Option: D

One gas bleaches the colour of flowers by reduction while the other gas by oxidation. The two gases are, respectively:

  1. $CO$ & $Cl _{2}$

  2. $H _{2}S$ & $Br _{2}$

  3. $SO _{2}$ & $Cl _{2}$

  4. $NH _{3}$ & $SO _{3}$


Correct Option: C
Explanation:
Sulphur dioxide and chlorine gas are the two gases that bleach the colour of flower by oxidation and reduction.

Chlorine reacts with water and produces a single oxygen atom also called nascent oxygen. This nascent oxygen when combines with any colour make it colourless.

$Cl _{2}+H _{2}O\rightarrow 2HCl+(O)$ (O is oxidizing)

Sulphur dioxide when reacts with coloured substances it releases oxygen from the substance resulting in loss of colour.

$SO _{2}+2H _{2}O\rightarrow H _{2}SO _{4}+2(H)$  (H is reducing)

Hence, the correct option is $C$

Oxidation number of $N$ in $HN _3$ ( hydrazoic acid) is:

  1. -1/3

  2. +3

  3. 0

  4. -3


Correct Option: A

Sulphur dioxide bleaches colouring matter by ___________.

  1. dehydration

  2. oxidation

  3. decomposition

  4. reduction


Correct Option: D
Explanation:

Sulphur dioxide bleaches coloring matter by reduction. Bleaching action of sulphur dioxide is not permanent.

$SO _2(g) + 2H _2O \rightarrow H _2SO _4(aq) + 2[H^+]$

When a piece of magnesium ribbon is placed in a beaker of copper sulfate solution, which one of the following is most likely to happen?

  1. A vigorous reaction immediately takes place, with the release of a lot of heat.

  2. The blue colour of the copper sulfate solution gets darker.

  3. The blue colour of the copper sulfate solution gets lighter.

  4. The magnesium catches fire.


Correct Option: C
Explanation:

$Mg+CuSO _4 \longrightarrow MgSO _4 + Cu$

            $(Blue)$
By adding$Magnesium$ to $CuSO _4$, it turns to $MgSO _4.$
Because $Mg$ is above $Cu$ in reactivity series. So $Mg$ displaces $Cu$ in $CuSO _4$. As reaction processes, $CuSO _4$ content decreases.
$\therefore$ Color blue fades. 

In the electrochemical series, metals are arranged in order of their tendency to :

  1. release halogens from their salts

  2. become reduced

  3. react with water

  4. lose electrons


Correct Option: D
Explanation:

In the electrochemical series, metals are arrange in order of their tendency to lose electrons. 

$\bullet$ Oxidation potential increases down the group.(tendency to lose electron increases).
$\bullet$ Reduction potentials decreases down the group. (tendency to gain $e^-$ decreases).

Which one of the following substances is a good oxidising agent?

  1. coke

  2. water

  3. Hydrogen peroxide

  4. Sulphur dioxide


Correct Option: C
Explanation:

A good oxidizing agent is one which can readily oxidize other chemical species and reduce itself. Therefore, the compound which is good oxidizing agent must have initial oxidation as high, so that it can reduce its oxidation state and get reduced and oxidize others.

$Coke:$ Reducing agent 
$Hydrogen$ $peroxide$ $(H _2O _2):$ because oxidation state of oxygen is -1 in hydrogen peroxide. So, it oxidizes to 0 & acts as a reducing agent & it reduces to -2 & acts as an oxidizing agent. But, generally, it acts as an oxidizing agent because a stable oxidation state of oxygen is -2(oxide form)
$H _2O:$ Reducing agent.
$SO _2$: sulfur in the +4 oxidation state, sulfur dioxide is a reducing agent. It is oxidized by halogens to give the sulfuryl halides, such as sulfuryl chloride

Hence, the correct option is $\text{C}$

What electrolyte is used to electroplate silver onto a spoon?

  1. pure water

  2. silver nitrate

  3. copper sulfate

  4. acidified water


Correct Option: B
Explanation:

$Ans:$ $Silver$ $Nitrate.$

In silver plating, the object (eg.spood) to be plated is made the cathode of an electrolytic cell. The anode is bar of silver metal and the electrodyte must contain $Ag^+$ ions (i.e. $AgNO _3$), so that when the direct current is passed through the cell, positive ($Ag^+$) ions move towards the negative anode (spoon) where they are neutralized by $e^-$ and stick to spoon as silver metal.

Oxalic acid dihydrate, $H _{2}C _{2}O _{4}\cdot 2H _{2}O(s)$ is often used as a primary reagent to standardise sodium hydroxide solution. Which of these facts are reasons to choose this substance as a primary standard?
I. It is diprotic.
II. It is a stable compound that can be weighed directly in air.
III. It is available in pure form.

  1. III only

  2. I and II only

  3. II and III only

  4. I, II and III


Correct Option: C
Explanation:

$\begin{array}{l} { H _{ 2 } }{ C _{ 2 } }{ O _{ 4 } }\cdot 2{ H _{ 2 } }O\left( s \right) \to oxadic\, acid\, dihydrate\,  \ Since,\, it\, is\, very\, stable\, and\, also\, available\, in\, pressure \ from\, in\, the\, laboratory\, i.e.\, it's\, always\, used\, as\, primary\, s\tan  dard\, in\, the\, lab. \ Hence,\, the\, option\, C\, is\, the\, correct\, answer. \end{array}$

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:

  1. $4, 1, 7$

  2. $7, 4, 1$

  3. $4, 1, 10$

  4. $1, 4, 10$


Correct Option: C

The mass of $H _{2}O _{2}$ that is completely oxidised by $30.2\ g$ of $KMnO _{4}$ (molar mass= $158\ g mol^{-1}$) in acidic medium is 

  1. $12\ g$

  2. $14\ g$

  3. $16\ g$

  4. $1\ g$


Correct Option: C

Which of the following is a non - redox process ?

  1. $ SO _{4}^{2-}\rightarrow SO _{3} $

  2. $ Cr _{2}O _{7}^{2-}\rightarrow CrO _{4}^{2-} $

  3. $ PO _{4}^{3-}\rightarrow P _{2}O _{7}^{4-} $

  4. $ C _{2}O _{4}^{-2}\rightarrow CO _{2} $


Correct Option: A,B,C
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 $

In $ 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

(C) In $ 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 $\text {C}$

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