QBManager

ELECTROCHEMISTRY

276231 For a galvanic cell ${Cr}^{3 +} / {Cr}^{3 +} / / {Cd}^{2} / Cd$ . calculated $Δ G^{0}$ for its all reaction will be $[E_{{Cr}^{3 +} / cr}^{0} = - 0.74 V, E_{{Cd}^{2} / Cd}^{0} = - 0.40 V]$

1

- 28.95 kJ / mol

2

- 125.4 kJ / mol

3

- 196.8 KJ / mol

4

- 87.6 kJ / mol

Explanation:

Given that,
$E_{cathode} = - 0.40$
$E_{anode} = - 0.74$
$E_{cell} = E_{cathode} - E_{anode}$
$\begin{aligned} = & - 0.40 - (- 0.74) \\ = & 0.34 \\ Δ G^{\circ} = - {nf}_{cell} \\ = & - 6 \times 96500 \times 0.34 \\ = & - 196869 J / {mol}_{mol} \\ = & - 196.869 kJ mol . \end{aligned}$

AP- EAPCET- 07-09-2021

ELECTROCHEMISTRY

276232 Among the following, number of metal/s which can be used as electrodes in the photoelectric cell is ——. (Integer answer)

1

Li

2

Na

3

Rb

4

Cs

Explanation:

Cesium is used as an electrode in photoelectric cells because its ionization potential is low. Due to which its valence shell electrons can be easily ejected out by the energy of light.
- Li is used to make electrochemical cells.
- Sodium is used to make a $Na / Pb$ alloy needed to make ${PbEt}_{4}$ and ${PbMe}_{4}$ .
- Liquid sodium metal is used as a coolant in fast breeder nuclear reactors.
Hence, Cs is only metal which used as electrodes in the photoelectric cell.

JEE Main 25.02.2021

ELECTROCHEMISTRY

276237 Which of the following acts as oxidising agent in hydrogen-oxygen fuel cell ?

1

H_{2}

2

O_{2}

3

KOH

4

C

Explanation:

A hydrogen-oxygen fuel cell is an electrochemical cell that uses a pair of redox reactions to turn the chemical energy of hydrogen which serves as a fuel, and oxygen as an oxidizing agent into electricity.
$\begin{aligned} 2 H_{2} + 4 {OH}^{-} ⟶ 4 H_{2} O + 4 e^{-} \\ \frac{O_{2} + 2 H_{2} O + 4 e^{-} ⟶ 4 {OH}^{-}}{2 H_{2} + O_{2} ⟶ 2 H_{2} O} \end{aligned}$

MHT CET-02.05.2019

ELECTROCHEMISTRY

276239 For a fuel cell, which statement is correct?

1 Hydrogen is oxidized to

H_{2} O

at cathode

2 Oxygen is reduced to

{OH}^{-}

at anode

3 Hydrogen is oxidized to

H_{2} O

at anode

4 Hydrogen is oxidized to

H^{+}

at anode

Explanation:

Galvanic cells which use energy for combustion of fuels like $H_{2}, {CH}_{4}, {CH}_{3} OH$ etc. as the source to produce electrical energy are called fuel cells. Fuel cell uses the reaction of $H_{2}$ and $O_{2}$ in gaseous state to form water.
At anode, hydrogen is oxidized to $H_{2} O$
(i) $2 H_{2} \to 4 H^{+} + 4 e^{-}$
(ii) $4 H^{+} + 4 {OH}^{-} \to 4 H_{2} O$
At cathode, oxygen is reduced to ${OH}^{-}$ ions
$O_{2} (g) + 2 H_{2} O (l) + 4 e^{-} \to 4 {OH}^{-} (aq$

CG PET -2017

ELECTROCHEMISTRY

276231 For a galvanic cell ${Cr}^{3 +} / {Cr}^{3 +} / / {Cd}^{2} / Cd$ . calculated $Δ G^{0}$ for its all reaction will be $[E_{{Cr}^{3 +} / cr}^{0} = - 0.74 V, E_{{Cd}^{2} / Cd}^{0} = - 0.40 V]$

1

- 28.95 kJ / mol

2

- 125.4 kJ / mol

3

- 196.8 KJ / mol

4

- 87.6 kJ / mol

Explanation:

Given that,
$E_{cathode} = - 0.40$
$E_{anode} = - 0.74$
$E_{cell} = E_{cathode} - E_{anode}$
$\begin{aligned} = & - 0.40 - (- 0.74) \\ = & 0.34 \\ Δ G^{\circ} = - {nf}_{cell} \\ = & - 6 \times 96500 \times 0.34 \\ = & - 196869 J / {mol}_{mol} \\ = & - 196.869 kJ mol . \end{aligned}$

AP- EAPCET- 07-09-2021

ELECTROCHEMISTRY

276232 Among the following, number of metal/s which can be used as electrodes in the photoelectric cell is ——. (Integer answer)

1

Li

2

Na

3

Rb

4

Cs

Explanation:

Cesium is used as an electrode in photoelectric cells because its ionization potential is low. Due to which its valence shell electrons can be easily ejected out by the energy of light.
- Li is used to make electrochemical cells.
- Sodium is used to make a $Na / Pb$ alloy needed to make ${PbEt}_{4}$ and ${PbMe}_{4}$ .
- Liquid sodium metal is used as a coolant in fast breeder nuclear reactors.
Hence, Cs is only metal which used as electrodes in the photoelectric cell.

JEE Main 25.02.2021

ELECTROCHEMISTRY

276237 Which of the following acts as oxidising agent in hydrogen-oxygen fuel cell ?

1

H_{2}

2

O_{2}

3

KOH

4

C

Explanation:

A hydrogen-oxygen fuel cell is an electrochemical cell that uses a pair of redox reactions to turn the chemical energy of hydrogen which serves as a fuel, and oxygen as an oxidizing agent into electricity.
$\begin{aligned} 2 H_{2} + 4 {OH}^{-} ⟶ 4 H_{2} O + 4 e^{-} \\ \frac{O_{2} + 2 H_{2} O + 4 e^{-} ⟶ 4 {OH}^{-}}{2 H_{2} + O_{2} ⟶ 2 H_{2} O} \end{aligned}$

MHT CET-02.05.2019

ELECTROCHEMISTRY

276239 For a fuel cell, which statement is correct?

1 Hydrogen is oxidized to

H_{2} O

at cathode

2 Oxygen is reduced to

{OH}^{-}

at anode

3 Hydrogen is oxidized to

H_{2} O

at anode

4 Hydrogen is oxidized to

H^{+}

at anode

Explanation:

Galvanic cells which use energy for combustion of fuels like $H_{2}, {CH}_{4}, {CH}_{3} OH$ etc. as the source to produce electrical energy are called fuel cells. Fuel cell uses the reaction of $H_{2}$ and $O_{2}$ in gaseous state to form water.
At anode, hydrogen is oxidized to $H_{2} O$
(i) $2 H_{2} \to 4 H^{+} + 4 e^{-}$
(ii) $4 H^{+} + 4 {OH}^{-} \to 4 H_{2} O$
At cathode, oxygen is reduced to ${OH}^{-}$ ions
$O_{2} (g) + 2 H_{2} O (l) + 4 e^{-} \to 4 {OH}^{-} (aq$

CG PET -2017

ELECTROCHEMISTRY

276231 For a galvanic cell ${Cr}^{3 +} / {Cr}^{3 +} / / {Cd}^{2} / Cd$ . calculated $Δ G^{0}$ for its all reaction will be $[E_{{Cr}^{3 +} / cr}^{0} = - 0.74 V, E_{{Cd}^{2} / Cd}^{0} = - 0.40 V]$

1

- 28.95 kJ / mol

2

- 125.4 kJ / mol

3

- 196.8 KJ / mol

4

- 87.6 kJ / mol

Explanation:

Given that,
$E_{cathode} = - 0.40$
$E_{anode} = - 0.74$
$E_{cell} = E_{cathode} - E_{anode}$
$\begin{aligned} = & - 0.40 - (- 0.74) \\ = & 0.34 \\ Δ G^{\circ} = - {nf}_{cell} \\ = & - 6 \times 96500 \times 0.34 \\ = & - 196869 J / {mol}_{mol} \\ = & - 196.869 kJ mol . \end{aligned}$

AP- EAPCET- 07-09-2021

ELECTROCHEMISTRY

276232 Among the following, number of metal/s which can be used as electrodes in the photoelectric cell is ——. (Integer answer)

1

Li

2

Na

3

Rb

4

Cs

Explanation:

Cesium is used as an electrode in photoelectric cells because its ionization potential is low. Due to which its valence shell electrons can be easily ejected out by the energy of light.
- Li is used to make electrochemical cells.
- Sodium is used to make a $Na / Pb$ alloy needed to make ${PbEt}_{4}$ and ${PbMe}_{4}$ .
- Liquid sodium metal is used as a coolant in fast breeder nuclear reactors.
Hence, Cs is only metal which used as electrodes in the photoelectric cell.

JEE Main 25.02.2021

ELECTROCHEMISTRY

276237 Which of the following acts as oxidising agent in hydrogen-oxygen fuel cell ?

1

H_{2}

2

O_{2}

3

KOH

4

C

Explanation:

A hydrogen-oxygen fuel cell is an electrochemical cell that uses a pair of redox reactions to turn the chemical energy of hydrogen which serves as a fuel, and oxygen as an oxidizing agent into electricity.
$\begin{aligned} 2 H_{2} + 4 {OH}^{-} ⟶ 4 H_{2} O + 4 e^{-} \\ \frac{O_{2} + 2 H_{2} O + 4 e^{-} ⟶ 4 {OH}^{-}}{2 H_{2} + O_{2} ⟶ 2 H_{2} O} \end{aligned}$

MHT CET-02.05.2019

ELECTROCHEMISTRY

276239 For a fuel cell, which statement is correct?

1 Hydrogen is oxidized to

H_{2} O

at cathode

2 Oxygen is reduced to

{OH}^{-}

at anode

3 Hydrogen is oxidized to

H_{2} O

at anode

4 Hydrogen is oxidized to

H^{+}

at anode

Explanation:

Galvanic cells which use energy for combustion of fuels like $H_{2}, {CH}_{4}, {CH}_{3} OH$ etc. as the source to produce electrical energy are called fuel cells. Fuel cell uses the reaction of $H_{2}$ and $O_{2}$ in gaseous state to form water.
At anode, hydrogen is oxidized to $H_{2} O$
(i) $2 H_{2} \to 4 H^{+} + 4 e^{-}$
(ii) $4 H^{+} + 4 {OH}^{-} \to 4 H_{2} O$
At cathode, oxygen is reduced to ${OH}^{-}$ ions
$O_{2} (g) + 2 H_{2} O (l) + 4 e^{-} \to 4 {OH}^{-} (aq$

CG PET -2017

ELECTROCHEMISTRY

276231 For a galvanic cell ${Cr}^{3 +} / {Cr}^{3 +} / / {Cd}^{2} / Cd$ . calculated $Δ G^{0}$ for its all reaction will be $[E_{{Cr}^{3 +} / cr}^{0} = - 0.74 V, E_{{Cd}^{2} / Cd}^{0} = - 0.40 V]$

1

- 28.95 kJ / mol

2

- 125.4 kJ / mol

3

- 196.8 KJ / mol

4

- 87.6 kJ / mol

Explanation:

Given that,
$E_{cathode} = - 0.40$
$E_{anode} = - 0.74$
$E_{cell} = E_{cathode} - E_{anode}$
$\begin{aligned} = & - 0.40 - (- 0.74) \\ = & 0.34 \\ Δ G^{\circ} = - {nf}_{cell} \\ = & - 6 \times 96500 \times 0.34 \\ = & - 196869 J / {mol}_{mol} \\ = & - 196.869 kJ mol . \end{aligned}$

AP- EAPCET- 07-09-2021

ELECTROCHEMISTRY

276232 Among the following, number of metal/s which can be used as electrodes in the photoelectric cell is ——. (Integer answer)

1

Li

2

Na

3

Rb

4

Cs

Explanation:

Cesium is used as an electrode in photoelectric cells because its ionization potential is low. Due to which its valence shell electrons can be easily ejected out by the energy of light.
- Li is used to make electrochemical cells.
- Sodium is used to make a $Na / Pb$ alloy needed to make ${PbEt}_{4}$ and ${PbMe}_{4}$ .
- Liquid sodium metal is used as a coolant in fast breeder nuclear reactors.
Hence, Cs is only metal which used as electrodes in the photoelectric cell.

JEE Main 25.02.2021

ELECTROCHEMISTRY

276237 Which of the following acts as oxidising agent in hydrogen-oxygen fuel cell ?

1

H_{2}

2

O_{2}

3

KOH

4

C

Explanation:

A hydrogen-oxygen fuel cell is an electrochemical cell that uses a pair of redox reactions to turn the chemical energy of hydrogen which serves as a fuel, and oxygen as an oxidizing agent into electricity.
$\begin{aligned} 2 H_{2} + 4 {OH}^{-} ⟶ 4 H_{2} O + 4 e^{-} \\ \frac{O_{2} + 2 H_{2} O + 4 e^{-} ⟶ 4 {OH}^{-}}{2 H_{2} + O_{2} ⟶ 2 H_{2} O} \end{aligned}$

MHT CET-02.05.2019

ELECTROCHEMISTRY

276239 For a fuel cell, which statement is correct?

1 Hydrogen is oxidized to

H_{2} O

at cathode

2 Oxygen is reduced to

{OH}^{-}

at anode

3 Hydrogen is oxidized to

H_{2} O

at anode

4 Hydrogen is oxidized to

H^{+}

at anode

Explanation:

Galvanic cells which use energy for combustion of fuels like $H_{2}, {CH}_{4}, {CH}_{3} OH$ etc. as the source to produce electrical energy are called fuel cells. Fuel cell uses the reaction of $H_{2}$ and $O_{2}$ in gaseous state to form water.
At anode, hydrogen is oxidized to $H_{2} O$
(i) $2 H_{2} \to 4 H^{+} + 4 e^{-}$
(ii) $4 H^{+} + 4 {OH}^{-} \to 4 H_{2} O$
At cathode, oxygen is reduced to ${OH}^{-}$ ions
$O_{2} (g) + 2 H_{2} O (l) + 4 e^{-} \to 4 {OH}^{-} (aq$

CG PET -2017