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ELECTROCHEMISTRY

275951 Which of the following electrolytes will have maximum flocculation value for $Fe (OH)_{3}$ sol ?

1

NaCl

2

{Na}_{2} S

3

{({NH}_{4})}_{3} {PO}_{4}

4

K_{2} {SO}_{4}

Explanation:

$Fe (OH)_{3}$ is a positively charge solution. Comparing the charge on negative ions, ${Cl}^{-}$ has least charge and hence least coagulating power and maximum coagulation values.
Flocculation value $\propto \frac{1}{Coagulating power}$
Among the given electrolytes, $NaCl$
has lowest coagulation power, so its flocculation value will be maximum.

JCECE - 2009

ELECTROCHEMISTRY

275952 Kohlrausch's law states that at

1 Infinite dilution, each ion makes definite contribution to conductance of an electrolyte whatever be the nature of the other ion of the electrolyte

2 Infinite dilution, each ion makes definite contribution to equivalent conductance of an electrolyte, whatever be the nature of the other ion of the electrolyte

3 Finite dilution, each ion makes definite contribution to equivalent conductance of an electrolyte, whatever be the nature of the other ion of the electrolyte

4 Infinite dilution each ion makes definite contribution to equivalent conductance of an electrolyte depending on the nature of the other ion of the electrolyte

Explanation:

At infinite dilution, when the dissociation of electrolyte is complete, each ion makes a definite contribution towards the molar conductivity irrespective of the nature of the other ion with which it is associated. According to Kohlrausch's law the molar conductivity of an electrolyte at infinite dilution can be expressed as the sum of the contributions from its individual ions.
$λ_{\infty} = λ_{a} + λ_{c}$
Where, $λ_{a}$ and $λ_{c}$ are know ionic conductance of anion and cation at infinite dilution respectively.

(AIPMT -2008)

ELECTROCHEMISTRY

275756 The molar conductivity is maximum for the solution of concentration

1

0.005 M

2

0.001 M

3

0.004 M

4

0.002 M

Explanation:

Molar conductance $\propto \frac{1}{Molarity}$
The molar conductivity of electrolytes increases with decreasing concentration, so the molar conductivity of the solution will be maximum at a concentration of $0.001 M$ .

Karnataka CET-17.06.2022

ELECTROCHEMISTRY

275958 The relationship between Gibb's free energy change $(Δ G)$ and emf (E) of a reversible electrochemical cell is given by

1

Δ G = nFE

2

Δ G = nF / E

3

Δ G = - nFE

4

Δ G = E / nF

Explanation:

If the free energy change $(Δ G)$ is negative then any redox reaction would occur spontaneously.
$Δ G = - {nFE}_{cell}$ where,
$n = No. of electrons$
F = value of Faraday
E = The emf of cell
$E = The emf of cell$
$Δ G$ can be negative if $E$ is positive .

J and K CET-(2008)

ELECTROCHEMISTRY

275951 Which of the following electrolytes will have maximum flocculation value for $Fe (OH)_{3}$ sol ?

1

NaCl

2

{Na}_{2} S

3

{({NH}_{4})}_{3} {PO}_{4}

4

K_{2} {SO}_{4}

Explanation:

$Fe (OH)_{3}$ is a positively charge solution. Comparing the charge on negative ions, ${Cl}^{-}$ has least charge and hence least coagulating power and maximum coagulation values.
Flocculation value $\propto \frac{1}{Coagulating power}$
Among the given electrolytes, $NaCl$
has lowest coagulation power, so its flocculation value will be maximum.

JCECE - 2009

ELECTROCHEMISTRY

275952 Kohlrausch's law states that at

1 Infinite dilution, each ion makes definite contribution to conductance of an electrolyte whatever be the nature of the other ion of the electrolyte

2 Infinite dilution, each ion makes definite contribution to equivalent conductance of an electrolyte, whatever be the nature of the other ion of the electrolyte

3 Finite dilution, each ion makes definite contribution to equivalent conductance of an electrolyte, whatever be the nature of the other ion of the electrolyte

4 Infinite dilution each ion makes definite contribution to equivalent conductance of an electrolyte depending on the nature of the other ion of the electrolyte

Explanation:

At infinite dilution, when the dissociation of electrolyte is complete, each ion makes a definite contribution towards the molar conductivity irrespective of the nature of the other ion with which it is associated. According to Kohlrausch's law the molar conductivity of an electrolyte at infinite dilution can be expressed as the sum of the contributions from its individual ions.
$λ_{\infty} = λ_{a} + λ_{c}$
Where, $λ_{a}$ and $λ_{c}$ are know ionic conductance of anion and cation at infinite dilution respectively.

(AIPMT -2008)

ELECTROCHEMISTRY

275756 The molar conductivity is maximum for the solution of concentration

1

0.005 M

2

0.001 M

3

0.004 M

4

0.002 M

Explanation:

Molar conductance $\propto \frac{1}{Molarity}$
The molar conductivity of electrolytes increases with decreasing concentration, so the molar conductivity of the solution will be maximum at a concentration of $0.001 M$ .

Karnataka CET-17.06.2022

ELECTROCHEMISTRY

275958 The relationship between Gibb's free energy change $(Δ G)$ and emf (E) of a reversible electrochemical cell is given by

1

Δ G = nFE

2

Δ G = nF / E

3

Δ G = - nFE

4

Δ G = E / nF

Explanation:

If the free energy change $(Δ G)$ is negative then any redox reaction would occur spontaneously.
$Δ G = - {nFE}_{cell}$ where,
$n = No. of electrons$
F = value of Faraday
E = The emf of cell
$E = The emf of cell$
$Δ G$ can be negative if $E$ is positive .

J and K CET-(2008)

ELECTROCHEMISTRY

275951 Which of the following electrolytes will have maximum flocculation value for $Fe (OH)_{3}$ sol ?

1

NaCl

2

{Na}_{2} S

3

{({NH}_{4})}_{3} {PO}_{4}

4

K_{2} {SO}_{4}

Explanation:

$Fe (OH)_{3}$ is a positively charge solution. Comparing the charge on negative ions, ${Cl}^{-}$ has least charge and hence least coagulating power and maximum coagulation values.
Flocculation value $\propto \frac{1}{Coagulating power}$
Among the given electrolytes, $NaCl$
has lowest coagulation power, so its flocculation value will be maximum.

JCECE - 2009

ELECTROCHEMISTRY

275952 Kohlrausch's law states that at

1 Infinite dilution, each ion makes definite contribution to conductance of an electrolyte whatever be the nature of the other ion of the electrolyte

2 Infinite dilution, each ion makes definite contribution to equivalent conductance of an electrolyte, whatever be the nature of the other ion of the electrolyte

3 Finite dilution, each ion makes definite contribution to equivalent conductance of an electrolyte, whatever be the nature of the other ion of the electrolyte

4 Infinite dilution each ion makes definite contribution to equivalent conductance of an electrolyte depending on the nature of the other ion of the electrolyte

Explanation:

At infinite dilution, when the dissociation of electrolyte is complete, each ion makes a definite contribution towards the molar conductivity irrespective of the nature of the other ion with which it is associated. According to Kohlrausch's law the molar conductivity of an electrolyte at infinite dilution can be expressed as the sum of the contributions from its individual ions.
$λ_{\infty} = λ_{a} + λ_{c}$
Where, $λ_{a}$ and $λ_{c}$ are know ionic conductance of anion and cation at infinite dilution respectively.

(AIPMT -2008)

ELECTROCHEMISTRY

275756 The molar conductivity is maximum for the solution of concentration

1

0.005 M

2

0.001 M

3

0.004 M

4

0.002 M

Explanation:

Molar conductance $\propto \frac{1}{Molarity}$
The molar conductivity of electrolytes increases with decreasing concentration, so the molar conductivity of the solution will be maximum at a concentration of $0.001 M$ .

Karnataka CET-17.06.2022

ELECTROCHEMISTRY

275958 The relationship between Gibb's free energy change $(Δ G)$ and emf (E) of a reversible electrochemical cell is given by

1

Δ G = nFE

2

Δ G = nF / E

3

Δ G = - nFE

4

Δ G = E / nF

Explanation:

If the free energy change $(Δ G)$ is negative then any redox reaction would occur spontaneously.
$Δ G = - {nFE}_{cell}$ where,
$n = No. of electrons$
F = value of Faraday
E = The emf of cell
$E = The emf of cell$
$Δ G$ can be negative if $E$ is positive .

J and K CET-(2008)

ELECTROCHEMISTRY

275951 Which of the following electrolytes will have maximum flocculation value for $Fe (OH)_{3}$ sol ?

1

NaCl

2

{Na}_{2} S

3

{({NH}_{4})}_{3} {PO}_{4}

4

K_{2} {SO}_{4}

Explanation:

$Fe (OH)_{3}$ is a positively charge solution. Comparing the charge on negative ions, ${Cl}^{-}$ has least charge and hence least coagulating power and maximum coagulation values.
Flocculation value $\propto \frac{1}{Coagulating power}$
Among the given electrolytes, $NaCl$
has lowest coagulation power, so its flocculation value will be maximum.

JCECE - 2009

ELECTROCHEMISTRY

275952 Kohlrausch's law states that at

1 Infinite dilution, each ion makes definite contribution to conductance of an electrolyte whatever be the nature of the other ion of the electrolyte

2 Infinite dilution, each ion makes definite contribution to equivalent conductance of an electrolyte, whatever be the nature of the other ion of the electrolyte

3 Finite dilution, each ion makes definite contribution to equivalent conductance of an electrolyte, whatever be the nature of the other ion of the electrolyte

4 Infinite dilution each ion makes definite contribution to equivalent conductance of an electrolyte depending on the nature of the other ion of the electrolyte

Explanation:

At infinite dilution, when the dissociation of electrolyte is complete, each ion makes a definite contribution towards the molar conductivity irrespective of the nature of the other ion with which it is associated. According to Kohlrausch's law the molar conductivity of an electrolyte at infinite dilution can be expressed as the sum of the contributions from its individual ions.
$λ_{\infty} = λ_{a} + λ_{c}$
Where, $λ_{a}$ and $λ_{c}$ are know ionic conductance of anion and cation at infinite dilution respectively.

(AIPMT -2008)

ELECTROCHEMISTRY

275756 The molar conductivity is maximum for the solution of concentration

1

0.005 M

2

0.001 M

3

0.004 M

4

0.002 M

Explanation:

Molar conductance $\propto \frac{1}{Molarity}$
The molar conductivity of electrolytes increases with decreasing concentration, so the molar conductivity of the solution will be maximum at a concentration of $0.001 M$ .

Karnataka CET-17.06.2022

ELECTROCHEMISTRY

275958 The relationship between Gibb's free energy change $(Δ G)$ and emf (E) of a reversible electrochemical cell is given by

1

Δ G = nFE

2

Δ G = nF / E

3

Δ G = - nFE

4

Δ G = E / nF

Explanation:

If the free energy change $(Δ G)$ is negative then any redox reaction would occur spontaneously.
$Δ G = - {nFE}_{cell}$ where,
$n = No. of electrons$
F = value of Faraday
E = The emf of cell
$E = The emf of cell$
$Δ G$ can be negative if $E$ is positive .

J and K CET-(2008)