0

Redox and stoichiometry - class-XI

Description: Redox and stoichiometry
Number of Questions: 15
Created by:
Tags: redox reactions chemistry oxidation- reduction reactions
Attempted 0/14 Correct 0 Score 0

The number of equivalents of $Na _{2}S _{2}O _{3}$ required for the volumetric estimation of one equivalent of $Cu^{2+}$ is:

  1. $1/3$

  2. $1$

  3. $3/2$

  4. $2/3$


Correct Option: B
Explanation:

Number of equivalents of reacting species in a chemical reaction are same.

A volume of $100 ml$ of $ H _{2}O _{2}$ is oxidized by 100 ml of $ 1 M \ KMnO _{4}$ in acidic medium $ (MnO _{4}^{-}$ reduced to $ Mn^{2+}).$ A volume of $100 ml$ of same $H _{2}O _{2}$ is oxidized by $'V'$ ml of $1 M \ KMnO _{4}$ in basic medium ($ MnO _{4}^{-}$ reduced to $ MnO _{2}).$ The value of $'V'$ is

  1. $500$

  2. $100$

  3. $33.33$

  4. $166.67$


Correct Option: D
Explanation:

In acidic medium, 

$ N _{1}V _{1} = N _{2}V _{2} $

$ n _1\times M _{1}\times V _{1} = n _2\times M _{2}\times V _{2} $

$ 5\times 1\times 100 =  2\times M _{2} \times 100$ 

$ M _{2} = 2.5\,M $

Now in basic medium,

$ N _{1}V _{1} = N _{2}V _{2} $

$ 2\times 2.5 \times 100=  3\times 1\times V $

$ \therefore V = \dfrac{200\times 2.5}{3} $

$V = 166.66\,ml $

A solid $PQ$ has rock salt type structure in which $Q$ atoms are the corners of the unit cell. If the body-centered atoms in all the unit cells are missing, the resulting stoichiometry will be

  1. $PQ$

  2. $P{Q} _{2}$

  3. ${ P } _{ 3 }{ Q } _{ 4 }$

  4. ${ P } _{ 4 }{ Q } _{ 3 }$


Correct Option: C

In a solid $AB$ of $NaCl$ structure, A atoms occupy the corners of the cubic unit cell. If all the corner atoms are removed then the formula of the unit cell will be

  1. $A _{4}B _{4}$

  2. $B$

  3. $A _{3}B _{4}$

  4. $AB$


Correct Option: C

Number of electrons involved in the electrodeposition of 63.5 g of Cu from solution of $ CuSO _{4} $ is:

  1. $ 6.022\times 10^{23} $

  2. $ 3.011\times 10^{23} $

  3. $ 12.044\times 10^{23} $

  4. $ 6.022\times 10^{23} $


Correct Option: C
Explanation:
$ Cu^{2+}+2e^-\rightarrow Cu $ 

1 mole Cu = 63.5 g
 
For deposition of 1 mole of Cu, 2 moles of $e^-$ is required.

$ \therefore$ Number of electrons $ =12.044\times 10^{23} $      
                                    
Hence, option C is correct.

Write balanced half reactions for the following redox reaction:
$Cr _2O _7^{2-} + Fe^{2+}\rightarrow  Cr^{3+} + Fe^{3+}$

  1. Reduction: $6e^-+ 14H^+ + Cr _2O _7^{2-} \rightarrow Cr^{3+} + 7H _2O $

    Oxidation: $ \ Fe^{2+} \rightarrow Fe^{3+} + e^{-}$

  2. Oxidation: $6e^-+ 14H^+ + Cr _2O _7^{2-} \rightarrow Cr^{3+} + 7H _2O $

    Reduction: $ \ Fe^{2+} \rightarrow Fe^{3+} + e^{-}$

  3. Reduction: $4e^-+ 14H^+ + Cr _2O _7^{2-} \rightarrow Cr^{3+} + 7H _2O $

    Oxidation: $ \ Fe^{2+} \rightarrow Fe^{3+} + e^{-}$

  4. Oxidation: $6e^-+ 14H^+ + Cr _2O _7^{2-} \rightarrow Cr^{3+} + 6H _2O $

    Reduction: $ \ Fe^{2+} \rightarrow Fe^{3+} + e^{-}$


Correct Option: A
Explanation:

Given Redox reaction is

$Cr _2O _7^{2-}+Fe^{+2} \longrightarrow Cr^{+3}+Fe^ {3+}$
Oxidation half reaction:- $Fe^{+2} \longrightarrow Fe^{+3}+e^-$
                                          [as oxidation number increases from $2$ to $3$]
Reduction half reaction:- $Cr _2O _7^{2-}+6e^-+14H^+ \longrightarrow 2Cr^{+3}+7H _2O$
                                          [as oxidation number decreases from $6$ to $3$]

Which particles are gained and lost during a redox reaction?

  1. Protons

  2. Neutrons

  3. Electrons

  4. Positrons


Correct Option: C
Explanation:

  • Indirect redox reactions only take place if in between the electrodes is an environment that is conducive for electric current (meaning that charged particles are present that can freely move around, like the ions of dissolved or molten salts). There is no direct contact between the particles of the oxidation and the reduction. The transfer of the electrons is realised via-via, mostly via metal wiring or other conductive material. 
Exteriorly the conducting wires take care for the electron transport from RED to OX, without direct contact between the reactants.

Two half reactions are given as follows:
$2e^{-} + H^{+} + H _5IO _6 \rightarrow IO^{3-} + 3H _2O $
$Cr \rightarrow Cr^{3+} + 3e^{-}$
Final balanced reaction is:

  1. $3H^{+} + 3H _5IO _6 + 2Cr \rightarrow 2Cr^{3+} + 3IO^{3-} + 9H _2O$

  2. $5H^{+} + 3H _5IO _6 + 2Cr \rightarrow 2Cr^{3+} + 3IO^{3-} + 10H _2O$

  3. $3H^{+} + 3H _5IO _6 + 4Cr \rightarrow 4Cr^{3+} + 3IO^{3-} + 9H _2O$

  4. $3H^{+} + 3H _5IO _6 + 3Cr \rightarrow 3Cr^{3+} + 3IO^{3-} + 9H _2O$


Correct Option: A
Explanation:

Multiply each given half reactions by a number to balance electrons on both sides.

$[2e^-+H^++H _5IO _6 \longrightarrow IO^{-3}+3H _2O] \times 3 \longrightarrow (1)$
$[Cr \longrightarrow Cr^{+3}+3e^- ] \times 2 \longrightarrow (2)$
Now adding equation (1) and (2) we get balanced equation
$3H^++3H _5IO _6+2Cr \longrightarrow 2Cr^{+3}+3IO^{3-}+9H _2O$ .

In the reaction, $FeS {2} + KMnO _{4} + H^{+} \rightarrow Fe^{3+} + SO _{2} + Mn^{2+} + H _{2}O$, the equivalent mass of $FeS _{2}$ would be equal to__________.

  1. $\text{molar mass}$

  2. $\dfrac {\text {molar mass}}{10}$

  3. $\dfrac {\text {molar mass}}{11}$

  4. $\dfrac {\text {molar mass}}{13}$


Correct Option: C
Explanation:

$Fe^{2+} \rightarrow Fe^{3+} + e^{-}; S _{2}^{2-} \rightarrow 2S^{4+} + 10e^{-}$
$\therefore FeS _{2} \rightarrow 2S^{4+} + Fe^{3+} + 11e^{-}$
Equivalent mass of $FeS _{2} = \dfrac {\text {Molar mass}}{11}$. 

How much volume of $0.1M$ $Zn{ \left( Mn{ O } _{ 4 } \right)  } _{ 2 }$ is required to react with $50ml$ of $0.2M$ ferrous oxalate. Given ${ MnO } _{ 4 }^{ - }$ reduces into ${ Mn }^{ 2+ }$ in acidic medium.

  1. $30ml$

  2. $60ml$

  3. $40ml$

  4. None


Correct Option: C

For the redox reaction $MnO^- _4+C _2O^{3+} _4+H^+\rightarrow Mn^{2+}+CO _2+H _2O$. The correct whole number stoichiometric coefficients of $MnO^- _4, C _2O^{3+} _4$ and $H^+$ are respectively.

  1. $2.5, 16$

  2. $16, 5.2$

  3. $5, 16.2$

  4. $2, 16.5$


Correct Option: A

0.2 g of a sample of $
{\text{H}} _{\text{2}} {\text{O}} _{\text{2}}
$ required 10 mL of 1 N $
{\text{KMnO}} _{\text{4}}
$ in a titration in the presence of $
{\text{H}} _{\text{2}} {\text{SO}} _{\text{4}}
$. Purity of $
{\text{H}} _{\text{2}} {\text{O}} _{\text{2}}
$ is:

  1. 25%

  2. 85%

  3. 65%

  4. 95%


Correct Option: B
Explanation:

b'

In 0.2 g of a sample of  H2O2,

Let \'\'x\'\' gm of pure H2O2 is present, then

Equivalents of H2O2 = Equivalents of KMnO4

Moles of H2O2 X V.F of H2O2moles of KMnO4 X V.F of KMnO4

  = Molarity x volume x V.F of KMnO4

 = Normality x V.F of KMnO4   [N =M. V.F]

   Thus,

(x/34) x 2 = 1 x 10/1000

X/17 = 1/100

x = 17/100

x = 0.17.

Thus Pure H2O2 in 0.2 gm sample is =0.17/0.2 x 100

= 85 %

Hence Option “B” is correct answer.

'

In alkaline medium, $ClO _2$ oxidizes $H _2O _2$ to $O _2$ and is itself reduced to ${ClO} _2^{\displaystyle-}$. Number of moles of $H _2O _2$ oxidized by 1 mole of $ClO _2$ is :

  1. 1

  2. 1.5

  3. 0.5

  4. 3.5


Correct Option: C
Explanation:

The balanced reaction is as follows:
$H _2O _2 + 2 ClO _2 \rightarrow 2ClO _2^ + O _2 + 2 H^{+}$
So, number of moles of $H _2O _2$ oxidized by 1 mole of $ClO _2$ is 0.5.

For the reaction of $MnO _2\,+\,C _2O _{4}^{2-}\,+\,H^+\,\rightarrow\,Mn^{2+}\,+\,CO _2\,+\,H _2O$ the correct whole number stoichiometric coefficients of $MnO _4$ , $C _2$$O _4$ and $H^+$ are respectively :

  1. 2 , 5 , 16

  2. 16 , 5 , 2

  3. 5 , 16 , 2

  4. 2 , 16 , 5


Correct Option: A
- Hide questions