3 Marks Question

Question 13 Marks

Calculate the emf of the cell $Mg ( s )\left| Mg ^{+2}(0.1 M )\right| \mid Cu ^{+2}\left(1 \times 10^{-3} M \mid Cu ( s )\right)$ Given, $\left.E _{C u^{2+} / C u}^{\ominus}=+0.34 V, E _{M g^{+2} / M g}^{\ominus}=-2.37 V\right)$

Answer

We have
$Mg(s)\left|Mg^{+2}(0.1 M)\right| \mid Cu^{+2}\left(1 \times 10^{-3} M \mid Cu(s)\right)$
Half cell reactions of this cell are:
At Cathode $($Reduction$)$ :
$C u^{2+}(a q)+2 e^{-} \rightarrow C u(s)$
At Anode $($Oxidation$)$ :
$M g(s) \rightarrow M g^{2+}(a q)+2 e^{-}$
For this cell, we have, $n =2$ moles of electrons.
$ ($Given $, E_{Cu 2+}^{\ominus} / Cu$
$\left.E_{\text {cell }}^{\ominus}=+0.34 V, E_{Mg^{+2} / Mg}^{\ominus}=-2.37 V\right)$
$E_{\text {cell }}^{\ominus}=E_{\left(C Cu^{2+} / Cu\right)}^{\ominus}-E_{\left(Mg^{2+} / Mg\right)}^{\ominus}-E_{\text {oxidation }}^{\ominus}$
$=0.34-(-2.37) V$
$=2.71 V$
According to Nernst equation
$E_{\text {cell }}=E_{\text {cell }}^{\ominus}-\frac{0.059}{n} \log \frac{\left[Mg^{2+}\right]}{\left[Cu^{2+}\right]}$
$E_{\text {cell }}=E_{\text {cell }}^{\ominus}-\frac{0.059}{2} \log \left[\frac{0.1}{10^{-3}}\right]$
$=2.71-0.0295 \log 10^2$
$=2.71-0.0295 \times 2$
$=2.651$
$\therefore E_{\text {cell }}=2.651 V$

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Question 23 Marks

Tert$-$Butylbromide reacts with aq. $NaOH$ by $S_N1$ mechanism while $n-$butylbromide reacts by $SN_2$ mechanism. Why?

Answer

Tert$-$butylbromide undergoes substitution by $S_N1$ unimolecular substitution mechanism because it is able to form a stable carbocation in the first step after cleavage of the halide group. The carbocation then reacts with the nucleophile $OH^-.$ On the other hand, primary halide $n-$butylbromide cannot form a stable carbocation so it undergoes $S_N2$ bimolecular substitution mechanism,which is a one$-$step substitution that involves the attack of $OH^-$ and simultaneous leaving of $X^-$ to form $n-$butyl alcohol.

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Question 33 Marks

Write the Nernst equation and calculate emf of the following cell at $298 K:$
$ Cr \left| Cr ^{3+}(0 \cdot 1 M ) \| Fe ^{2+}(0 \cdot 01 M )\right| Fe$
$\text { Given : } E _{ Cr ^{3+} / Cr }^{\ominus}=-0 \cdot 75 V$
$E _{ Fe ^{2+} / Fe }^{\ominus}=-0 \cdot 45 V (\log 10=1)$

Answer

$E_{\text {cell }}=E_{\text {cell }}^0-\frac{0.059}{6} \log \frac{\left[ Cr ^{3+}\right]^2}{\left[ Fe ^{2+}\right]^3}$
$E_{\text {cell }}^0=-0.45-(-0.75) =0.30 V$
$E_{\text {cell }}=0.3-\frac{0.059}{6} \log \frac{(0-1)^2}{(0.01)^3}$
$=0 \cdot 3-.00985 \log \frac{\left(10^{-1}\right)^2}{\left(10^{-2}\right)^3}$
$=0.3-.00985 \times 4 \log 10$
$=0.3-0.0394$
$=0.2606 V$

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Question 43 Marks

Predict the products formed when cyclohexanecarbaldehyde reacts with following reagents.
$i. \ MgBr$ and then $H_3O^+$
$ii.$ Tollens' reagent
$iii.$ Semicarbazide and weak acid
$iv.$ Excess ethanol and acid
$v. \ Zinc$ amalgam and dilute hydrochloric acid

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Question 53 Marks

How can phenol be converted to aspirin?

Answer

Phenol is converted into salicylic acid. The reaction is usually carried out by allowing sodium phenoxide to absorb carbon dioxide and then heating the product to $400 K$ and $4-7 \ atm$ pressure. The first unstable intermediate is formed which undergoes a proton shift to form sodium salicylate. The subsequent acidification of sodium salicylate gives.

Then aspirin is obtained by acetylating salicylic acid with acetic anhydride and conc. $H_2SO_4$

The preparation of Aspirin from salicylic acid is an example of an electrophilic substitution reaction in which carbon dioxide is an electrophile.

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Question 63 Marks

Write the mechanism of acid dehydration of ethanol to yield ethane.

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Question 73 Marks

$i.$ Determine the units of rate constant for first and zero order reaction.
$ii.$ Show that time required for the completion of $99\%$ of the first order reaction is twice the 90\% of completion of the reaction.

Answer

$i. K =( mol )^{1- n } L ^{ n -1} s^{-1}$
For zero order, $n =0$
So, $K=( mol )^{1-0} L^{0-1} s^{-1}= s ^{-1} mol L ^{-1}$
For first order, $n =1$
$K=(mol)^{1-n} L^{n-1} s^{-1}$
$\text { So, } K=(mol)^{1-1} L^{1-1} s^{-1}$
$=s^{-1}$
$ii.$ For a first order reaction,
$t=\frac{2.303}{K} \log \frac{[A]_0}{[A]}$
${[A]_0=a,[A]=a-\frac{a \times 99}{100}=0.01 a}$
$t(99 \%)=\frac{2.303}{K} \log \frac{a}{0.01 a}$
$=\frac{2.303}{K} \log 100$
$=\frac{2.303}{K} \times 2 \ldots \text { (i) }$
For $90 \%$ completion of reaction,
${[A]=a-\frac{a \times 99}{100}=0.1 a}$
$t(90 \%)=\frac{2.303}{K} \log \frac{a}{0.1 a}$
$=\frac{2.303}{K} \times 1 \ldots .(i)$
Dividing equation $(i)$ by equation $(ii)$, we get
$t(99 \%)=2 \times t(90 \%)$
Hence, the time taken to complete $9\%$ of the first order reaction is twice the time required for the completion of $90\%$ of the reaction.

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Question 83 Marks

How much charge is required for the following reductions:
$i. 1$ mol of $Al ^{3+}$ to $Al\ ?$
$ii. 1$ mol of $Cu ^{2+}$ to $Cu\ ?$
$iii. 1$ mol of $MnO _4^{-}$to $Mn ^{2+}\ ?$

Answer

$i. A l^{3+}+3 e^{-} \rightarrow A l$
Therefore, Required charge $=3 F$
$=3 \times 96487 C$
$=289461 C$
$ii. C u^{2+}+2 e^{-} \rightarrow C u$
Therefore, Required charge $=2 F$
$=2 \times 96487 C$
$=192974 C$
$iii. MnO _4^{-} \rightarrow Mn ^{2+}$
i.e., $M n^{7+}+5 e^{-} \rightarrow M n^{2+}$
Therefore, Required charge $=5 F$
$=5 \times 96487 C$
$=482435 C$

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