3 Marks Question

Question 13 Marks

On the basis of the following data, explain why $Co -(III)$ is not stable in aqueous solution?
$Co^{3+}+e^{-} \rightarrow Co^2+, E^o=+1.82 V$
$2 H_2 O \rightarrow O_2+4 H^{+}+4 e^{-}, E^o=1.23 V$

Answer

Adding the two half reactions, $e.m.f.$ comes out to positive.
$emf=E_{\text {cathode }}-E_{\text {anode }}$
$emf=1.82-1.23=+0.59$
This means that $Co (III)$ in aqueous solution has the tendency to change to $Co (II).$
Hence, $Co (III)$ is not stable in aqueous solution.

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Question 23 Marks

How the following conversions can be carried out?
i. 2-Bromopropane to 1-bromopropane
ii. Chloroethane to butane
iii. Benzene to diphenyl

Answer

i. 2-Bromopropane to 1-bromopropane
a. alc KOH heat
b. HBr peroxide.
ii. Chloroethane to butane
Na wurtz reaction
iii. Benzene to diphenyl
a. Bromine, ferric bromide
b. Na Dry ether fitting reaction

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Question 33 Marks

How long will it take an electric current of $0.15 A$ to deposit all the copper from $500\ ml$ of $0.15 M$ copper sulphate solution?

Answer

You have mols in
$500 ml$ of $0.15\ M\ CuSO _4$ solution contains
$\frac{500 \times 0.15}{1000}=0.075\ mole$ of $Cu$
Mass of $Cu =0.075 \times 63.5=4.7625 g$
Eq. Wt. of $C u=\frac{635}{2}=31.75$
$m=Z \times I \times t$
$4.7625=\frac{31.75}{96500} \times 0.15 \times t$
$t=\frac{4.7625 \times 96500}{31.75 \times 0.15}$
$=96500 \sec$
$=\frac{96500}{60 \times 60}$
$=26.80$ hours.

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Question 43 Marks

An organic compound $(A)$ $($molecular formula $C_8H_{16}0_2)$ was hydrolysed with dilute sulphuric acid to give a carboxylic acid $(B)$ and an alcohol $(C).$ Oxidation of $(C)$ with chromic acid produced $(B). (C)$ on dehydration gives but$-1-$ene. Write equations for the reactions involved.

Answer

The relevant equations for all the reactions involved may be explained as follows$:$

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Question 53 Marks

Name the reagents used in the following reactions$:$
$i.$ Oxidation of a primary alcohol to carboxylic acid.
$ii.$ Oxidation of a primary alcohol to aldehyde.
$iii.$ Bromination of phenol to $2,4,6-$tribromophenol.
$iv.$ Benzyl alcohol to benzoic acid.
$v.$ Dehydration of propan$-2-$ol to propene.
$vi.$ Butan$-2-$one to butan$-2-$ol.

Answer

$i.$ Acidified potassium permanganate
$ii.$ Pyridinium chlorochromate $(PCC)$
$iii.$ Bromine water
$iv.$ Acidified potassium permanganate
$v. 85\%$ phosphoric acid
$vi. NaBH_4$ or $LiAlH_4$

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Question 63 Marks

How can diethyl ether be prepared from
i. ethyl iodide
ii. ethyl alcohol?
Why is the boiling point of an ether lower than that of the isomeric alcohols.

Answer

i. $\underset{\text { ethanol (excess) }}{2 CH _3 CH _2 OH } \xrightarrow{ Conc . H _2 SO _4}{ }_4 CH _3 CH _2-\underset{\text { ethoxyethane }}{ O - CH _2 CH _3}+ H _2 O$
ii. $2 R^{\prime}- OH +2 Na \rightarrow 2 R ^{\prime}- O - Na a^{+}+ H _2$ Alcohol
iii. $R^{\prime}-O^{-} N a^{+}+R-X \xrightarrow{S_{N^2}} R^{\prime}-\underset{\text { Ether }}{O}-R+N a^{+} X^{-}$
Boiling points: Ethers are isomeric with monohydric alcohols but their boiling points are much lower than those of the isomeric alcohols. This is due to the reason that unlike alcohols, ethers do not form hydrogen bonds. As a result ethers do not show molecular association and hence have lower boiling point than corresponding alcohols.

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Question 73 Marks

The decomposition of $NH _3$ on platinum surface is zero order reaction. What are the rates of production of $N _2$ and $H_2$ if $k=2.5 \times 10^{-4} mol^{-1} Ls ^{-1}$ ?

Answer

The decomposition of on platinum surface is represented by the following equation.
$
2 NH_3(g) \rightarrow N_2(g)+3 H_2(g)
$
Therefore, Rate $=-\frac{1}{2} \frac{d\left[N H_3\right]}{d t}=\frac{d\left[N_2\right]}{d t}=\frac{1}{3} \frac{d\left[H_2\right]}{d t}$
However, it is given that the reaction is of zero order.
$
-\frac{1}{2} \frac{d\left[N H_3\right]}{d t}=\frac{d\left[N_2\right]}{d t}=\frac{1}{3} \frac{d\left[H_2\right]}{d t}=k
$
Therefore, $k=2.5 \times 10^{-4} mol L^{-1} s^{-1}$
Hence, the rate of production of $N_2$ is
$
\frac{d\left[N_2\right]}{d t}=2.5 \times 10^{-4} mol L^{-1} s^{-1}
$
And, the rate of production of $H _2$ is
$
\frac{d\left[H_2\right]}{d t}=3 \times 2.5 \times 10^{-4} mol L^{-1} s^{-1}=7.5 \times 10^{-4} mol L^{-1} s^{-1}
$

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Question 83 Marks

$a.$ Write the formulation for the galvanic cell in which the reaction $C u(s)+2 A g^{+}(a q) \rightarrow C u^{2+}(a q)+2 A g(s)$ takes place.
Identify the cathode and the anode reactions in it.
$b.$ Write Nernst equation and calculate the emf of the following cell: $Sn ( s )\left| Sn ^{2+}(0.04 M ) \| H ^{+}(0.02 M )\right|$ $H _2(g) \mid \operatorname{Pt}( s )$
$\left(\right.$ Given $\left.E_{S n^{2+} / S n}=-0.14 V\right)$

Answer

$a.$ We have
$E_{\left(C u^{2+} / Cu\right)}^{\Theta}=0.34 V \text { and } E_{\left(Ag^{+} / Ag\right)}^{\Theta}=0.80 V$
Standard emf of Cu is less than Ag , therefore it is strong reducing agent and is oxidised. Therefore Cu acts as Anode and Ag acts as Cathode.
Half cell reactions are:
At Cathode (Reduction):
$2 Ag^{+}(a q)+2 e^{-} \rightarrow 2 Ag(s)$
At Anode (Oxidation):
$C u(s) \rightarrow C u^{2+}(a q)+2 e^{-}$
$b.$ The reactions are :
At Anode:
$S n(s) \rightarrow S n^{2+}(a q)+2 e^{-}$
At Cathode:
$2 H^{+}(a q)+2 e^{-} \rightarrow H_2(g)$
Full cell reaction:
$Sn(s)+2 H^{+}(a q) \rightarrow Sn^{2+}(a q)+H_2(g)$
Standard emf of the cell is:
$E_{\text {cell }}^0=E_{H^{+} / H_2}^0-E_{S n^{2+} / S n}^0$
$=0-(-0.14) V$
$=+0.14 V$
For this reaction $n =2$ moles of electrons. Using Nernst equation,
$E_{\text {cell }}=0.14-\frac{0.0591}{2} \log \frac{\left[Sn^2+\right]}{\left[H^{+}\right]^2}$
$=0.14-\frac{0.0591}{2} \log \frac{0.04}{(0.02)^2}$
$=0.14-\frac{0.0591}{2} \log \frac{4}{100} \times \frac{100}{2} \times \frac{100}{2}$
$=0.14 V-0.0591 V$
$=0.0809 V$

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