Tag: electrochemistry

Questions Related to electrochemistry

Which of the following is correct regarding current carrying ions in the solution of $C _{2}H _{5}COOH$ upon dilution?

conductivity and its types electrochemistry

  1. The number of ions in $1 cm^{3}$, as well as in total volume increases.

  2. The number of ions in $1 cm^{3}$ decreases, whereas that in the total volume remains constant.

  3. The number of ions in $1 cm^{3}$ decreases, but that in the total volume increases.

  4. The number of ions in $1 cm^{3}$, as well as in total volume decreases.

Show answer
Correct Option: C
Explanation:

For weak electrolytes, as the concentration decreases, the percentage dissociation increases and the total number of ions increases. But as the volume increases, number of ions per unit volume decreases.


Hence, option C is correct.

The resistance of $N/2$ solution of an electrolyte in a cell was found to be $45$ ohm. The equivalent conductivity of a solution, if the electrodes in the cell are $2.2$ cm apart and have an area of $3.8 cm^2$ will be:

conductivity and its types electrochemistry

  1. $52.72\ S cm^2 eq^{-1}$

  2. $22.57\ S cm^2 eq^{-1}$

  3. $27.52\ S cm^2 eq^{-1}$

  4. $25.72\ S cm^2 eq^{-1}$

Show answer
Correct Option: D
Explanation:
Cell constant = $\dfrac{l}{a}=\dfrac{2.2}{3.8}=0.579 cm^{-1} $

Specific conductance = cell constant $\times$ conductance

Specific conductance = $\dfrac{0.579}{45} $

Equivalent conductance= $\dfrac{0.579}{45} \times \dfrac{1000}{0.5} $

Equivalent conductance = $25.73 Scm^{2} eq^{-1} $

Hence, option D is correct.

The resistance of 0.1 N solution of a salt is found to be $2.5\times10^{3}$. The equivalent conductance of the solution is: (cell constant=1.15 $cm^{-1}$)

conductivity and its types electrochemistry

  1. 3.6

  2. 4.6

  3. 5.6

  4. 6.6

Show answer
Correct Option: B
Explanation:

The relationship between the specific conductance, resistance and the cell constant is $ \kappa =\cfrac { 1 }{ R } \times \cfrac { l }{ a }$.
Substitute $ R=2.5\times { 10 }^{ 3 }\quad ohm $ and $ \cfrac { l }{ a } =1.15\quad {cm }^{ -1 } $.
Hence $ \kappa =\cfrac { 1 }{ 2.5\times { 10 }^{ 3 }\quad ohm } \times1.15\quad { cm }^{ -1 }=\cfrac { 1.15 }{ 2.5\times { 10 }^{ 3 } } \quad { ohm }^{ -1 }\quad { cm }^{ -1 } $.
The relationship between the equivalent conductance and specifc conductance is $  { \Lambda  } _{ eq }=\cfrac { \kappa \times 1000 }{ M }  $.
Substitute $ M=0.1\quad N $ and $ \kappa=\cfrac { 1.15 }{ 2.5\times { 10 }^{ 3 } } \quad { ohm }^{ -1 }\quad { cm }^{ -1 } $.
Hence $ { \Lambda  } _{ eq }=\cfrac { \cfrac { 1.15 }{ 2.5\times { 10 }^{ 3 } } \times 1000 }{ 0.1 } \quad =\quad 4.6\quad \quad { ohm }^{ -1 }\quad { cm }^{ 2 }\quad { equiv }^{ -1 } $.

At 291 K, the equivalent conductivities at infinite dilution of $NH _4Cl$, $NaOH $ and $NaCl $ are $129.8$, $217.5$ and $108.9$ S $cm^2 eq^{-1}$ respectively. The equivalent conductivity at infinite dilution of $NH _4OH$ is:

conductivity and its types electrochemistry

  1. $208.4 S cm^2 eq^{-1}$

  2. $238.4 S cm^2 eq^{-1}$

  3. $283.4 S cm^2 eq^{-1}$

  4. None of these

Show answer
Correct Option: B
Explanation:
Equivalent conductance of $NH _4OH $ = Equivalent conductance of $NH _4Cl $+$NaOH$-$NaCl$

                                                                $= 129.8+218.4-108.9$

Equivalent conductance of $NH _4OH = 239.3$

Therefore, option B is the correct answer.

The correct order of equivalent conductivity at infinite dilution of $LiCl,\ NaCl$ and $KCl$ is:

conductivity and its types electrochemistry

  1. $LiCl > NaCl > KCl$

  2. $KCl > NaCl > LiCl$

  3. $NaCl > KCl > LiCl$

  4. $LiCl > KCl > NaCl$

Show answer
Correct Option: B
Explanation:

The correct order of equivalent conductivity at infinite dilution of $LiCl,\ NaCl$ and $KCl$ is $KCl > NaCl > LiCl$.

Ionic mobility depends upon size of the ion. The ionic size in case of hydrated cation, is $K^+(aq) < Na^+(aq) < Li^+(aq)$.

As the size of the hydrated ion increases, the equivalent conductivity at infinite dilution decreases.

$\lambda {eq}$ x Normality = _________

conductivity and its types electrochemistry

  1. K x 10$^3$

  2. K x 10$^4$

  3. K x 10$^5$

  4. K x 10$^6$

Show answer
Correct Option: A
Explanation:

As we know,
                     $\lambda _{eq} = k\times1000/N$

              Then, $\lambda _{eq} \times normality$ = K x 10$^3$


   So. the correct option is $A$

Equivalent conductance of an electrolyte containing $NaF$ at infinite dilution is $90.1\ Ohm^{-1} cm^{2}$. If $NaF$ is replaced by $KF$ what is the value of equivalent conductance?

conductivity and its types electrochemistry

  1. $90.1\ Ohm^{-1} cm^{2}$

  2. $11.2\ Ohm^{-1} cm^{2}$

  3. $0$

  4. $222.4\ Ohm^{-1} cm^{2}$

Show answer
Correct Option: A
Explanation:

At infinite dilution the equivalent conductance of strong electrolytes furnishing same number of ions is same.

For this case, Both NaF and KF are strong electrolyte and also furnish 2 ions in the solution.
Hence, Both will have same equivalent conductance.
So, option A is correct.

The resistance of $N/10$ solution is found to be $2.5\times 10^{3}ohm$. The equivalent conductance of the solution is (cell constant $= 1.25\ cm^{-1})$.

conductivity and its types electrochemistry

  1. $2.5\ ohm^{-1} cm^{2} equiv^{-1}$

  2. $5\ ohm^{-1} cm^{2} equiv^{-1}$

  3. $2.5\ ohm^{-1} cm^{-2} equiv^{-1}$

  4. $5\ ohm^{-1} cm^{-2} equiv^{-1}$

Show answer
Correct Option: B
Explanation:

Given Data : 1)  Resistance = $R$ = $2.5 \times 10^3$

                     2) Cell constant = $k$ = $\cfrac{l}{a}$= $1.15$$cm^2$
                     3)  Normality =$N$= $0.1 N$
To Find : Equivalent conductance = $\Lambda$$ _e$$ _q$

The relation between $K$ ,$R$ and $ k$ is,
$K$ = $\cfrac{1}{R} \times \cfrac{l}{a}$
where $K$ is specific conductance.
$\therefore$ Substituting the given values we get,

$K$ = $\cfrac{1}{2.5×10^3}$ × $1.15cm^-$$^1$

     = $\cfrac{1.15}{2.5×10^3}$ $ohm$$^-$$^1$$cm$$^-$$^1$

The relation between $\Lambda$$ _e$$ _q$ and $K$ is,

$\Lambda$$ _e$$ _q$ = $\cfrac{K×1000}{N}$

Substituting the value of $M$ and $K$ we get,

$\Lambda _{eq}= \cfrac{1.15}{2.5\times10^3\times0.1} \times 1000$

       = $4.6$$ohm^-$$^1$$cm$$^2$$equi$$^{-1}$      [Note:$ \text {Normality= no. of equiv.} /cm^3$]

Here appproximation is taken,

      $\approx$ $5$ $ohm$$^-$$^1$$cm$$^2$$equi$$^-$$^1$

Hence the correct option is 'B'.

What are the units of equivalent conductivity of a solution?

conductivity and its types electrochemistry

  1. $mho\ cm^{-1}$

  2. $ohm\ cm^{-1} g\ equiv^{-1}$

  3. $mho\ cm^{-2}g\ equiv^{-1}$

  4. $mho\ cm^{2}g\ equiv^{-1}$

Show answer
Correct Option: A
Explanation:
$\rightarrow$ If 1 $m^{3}$ of a solution of an electrolyte placed between two large electrodes 1m apart. The cross-sectional area of the solution will be $1m^{2}$.
$\rightarrow$ The conductance of solution will evidently be its specific conductance because we have a one-meter cube of the solution.
$\rightarrow$ If $1m^{3}$ of the solution contains 1 gram equivalent of the electrolyte.
Hence, $Conductance(C)=Specific Conductance(K)=Equivalent Conductance(\wedge)$
$\rightarrow$ As unit of Specific conductance is $Sm^{-1}$ or $Scm^{-1}$, so the unit of Equivalent Conductance is
$Scm^{-1}$ or $ mhocm^{-1}$.
$\rightarrow$ $mho$ is reciprocal of $ohm$ & reciprocal of $ohm$ i.e. $ohm^{-1}$ is also called $mho$ or $Siemen(S)$.

The equivalent conductances of $NaCl$ at concentration $c$ and at infinite dilution are $\lambda _{c}$ and $\lambda _{\infty}$ respectively. The correct relationship between $\lambda _{c}$ and $\lambda _{\infty}$ is given as: (where the constant $b$ is positive).

conductivity and its types electrochemistry

  1. $\lambda _{c} = \lambda _{\infty} - b\sqrt {c}$

  2. $\lambda _{c} = \lambda _{\infty} + b\sqrt {c}$

  3. $\lambda _{c} = \lambda _{\infty} + bc$

  4. $\lambda _{c} = \lambda _{\infty} - bc$

Show answer
Correct Option: A
Explanation:

According to debye huckel onsager equation we have,

equivalent conductance of NaCl is given by
$\lambda _C$ = $\lambda _{\infty}$ - $b C^{1/2}$
where b is positive.