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Multiple choice redox and stoichiometry applications of redox reaction oxidation- reduction reactions redox reactions chemistry

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. $3H^{+} + 3H _5IO _6 + 3Cr \rightarrow 3Cr^{3+} + 3IO^{3-} + 9H _2O$
Reveal answer Fill a bubble to check yourself
A Correct answer
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
$3H^++3H _5IO _6+2Cr \longrightarrow 2Cr^{+3}+3IO^{3-}+9H _2O$ .

Multiple choice redox and stoichiometry applications of redox reaction oxidation- reduction reactions redox reactions chemistry

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. $\dfrac {\text {molar mass}}{13}$
Reveal answer Fill a bubble to check yourself
C Correct answer
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 $FeS _{2} = \dfrac {\text {Molar mass}}{11}$. 

Multiple choice redox and stoichiometry applications of redox reaction oxidation- reduction reactions redox reactions chemistry

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%
Reveal answer Fill a bubble to check yourself
B Correct answer
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 H2O2 = moles 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.

'

Multiple choice redox and stoichiometry applications of redox reaction oxidation- reduction reactions redox reactions chemistry

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
Reveal answer Fill a bubble to check yourself
C Correct answer
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.

Multiple choice redox and stoichiometry applications of redox reaction oxidation- reduction reactions redox reactions chemistry

Give the products available on the cathode and the anode respectively during the electrolysis of an aqueous solution of $MgSO _4$ between inert electrodes.

  1. $O _2$ and $H _2$
  2. $H _2$ and $O _2$
  3. $O _2$ and $Mg$

  4. none of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The electrolysis of an aqueous solution of magnesium using inert electrodes produces hydrogen at the cathode and oxygen at the anode and neutral solution of magnesium sulphate remains unaltered by the electrolysis.

Multiple choice redox and stoichiometry applications of redox reaction oxidation- reduction reactions redox reactions chemistry

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
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Balance redox half-reactions: MnO4- + 8H+ + 5e- → Mn^2+ + 4H2O (reduction, gain 5e-); C2O4^2- → 2CO2 + 2e- (oxidation, lose 2e-). LCM of 5 and 2 is 10, so 2 MnO4- (gains 10e-) and 5 C2O4^2- (loses 10e-). H+ balance requires 16 H+ for 2 MnO4-.

Multiple choice chemistry amines chemical reactions of diazonium salts chemical properties of diazonium salts diazonium salts

Sandmeyer reaction is better than Gattermann reaction because:

  1. sandmeyer reaction uses salt of copper while Gattermann reaction uses copper powder.

  2. only one salt is produced in Sandmeyer reaction while Gattermann reaction produces two salts.

  3. sandmeyer reaction gives better yield than Gattermann reaction.

  4. all of the above.
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Sandmeyer gives a higher yield.So,it is better than Gattermann reaction.

Multiple choice chemistry amines chemical reactions of diazonium salts chemical properties of diazonium salts diazonium salts

Which of the reaction would you choose for the most beneficial conversion shown below?
$ArN _2^+X \rightarrow ArCl$

  1. Sandmeyer Reaction

  2. Gattermann Reaction
  3. Directly reacting with $HCl$

  4. Both A and B
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Sandmeyer and Gattermann reactions are two processes used mostly for the above conversion.But Sandmeyer is more beneficial as it gives a higher yield.