Tag: electrochemistry

Questions Related to electrochemistry

Equivalent conductance of $BaCl _2, H _2SO _4$ and $HCl$ are $x _1, x _2$ and $x _3 S cm^2 equiv^{-1}$ at infinite dilution.If specific conductance of saturated $BaSO _4$ solution is of $y S cm^1$ then $K _{sp}$ of $BaSO _4$ is:

  1. $\frac{10^3y}{2(x _1 + x _2 - 2x _3)}$

  2. $\frac{10^6y^2}{4(x _1 + x _2 - 2x _3)^2}$

  3. $\frac{10^6y^2}{2(x _1 + x _2 - x _3)^2}$

  4. $\frac{x _1 + x _2 - 2x _3}{10^6y^2}$


Correct Option: A
Explanation:

${ \Lambda  } _{ eq }\quad (Ba{ SO } _{ 4 })={ \Lambda  } _{ eq }\left( Ba{ Cl } _{ 2 } \right) +{ \Lambda  } _{ eq }\left( { H } _{ 2 }{ SO } _{ 4 } \right) -2{ \Lambda  } _{ eq }\left( HCl \right) $

                            $=\left( { x } _{ 1 }+{ x } _{ 2 }-{ 2x } _{ 3 } \right) S{ cm }^{ 2 }{ eq }^{ -1 }$
$\because { \Lambda  } _{ eq }=K\times \cfrac { 1000 }{ N } \quad \Rightarrow \quad N=\left( \cfrac { y\times { 10 }^{ 3 } }{ { x } _{ 1 }+{ x } _{ 2 }-{ 2x } _{ 3 } }  \right) \quad \left( \because { n } _{ f }=2 \right) $
$\therefore \quad M=\left{ \cfrac { y\times { 10 }^{ 3 } }{ 2\left( { x } _{ 1 }+{ x } _{ 2 }-{ 2x } _{ 3 } \right)  }  \right} $
$\therefore \quad { K } _{ sp }={ \left( M \right)  }^{ 2 }=\left{ \cfrac { { y }^{ 2 }\times { 10 }^{ 6 } }{ 4{ \left( { x } _{ 1 }+{ x } _{ 2 }-{ 2x } _{ 3 } \right)  }^{ 2 } }  \right} $

For $HCl$ solution at ${25}^{o}C$ equivalent conductance at infinite dilution is $425 \ {ohm}^{-1}{cm}^{2}{equiv}^{-1}$. The specific conductance of a solution of $HCl$ is $3.825$ ${ohm}^{-1}{cm}^{-1}$. If the apparent degree of dissociation is $90$% the normality of the solution is :

  1. $0.90N$

  2. $1.0N$

  3. $10\ N$

  4. $1.2N$


Correct Option: C

The equivalent conductivity of $0.1 N \ CHNCH _{3}COOH$ at $25^{0}C$ is 80 and at infinite dilution it is 400, the degree of dissociation of $CH _{3}COOH$ is :

  1. 1

  2. 0.2

  3. 0.1

  4. 0.5


Correct Option: B
Explanation:

Given:-

$\wedge _{eq}(CH _3COOH)= 80 S cm^{-2} eq^{-1}$

$\wedge^{\infty} _{eq}(CH _3COOH)$ at infinite dilution= $400S cm^2 eq^{-1}$

$\alpha \longrightarrow$ Degree of dissociation

$\alpha= \cfrac {\wedge^m _{eq}}{\wedge^{\infty} _{eq}}=\cfrac {80}{400}= 0.2$

$\alpha= 0.2$

In infinite dilusions, the equivalent conductances of $Ba^{2+}$ and $Cl^{-}$ are $127$ and $76 ohm^{-1} \, cm^{-1} \, eqvt^{-1}$. The equivalent conductivity of $BaCl _2$ at indefinite dilution is?

  1. $101.5$

  2. $203.5$

  3. $139.5$

  4. $279.5$


Correct Option: C
Explanation:

The equivalent conductance of BaCl2 at infinite dilution, 


λ of BaCl2=1/2 λ of Ba2+ + λof Cl
       

 =127/2+76 

=139.

Option C is correct answer

${\text{N}}{{\text{a}} _{\text{3}}}{\text{Al}}{{\text{F}} _{\text{6}}}\,\,$ is added to $\,{\text{A}}{{\text{l}} _{\text{2}}}{{\text{O}} _{\text{3}}}$

  1. Improve the electrical conductivity of the cell

  2. Increases rate of production

  3. Increases the melting point

  4. Decrease the electrical conductivity


Correct Option: A

The equivalent conductance of $0.02$ M acctic acid  $1.62.*{10^{ - 3}}$.  Degree of ironisation $'a'$ of $C{H _3}COOH$ is:
$({x _H} = 349.83oh{m^{ - 1}}and\lambda C{H _3}CO{O^ - } = 40.89ohm{s^{ - 1}}$

  1. $0.01$

  2. $0.02$

  3. $0.03$

  4. $0.04$


Correct Option: B

Equivalent conductivity of $BaCl _2,H _2SO _4$ and HCI, are $x _1,x _2$ and $x _3scm^{-1}eq^{-1}$ at infinite dilution. If conductivity of saturated $BaSO _4$ solution is x $Scm^{-1}$, then $K _{sp}$ of $BaSO _4$ is

  1. $\dfrac {500x} {(x _1+x _2-2x _3)}$

  2. $\dfrac {10^6x^2} {(x _1+x _2-2x _3)^3}$

  3. $\dfrac {2.5\times10^5 x^2} {x _1-2x _2-x _3)^2}$

  4. $\dfrac {0.25 x^2} {x _1 + x _2-x _3)^2}$


Correct Option: A

Molar conductance of $C{a^{2 + }}$ and $C{l^ - }$ are  $120\,c{m^2}\,$ $mo{l^{ - 1}}$ and $77\,S\,c{m^2}\,mo{l^{ - 1}}$ respectively. What is the equivalent conductance of $CaC{l _2}$ ?

  1. $98.5\,Sc{m^2}e{q^{ - 1}}$

  2. $137\,Sc{m^2}e{q^{ - 1}}$

  3. $197\,Sc{m^2}e{q^{ - 1}}$

  4. $247\,Sc{m^2}e{q^{ - 1}}$


Correct Option: D

At infinite dilution equivalent conductance of ${B^{ + 2}}$ & CI ions are 127  &  76$oh{m^{ - 1}}$ $c{m^{ - 1}}$` $e{q^{ - 1}}$ respectively. Equivalent conductance  of $BaC{I _2}$ at infinite diluition is : 

  1. 139.5

  2. 101.5

  3. 203

  4. 279


Correct Option: C

Equivalent constant of standard $BaSO _{4}$ is $400ohm^{-1}\ cm^{2}$ equiv$^{-1}$ and specific conduction is $8\times 10^{-5}\ ohm^{-1}\ cn^{-1}$. Hence $K _{SP}$ of $BaSO _{4}$ is

  1. $4\times 10^{-8}M^{2}$

  2. $1\times 10^{-8}M^{2}$

  3. $2\times 10^{-4}M^{2}$

  4. $1\times 10^{-4}M^{2}$


Correct Option: B