3 Marks Question - Chemistry STD 12 Science Questions - Vidyadip

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Question 13 Marks

How is the molar conductivity of a weak electrolyte at infinite dilution determined?

Answer

The value of limiting molar conductivity of weak electrolytes is not obtained from the graph between $\Lambda_{ m }$ and $C ^{\frac{1}{2}}$, hence it can be calculated as follows :
For example :
$C H _3 C O O H$ (Acetic acid): To determine this, the limiting molar conductivities of $CH _3 COONa , HCl$ and NaCl are used which are already known, because all of them are strong electrolytes.
Hence, $\Lambda_{ m }^{\circ}\left( CH _3 COOH \right)=\lambda_{\left( CH _3 COO ^{-}\right)}^{\circ}+\lambda_{\left( H ^{+}\right)}^{\circ}$
Acetate ion $\left( CH _3 COO ^{-}\right)$is indicated as $Ac ^{-}$.
Therefore, from $NaCl , HCl$ and $CH _3 COONa$
$
\begin{aligned}
\Lambda_{m(NaAc)}^{\circ} & =\lambda_{Ac^{+}}^{\circ}+\lambda_{Na^{+}}^{\circ} \\
\Lambda_{m(HCl)}^{\circ} & =\lambda_{H^{+}}^{\circ}+\lambda_{Cl^{-}}^{\circ} \\
\text { and } \Lambda_{m(NaCl)}^{\circ} & =\lambda_{Na^{+}}^{\circ}+\lambda_{Cl^{-}}^{\circ}
\end{aligned}
$
$\begin{aligned} \Lambda_{ m \left( CH H _3 COOH \right)}^{\circ} & =\Lambda_{ m ( HCl )}^{\circ}+\Lambda_{ m ( NaAc )}^{\circ}-\Lambda_{ m ( NaCl )}^{\circ} \\ \Lambda_{ m ( HAc )}^{\circ} & =\lambda_{ H ^{+}}^{\circ}+\lambda_{ Cl ^{-}}^{\circ}+\lambda_{ Na ^{+}}^{\circ}+\lambda_{ Ac ^{-}}^{\circ}-\lambda_{ Na ^{+}}^{\circ}-\lambda_{ Cl ^{-}}^{\circ} \\ \Lambda_{ m ( HAc )}^{\circ} & =\lambda_{\left( Ac ^{-}\right)}^{\circ}+\lambda_{ H ^{+}}^{\circ} \\ \text { or } \quad \Lambda_{ m ( HAc )}^{\circ} & =\lambda_{\left( CH _3 COO ^{-}\right)}^{\circ}+\lambda_{ H ^{+}}^{\circ}\end{aligned}$

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Question 23 Marks

How much electricity will be required to obtain 1.5 gram on copper in 3 hours in an electrolytic cell of aqueous solution of copper sulphate? (Atomic mass of Cu = 63.5 )

Answer

$Z =\frac{ E }{ F }=\frac{\text { Equivalent weight }}{\text { Faraday }}$
Equivalent weight of copper $=\frac{63.5}{2}=31.75$
Because Cu is formed from $Cu ^{2+}$ for which 2 electrons are important. Hence, $\quad Z=\frac{31.75}{96500}=3.29 \times 10^{-4}$
According to first law of Faraday,
$
\begin{array}{c}
W=ZIt \\
I=\frac{W}{Zt} \\
W=1.5 g, \quad Z=3.29 \times 10^{-4}, t=3 \times 60 \times 60 \text { second }
\end{array}
$
Hence, $\quad I=\frac{1.5}{3 \times 60 \times 60 \times 3.29 \times 10^{-4}}$
$
I=0.422 \text { Ampere }
$

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Question 33 Marks

In which of the following pairs will more electric current flow and why?
(a) KCl solution $25^{\circ} C$ and KCl solution at $50^{\circ} C$.
(b) $0.2 \quad M \quad CH _3 COOH$ solution and 2 M $CH _3 COOH$ solution.
(c) $NH _4 OH$ solution at $25^{\circ} C$ and at wire at $50^{\circ} C$.
(d) Al wire at $25^{\circ} C$ and Al wire at $50^{\circ} C$.

Answer

(a) KCl solution at $50^{\circ} C$, because on increasing the temperature, the speed of ion increases.
(b) $0.2 M CH _3 COOH$ solution because ionization of weak electrolytes is more in dilute solution.
(c) $NH _4 OH$ solution at $75^{\circ} C$, because it weakens on increasing the temperature, ionization of eletrolyte is more.
(d) Al wire at $25^{\circ} C$ because the conductivity of metal decreases with increasing temperature.

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Question 43 Marks

Explain the products obtained from electrolysis of aqueous solution of sodium chloride on the basis of electrode potential values.

Answer

Electrolysis of aqueous solution of sodium chloride produces $NaOH , Cl _2$ gas and $H _2$ gas.
Along with $Na ^{+}$and $Cl ^{-}$, it also contains $H ^{+}$and $OH ^{-}$ions and solvent molecules $\left( H _2 O \right)$.
The possible reactions ocuring at cathode are :
$Na^{+}(aq)+e^{-} \rightarrow Na(s) ; E_{\text {cell }}^{\ominus}=-2.71 V \ldots \ldots(1)$
$H^{+}(aq)+e^{-} \rightarrow \frac{1}{2} H_2(g) ; E_{\text {cell }}^{\ominus}=0.00 V \ldots \ldots(2)$
The value of $E_{\text {cell }}^{\circ}$ is higher for reaction $(2)$ so this will take place at cathode.
$H ^{+}$ is obtained from dissociation of $H _2 O$, hence, reaction taking place at cathode is as follows:

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Question 53 Marks

Calculate the charge of one mole of electrons.

Answer

Charge on one electron $=1.6021 \times 10^{-19} C$
$($coulomb$)$
$ 1\ mole$ electron $=$ Avogadro number $\left( N _{ A }\right)$
$=6.022 \times 10^{23} mol^{-1} .$
Hence, charge of one mole of electrons
$=$ Charge on one electron $\times$ Avogadro number $\left( N _{ A }\right)$
$=1.6021 \times 10^{-19} C \times 6.022 \times 10^{23} mol^{-1} mol$
$=96478\ C\ mol ^{-1} \approx 96500\ C\ mol ^{-1}$
$=1$ Faraday $(1 F)$

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Question 63 Marks

Explain the structure and function of salt bridge of molten galvanic cell.

Answer

A salt bridge is an U-shaped glass tube containing saturated agar-agar gel with $KCl , KNO _3, NH _4 Cl$ etc. It completes the electrical circuit and maintains the electrical neutrality of both the earth cells. After the release of electrons, the excess positive charge $\left( Zn ^{2+}\right)$ present at anode is neutralized. Anions $\left( Cl ^{-}, NO _3{ }^{-}\right.$moves through the salt bridge towards anode and similarly, to neutralize excess negative charge $\left( SO _4{ }^{2-}\right)$ on the cathode, the cations $\left( K ^{+}\right.$, $NH _4{ }^{+}$) present in salt bridge moves towards the cathode. Before the movement of mentioned ions, the reaction and the flow of electricity stops.

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Question 73 Marks

What will be the amount of $H_2$ liberated at the cathode when a current of $1$ ampere is passed for $3000$ second in water acidified with $\ce{HCl}$?

Answer

Quantity of electric current $= Q = It$
$I =1$ Ampere, $t=3000$ second
$Q=1 \times 3000=3000 C$
Reaction at cathode :
$H_{(aq)}^{+}+e^{-} \rightarrow \frac{1}{2} H_2$
Hence, according to the reaction, $\frac{1}{2}$ mole $H _2 ( 1 g$ of $H _2 )$ can be obtained.
Amount of electricity required $=96500$ coulomb $($for $1$ nole electron$)$
Since, $H _2$ obtained from $96500 C =1$ gram
Hence, $H _2$ obtained from $3000 C =\frac{ l }{96500} \times 3000$
$=0.0310 g$

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Question 83 Marks

If the price of electricity require to obtain $1$ gram $M g$ from the solution of $M g ^{ 2 + }$ ion is $Rs.5 . 0 0$, then what will the price of electricity required to obtain $10$ gram of $Al$ from $Al ^{3+} ? \ ($Atomic mass of $Al = 27$ and $M g =24)$

Answer

According to Faraday's first law,
For $Al$ :
$W=Z Q$
$Z=\frac{E}{F}=\frac{M}{n F}$
Hence, $W =\frac{ M }{ nF } Q\ [ W =10 g$ and $n =3$ because $Al$ forms from $Al ^{3+}]$
$Q=\frac{WnF}{M}=\frac{10 \times 3 \times 96500}{27}=107222$ coulomb $(C)$
Similarly for $Mg , Q =\frac{ WnF }{ M }$
$[W =1 g$ and $n =2$ because $Mg$ forms from $Mg ^{2+}]$
$Q=\frac{1 \times 2 \times 96500}{24}=8041$ coulomb
As per question$, 8041 C$ electricity $= Rs. 5.00$
then$, 107222 C$ electricity $=\frac{5}{8041} \times 107222$
$ =66.67 Rs .$

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Question 93 Marks

How many gram of $Al$ metal will be obtained at the cathode by passing $0.1$ Faraday electric current in $\text{AlCl} _3$ solution? $($Atomic mass of $Al =27)$

Answer

The reaction taking place at cathode is :
$Al_{(aq)}^{3+}+3 e^{-} \rightarrow Al_{(s)}$
According to Faraday's first law,
$W=Z Q$
$Z=\frac{M}{n F}$
Hence, $ W =\frac{ M }{ nF } Q$
For $Al , M =27, n =3, F=96500$ coulomb,
$Q=0.1$
Faraday $=9650 C$
$W =\frac{27 \times 9650}{3 \times 96500}=0.9$ gram

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Question 103 Marks

If 200 mol of $H _2$ (at NTP) and 0.60 g of Cu are obtained by passing electric current in the solution of acidified water and $CuSO _4$ respectively in two electrolytic cell connected in series, then what will be the equivalent weight of Cu ?

Answer

Molecular weight of $H _2=2$ and equivalent weight $=1$
At NTP, $22.41(22400 ml ) H _2=2$ gram
Hence, for $200 ml H _2=\frac{2 \times 200}{22400}=0.01785$ gram
Obtained weight of $Cu =0.60$ gram
According to Faraday's second law,
$
\begin{aligned}
\frac{\text { Weight of copper }}{\text { Weight of } H_2} & =\frac{\text { Equivalent weight of } Cu}{\text { Equivalent weight of } H_2} \\
\frac{0.60}{0.01785} & =\frac{\text { Equivalent weight of } Cu}{1}
\end{aligned}
$
Equivalent weight of copper $=\frac{0.60}{0.01785}=33.6$

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Question 113 Marks

To coat a layer of $0.004\ mm$ thick Ag on the surface of copper metal whose area is $60 \ cm^2$, it will be necessary to flow a current of $5.0$ ampere in $A g N O _3$ solution, for how much time should it flow? $($Density of silver $=10.5 g \ cm ^{-3}$ and atomic mass of $Ag =108)$

Answer

Amount of Ag coated on copper $(W)$
$=$ Volume $\times$ density
$=$ Surface area $\times$ thickness $\times$ density
$=60 \times 0.004 \times 10.5$
$W=2.52$ gram
According to Faraday's first law,
$W=ZIt$
$Z=\frac{E}{F}=\frac{\text { Equivalent weight }}{\text { Faraday }}$
One electron is being used in the reaction :
$Ag_{(aq)}^{+}+e^{-} \rightarrow Ag_{(s)}$
Hence, equivalent weight of $Ag =\frac{108}{1}=108$
Hence,
$Z=\frac{108}{96500}$
$t=\frac{W}{Z I}=\frac{2.52 \times 96500}{108 \times 5}$
$t=450.3$ seconds
Time

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Question 123 Marks

When a current of 50 ampere is passed for 2 hours in the molton NaCl :
(a) How many gram of chlorine gas will be obtained at anode?
(b) What will be the volume of chlorine gas obtained at standard temperature and pressure (NTP)?

Answer

When electricity is passed through molten NaCl , the reaction occuring at anode is :
(a) $2 Cl ^{-} \rightarrow Cl _2+2 e ^{-}$ Molecular weight of $Cl _2=35.5 \times 2=71$ gram
Charge $\quad Q = It$
$I =50$ ampere and $t=2 \times 60 \times 60=7200$ seconds Hence, $Q=50 \times 7200=360000 C$
According to the reaction,
$
\begin{aligned}
2 \times 96500 \text { coulomb, } Cl_2 & =71 \text { gram } Cl_2(1 \text { mole }) \\
360000 \text { coulomb } Cl_2 & =\frac{360000}{2 \times 96500} \times 71 \\
& =132.4 \text { gram chlorine }
\end{aligned}
$
(b) $2 \times 96500 C , Cl _2=1 mole Cl _2$ Hence, 360000 coulomb $Cl _2=\frac{3,60,000}{2 \times 96500}$
$
=1.8652 mol \text { chlorine }
$
One mole $Cl _2$ at $NTP =22.4 l$
Therefore, $1.8652{mole Cl _2}^2=22.4 \times 1.8652$ $=41.78 l$ chlorine.

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Question 133 Marks

In the button cells widely used in watches and other devices the following reaction takes place :
$Zn _{( s )}+ Ag _2 O _{( s )}+ H _2 O _{( l )} \rightarrow Zn _{( aq )}^{2+}+2 Ag _{( s )}+2 OH _{( aq )}^{-}$
Determine $\Delta_{ r } G ^{\circ}$ and $E ^{\circ}$ for the reaction.

Answer

Cell reaction :
$Zn ( s )+ Ag _2 O ( s )+ H _2 O ( l ) \rightarrow Zn ^{2+}( aq )+2 Ag ( s )+2 OH ^{-}( aq )$
From active part : $E _{ Zn ^{+2} / Zn }^{\circ}=-0.76 V \quad( n =2)$
$E _{ Ag ^{+} / Ag }=0.80 V$
In this cell, Zn is of anode and Ag works as cathode.
Hence,
$E _{ cell }^{\circ}= E _{ cathode }^{\circ}- E _{ anode }^{\circ}$
$E_{\text {cell }}^{\circ}=0.80-(-0.76)$
$E _{ cell }^{\circ}=1.56 V$
$\Delta_{ r } G ^{\circ}=- nFE E_{\text {cell }}^{\circ}$
$\Delta_{ r } G ^{\circ}=-2 \times 96500 \times 1.56$
$\Delta_{ r } G ^{\circ}=-301080 eV =-3.01 \times 10^5 J$

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Question 143 Marks

Depict the galvanic cell in which the reaction
$Zn ( s )+2 Ag ^{+}( aq ) \rightarrow Zn ^{2+}( aq )+2 Ag ( s )$,
takes place. Further show :
(i) Which of the electrode is negatively charged?
(ii) The carriers of the current in the cell.
(iii) Individual reaction at each electrode.

Answer

On the basis of the given reaction, a galvanic (electrochemical cell) can be shown as follows :
$\qquad\qquad\qquad$$Zn _{( s )}\left| Zn _{( aq )}^{2+}\right|\left| Ag _{( aq )}^{+}\right| Ag$
(i) $Zn \mid Zn ^{+2}$ in this cell electrode is negatively charged, hence it will be anode.
(ii) The carriers of electric current in the cell are electrons and the flow of current is from the silver electrode towards the zinc electrode because the flow of electric current is in opposite direction to the flow of electrons.
(iii) The reaction taking place at the cathode is as follows :
$\qquad\qquad\qquad$$2 Ag ^{+}+2 e ^{-} \rightarrow 2 Ag$
And reaction taking place at anode :
$\qquad\qquad\qquad$$Zn \rightarrow Zn ^{2+}+2 e ^{-}$

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Question 153 Marks

A solution Ni(NO₃)₂ is electrolysed between platinum electrodes a current of 5 amperes for 20 minutes. What mass of Ni is deposited at the cathode?

Answer

Amolunt of electricity flow
$Q = It$
I = 5 Ampere, 1 = time (in second) = 20 × 60s
$Q =5 \times 20 \times 60=6000 C$
The reaction taking place at cathode :
$Ni ^{2+}+2 e ^{-} \rightarrow Ni$
Atomic mass of Ni = 58.7 (1 mole)
According to the reaction, 2F or 2 × 96500 C of electric current is passed to get 58.7 g Ni.
Hence, the amount of Ni obtained by passing 6000 C eletric current :
$=\frac{58.7 \times 6000}{2 \times 96500}=1.824 g$
Therefore, the amount of Ni deposited at cathode = 1.824 g.

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Question 163 Marks

How much electricity is required in coulomb for the oxidation of :
(i) 1 mol of H₂O to O₂?
(ii) 1 mol of FeO to Fe₂O₃?

Answer

(i) The reaction of formation of O₂ from H₂O is of follows :
$2 H _2 O \rightarrow O _2+4 H ^{+}+4 e ^{-}$
Here, 4 moles of electron are released from 2 moles of H₂O, hence 2 moles of electron will be released from 1 mole of H₂O, hence, the amount of electricity required for oxidation of 1 mole of H₂O = 2F.
2F = 2 x 96,500 Coulomb.
Amount of electricity required = 1,93,000 coulomb.
(ii) Formation of Fe₂O₃ from FeO, the reaction is as follows :
$\underset{+2}{2 FeO } \rightarrow \underset{+3}{ Fe _2 O _3}+2 e ^{-}$
Here, Fe³⁺ is being formed from Fe²⁺ , hence, for 1 mole of FeO to be oxidised to $Fe _2 O _3$ the required amount of electricity = 1F = 96 ,500 coulomb.

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Question 173 Marks

How much electricity in terms of Faraday is required to produce?
(i) 20.0 g of Ca from molten CaCl₂?
(ii) 40.0 g of Al from molten Al₂O₃?

Answer

(i) To obtain Ca from molten CaCl₂, the following reaction will take place at the cathode :
$Ca ^{2+}+2 e ^{-} \rightarrow Ca$
1 mole of Ca = 40g (atomic mass)
hence, to obtain 40 g Ca, according to the reaction, Required amount of electricity = 2F
So, to obtain 20 g Ca, 1F amount of electricity is required.
(ii) To obtain Al from molten Al₂O₃, the reaction will take place at cathode : $Al ^{3+}+3 e ^{-} \rightarrow Al$
Atomic mass of 1 mole Al = 27 g
Hence, according to the reaction, 3F amount of electricity is required to obtain 27 g Al.
So, for 40 g of $Al =\frac{3 F \times 40}{27}=4.44 F$ electricity is required.

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Question 183 Marks

How much charge is required for the following reductions :
(i) 1 mol of Al³⁺ to Al?
(ii) 1 mol of Cu²⁺ to Cu?
(iii) 1 mol of MnO₄^- to Mn²⁺?

Answer

(i) Electrode reaction is Al³⁺ + 3e^-→ Al.
Hence, for the reduction 1 mol of Al³⁺, 3F charge will be required because 3 moles of electron are used in this reaction and 3 Faraday 3 × 96500 Coulomb (C) = 289500 С.
(ii) Reaction : $Cu ^{2+}+2 e \rightarrow Cu$
For the reduction of 1 mole of Cu²⁺, 2F charge will be required and 2F = 2 × 96500 C = 193000 С.
(iii) Reaction : $MnO _4^{-} \rightarrow Mn ^{2+}$
In MnO₄^-, the oxidation number of Mn is +7 and it is forming Mn²⁺ hence 5 electrons are being used in it.
Therefore, for the reduction of 1 mole of MnO₄^- to Mn²⁺ , 5F charge will be required and 5F = 5 × 96500 C = 482500 С.

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3 Marks Question - Chemistry STD 12 Science Questions - Vidyadip