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CHEMISTRY(KCET)

285388 For spontaneity of a cell, which is correct?

1

Δ G = + ve, Δ E = + ve

2

Δ G = - v e

3

Δ G = 0, Δ E = 0

4

Δ G = - ve, Δ E = 0

Explanation:

(b) For spontaneity of a cell, $Δ G = - ve, E_{cell} = + ve$

Karnataka CET 2022

CHEMISTRY(KCET)

285389 Consider the following electrodes:

$P = {Zn}^{2 +} (0.0001 M) / Zn$ ;
$Q = {Zn}^{2 +} (0.1 M) / Zn$ ;
$R = {Zn}^{2 +} (0.01 M) / Zn$ ;
$S = {Zn}^{2 +} (0.001 M) / Zn$
$E_{Zn / {Zn}^{2 +}}^{\circ} = - 0.76 V$ electrode potentials of the above electrode in volts are in the order

1

P > S > R > Q

2

S > R > Q > P

3

Q > R > S > P

4

P > Q > R > S

Explanation:

(c) ${Zn}^{2 +}_{(a q)} + 2 e^{-} \to {Zn}_{(s)}$
$E_{red} = E_{red}^{\circ} - \frac{0.059}{n} \log \frac{1}{[{Zn}^{2 +}]}$
$E_{red} = + 0.76 + \frac{0.059}{2} \log [{Zn}^{2 +}];$
as $[{Zn}^{2 +}] ↑, E_{red} ↑$
So, the correct order is $Q > R > S > P$ .

Karnataka CET 2021

CHEMISTRY(KCET)

285390 The resistance of 0.01 M KCl solution at 298 K is $1500 Ω$ . If the conductivity of 0.01 M KCl solution at 298 K is $0.1466 \times 10^{- 3} S {cm}^{- 1}$ . The cell constant of the conductivity cell in $({cm}^{- 1})$ is

1 0.219

2 0.291

3 0.301

4 0.194

Explanation:

(a) $G^{*} = κ R [G^{*} =$ cell constant $]$
$= 0.1466 \times 10^{- 3} \times 1500 = 0.219 {cm}^{- 1}$ .

Karnataka CET 2021

CHEMISTRY(KCET)

285391 $H_{2} (g) + 2 AgCl (s) ⇌ 2 Ag (s) + 2 HCl (aq)$ $E_{cell}^{\circ}$ at $25^{\circ} C$ for the cell is 0.22 V . The equilibrium constant at $25^{\circ} C$ is

1

2.8 \times 10^{7}

2

5.2 \times 10^{8}

3

2.8 \times 10^{5}

4

5.2 \times 10^{4}

Explanation:

(a) At equilibrium, $E_{cell} = 0$ .
Number of electrons involved in the green reaction is 2 .
$\log K_{c} = \frac{E_{cell}^{0} \times n}{0.059} = \frac{0.22 \times 2}{0.059} = 7.46$
$K_{c} = Antilog (7.46) = 2.8 \times 10^{7}$

Karnataka CET 2021

CHEMISTRY(KCET)

285388 For spontaneity of a cell, which is correct?

1

Δ G = + ve, Δ E = + ve

2

Δ G = - v e

3

Δ G = 0, Δ E = 0

4

Δ G = - ve, Δ E = 0

Explanation:

(b) For spontaneity of a cell, $Δ G = - ve, E_{cell} = + ve$

Karnataka CET 2022

CHEMISTRY(KCET)

285389 Consider the following electrodes:

$P = {Zn}^{2 +} (0.0001 M) / Zn$ ;
$Q = {Zn}^{2 +} (0.1 M) / Zn$ ;
$R = {Zn}^{2 +} (0.01 M) / Zn$ ;
$S = {Zn}^{2 +} (0.001 M) / Zn$
$E_{Zn / {Zn}^{2 +}}^{\circ} = - 0.76 V$ electrode potentials of the above electrode in volts are in the order

1

P > S > R > Q

2

S > R > Q > P

3

Q > R > S > P

4

P > Q > R > S

Explanation:

(c) ${Zn}^{2 +}_{(a q)} + 2 e^{-} \to {Zn}_{(s)}$
$E_{red} = E_{red}^{\circ} - \frac{0.059}{n} \log \frac{1}{[{Zn}^{2 +}]}$
$E_{red} = + 0.76 + \frac{0.059}{2} \log [{Zn}^{2 +}];$
as $[{Zn}^{2 +}] ↑, E_{red} ↑$
So, the correct order is $Q > R > S > P$ .

Karnataka CET 2021

CHEMISTRY(KCET)

285390 The resistance of 0.01 M KCl solution at 298 K is $1500 Ω$ . If the conductivity of 0.01 M KCl solution at 298 K is $0.1466 \times 10^{- 3} S {cm}^{- 1}$ . The cell constant of the conductivity cell in $({cm}^{- 1})$ is

1 0.219

2 0.291

3 0.301

4 0.194

Explanation:

(a) $G^{*} = κ R [G^{*} =$ cell constant $]$
$= 0.1466 \times 10^{- 3} \times 1500 = 0.219 {cm}^{- 1}$ .

Karnataka CET 2021

CHEMISTRY(KCET)

285391 $H_{2} (g) + 2 AgCl (s) ⇌ 2 Ag (s) + 2 HCl (aq)$ $E_{cell}^{\circ}$ at $25^{\circ} C$ for the cell is 0.22 V . The equilibrium constant at $25^{\circ} C$ is

1

2.8 \times 10^{7}

2

5.2 \times 10^{8}

3

2.8 \times 10^{5}

4

5.2 \times 10^{4}

Explanation:

(a) At equilibrium, $E_{cell} = 0$ .
Number of electrons involved in the green reaction is 2 .
$\log K_{c} = \frac{E_{cell}^{0} \times n}{0.059} = \frac{0.22 \times 2}{0.059} = 7.46$
$K_{c} = Antilog (7.46) = 2.8 \times 10^{7}$

Karnataka CET 2021

CHEMISTRY(KCET)

285388 For spontaneity of a cell, which is correct?

1

Δ G = + ve, Δ E = + ve

2

Δ G = - v e

3

Δ G = 0, Δ E = 0

4

Δ G = - ve, Δ E = 0

Explanation:

(b) For spontaneity of a cell, $Δ G = - ve, E_{cell} = + ve$

Karnataka CET 2022

CHEMISTRY(KCET)

285389 Consider the following electrodes:

$P = {Zn}^{2 +} (0.0001 M) / Zn$ ;
$Q = {Zn}^{2 +} (0.1 M) / Zn$ ;
$R = {Zn}^{2 +} (0.01 M) / Zn$ ;
$S = {Zn}^{2 +} (0.001 M) / Zn$
$E_{Zn / {Zn}^{2 +}}^{\circ} = - 0.76 V$ electrode potentials of the above electrode in volts are in the order

1

P > S > R > Q

2

S > R > Q > P

3

Q > R > S > P

4

P > Q > R > S

Explanation:

(c) ${Zn}^{2 +}_{(a q)} + 2 e^{-} \to {Zn}_{(s)}$
$E_{red} = E_{red}^{\circ} - \frac{0.059}{n} \log \frac{1}{[{Zn}^{2 +}]}$
$E_{red} = + 0.76 + \frac{0.059}{2} \log [{Zn}^{2 +}];$
as $[{Zn}^{2 +}] ↑, E_{red} ↑$
So, the correct order is $Q > R > S > P$ .

Karnataka CET 2021

CHEMISTRY(KCET)

285390 The resistance of 0.01 M KCl solution at 298 K is $1500 Ω$ . If the conductivity of 0.01 M KCl solution at 298 K is $0.1466 \times 10^{- 3} S {cm}^{- 1}$ . The cell constant of the conductivity cell in $({cm}^{- 1})$ is

1 0.219

2 0.291

3 0.301

4 0.194

Explanation:

(a) $G^{*} = κ R [G^{*} =$ cell constant $]$
$= 0.1466 \times 10^{- 3} \times 1500 = 0.219 {cm}^{- 1}$ .

Karnataka CET 2021

CHEMISTRY(KCET)

285391 $H_{2} (g) + 2 AgCl (s) ⇌ 2 Ag (s) + 2 HCl (aq)$ $E_{cell}^{\circ}$ at $25^{\circ} C$ for the cell is 0.22 V . The equilibrium constant at $25^{\circ} C$ is

1

2.8 \times 10^{7}

2

5.2 \times 10^{8}

3

2.8 \times 10^{5}

4

5.2 \times 10^{4}

Explanation:

(a) At equilibrium, $E_{cell} = 0$ .
Number of electrons involved in the green reaction is 2 .
$\log K_{c} = \frac{E_{cell}^{0} \times n}{0.059} = \frac{0.22 \times 2}{0.059} = 7.46$
$K_{c} = Antilog (7.46) = 2.8 \times 10^{7}$

Karnataka CET 2021

CHEMISTRY(KCET)

285388 For spontaneity of a cell, which is correct?

1

Δ G = + ve, Δ E = + ve

2

Δ G = - v e

3

Δ G = 0, Δ E = 0

4

Δ G = - ve, Δ E = 0

Explanation:

(b) For spontaneity of a cell, $Δ G = - ve, E_{cell} = + ve$

Karnataka CET 2022

CHEMISTRY(KCET)

285389 Consider the following electrodes:

$P = {Zn}^{2 +} (0.0001 M) / Zn$ ;
$Q = {Zn}^{2 +} (0.1 M) / Zn$ ;
$R = {Zn}^{2 +} (0.01 M) / Zn$ ;
$S = {Zn}^{2 +} (0.001 M) / Zn$
$E_{Zn / {Zn}^{2 +}}^{\circ} = - 0.76 V$ electrode potentials of the above electrode in volts are in the order

1

P > S > R > Q

2

S > R > Q > P

3

Q > R > S > P

4

P > Q > R > S

Explanation:

(c) ${Zn}^{2 +}_{(a q)} + 2 e^{-} \to {Zn}_{(s)}$
$E_{red} = E_{red}^{\circ} - \frac{0.059}{n} \log \frac{1}{[{Zn}^{2 +}]}$
$E_{red} = + 0.76 + \frac{0.059}{2} \log [{Zn}^{2 +}];$
as $[{Zn}^{2 +}] ↑, E_{red} ↑$
So, the correct order is $Q > R > S > P$ .

Karnataka CET 2021

CHEMISTRY(KCET)

285390 The resistance of 0.01 M KCl solution at 298 K is $1500 Ω$ . If the conductivity of 0.01 M KCl solution at 298 K is $0.1466 \times 10^{- 3} S {cm}^{- 1}$ . The cell constant of the conductivity cell in $({cm}^{- 1})$ is

1 0.219

2 0.291

3 0.301

4 0.194

Explanation:

(a) $G^{*} = κ R [G^{*} =$ cell constant $]$
$= 0.1466 \times 10^{- 3} \times 1500 = 0.219 {cm}^{- 1}$ .

Karnataka CET 2021

CHEMISTRY(KCET)

285391 $H_{2} (g) + 2 AgCl (s) ⇌ 2 Ag (s) + 2 HCl (aq)$ $E_{cell}^{\circ}$ at $25^{\circ} C$ for the cell is 0.22 V . The equilibrium constant at $25^{\circ} C$ is

1

2.8 \times 10^{7}

2

5.2 \times 10^{8}

3

2.8 \times 10^{5}

4

5.2 \times 10^{4}

Explanation:

(a) At equilibrium, $E_{cell} = 0$ .
Number of electrons involved in the green reaction is 2 .
$\log K_{c} = \frac{E_{cell}^{0} \times n}{0.059} = \frac{0.22 \times 2}{0.059} = 7.46$
$K_{c} = Antilog (7.46) = 2.8 \times 10^{7}$

Karnataka CET 2021

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