Tag: electrochemistry

Questions Related to electrochemistry

The equivalent conductances at infinite dilution of $HCl$ and $NaCl$ are $426.15$ and $126.15\ mho\ cm^{2}g\ eq^{-1}$ respectively. If can be said that the mobility of:

conductivity and its types electrochemistry

  1. $H^{+}$ ions is much more than that of $Cl^{-}$ ions

  2. $Cl^{-}$ ions is much more than that of $H^{+}$ ions

  3. $H^{+}$ ions is much more than that of $Na^{+}$ ions

  4. $Na^{+}$ ions is much more than that of $H^{+}$ ions

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Correct Option: C
Explanation:
$\rightarrow$ The equivalent conductance of $HCl$ is more than $NaCl$ , it is because of difference in speed of ions.
$\rightarrow\,HCl$ has conductance at infinite dilution more than three times as high as that of $NaCl$.
$\rightarrow$ As Chlorine ion is common in both of them, it seems that speed of $H^{+}$ ion is much more than that of $Na^{+}$ ions.

The resistance of $0.01\ N$ solution at $25^{\circ}$ is $200\ ohm$. Cell constant of the conductivity cell is unity. Calculate the equivalent conductance of the solution.

conductivity and its types electrochemistry

  1. $200\ ohm^{-1}cm^{2} eq^{-1}$.

  2. $300\ ohm^{-1}cm^{2} eq^{-1}$.

  3. $400\ ohm^{-1}cm^{2} eq^{-1}$.

  4. $500\ ohm^{-1}cm^{2} eq^{-1}$.

Show answer
Correct Option: D
Explanation:

since, we have

$conductance * cell constant$ =  specific conductance
cell consyant = 1
conductance = specific conductance= $\dfrac{1}{200}$

equivalent conductance= $\dfrac{K*1000}{N}$ $Scm^{2}eq^{-1}$

equivalent conductance= $\dfrac{1*1000}{200*0.01}$

equivalent conductance= $500$ $ Scm^{2}eq^{-1}$
 

The specific conductance $(K)$ of an electrolyte of $0.1\ N$ concentration is related to equivalent conductance $(\wedge _{e})$ by the following formula.

conductivity and its types electrochemistry

  1. $\wedge _{e} = K$

  2. $\wedge _{e} = 10 K$

  3. $\wedge _{e} = 100 K$

  4. $\wedge _{e} = 10000 K$

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Correct Option: D
Explanation:

we have,

equivalent conductance = $\dfrac{K*1000}{N}$

equivalent conductance= $\dfrac{K*1000}{0.1 N}$

equivalent conductance= $10000K$

The gas leaked from a storage tank of the Union Carbide plant in Bhopal gas tragedy was

conductivity and its types electrochemistry

  1. methyl isocyanate

  2. methylamine

  3. ammonia

  4. phosgene

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Correct Option: A

The specific conductivity of $0.1$ $N$ $KCl$ solution at $20^0$C is $0.0212$ $ohm^{-1} cm^{-1}$. The solution was found to offer resistance of $55$ ohms. Find the cell constant of the conductivity cell.

conductivity and its types electrochemistry

  1. $2.25$

  2. $1.166$

  3. $1.936$

  4. none of these

Show answer
Correct Option: B
Explanation:
Specific conductance $K=0.0212 \Omega ^{-1}cm^{-1}$

Normality = $0.1 N$, Resistance = $55\Omega $

Cell constant $G^*$ = conductivity $\times $ Resistance

$G^*=K\times R$

$=0.0212\times 55$

$=1.166 cm^{-1}$

Option B

What would be the equivalent conductivity of a cell in which $0.5$ N salt solution offers a resistance of $40$ ohm whose electrodes are $2$ cm apart and $5$ $cm^{2}$ in area?

conductivity and its types electrochemistry

  1. $10$ $ohm^{-1}$ $cm^2$ $eq^{-1}$

  2. $20$ $ohm^{-1}$ $cm^{2}$ $eq^{-1}$

  3. $30$ $ohm^{-1}$ $cm^{2}$ $eq^{-1}$

  4. $25$ $ohm^{-1}$ $cm^{2}$ $eq^{-1}$

Show answer
Correct Option: B
Explanation:
Given,
$R=40 ohm ,l=2cm,A=2cm^{2}$

We know the relation,

$\kappa=\dfrac{1}{R}\times\dfrac{l}{A}=\dfrac{1}{40}\times\dfrac{2}{5}=\dfrac{1}{100} ohm^{-1} m^{-1}$


Now we also know the relation,

$\Lambda _{eq}=\dfrac{\kappa\times1000}{N}=\dfrac{1}{100}\times\dfrac{1000}{0.5}=20 ohm^{-1} cm^{2} eq^{-1}$

Hence, option B is correct.

The equivalent conductivity of $N/10$ solution of acetic acid at $25^o$C is $14.3$ $ohm^{-1}$ $cm^2$ $equiv^{-1}$. What will be the degree of dissociation of acetic acid?
$(\Lambda _{\infty CH _3COOH}=390.71$ $ohm^{-1}$ $cm^2$ $equiv^{-1}$).

conductivity and its types electrochemistry

  1. $3.66\%$

  2. $3.9\%$

  3. $2.12\%$

  4. $0.008\%$

Show answer
Correct Option: A
Explanation:

The equivalent conductivity of an electrolyte is defined as the conductivity of a volume of solution containing one equivalent weight of dissolved substance when placed between two parallel electrodes 1 cm apart, and large enough to contain between them all of the solution.
Given,
$\Lambda^{\infty} _{m(CH _3COOH)}=390.71 ohm^{-1} cm^{2} eq^{-1}$
$\Lambda _{m(CH _3COOH)}=14.3 ohm^{-1} cm^{2} eq^{-1}$
Degree of dissociation is given by,
$\alpha=\dfrac{\Lambda _m}{\Lambda^{0} _m}=\dfrac{14.3}{390.71}=0.0366 \implies$3.66%


Which of the following statements is true?

conductivity and its types electrochemistry

  1. When an aqueous solution of NaCl is electrolysed, sodium metal is deposited at cathode

  2. There is no difference between specific conductivity and molar conductivity

  3. Silver nitrate solution can be stored in a copper container

  4. The addition of liquid bromine to iodide solution turns it violet

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Correct Option: D
Explanation:

When an aqueous solution of NaCl is electrolysed, hydrogen is liberated at cathode. Specific conductivity and molar conductivity are different terms. Silver nitrate solution cannot be stored in a copper container as silver will get precipitated because of high reactivity of Cu than Ag.
The addition of liquid bromine to iodide solution turns it violet.

$Br _{2(l)}+2I^{-} _{(aq)}\rightarrow 2Br^{-} _{(aq)}+I _{2(aq)}$
Here, $Br _{2(l)}$ is Reddish-brown in color and $I _{2(aq)}$is violet in color

The ionic conductivity of $B{a^{2 + }}$ and $C{l^ - }$ at infinite dilution are 127 and 76 respectively. The equivalent conductivity of $BaC{l _2}$ at infinite dilution (in $oh{m^ - }\,c{m^2}\,e{q^{ - 1}}$) would be:

conductivity and its types electrochemistry

  1. 279

  2. 280

  3. 139.5

  4. 102

Show answer
Correct Option: C
Explanation:

Given data,


$\lambda (Ba^{2+}) = 127$


$\lambda(Cl) = 76$

$\wedge _m^{\infty} = \lambda (Ba^{2+}) + 2\lambda(Cl)$

$= 127 + 2\times 76$

$\wedge _m^{\infty} = 279$

We know that-

$\wedge _{eq}^{\infty} = \dfrac{\lambda _m^{\infty}(Ba^{2+})}{2} + \lambda _m^{\infty}(Cl^-)$ 

$= \dfrac{127}{2} + 76$

$\wedge _{eq}^{\infty} = 139.5\ {ohm}^{-1}{cm}^2$

Hence, option C is correct.

The correct order of equivalent conductance 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:

Since, the anion are same, comparing the size of cation.

$Li^+<Na^+<K^+$
but due to hydration of cation, the radii order is inverted.
$Li^+>Na^+>K^+$
$Conductance \propto\cfrac{1}{radii}$
$\therefore$ The order is $KCl>NaCl>LiCl$