QBManager

CHEMISTRY(KCET)

285383 During the electrolysis of brine, by using inert electrodes,

1

O_{2}

liberates at anode

2

H_{2}

liberates at anode

3 Na deposits on cathode

4

{Cl}_{2}

liberates at anode

Explanation:

(d) Brine solution $= NaCl$ aqueous solution.
We have inert electrodes.
$\Rightarrow$ Electrode reactions:-
Anode (oxidation):
$2 {Cl}^{-}$ (aq.) $\to {Cl}_{2}$ (g) $+ 2 e^{-}$
$2 H_{2} O (l) \to O_{2} (g) + 4 H^{+} + 4 e^{-}$
${Cl}_{2}$ gas is liberated due to over potential of $O_{2} . (g)$ .
Cathode (reduction):
${Na}^{+}$ (aq.) $+ e^{-} \to Na$ (s)
$2 H^{+}$ (aq.) $+ 2 e^{-} \to H_{2}$ (g)
$H_{2}$ gas is liberated at cathode due to higher SEP of $H_{2}$ .

Karnataka CET 2023

CHEMISTRY(KCET)

285384 Consider the following 4 electrodes
$A : {Ag}^{+} (0.0001 M) / {Ag}_{(s)}$ ;
$B : {Ag}^{+} (0.1 M) / {Ag}_{(s)}$
C: ${Ag}^{+} (0.01 M / {Ag}_{(s)}$ ;
$D : {Ag}^{+} (0.001 M) / {Ag}_{(s)}; E_{{Ag}^{+} / Ag}^{0} = + 0.80 V$
Then reduction potential in volts of the electrodes in the order.

1 B

>

C

>

D

>

A

2 C

>

D

>

A

>

B

3 A

>

D

>

C

>

B

4 A

>

B

>

C

>

D

Explanation:

(a) $E = E^{\circ} + \frac{0.059}{n} \log [M^{n +}]$
$\Rightarrow$ A higher value of ionic concentration means higher reduction potential.
$\Rightarrow$ Order of ionic concentration
$= B > C > D > A$
$\Rightarrow$ Order of reduction potentials
$= B > C > D > A$

Karnataka CET 2023

CHEMISTRY(KCET)

285385 Specific conductance of 0.1 M HNO 3 is $6.3 \times 10^{- 2} {ohm}^{- 1} {cm}^{- 1}$ . The molar conductance of the solution is

1

6.300 . {ohm}^{- 1} {cm}^{2} {mol}^{- 1}

2

63.0 {ohm}^{- 1} {cm}^{2} {mol}^{- 1}

3

630 {ohm}^{- 1} {cm}^{2} {mol}^{- 1}

4

315 {ohm}^{- 1} {cm}^{2} {mol}^{- 1}

Explanation:

(c) $λ_{m} = \frac{κ \times 1000}{M}$
$= \frac{6.3 \times 10^{- 2} {ohm}^{- 1} {cm}^{- 1} \times 1000}{0.1 {molcm}^{- 3}}$
$= 630 {ohm}^{- 1} {cm}^{2} {mol}^{- 1}$

Karnataka CET 2022

CHEMISTRY(KCET)

285386 In fuel cells, are used as catalysts.

1 zinc-mercury

2 lead-manganese

3 platinum-palladium

4 nickel-cadmium

Karnataka CET 2022

CHEMISTRY(KCET)

285387 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:

(b) $A_{m} = \frac{κ \times 1000}{c}$ .Thus, $A_{m} \propto \frac{1}{c}$
Lower the value of concentration, higher is the molar conductivity.

Karnataka CET 2022

CHEMISTRY(KCET)

285383 During the electrolysis of brine, by using inert electrodes,

1

O_{2}

liberates at anode

2

H_{2}

liberates at anode

3 Na deposits on cathode

4

{Cl}_{2}

liberates at anode

Explanation:

(d) Brine solution $= NaCl$ aqueous solution.
We have inert electrodes.
$\Rightarrow$ Electrode reactions:-
Anode (oxidation):
$2 {Cl}^{-}$ (aq.) $\to {Cl}_{2}$ (g) $+ 2 e^{-}$
$2 H_{2} O (l) \to O_{2} (g) + 4 H^{+} + 4 e^{-}$
${Cl}_{2}$ gas is liberated due to over potential of $O_{2} . (g)$ .
Cathode (reduction):
${Na}^{+}$ (aq.) $+ e^{-} \to Na$ (s)
$2 H^{+}$ (aq.) $+ 2 e^{-} \to H_{2}$ (g)
$H_{2}$ gas is liberated at cathode due to higher SEP of $H_{2}$ .

Karnataka CET 2023

CHEMISTRY(KCET)

285384 Consider the following 4 electrodes
$A : {Ag}^{+} (0.0001 M) / {Ag}_{(s)}$ ;
$B : {Ag}^{+} (0.1 M) / {Ag}_{(s)}$
C: ${Ag}^{+} (0.01 M / {Ag}_{(s)}$ ;
$D : {Ag}^{+} (0.001 M) / {Ag}_{(s)}; E_{{Ag}^{+} / Ag}^{0} = + 0.80 V$
Then reduction potential in volts of the electrodes in the order.

1 B

>

C

>

D

>

A

2 C

>

D

>

A

>

B

3 A

>

D

>

C

>

B

4 A

>

B

>

C

>

D

Explanation:

(a) $E = E^{\circ} + \frac{0.059}{n} \log [M^{n +}]$
$\Rightarrow$ A higher value of ionic concentration means higher reduction potential.
$\Rightarrow$ Order of ionic concentration
$= B > C > D > A$
$\Rightarrow$ Order of reduction potentials
$= B > C > D > A$

Karnataka CET 2023

CHEMISTRY(KCET)

285385 Specific conductance of 0.1 M HNO 3 is $6.3 \times 10^{- 2} {ohm}^{- 1} {cm}^{- 1}$ . The molar conductance of the solution is

1

6.300 . {ohm}^{- 1} {cm}^{2} {mol}^{- 1}

2

63.0 {ohm}^{- 1} {cm}^{2} {mol}^{- 1}

3

630 {ohm}^{- 1} {cm}^{2} {mol}^{- 1}

4

315 {ohm}^{- 1} {cm}^{2} {mol}^{- 1}

Explanation:

(c) $λ_{m} = \frac{κ \times 1000}{M}$
$= \frac{6.3 \times 10^{- 2} {ohm}^{- 1} {cm}^{- 1} \times 1000}{0.1 {molcm}^{- 3}}$
$= 630 {ohm}^{- 1} {cm}^{2} {mol}^{- 1}$

Karnataka CET 2022

CHEMISTRY(KCET)

285386 In fuel cells, are used as catalysts.

1 zinc-mercury

2 lead-manganese

3 platinum-palladium

4 nickel-cadmium

Karnataka CET 2022

CHEMISTRY(KCET)

285387 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:

(b) $A_{m} = \frac{κ \times 1000}{c}$ .Thus, $A_{m} \propto \frac{1}{c}$
Lower the value of concentration, higher is the molar conductivity.

Karnataka CET 2022

CHEMISTRY(KCET)

285383 During the electrolysis of brine, by using inert electrodes,

1

O_{2}

liberates at anode

2

H_{2}

liberates at anode

3 Na deposits on cathode

4

{Cl}_{2}

liberates at anode

Explanation:

(d) Brine solution $= NaCl$ aqueous solution.
We have inert electrodes.
$\Rightarrow$ Electrode reactions:-
Anode (oxidation):
$2 {Cl}^{-}$ (aq.) $\to {Cl}_{2}$ (g) $+ 2 e^{-}$
$2 H_{2} O (l) \to O_{2} (g) + 4 H^{+} + 4 e^{-}$
${Cl}_{2}$ gas is liberated due to over potential of $O_{2} . (g)$ .
Cathode (reduction):
${Na}^{+}$ (aq.) $+ e^{-} \to Na$ (s)
$2 H^{+}$ (aq.) $+ 2 e^{-} \to H_{2}$ (g)
$H_{2}$ gas is liberated at cathode due to higher SEP of $H_{2}$ .

Karnataka CET 2023

CHEMISTRY(KCET)

285384 Consider the following 4 electrodes
$A : {Ag}^{+} (0.0001 M) / {Ag}_{(s)}$ ;
$B : {Ag}^{+} (0.1 M) / {Ag}_{(s)}$
C: ${Ag}^{+} (0.01 M / {Ag}_{(s)}$ ;
$D : {Ag}^{+} (0.001 M) / {Ag}_{(s)}; E_{{Ag}^{+} / Ag}^{0} = + 0.80 V$
Then reduction potential in volts of the electrodes in the order.

1 B

>

C

>

D

>

A

2 C

>

D

>

A

>

B

3 A

>

D

>

C

>

B

4 A

>

B

>

C

>

D

Explanation:

(a) $E = E^{\circ} + \frac{0.059}{n} \log [M^{n +}]$
$\Rightarrow$ A higher value of ionic concentration means higher reduction potential.
$\Rightarrow$ Order of ionic concentration
$= B > C > D > A$
$\Rightarrow$ Order of reduction potentials
$= B > C > D > A$

Karnataka CET 2023

CHEMISTRY(KCET)

285385 Specific conductance of 0.1 M HNO 3 is $6.3 \times 10^{- 2} {ohm}^{- 1} {cm}^{- 1}$ . The molar conductance of the solution is

1

6.300 . {ohm}^{- 1} {cm}^{2} {mol}^{- 1}

2

63.0 {ohm}^{- 1} {cm}^{2} {mol}^{- 1}

3

630 {ohm}^{- 1} {cm}^{2} {mol}^{- 1}

4

315 {ohm}^{- 1} {cm}^{2} {mol}^{- 1}

Explanation:

(c) $λ_{m} = \frac{κ \times 1000}{M}$
$= \frac{6.3 \times 10^{- 2} {ohm}^{- 1} {cm}^{- 1} \times 1000}{0.1 {molcm}^{- 3}}$
$= 630 {ohm}^{- 1} {cm}^{2} {mol}^{- 1}$

Karnataka CET 2022

CHEMISTRY(KCET)

285386 In fuel cells, are used as catalysts.

1 zinc-mercury

2 lead-manganese

3 platinum-palladium

4 nickel-cadmium

Karnataka CET 2022

CHEMISTRY(KCET)

285387 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:

(b) $A_{m} = \frac{κ \times 1000}{c}$ .Thus, $A_{m} \propto \frac{1}{c}$
Lower the value of concentration, higher is the molar conductivity.

Karnataka CET 2022

CHEMISTRY(KCET)

285383 During the electrolysis of brine, by using inert electrodes,

1

O_{2}

liberates at anode

2

H_{2}

liberates at anode

3 Na deposits on cathode

4

{Cl}_{2}

liberates at anode

Explanation:

(d) Brine solution $= NaCl$ aqueous solution.
We have inert electrodes.
$\Rightarrow$ Electrode reactions:-
Anode (oxidation):
$2 {Cl}^{-}$ (aq.) $\to {Cl}_{2}$ (g) $+ 2 e^{-}$
$2 H_{2} O (l) \to O_{2} (g) + 4 H^{+} + 4 e^{-}$
${Cl}_{2}$ gas is liberated due to over potential of $O_{2} . (g)$ .
Cathode (reduction):
${Na}^{+}$ (aq.) $+ e^{-} \to Na$ (s)
$2 H^{+}$ (aq.) $+ 2 e^{-} \to H_{2}$ (g)
$H_{2}$ gas is liberated at cathode due to higher SEP of $H_{2}$ .

Karnataka CET 2023

CHEMISTRY(KCET)

285384 Consider the following 4 electrodes
$A : {Ag}^{+} (0.0001 M) / {Ag}_{(s)}$ ;
$B : {Ag}^{+} (0.1 M) / {Ag}_{(s)}$
C: ${Ag}^{+} (0.01 M / {Ag}_{(s)}$ ;
$D : {Ag}^{+} (0.001 M) / {Ag}_{(s)}; E_{{Ag}^{+} / Ag}^{0} = + 0.80 V$
Then reduction potential in volts of the electrodes in the order.

1 B

>

C

>

D

>

A

2 C

>

D

>

A

>

B

3 A

>

D

>

C

>

B

4 A

>

B

>

C

>

D

Explanation:

(a) $E = E^{\circ} + \frac{0.059}{n} \log [M^{n +}]$
$\Rightarrow$ A higher value of ionic concentration means higher reduction potential.
$\Rightarrow$ Order of ionic concentration
$= B > C > D > A$
$\Rightarrow$ Order of reduction potentials
$= B > C > D > A$

Karnataka CET 2023

CHEMISTRY(KCET)

285385 Specific conductance of 0.1 M HNO 3 is $6.3 \times 10^{- 2} {ohm}^{- 1} {cm}^{- 1}$ . The molar conductance of the solution is

1

6.300 . {ohm}^{- 1} {cm}^{2} {mol}^{- 1}

2

63.0 {ohm}^{- 1} {cm}^{2} {mol}^{- 1}

3

630 {ohm}^{- 1} {cm}^{2} {mol}^{- 1}

4

315 {ohm}^{- 1} {cm}^{2} {mol}^{- 1}

Explanation:

(c) $λ_{m} = \frac{κ \times 1000}{M}$
$= \frac{6.3 \times 10^{- 2} {ohm}^{- 1} {cm}^{- 1} \times 1000}{0.1 {molcm}^{- 3}}$
$= 630 {ohm}^{- 1} {cm}^{2} {mol}^{- 1}$

Karnataka CET 2022

CHEMISTRY(KCET)

285386 In fuel cells, are used as catalysts.

1 zinc-mercury

2 lead-manganese

3 platinum-palladium

4 nickel-cadmium

Karnataka CET 2022

CHEMISTRY(KCET)

285387 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:

(b) $A_{m} = \frac{κ \times 1000}{c}$ .Thus, $A_{m} \propto \frac{1}{c}$
Lower the value of concentration, higher is the molar conductivity.

Karnataka CET 2022

CHEMISTRY(KCET)

285383 During the electrolysis of brine, by using inert electrodes,

1

O_{2}

liberates at anode

2

H_{2}

liberates at anode

3 Na deposits on cathode

4

{Cl}_{2}

liberates at anode

Explanation:

(d) Brine solution $= NaCl$ aqueous solution.
We have inert electrodes.
$\Rightarrow$ Electrode reactions:-
Anode (oxidation):
$2 {Cl}^{-}$ (aq.) $\to {Cl}_{2}$ (g) $+ 2 e^{-}$
$2 H_{2} O (l) \to O_{2} (g) + 4 H^{+} + 4 e^{-}$
${Cl}_{2}$ gas is liberated due to over potential of $O_{2} . (g)$ .
Cathode (reduction):
${Na}^{+}$ (aq.) $+ e^{-} \to Na$ (s)
$2 H^{+}$ (aq.) $+ 2 e^{-} \to H_{2}$ (g)
$H_{2}$ gas is liberated at cathode due to higher SEP of $H_{2}$ .

Karnataka CET 2023

CHEMISTRY(KCET)

285384 Consider the following 4 electrodes
$A : {Ag}^{+} (0.0001 M) / {Ag}_{(s)}$ ;
$B : {Ag}^{+} (0.1 M) / {Ag}_{(s)}$
C: ${Ag}^{+} (0.01 M / {Ag}_{(s)}$ ;
$D : {Ag}^{+} (0.001 M) / {Ag}_{(s)}; E_{{Ag}^{+} / Ag}^{0} = + 0.80 V$
Then reduction potential in volts of the electrodes in the order.

1 B

>

C

>

D

>

A

2 C

>

D

>

A

>

B

3 A

>

D

>

C

>

B

4 A

>

B

>

C

>

D

Explanation:

(a) $E = E^{\circ} + \frac{0.059}{n} \log [M^{n +}]$
$\Rightarrow$ A higher value of ionic concentration means higher reduction potential.
$\Rightarrow$ Order of ionic concentration
$= B > C > D > A$
$\Rightarrow$ Order of reduction potentials
$= B > C > D > A$

Karnataka CET 2023

CHEMISTRY(KCET)

285385 Specific conductance of 0.1 M HNO 3 is $6.3 \times 10^{- 2} {ohm}^{- 1} {cm}^{- 1}$ . The molar conductance of the solution is

1

6.300 . {ohm}^{- 1} {cm}^{2} {mol}^{- 1}

2

63.0 {ohm}^{- 1} {cm}^{2} {mol}^{- 1}

3

630 {ohm}^{- 1} {cm}^{2} {mol}^{- 1}

4

315 {ohm}^{- 1} {cm}^{2} {mol}^{- 1}

Explanation:

(c) $λ_{m} = \frac{κ \times 1000}{M}$
$= \frac{6.3 \times 10^{- 2} {ohm}^{- 1} {cm}^{- 1} \times 1000}{0.1 {molcm}^{- 3}}$
$= 630 {ohm}^{- 1} {cm}^{2} {mol}^{- 1}$

Karnataka CET 2022

CHEMISTRY(KCET)

285386 In fuel cells, are used as catalysts.

1 zinc-mercury

2 lead-manganese

3 platinum-palladium

4 nickel-cadmium

Karnataka CET 2022

CHEMISTRY(KCET)

285387 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:

(b) $A_{m} = \frac{κ \times 1000}{c}$ .Thus, $A_{m} \propto \frac{1}{c}$
Lower the value of concentration, higher is the molar conductivity.

Karnataka CET 2022