QBManager

ELECTROCHEMISTRY

276128 Which of the following relation represents correct relation between standard electrode potential and equilibrium constant?
I. $\log K = \frac{{nFE}^{0}}{2.303 RT}$
II. $K = e^{\frac{{nFE}^{0}}{RT}}$
III. $\log K = \frac{- {nFE}^{0}}{2.303 RT}$
IV. $\log K = 0.4342 \frac{{nFE}^{0}}{RT}$
Choose the correct statement (s).

1 I, II and III are correct

2 II and III are correct

3 I, II and IV are correct

4 I and IV are correct

Explanation:

$Δ G^{\circ} = - 2.303 RT \log K$
$- {nFE}^{\circ} = - 2.303 RT \log K$
$\begin{array}{r} \log K = \frac{1}{2.303} \times \frac{nFE}{RT} \\ \log K = 0.4342 \frac{nFE}{RT} \end{array}$
$\begin{aligned} \ln K = \frac{{nFE}^{\circ}}{RT} \\ K = e^{\frac{{nFE}^{\circ}}{RT}} \end{aligned}$
Hence, equation (I), (II) and (IV) are correct.

BITSAT - 2017

ELECTROCHEMISTRY

276129 $λ^{\circ}$ for ${NH}_{4} Cl, NaOH$ and $NaCl$ are 130,248 and $126.5 {ohm}^{- 1} {cm}^{2} {mol}^{- 1}$ respectively. The $λ^{\circ}_{m}$ of ${NH}_{4} OH$ will be

1 251.5

2 244.5

3 130

4 504.5

Explanation:

$\begin{aligned} Given: \\ λ^{\circ}_{m ({NH}_{4} Cl)} = 130 {ohm}^{- 1} {cm}^{2} {mol}^{- 1} \\ λ_{m (NaOH)}^{\circ} = 248 {ohm}^{- 1} {cm}^{2} {mol}^{- 1} \\ λ^{\circ}_{m (NaCl)} = 126.5 {ohm}^{- 1} {cm}^{2} {mol}^{- 1} \\ λ_{m ({NH}_{4} OH)}^{\circ} = λ^{\circ}_{m ({NH}_{4} Cl)} + λ^{\circ}_{m (NaOH)} - λ^{\circ}_{m (NaCl)} \\ λ_{m ({NH}_{4} OH)}^{\circ} = 130 + 248 - 126.5 \\ λ^{\circ}_{m ({NH}_{4} OH)} = 251.5 {ohm}^{- 1} {cm}^{2} {mol}^{- 1} \end{aligned}$

COMEDK 2015

ELECTROCHEMISTRY

276130 For the reaction,
$2 {SO}_{2} (g) + O_{2} (g) ⇌ 2 {SO}_{3} (g)$ at $300 K$ , the value of $◻ G^{0}$ is -690.9 R. The equilibrium constant value for the reaction at that temperature is ( $R$ is gas constant)

1

10 {atm}^{- 1}

2

10 atm

3 10

4 1

Explanation:

For the reaction
$2 {SO}_{2} + O_{2} ⇌ 2 {SO}_{3}$
Given that,
$Δ G^{\circ} = - 690.9 R,$
$T = 300 K$
$Δ G^{\circ} = - RT \ln K$
$- 690.9 R = - R \times 300 \times \ln K$
$\ln K = \frac{690.9}{300} = 2.303$
$2.303 \log K = 2.303$
$\log K = 1$
$K = 10^{1} = 10$
Unit of $K = (atm)^{Δ n} = (atm)^{2 - 3}$
Unit of $K = (atm)^{- 1}$
$K = 10 {atm}^{- 1}$

WB-JEE-2015

ELECTROCHEMISTRY

276131 If for the cell reaction, $Zn + {Cu}^{2 +} ⇌ Cu + {Zn}^{2 +}$ Entropy change $Δ S^{\circ}$ is $96.5 J {mol}^{- 1} K^{- 1}$ , then temperature coefficient of the emf of a cell is

1

5 \times 10^{- 4} {VK}^{- 1}

2

1 \times 10^{- 3} {VK}^{- 1}

3

2 \times 10^{- 3} {VK}^{- 1}

4

9.65 \times 10^{- 4} {VK}^{- 1}

Explanation:

We know that
$Δ G = Δ H - nFT {(\frac{dE}{dT})}_{P}$
and from Gibbs free energy
$Δ G = Δ H - T Δ S$
from (1) and (2) we get
$\begin{aligned} {(\frac{dE}{dT})}_{P} = \frac{Δ S}{nF} \\ {(\frac{dE}{dT})}_{P} = \frac{96.5}{2 \times 96500} \\ = 5 \times 10^{- 4} {VK}^{- 1} \end{aligned}$

VITEEE-2013

ELECTROCHEMISTRY

276128 Which of the following relation represents correct relation between standard electrode potential and equilibrium constant?
I. $\log K = \frac{{nFE}^{0}}{2.303 RT}$
II. $K = e^{\frac{{nFE}^{0}}{RT}}$
III. $\log K = \frac{- {nFE}^{0}}{2.303 RT}$
IV. $\log K = 0.4342 \frac{{nFE}^{0}}{RT}$
Choose the correct statement (s).

1 I, II and III are correct

2 II and III are correct

3 I, II and IV are correct

4 I and IV are correct

Explanation:

$Δ G^{\circ} = - 2.303 RT \log K$
$- {nFE}^{\circ} = - 2.303 RT \log K$
$\begin{array}{r} \log K = \frac{1}{2.303} \times \frac{nFE}{RT} \\ \log K = 0.4342 \frac{nFE}{RT} \end{array}$
$\begin{aligned} \ln K = \frac{{nFE}^{\circ}}{RT} \\ K = e^{\frac{{nFE}^{\circ}}{RT}} \end{aligned}$
Hence, equation (I), (II) and (IV) are correct.

BITSAT - 2017

ELECTROCHEMISTRY

276129 $λ^{\circ}$ for ${NH}_{4} Cl, NaOH$ and $NaCl$ are 130,248 and $126.5 {ohm}^{- 1} {cm}^{2} {mol}^{- 1}$ respectively. The $λ^{\circ}_{m}$ of ${NH}_{4} OH$ will be

1 251.5

2 244.5

3 130

4 504.5

Explanation:

$\begin{aligned} Given: \\ λ^{\circ}_{m ({NH}_{4} Cl)} = 130 {ohm}^{- 1} {cm}^{2} {mol}^{- 1} \\ λ_{m (NaOH)}^{\circ} = 248 {ohm}^{- 1} {cm}^{2} {mol}^{- 1} \\ λ^{\circ}_{m (NaCl)} = 126.5 {ohm}^{- 1} {cm}^{2} {mol}^{- 1} \\ λ_{m ({NH}_{4} OH)}^{\circ} = λ^{\circ}_{m ({NH}_{4} Cl)} + λ^{\circ}_{m (NaOH)} - λ^{\circ}_{m (NaCl)} \\ λ_{m ({NH}_{4} OH)}^{\circ} = 130 + 248 - 126.5 \\ λ^{\circ}_{m ({NH}_{4} OH)} = 251.5 {ohm}^{- 1} {cm}^{2} {mol}^{- 1} \end{aligned}$

COMEDK 2015

ELECTROCHEMISTRY

276130 For the reaction,
$2 {SO}_{2} (g) + O_{2} (g) ⇌ 2 {SO}_{3} (g)$ at $300 K$ , the value of $◻ G^{0}$ is -690.9 R. The equilibrium constant value for the reaction at that temperature is ( $R$ is gas constant)

1

10 {atm}^{- 1}

2

10 atm

3 10

4 1

Explanation:

For the reaction
$2 {SO}_{2} + O_{2} ⇌ 2 {SO}_{3}$
Given that,
$Δ G^{\circ} = - 690.9 R,$
$T = 300 K$
$Δ G^{\circ} = - RT \ln K$
$- 690.9 R = - R \times 300 \times \ln K$
$\ln K = \frac{690.9}{300} = 2.303$
$2.303 \log K = 2.303$
$\log K = 1$
$K = 10^{1} = 10$
Unit of $K = (atm)^{Δ n} = (atm)^{2 - 3}$
Unit of $K = (atm)^{- 1}$
$K = 10 {atm}^{- 1}$

WB-JEE-2015

ELECTROCHEMISTRY

276131 If for the cell reaction, $Zn + {Cu}^{2 +} ⇌ Cu + {Zn}^{2 +}$ Entropy change $Δ S^{\circ}$ is $96.5 J {mol}^{- 1} K^{- 1}$ , then temperature coefficient of the emf of a cell is

1

5 \times 10^{- 4} {VK}^{- 1}

2

1 \times 10^{- 3} {VK}^{- 1}

3

2 \times 10^{- 3} {VK}^{- 1}

4

9.65 \times 10^{- 4} {VK}^{- 1}

Explanation:

We know that
$Δ G = Δ H - nFT {(\frac{dE}{dT})}_{P}$
and from Gibbs free energy
$Δ G = Δ H - T Δ S$
from (1) and (2) we get
$\begin{aligned} {(\frac{dE}{dT})}_{P} = \frac{Δ S}{nF} \\ {(\frac{dE}{dT})}_{P} = \frac{96.5}{2 \times 96500} \\ = 5 \times 10^{- 4} {VK}^{- 1} \end{aligned}$

VITEEE-2013

ELECTROCHEMISTRY

276128 Which of the following relation represents correct relation between standard electrode potential and equilibrium constant?
I. $\log K = \frac{{nFE}^{0}}{2.303 RT}$
II. $K = e^{\frac{{nFE}^{0}}{RT}}$
III. $\log K = \frac{- {nFE}^{0}}{2.303 RT}$
IV. $\log K = 0.4342 \frac{{nFE}^{0}}{RT}$
Choose the correct statement (s).

1 I, II and III are correct

2 II and III are correct

3 I, II and IV are correct

4 I and IV are correct

Explanation:

$Δ G^{\circ} = - 2.303 RT \log K$
$- {nFE}^{\circ} = - 2.303 RT \log K$
$\begin{array}{r} \log K = \frac{1}{2.303} \times \frac{nFE}{RT} \\ \log K = 0.4342 \frac{nFE}{RT} \end{array}$
$\begin{aligned} \ln K = \frac{{nFE}^{\circ}}{RT} \\ K = e^{\frac{{nFE}^{\circ}}{RT}} \end{aligned}$
Hence, equation (I), (II) and (IV) are correct.

BITSAT - 2017

ELECTROCHEMISTRY

276129 $λ^{\circ}$ for ${NH}_{4} Cl, NaOH$ and $NaCl$ are 130,248 and $126.5 {ohm}^{- 1} {cm}^{2} {mol}^{- 1}$ respectively. The $λ^{\circ}_{m}$ of ${NH}_{4} OH$ will be

1 251.5

2 244.5

3 130

4 504.5

Explanation:

$\begin{aligned} Given: \\ λ^{\circ}_{m ({NH}_{4} Cl)} = 130 {ohm}^{- 1} {cm}^{2} {mol}^{- 1} \\ λ_{m (NaOH)}^{\circ} = 248 {ohm}^{- 1} {cm}^{2} {mol}^{- 1} \\ λ^{\circ}_{m (NaCl)} = 126.5 {ohm}^{- 1} {cm}^{2} {mol}^{- 1} \\ λ_{m ({NH}_{4} OH)}^{\circ} = λ^{\circ}_{m ({NH}_{4} Cl)} + λ^{\circ}_{m (NaOH)} - λ^{\circ}_{m (NaCl)} \\ λ_{m ({NH}_{4} OH)}^{\circ} = 130 + 248 - 126.5 \\ λ^{\circ}_{m ({NH}_{4} OH)} = 251.5 {ohm}^{- 1} {cm}^{2} {mol}^{- 1} \end{aligned}$

COMEDK 2015

ELECTROCHEMISTRY

276130 For the reaction,
$2 {SO}_{2} (g) + O_{2} (g) ⇌ 2 {SO}_{3} (g)$ at $300 K$ , the value of $◻ G^{0}$ is -690.9 R. The equilibrium constant value for the reaction at that temperature is ( $R$ is gas constant)

1

10 {atm}^{- 1}

2

10 atm

3 10

4 1

Explanation:

For the reaction
$2 {SO}_{2} + O_{2} ⇌ 2 {SO}_{3}$
Given that,
$Δ G^{\circ} = - 690.9 R,$
$T = 300 K$
$Δ G^{\circ} = - RT \ln K$
$- 690.9 R = - R \times 300 \times \ln K$
$\ln K = \frac{690.9}{300} = 2.303$
$2.303 \log K = 2.303$
$\log K = 1$
$K = 10^{1} = 10$
Unit of $K = (atm)^{Δ n} = (atm)^{2 - 3}$
Unit of $K = (atm)^{- 1}$
$K = 10 {atm}^{- 1}$

WB-JEE-2015

ELECTROCHEMISTRY

276131 If for the cell reaction, $Zn + {Cu}^{2 +} ⇌ Cu + {Zn}^{2 +}$ Entropy change $Δ S^{\circ}$ is $96.5 J {mol}^{- 1} K^{- 1}$ , then temperature coefficient of the emf of a cell is

1

5 \times 10^{- 4} {VK}^{- 1}

2

1 \times 10^{- 3} {VK}^{- 1}

3

2 \times 10^{- 3} {VK}^{- 1}

4

9.65 \times 10^{- 4} {VK}^{- 1}

Explanation:

We know that
$Δ G = Δ H - nFT {(\frac{dE}{dT})}_{P}$
and from Gibbs free energy
$Δ G = Δ H - T Δ S$
from (1) and (2) we get
$\begin{aligned} {(\frac{dE}{dT})}_{P} = \frac{Δ S}{nF} \\ {(\frac{dE}{dT})}_{P} = \frac{96.5}{2 \times 96500} \\ = 5 \times 10^{- 4} {VK}^{- 1} \end{aligned}$

VITEEE-2013

ELECTROCHEMISTRY

276128 Which of the following relation represents correct relation between standard electrode potential and equilibrium constant?
I. $\log K = \frac{{nFE}^{0}}{2.303 RT}$
II. $K = e^{\frac{{nFE}^{0}}{RT}}$
III. $\log K = \frac{- {nFE}^{0}}{2.303 RT}$
IV. $\log K = 0.4342 \frac{{nFE}^{0}}{RT}$
Choose the correct statement (s).

1 I, II and III are correct

2 II and III are correct

3 I, II and IV are correct

4 I and IV are correct

Explanation:

$Δ G^{\circ} = - 2.303 RT \log K$
$- {nFE}^{\circ} = - 2.303 RT \log K$
$\begin{array}{r} \log K = \frac{1}{2.303} \times \frac{nFE}{RT} \\ \log K = 0.4342 \frac{nFE}{RT} \end{array}$
$\begin{aligned} \ln K = \frac{{nFE}^{\circ}}{RT} \\ K = e^{\frac{{nFE}^{\circ}}{RT}} \end{aligned}$
Hence, equation (I), (II) and (IV) are correct.

BITSAT - 2017

ELECTROCHEMISTRY

276129 $λ^{\circ}$ for ${NH}_{4} Cl, NaOH$ and $NaCl$ are 130,248 and $126.5 {ohm}^{- 1} {cm}^{2} {mol}^{- 1}$ respectively. The $λ^{\circ}_{m}$ of ${NH}_{4} OH$ will be

1 251.5

2 244.5

3 130

4 504.5

Explanation:

$\begin{aligned} Given: \\ λ^{\circ}_{m ({NH}_{4} Cl)} = 130 {ohm}^{- 1} {cm}^{2} {mol}^{- 1} \\ λ_{m (NaOH)}^{\circ} = 248 {ohm}^{- 1} {cm}^{2} {mol}^{- 1} \\ λ^{\circ}_{m (NaCl)} = 126.5 {ohm}^{- 1} {cm}^{2} {mol}^{- 1} \\ λ_{m ({NH}_{4} OH)}^{\circ} = λ^{\circ}_{m ({NH}_{4} Cl)} + λ^{\circ}_{m (NaOH)} - λ^{\circ}_{m (NaCl)} \\ λ_{m ({NH}_{4} OH)}^{\circ} = 130 + 248 - 126.5 \\ λ^{\circ}_{m ({NH}_{4} OH)} = 251.5 {ohm}^{- 1} {cm}^{2} {mol}^{- 1} \end{aligned}$

COMEDK 2015

ELECTROCHEMISTRY

276130 For the reaction,
$2 {SO}_{2} (g) + O_{2} (g) ⇌ 2 {SO}_{3} (g)$ at $300 K$ , the value of $◻ G^{0}$ is -690.9 R. The equilibrium constant value for the reaction at that temperature is ( $R$ is gas constant)

1

10 {atm}^{- 1}

2

10 atm

3 10

4 1

Explanation:

For the reaction
$2 {SO}_{2} + O_{2} ⇌ 2 {SO}_{3}$
Given that,
$Δ G^{\circ} = - 690.9 R,$
$T = 300 K$
$Δ G^{\circ} = - RT \ln K$
$- 690.9 R = - R \times 300 \times \ln K$
$\ln K = \frac{690.9}{300} = 2.303$
$2.303 \log K = 2.303$
$\log K = 1$
$K = 10^{1} = 10$
Unit of $K = (atm)^{Δ n} = (atm)^{2 - 3}$
Unit of $K = (atm)^{- 1}$
$K = 10 {atm}^{- 1}$

WB-JEE-2015

ELECTROCHEMISTRY

276131 If for the cell reaction, $Zn + {Cu}^{2 +} ⇌ Cu + {Zn}^{2 +}$ Entropy change $Δ S^{\circ}$ is $96.5 J {mol}^{- 1} K^{- 1}$ , then temperature coefficient of the emf of a cell is

1

5 \times 10^{- 4} {VK}^{- 1}

2

1 \times 10^{- 3} {VK}^{- 1}

3

2 \times 10^{- 3} {VK}^{- 1}

4

9.65 \times 10^{- 4} {VK}^{- 1}

Explanation:

We know that
$Δ G = Δ H - nFT {(\frac{dE}{dT})}_{P}$
and from Gibbs free energy
$Δ G = Δ H - T Δ S$
from (1) and (2) we get
$\begin{aligned} {(\frac{dE}{dT})}_{P} = \frac{Δ S}{nF} \\ {(\frac{dE}{dT})}_{P} = \frac{96.5}{2 \times 96500} \\ = 5 \times 10^{- 4} {VK}^{- 1} \end{aligned}$

VITEEE-2013

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