3 Marks Question

Question 13 Marks

Zinc rod is dipped in 0.1 M solution of $ZnSO _4$
The salt is $95 \%$ dissociated at is dilution at 298 K . Calculate the electrode potential. Given
$E ^0\left( Zn ^{2+} / Zn \right)=-0.76$

Answer

$\begin{array}{l}{\left[Z n^{2+}\right]=0.1 \times \frac{95}{100}=0.095 M} \\ Z n^{2+}+2 e^{-} \rightarrow Z n \\ E_{\left(Z n^{+} / Z n\right)}=E_{\left(Z n^{+} / Z n\right)}^0-\frac{0.0591}{2} \log \frac{1}{\left[Z n^{2+}\right]} \\ =-0.76 V-\frac{0.0591}{2} \log \frac{1}{0.095} \\ =-0.76 V-\frac{0.0591}{2}[\log 1000-\log 95] \\ =-0.76-\frac{0.0591}{2}[3.000-1.9777] \\ =-0.76 V-\frac{0.0591}{2} \times 1.0223 \\ =-0.76 V-\frac{0.0604}{2}=0.76-0.0302 \\ =-0.7902 V\end{array}$

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Question 23 Marks

Give structures of the products you would except when each of the following alcohols
i. Butan-1-ol
ii. 2-Methylbutan-2-ol react with
a. HCI - ZnCl₂
b. HBr and
c. SOCI₂

Answer

i. a. With HCl - ZnCl₂ (lucas reagent) 2-Methylbutane-2-ol
ii. Being a 3º alcohol, reacts with Lucas reagent to produce turbidity immediately due to the formation of insoluble tert-alkyl chloride while butane-l-ol (i) being a 1ºalcohol does not react with Lucas reagent at room temperature.

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Question 33 Marks

Answer

is more reactive towards S_N2 because it is a primary halide.

is more reactive towards nucleophilic substitution reaction because it is more reactive due to the presence of electron with drawing - NO₂ group.

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Question 43 Marks

The following results have been obtained during the kinetic studies of the reaction.
2A + BC → D

Experiment	[A]/mol L^-1	[B]/mol L^-1	Initial rate of formation of D/mol L^-1 min^-1
I	0.1	0.1	6.0 x 10^-3
II	0.3	0.2	7.2 x 10^-2
III	0.3	0.4	2.88 × 10^-1
IV	0.4	0.1	2.40 x 10^-2

Determine the rate law and the rate constant for the reaction.

Answer

From experiments I and IV, it may be noted that $[B]$ is same but $[A]$ has been made four times, the rate of reaction has also becomes four times. This means w.r.t A
$
\text { Rate } \propto[A]
$
From experiments II and III it may be noted that [A] is kept same an [B] has been doubled, the rate of reaction has become four times. This means w.r.t B,
$
\text { Rate } \propto[B]^2
$
Combining (i) and (ii), we get the rate law for the given reaction as:
$
\text { Rate }=k[A][B]^2
$
Thus order w.r.t $A =1$
Order w.r.t B $=2$
The overall order of the reaction
$
=1+2=3
$
The rate constant and its units can be calculated from the data of each experiment using the expression. $
\begin{array}{l}
k=\frac{\text { rate }}{[A][B]^2} \\
=\frac{mol L^{-1} \min ^{-1}}{\left(mol L^{-1}\right)\left(mol L^{-1}\right)^2} \\
=mol^{-2} L^{-2} \min ^{-1}
\end{array}
$
Expt. $k=$ mol $^{-2} L^{-2} \min ^{-1}$
i. $\frac{6.0 \times 10^{-3}}{0.1 \times(0.1)^2}=6.0$
ii. $\frac{7.2 \times 10^{-2}}{0.3 \times(0.2)^2}=6.0$
iii. $\frac{2.88 \times 10^{-1}}{0.3 \times(0.4)^2}=6.0$
iv. $\frac{2.4 \times 10^{-2}}{0.4 \times(0.1)^2}=6.0$
Rate constant $k=6.0 mol^{-2} L^{-2} \min ^{-1}$

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Question 53 Marks

Calculate the emf of the following cell at 298 K .
$
2 Cr(s)+3 Fe^{2+}(0.1 M) \rightarrow 2 Cr^{3+}(0.01 M)+3 Fe(s)
$
Given, $E_{c r^{3+} / c r}^o=-0.74 V, E_{F e^{2+} / F e}^o=-0.44 V$

Answer

Since oxidation of Cr is taking place in the given reaction, the chromium electrode is anode and as Fe is reduced in the reaction, Fe electrode is the cathode. The half-cell reactions are as follows.
At anode $\left.Cr \rightarrow Cr ^{3+}+3 e ^{-}\right] \times 2$
At cathode $\left.Fe ^{2+}+2 e ^{-} \rightarrow Fe \right] \times 3$
Overall reaction
$
2 Cr+3 Fe^{2+} \rightarrow 2 Cr^{3+}+3 Fe
$
$
\begin{array}{l}
\text { Eo }=E_{\text {cathode }}-E_{\text {anode }}=-0.44-(-0.74)=0.3 V \\
E=E^{o}-\frac{0.0591}{n} \log \frac{\left[Cr^{3+}\right]^2}{\left[Fe^{2+}\right]^3}
\end{array}
$
Here, $n =$ number of electrons transferred, i.e. equal to 6 .
$
\begin{array}{l}
=0.3-\frac{0.0591}{6} \log \frac{[0.01]^2}{[0.1]^3} \\
=0.309 \approx 0.31
\end{array}
$

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Question 63 Marks

Calculate the e.m.f at $25^{\circ} c$ for the following cell:
$
N i(s)\left|N i^{2+}(0.01 M) \| C u^{2+}(0.1 M)\right| C u(s)
$
Given: $E ^0 Ni ^{+2} / Ni =-0.25 V$
$
E_{C u^{2+} / C u}^0=+0.34 V
$
$\left[1 F=96,500 Cmol ^{-1}\right]$. Calculate the maximum work that can be accomplished by the operation of this cell

Answer

Cell reaction is
$
N i(s)\left|N i^{2+}(0.01 M)\right|\left|C u^{2+}(0.1 M)\right| C u(s)
$
At anode : $N i(s) \rightarrow N i^{2+}(a q)+2 e^{-}$
At cathode: $Cu ^{2+}(a q)+2 e \rightarrow Cu (s)$
Net cell reaction
$
\begin{array}{l}
Ni(s)+Cu^{2+}(a q) \rightarrow N i^{2+}(a q)+Cu(s) \\
E_{\text {cell }}=E_{\text {cell }}^0-\frac{0.0591}{2} \log \frac{\left[Ni^{2+}\right]}{\left[Cu^{2+}\right]} \\
=[+0.34 V-(-0.25 V)]-\frac{0.0591}{2} \log \frac{1}{10} \\
=0.59 V-\frac{0.0591}{2} \times-1 \\
=0.59+0.0295 \\
=0.6195 V
\end{array}
$
$
\begin{array}{l}
\Delta G=-n E F \\
=-2 \times 0.6195 V \times 96500 C / mol \\
=-119563.5 J / mol \\
\Delta G=-119.5635 KJ / mol \\
-\Delta G=W_{\max }=119.5635 KJ / mol
\end{array}
$

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Question 73 Marks

Draw structures of the following derivatives:
i. 2,4-Dinitrophenylhydrazone of benzaldehyde
ii. Cyclopropanone oxime
iii. Acetaldehydedimethylacetal

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Question 83 Marks

Explain how does the - OH group attached to a carbon of benzene ring activate it towards electrophilic substitution?

Answer

The -OH group is an electron-donating group. Thus, it increases the electron density in the benzene ring as shown in the given resonance structure of phenol.

As a result, the benzene ring is activated towards electrophilic substitution.

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Chemistry 3 Marks Question

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