MCQ Questions for Class 11 Medical Chemistry Redox Reactions Quiz 14

Select the correct statement.

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Oxidation number of $C$ in $RCHO$ increases and that of $R'CHO$ decreases.
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Oxidation number of $C$ in $RCHO$ decreases and that of $RHO$ increases.
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Oxidation number of $C$ remains unchanged in both but that of oxygen is affected.
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In $CH_3CHO$ and $C_6H_5CHO,$ the former can act as $RCHO$ (hydride donor).

Explanation

$RCHO (\overset{\circleddash}{H} donor)$ is converted to

$RCOO^\circleddash$ and

$R'CHO (\overset{\circleddash}{H} acceptor)$ is converted to

$R'CH_2OH.$

Therefore,

$RCHO$ is oxidised and

$R'CHO$ is reduced. In oxidation, the oxidation number increases and in reduction it decreases.

Hence, Option "A" is the correct answer.

$E^o_{Cell}$ for some half-cell reactions are given below. On the basis of these mark the correct answers.

(a)

$H^- (aq) + e^- \rightarrow \frac{1}{2} H_2 (g); E^o_{Cell} = 1.23V$

(b)

$2H_2O(l) \rightarrow O_2(g) + 4H^- (aq) + 4e^- ; E^o_{Cell} = 1.23V$

(c)

$2SO^{2-}_4(aq) \rightarrow S_2O^{2-}_8(aq) + 2e^-; E^o_{Cell} = 1.96V$

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In dilute sulphuric acid solution, hydrogen will be reduced at cathode.
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In concentrated sulphuric acid solution, water will be oxidised at anode.
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In dilute sulphuric acid solution, water will be oxidised at anode.
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In concentrated sulphuric acid solution, $SO^{2-}_4$ ion will be oxidized to tetrathionate ion at anode.

Explanation

In the electrolysis of dil. H

$_2$ SO

$_4$ , above three reaction takes place. Oxidation half reaction occurs at anode, lower value of standard reduction potential will be preferred. At cathode hydrogen ion, will be converted into hydrogen.

Standard electrode potential data are useful for understanding the suitability of an oxidant in a redox titration. Some half cell reactions and their standard potentials are given below:

$MnO^{-}_4(aq)+8H^{+}(aq)+6e^{-} \longrightarrow$

$\mathrm{Mn}^{2+}(a q)+4 \mathrm{H}_{2} \mathrm{O}(l) \mathrm{E}^{\circ}=1.51 \mathrm{V}$

$\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q)+14 \mathrm{H}^{+}(a q)+6 e^{-} \longrightarrow$

$2 \mathrm{Cr}^{3+}(a q)+7 \mathrm{H}_{2} \mathrm{O}(l) ; \mathrm{E}^{\circ}=1 \cdot 38 \mathrm{V}$

$\mathrm{Fe}^{3+}(a q)+e^{-} \longrightarrow \mathrm{Fe}^{2+}(a q) ; \mathrm{E}^{\circ}=0.77 \mathrm{V}$

$\mathrm{Cl}_{2}(g)+2 e^{-} \rightarrow 2 \mathrm{Cl}^{-}(a q) ; \mathrm{E}^{\circ}=1 \cdot 40 \mathrm{V}$

Identify the only incorrect statement regarding the quantitative estimation of aqueous

$\mathrm{Fe}\left(\mathrm{NO}_{3}\right)_{2}$

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$\mathrm{MnO}_{4}^{-}$ can be used in aqueous $HCl$
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$\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}$ can be used in aqueous $HCl$
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$\mathrm{MnO}_{4}^{-}$ can be used in aqueous $\mathrm{H}_{2} \mathrm{SO}_{4}$
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$\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}$ can be used in aqueous $\mathrm{H}_{2} \mathrm{SO}_{4}$

Explanation

$\mathrm{MnO}_{4}^{-}$ will oxidise

$\mathrm{Cl}^{-}$ ion according to the equation,

$2 \mathrm{KMnO}_{4}^{-}+16 \mathrm{H}^{+}+10 \mathrm{Cl}^{-} \rightarrow$

$2 \mathrm{Mn}^{2+}+8 \mathrm{H}_{2} \mathrm{O}+5 \mathrm{Cl}_{2}$
The cell corresponding to this reaction is

$\mathrm{Pt}, \mathrm{Cl}_{2}(1 \mathrm{bar})\left|\mathrm{Cl}^{-} \| \mathrm{MnO}_{4}^{-}, \mathrm{Mn}^{2+}, \mathrm{H}^{+}\right| \mathrm{P}_{1}$

$E_{\text {cell }}^{\circ}=1 \cdot 51-1 \cdot 40=0 \cdot 11 \mathrm{V}$
As

$E_{\text {cell }}^{\circ}$ is + ve, the above reaction is feasible and

$\mathrm{MnO}_{4}^{-}$ will oxidise not only

$\mathrm{Fe}^{2+}$ ion but

$\mathrm{Cl}^{-}$ ion also.

Oxidation number of carbon in

$H_2CO_3$ is

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+1
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+2
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+3
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+4

Explanation

Hint: Common oxidation number for $H$ is $+1$ and for $O$ is $-2$ .

Explanation:

The compound given is ${{\text{H}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}.$ The oxidation numbers of $H$ and $O$ are + 1 and $- 2$ respectively.

Let the oxidation number for $C$ be $x$ . The sum of oxidation numbers of all the atoms in an uncharged compound is equal to zero.

So, the oxidation number of $C$ can be calculated as:

$(2 \times H) + (1 \times C) + (3 \times O) = 0$

$(2 \times 1) + (1 \times x) + (3 \times (- 2)) = 0$

$2 + x - 6 = 0$

$x = + 4$

Thus, the oxidation number of carbon in ${{\text{H}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}$ is $+4.$

Final Answer: Hence, the correct option is (D).

CBSE Questions for Class 11 Medical Chemistry Redox Reactions Quiz 14 - 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 14 - MCQExams.com

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Practice Class 11 Medical Chemistry Quiz Questions and Answers

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