QBManager

ELECTROCHEMISTRY

276265 In electrolysis of $NaCl$ when $Pt$ electrode is taken then $H_{2}$ is liberated at cathode while with Hg electrode if forms sodium amalgam

1

Hg

is more inert that

Pt

2 more voltage is required to reduce

H^{+}

at

Hg

than at

Pt

3

Na

is dissolved in

Hg

while it does not dissolved in

Pt

4 concentration of

H^{+}

ions is larger when

Pt

electrode is taken

Explanation:

Sodium chloride in water dissociates as
$NaCl \to {Na}^{+} + {Cl}^{-}$
$H_{2} O \to H^{+} + {OH}^{-}$
When electric current is passed through this solution using platinum electrodes ${Na}^{+}$ and $H^{+}$ move towards cathode.
More voltage is required to reduce $H^{+}$ at $Hg$ than at $Pt$ .

CG PET -2007

ELECTROCHEMISTRY

276266 On the basis of the information available from the reaction $\frac{4}{3} Al + O_{2} \to \frac{2}{3} {Al}_{2} O_{3}, Δ G = - 827 kJ$ ${mol}^{- 1}$ of $O_{2}$ , the minimum emf required to carry out an electrolysis of ${Al}_{2} O_{3}$ is $(F = 96500 C$ ${mol}^{- 1}$ )

1

2.14 V

2

4.28 V

3

6.42 V

4

8.56 V

Explanation:

Given that,
$Δ G = 827 kJ {mol}^{- 1}$
and $\frac{4}{3} Al + O_{3} \to \frac{2}{3} {Al}_{2} O_{3}$
For $O_{2} Δ G = - {nFE}^{\circ} n = 4$
$- 827 \times 10^{3} J = - 4 \times E^{\circ} \times 96500$
$E = \frac{Δ G}{nF} = \frac{827 \times 10^{3}}{4 \times 96500} = 4.28$
$E = \frac{4.28}{2} = 2.14 V$

CG PET -2007

ELECTROCHEMISTRY

276267 The standard emf of a galvanic cell involving cell reaction with $n = 2$ is found to be $0.295 V$ at $25^{\circ} C$ . The equilibrium constant of the reaction would be

1

2.0 \times 10^{11}

2

4.0 \times 10^{12}

3

1.0 \times 10^{2}

4

1.0 \times 10^{10}

(Given,

F = 96500 C {mol}^{- 1}

R = 8.314 {JK}^{- 1} {mol}^{- 1}

)

Explanation:

Given that,
$E_{cell}^{\circ} = 0.295 V$
$R = 8.314 {Jk}^{- 1}$
$T = 25^{\circ} C + 273 = 298 k$
$F = 96500 C {mol}^{- 1}$ and $n = 2$
By Nernst equation,
$E_{cell} = E_{cell}^{\circ} - \frac{2.303 RT}{nF} \log_{10} k_{eq}$
At equilibrium $E_{cell} = 0$
$\begin{aligned} E_{cell}^{o} - \frac{2.303 RT}{nF} \log k_{eq} \\ 0.295 = \frac{2.303 \times 8.314 \times 298}{2 \times 96500} \log k_{eq} \\ \log k_{eq} = \frac{2 \times 96500 \times 0.295}{2.303 \times 8.314 \times 298} = 9.97 \approx 10 \\ K_{eq} = antilog 10 = 1 \times 10^{10} \end{aligned}$

CG PET -2007

ELECTROCHEMISTRY

276259 In a electrochemical cell, the reaction will be feasible when

1

Δ G = - ve, E = + ve

2

Δ G = + ve, E = - ve

3

Δ G = 0, E = - ve

4

Δ G = 0, E = 0

Explanation:

Exp:For a spontaneous reaction, $Δ G$ should be $- ve$ which is possible only when $E$ is +ve as $Δ G = - nFE$ . then $Δ G = - ve$ and $E = + ve$ .

COMEDK 2014

ELECTROCHEMISTRY

276265 In electrolysis of $NaCl$ when $Pt$ electrode is taken then $H_{2}$ is liberated at cathode while with Hg electrode if forms sodium amalgam

1

Hg

is more inert that

Pt

2 more voltage is required to reduce

H^{+}

at

Hg

than at

Pt

3

Na

is dissolved in

Hg

while it does not dissolved in

Pt

4 concentration of

H^{+}

ions is larger when

Pt

electrode is taken

Explanation:

Sodium chloride in water dissociates as
$NaCl \to {Na}^{+} + {Cl}^{-}$
$H_{2} O \to H^{+} + {OH}^{-}$
When electric current is passed through this solution using platinum electrodes ${Na}^{+}$ and $H^{+}$ move towards cathode.
More voltage is required to reduce $H^{+}$ at $Hg$ than at $Pt$ .

CG PET -2007

ELECTROCHEMISTRY

276266 On the basis of the information available from the reaction $\frac{4}{3} Al + O_{2} \to \frac{2}{3} {Al}_{2} O_{3}, Δ G = - 827 kJ$ ${mol}^{- 1}$ of $O_{2}$ , the minimum emf required to carry out an electrolysis of ${Al}_{2} O_{3}$ is $(F = 96500 C$ ${mol}^{- 1}$ )

1

2.14 V

2

4.28 V

3

6.42 V

4

8.56 V

Explanation:

Given that,
$Δ G = 827 kJ {mol}^{- 1}$
and $\frac{4}{3} Al + O_{3} \to \frac{2}{3} {Al}_{2} O_{3}$
For $O_{2} Δ G = - {nFE}^{\circ} n = 4$
$- 827 \times 10^{3} J = - 4 \times E^{\circ} \times 96500$
$E = \frac{Δ G}{nF} = \frac{827 \times 10^{3}}{4 \times 96500} = 4.28$
$E = \frac{4.28}{2} = 2.14 V$

CG PET -2007

ELECTROCHEMISTRY

276267 The standard emf of a galvanic cell involving cell reaction with $n = 2$ is found to be $0.295 V$ at $25^{\circ} C$ . The equilibrium constant of the reaction would be

1

2.0 \times 10^{11}

2

4.0 \times 10^{12}

3

1.0 \times 10^{2}

4

1.0 \times 10^{10}

(Given,

F = 96500 C {mol}^{- 1}

R = 8.314 {JK}^{- 1} {mol}^{- 1}

)

Explanation:

Given that,
$E_{cell}^{\circ} = 0.295 V$
$R = 8.314 {Jk}^{- 1}$
$T = 25^{\circ} C + 273 = 298 k$
$F = 96500 C {mol}^{- 1}$ and $n = 2$
By Nernst equation,
$E_{cell} = E_{cell}^{\circ} - \frac{2.303 RT}{nF} \log_{10} k_{eq}$
At equilibrium $E_{cell} = 0$
$\begin{aligned} E_{cell}^{o} - \frac{2.303 RT}{nF} \log k_{eq} \\ 0.295 = \frac{2.303 \times 8.314 \times 298}{2 \times 96500} \log k_{eq} \\ \log k_{eq} = \frac{2 \times 96500 \times 0.295}{2.303 \times 8.314 \times 298} = 9.97 \approx 10 \\ K_{eq} = antilog 10 = 1 \times 10^{10} \end{aligned}$

CG PET -2007

ELECTROCHEMISTRY

276259 In a electrochemical cell, the reaction will be feasible when

1

Δ G = - ve, E = + ve

2

Δ G = + ve, E = - ve

3

Δ G = 0, E = - ve

4

Δ G = 0, E = 0

Explanation:

Exp:For a spontaneous reaction, $Δ G$ should be $- ve$ which is possible only when $E$ is +ve as $Δ G = - nFE$ . then $Δ G = - ve$ and $E = + ve$ .

COMEDK 2014

ELECTROCHEMISTRY

276265 In electrolysis of $NaCl$ when $Pt$ electrode is taken then $H_{2}$ is liberated at cathode while with Hg electrode if forms sodium amalgam

1

Hg

is more inert that

Pt

2 more voltage is required to reduce

H^{+}

at

Hg

than at

Pt

3

Na

is dissolved in

Hg

while it does not dissolved in

Pt

4 concentration of

H^{+}

ions is larger when

Pt

electrode is taken

Explanation:

Sodium chloride in water dissociates as
$NaCl \to {Na}^{+} + {Cl}^{-}$
$H_{2} O \to H^{+} + {OH}^{-}$
When electric current is passed through this solution using platinum electrodes ${Na}^{+}$ and $H^{+}$ move towards cathode.
More voltage is required to reduce $H^{+}$ at $Hg$ than at $Pt$ .

CG PET -2007

ELECTROCHEMISTRY

276266 On the basis of the information available from the reaction $\frac{4}{3} Al + O_{2} \to \frac{2}{3} {Al}_{2} O_{3}, Δ G = - 827 kJ$ ${mol}^{- 1}$ of $O_{2}$ , the minimum emf required to carry out an electrolysis of ${Al}_{2} O_{3}$ is $(F = 96500 C$ ${mol}^{- 1}$ )

1

2.14 V

2

4.28 V

3

6.42 V

4

8.56 V

Explanation:

Given that,
$Δ G = 827 kJ {mol}^{- 1}$
and $\frac{4}{3} Al + O_{3} \to \frac{2}{3} {Al}_{2} O_{3}$
For $O_{2} Δ G = - {nFE}^{\circ} n = 4$
$- 827 \times 10^{3} J = - 4 \times E^{\circ} \times 96500$
$E = \frac{Δ G}{nF} = \frac{827 \times 10^{3}}{4 \times 96500} = 4.28$
$E = \frac{4.28}{2} = 2.14 V$

CG PET -2007

ELECTROCHEMISTRY

276267 The standard emf of a galvanic cell involving cell reaction with $n = 2$ is found to be $0.295 V$ at $25^{\circ} C$ . The equilibrium constant of the reaction would be

1

2.0 \times 10^{11}

2

4.0 \times 10^{12}

3

1.0 \times 10^{2}

4

1.0 \times 10^{10}

(Given,

F = 96500 C {mol}^{- 1}

R = 8.314 {JK}^{- 1} {mol}^{- 1}

)

Explanation:

Given that,
$E_{cell}^{\circ} = 0.295 V$
$R = 8.314 {Jk}^{- 1}$
$T = 25^{\circ} C + 273 = 298 k$
$F = 96500 C {mol}^{- 1}$ and $n = 2$
By Nernst equation,
$E_{cell} = E_{cell}^{\circ} - \frac{2.303 RT}{nF} \log_{10} k_{eq}$
At equilibrium $E_{cell} = 0$
$\begin{aligned} E_{cell}^{o} - \frac{2.303 RT}{nF} \log k_{eq} \\ 0.295 = \frac{2.303 \times 8.314 \times 298}{2 \times 96500} \log k_{eq} \\ \log k_{eq} = \frac{2 \times 96500 \times 0.295}{2.303 \times 8.314 \times 298} = 9.97 \approx 10 \\ K_{eq} = antilog 10 = 1 \times 10^{10} \end{aligned}$

CG PET -2007

ELECTROCHEMISTRY

276259 In a electrochemical cell, the reaction will be feasible when

1

Δ G = - ve, E = + ve

2

Δ G = + ve, E = - ve

3

Δ G = 0, E = - ve

4

Δ G = 0, E = 0

Explanation:

Exp:For a spontaneous reaction, $Δ G$ should be $- ve$ which is possible only when $E$ is +ve as $Δ G = - nFE$ . then $Δ G = - ve$ and $E = + ve$ .

COMEDK 2014

ELECTROCHEMISTRY

276265 In electrolysis of $NaCl$ when $Pt$ electrode is taken then $H_{2}$ is liberated at cathode while with Hg electrode if forms sodium amalgam

1

Hg

is more inert that

Pt

2 more voltage is required to reduce

H^{+}

at

Hg

than at

Pt

3

Na

is dissolved in

Hg

while it does not dissolved in

Pt

4 concentration of

H^{+}

ions is larger when

Pt

electrode is taken

Explanation:

Sodium chloride in water dissociates as
$NaCl \to {Na}^{+} + {Cl}^{-}$
$H_{2} O \to H^{+} + {OH}^{-}$
When electric current is passed through this solution using platinum electrodes ${Na}^{+}$ and $H^{+}$ move towards cathode.
More voltage is required to reduce $H^{+}$ at $Hg$ than at $Pt$ .

CG PET -2007

ELECTROCHEMISTRY

276266 On the basis of the information available from the reaction $\frac{4}{3} Al + O_{2} \to \frac{2}{3} {Al}_{2} O_{3}, Δ G = - 827 kJ$ ${mol}^{- 1}$ of $O_{2}$ , the minimum emf required to carry out an electrolysis of ${Al}_{2} O_{3}$ is $(F = 96500 C$ ${mol}^{- 1}$ )

1

2.14 V

2

4.28 V

3

6.42 V

4

8.56 V

Explanation:

Given that,
$Δ G = 827 kJ {mol}^{- 1}$
and $\frac{4}{3} Al + O_{3} \to \frac{2}{3} {Al}_{2} O_{3}$
For $O_{2} Δ G = - {nFE}^{\circ} n = 4$
$- 827 \times 10^{3} J = - 4 \times E^{\circ} \times 96500$
$E = \frac{Δ G}{nF} = \frac{827 \times 10^{3}}{4 \times 96500} = 4.28$
$E = \frac{4.28}{2} = 2.14 V$

CG PET -2007

ELECTROCHEMISTRY

276267 The standard emf of a galvanic cell involving cell reaction with $n = 2$ is found to be $0.295 V$ at $25^{\circ} C$ . The equilibrium constant of the reaction would be

1

2.0 \times 10^{11}

2

4.0 \times 10^{12}

3

1.0 \times 10^{2}

4

1.0 \times 10^{10}

(Given,

F = 96500 C {mol}^{- 1}

R = 8.314 {JK}^{- 1} {mol}^{- 1}

)

Explanation:

Given that,
$E_{cell}^{\circ} = 0.295 V$
$R = 8.314 {Jk}^{- 1}$
$T = 25^{\circ} C + 273 = 298 k$
$F = 96500 C {mol}^{- 1}$ and $n = 2$
By Nernst equation,
$E_{cell} = E_{cell}^{\circ} - \frac{2.303 RT}{nF} \log_{10} k_{eq}$
At equilibrium $E_{cell} = 0$
$\begin{aligned} E_{cell}^{o} - \frac{2.303 RT}{nF} \log k_{eq} \\ 0.295 = \frac{2.303 \times 8.314 \times 298}{2 \times 96500} \log k_{eq} \\ \log k_{eq} = \frac{2 \times 96500 \times 0.295}{2.303 \times 8.314 \times 298} = 9.97 \approx 10 \\ K_{eq} = antilog 10 = 1 \times 10^{10} \end{aligned}$

CG PET -2007

ELECTROCHEMISTRY

276259 In a electrochemical cell, the reaction will be feasible when

1

Δ G = - ve, E = + ve

2

Δ G = + ve, E = - ve

3

Δ G = 0, E = - ve

4

Δ G = 0, E = 0

Explanation:

Exp:For a spontaneous reaction, $Δ G$ should be $- ve$ which is possible only when $E$ is +ve as $Δ G = - nFE$ . then $Δ G = - ve$ and $E = + ve$ .

COMEDK 2014

Question Bank Series

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