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CHXII03:ELECTROCHEMISTRY

330164 How many Faraday of electricity is required to deposit $0.8 g$ of calcium at cathode by the electrolysis of ${CaCl}_{2}$ ?
(Molar mass of $Ca = 40 g {mol}^{- 1}$ )

1

4 F

2

0.04 F

3

2.5 F

4

2 F

Explanation:

${Ca}^{2 +} + 2 e^{-} \to Ca$
$40 g$ of $Ca$ requires $2 F$ of electricity
$∴ 0.8 g$ of $Ca$ requires $= \frac{2 F \times 0.8 g}{40 g} = 0.04 F$ of electricity.

MHTCET - 2021

CHXII03:ELECTROCHEMISTRY

330165 When during electrolysis of a solution of $A g N O_{3}, 9650$ coulombs of charge pass through the electroplating bath, the mass of silver deposited on the cathode will be:

1 108 g

2 10.8 g

3 1.08 g

4 21.6 g

Explanation:

$w = \frac{E . Q .}{96500} = \frac{108 \times 9650}{96500} = 10 .8 g$

CHXII03:ELECTROCHEMISTRY

330166 A current of $96.5 A$ is passed for $18 \min$ between nickel electrodes in $500 mL$ solution of $2 M Ni {(N O_{3})}_{2}$ . The molarity of solution after electrolysis would be

1

0.46 M

2

0.92 M

3

0.625 M

4

1.25 M

Explanation:

Moles of $Ni {({NO}_{3})}_{2}$ in $500 mL$ of
$2 M Ni {({NO}_{3})}_{2}$ is $\frac{2 \times 500}{1000} = 1 mol$
Charge $=$ it $= 96.5 \times 18 \times 60 = 104220 C$
${Ni}^{2 +} + 2 e^{-} \to Ni$
$∵ 2 \times 96500 C$ deposits 1 mole of $Ni {(N O_{3})}_{2}$
$∴ 104220 C$ will deposit $= \frac{104220}{2 \times 96500} = 0.54 mol$
So, moles of $Ni {({NO}_{3})}_{2}$ left
$= 1.0 - 0.54 = 0.46 mol$
Thus, molarity of $Ni {({NO}_{3})}_{2}$ , solution
$= 0.46 \times \frac{1000}{500} = 0.92 mol / L$

AIIMS - 2007

CHXII03:ELECTROCHEMISTRY

330167 In the electrolysis of acidulated water, it is desired to obtain 1.12 cc of hydrogen per second under S.T.P condition. The current to be passed is

1 9.65 A

2 19.3 A

3 0.965 A

4 1.93 A

Explanation:

No. of moles of $H_{2} = \frac{1 .12}{22400}$
$∴$ No. of equivalents of hydrogen
$= \frac{1 .12 \times 2}{22400} = 1 0^{- 4}$
No. of Faradays required $= 1 0^{- 4}$
$∴$ Current to be passed in 1 sec. $= 96500 \times 1 0^{- 4}$
= 9.65 A

KCET - 2009

CHXII03:ELECTROCHEMISTRY

330168 The number of Faradays needed to reduce $4 g$ equivalents of ${Cu}^{2 +}$ to $Cu$ metal will be

1 1

2 2

3

1 / 2

4 4

Explanation:

$\begin{matrix} C u^{2 +} & + & 2 e^{-} & \to & Cu \\ 1 mol & 2 F & 1 mol \\ \frac{1}{2} mol & 1 F & \frac{1}{2} mol \\ 1 g equiv . & 1 F & 1 g equiv . \end{matrix}$
Thus, to reduce $4 g$ equivalents of $C u^{2 +}$ into $C u$ four Faradays are required.

CHXII03:ELECTROCHEMISTRY

330164 How many Faraday of electricity is required to deposit $0.8 g$ of calcium at cathode by the electrolysis of ${CaCl}_{2}$ ?
(Molar mass of $Ca = 40 g {mol}^{- 1}$ )

1

4 F

2

0.04 F

3

2.5 F

4

2 F

Explanation:

${Ca}^{2 +} + 2 e^{-} \to Ca$
$40 g$ of $Ca$ requires $2 F$ of electricity
$∴ 0.8 g$ of $Ca$ requires $= \frac{2 F \times 0.8 g}{40 g} = 0.04 F$ of electricity.

MHTCET - 2021

CHXII03:ELECTROCHEMISTRY

330165 When during electrolysis of a solution of $A g N O_{3}, 9650$ coulombs of charge pass through the electroplating bath, the mass of silver deposited on the cathode will be:

1 108 g

2 10.8 g

3 1.08 g

4 21.6 g

Explanation:

$w = \frac{E . Q .}{96500} = \frac{108 \times 9650}{96500} = 10 .8 g$

CHXII03:ELECTROCHEMISTRY

330166 A current of $96.5 A$ is passed for $18 \min$ between nickel electrodes in $500 mL$ solution of $2 M Ni {(N O_{3})}_{2}$ . The molarity of solution after electrolysis would be

1

0.46 M

2

0.92 M

3

0.625 M

4

1.25 M

Explanation:

Moles of $Ni {({NO}_{3})}_{2}$ in $500 mL$ of
$2 M Ni {({NO}_{3})}_{2}$ is $\frac{2 \times 500}{1000} = 1 mol$
Charge $=$ it $= 96.5 \times 18 \times 60 = 104220 C$
${Ni}^{2 +} + 2 e^{-} \to Ni$
$∵ 2 \times 96500 C$ deposits 1 mole of $Ni {(N O_{3})}_{2}$
$∴ 104220 C$ will deposit $= \frac{104220}{2 \times 96500} = 0.54 mol$
So, moles of $Ni {({NO}_{3})}_{2}$ left
$= 1.0 - 0.54 = 0.46 mol$
Thus, molarity of $Ni {({NO}_{3})}_{2}$ , solution
$= 0.46 \times \frac{1000}{500} = 0.92 mol / L$

AIIMS - 2007

CHXII03:ELECTROCHEMISTRY

330167 In the electrolysis of acidulated water, it is desired to obtain 1.12 cc of hydrogen per second under S.T.P condition. The current to be passed is

1 9.65 A

2 19.3 A

3 0.965 A

4 1.93 A

Explanation:

No. of moles of $H_{2} = \frac{1 .12}{22400}$
$∴$ No. of equivalents of hydrogen
$= \frac{1 .12 \times 2}{22400} = 1 0^{- 4}$
No. of Faradays required $= 1 0^{- 4}$
$∴$ Current to be passed in 1 sec. $= 96500 \times 1 0^{- 4}$
= 9.65 A

KCET - 2009

CHXII03:ELECTROCHEMISTRY

330168 The number of Faradays needed to reduce $4 g$ equivalents of ${Cu}^{2 +}$ to $Cu$ metal will be

1 1

2 2

3

1 / 2

4 4

Explanation:

$\begin{matrix} C u^{2 +} & + & 2 e^{-} & \to & Cu \\ 1 mol & 2 F & 1 mol \\ \frac{1}{2} mol & 1 F & \frac{1}{2} mol \\ 1 g equiv . & 1 F & 1 g equiv . \end{matrix}$
Thus, to reduce $4 g$ equivalents of $C u^{2 +}$ into $C u$ four Faradays are required.

CHXII03:ELECTROCHEMISTRY

330164 How many Faraday of electricity is required to deposit $0.8 g$ of calcium at cathode by the electrolysis of ${CaCl}_{2}$ ?
(Molar mass of $Ca = 40 g {mol}^{- 1}$ )

1

4 F

2

0.04 F

3

2.5 F

4

2 F

Explanation:

${Ca}^{2 +} + 2 e^{-} \to Ca$
$40 g$ of $Ca$ requires $2 F$ of electricity
$∴ 0.8 g$ of $Ca$ requires $= \frac{2 F \times 0.8 g}{40 g} = 0.04 F$ of electricity.

MHTCET - 2021

CHXII03:ELECTROCHEMISTRY

330165 When during electrolysis of a solution of $A g N O_{3}, 9650$ coulombs of charge pass through the electroplating bath, the mass of silver deposited on the cathode will be:

1 108 g

2 10.8 g

3 1.08 g

4 21.6 g

Explanation:

$w = \frac{E . Q .}{96500} = \frac{108 \times 9650}{96500} = 10 .8 g$

CHXII03:ELECTROCHEMISTRY

330166 A current of $96.5 A$ is passed for $18 \min$ between nickel electrodes in $500 mL$ solution of $2 M Ni {(N O_{3})}_{2}$ . The molarity of solution after electrolysis would be

1

0.46 M

2

0.92 M

3

0.625 M

4

1.25 M

Explanation:

Moles of $Ni {({NO}_{3})}_{2}$ in $500 mL$ of
$2 M Ni {({NO}_{3})}_{2}$ is $\frac{2 \times 500}{1000} = 1 mol$
Charge $=$ it $= 96.5 \times 18 \times 60 = 104220 C$
${Ni}^{2 +} + 2 e^{-} \to Ni$
$∵ 2 \times 96500 C$ deposits 1 mole of $Ni {(N O_{3})}_{2}$
$∴ 104220 C$ will deposit $= \frac{104220}{2 \times 96500} = 0.54 mol$
So, moles of $Ni {({NO}_{3})}_{2}$ left
$= 1.0 - 0.54 = 0.46 mol$
Thus, molarity of $Ni {({NO}_{3})}_{2}$ , solution
$= 0.46 \times \frac{1000}{500} = 0.92 mol / L$

AIIMS - 2007

CHXII03:ELECTROCHEMISTRY

330167 In the electrolysis of acidulated water, it is desired to obtain 1.12 cc of hydrogen per second under S.T.P condition. The current to be passed is

1 9.65 A

2 19.3 A

3 0.965 A

4 1.93 A

Explanation:

No. of moles of $H_{2} = \frac{1 .12}{22400}$
$∴$ No. of equivalents of hydrogen
$= \frac{1 .12 \times 2}{22400} = 1 0^{- 4}$
No. of Faradays required $= 1 0^{- 4}$
$∴$ Current to be passed in 1 sec. $= 96500 \times 1 0^{- 4}$
= 9.65 A

KCET - 2009

CHXII03:ELECTROCHEMISTRY

330168 The number of Faradays needed to reduce $4 g$ equivalents of ${Cu}^{2 +}$ to $Cu$ metal will be

1 1

2 2

3

1 / 2

4 4

Explanation:

$\begin{matrix} C u^{2 +} & + & 2 e^{-} & \to & Cu \\ 1 mol & 2 F & 1 mol \\ \frac{1}{2} mol & 1 F & \frac{1}{2} mol \\ 1 g equiv . & 1 F & 1 g equiv . \end{matrix}$
Thus, to reduce $4 g$ equivalents of $C u^{2 +}$ into $C u$ four Faradays are required.

CHXII03:ELECTROCHEMISTRY

330164 How many Faraday of electricity is required to deposit $0.8 g$ of calcium at cathode by the electrolysis of ${CaCl}_{2}$ ?
(Molar mass of $Ca = 40 g {mol}^{- 1}$ )

1

4 F

2

0.04 F

3

2.5 F

4

2 F

Explanation:

${Ca}^{2 +} + 2 e^{-} \to Ca$
$40 g$ of $Ca$ requires $2 F$ of electricity
$∴ 0.8 g$ of $Ca$ requires $= \frac{2 F \times 0.8 g}{40 g} = 0.04 F$ of electricity.

MHTCET - 2021

CHXII03:ELECTROCHEMISTRY

330165 When during electrolysis of a solution of $A g N O_{3}, 9650$ coulombs of charge pass through the electroplating bath, the mass of silver deposited on the cathode will be:

1 108 g

2 10.8 g

3 1.08 g

4 21.6 g

Explanation:

$w = \frac{E . Q .}{96500} = \frac{108 \times 9650}{96500} = 10 .8 g$

CHXII03:ELECTROCHEMISTRY

330166 A current of $96.5 A$ is passed for $18 \min$ between nickel electrodes in $500 mL$ solution of $2 M Ni {(N O_{3})}_{2}$ . The molarity of solution after electrolysis would be

1

0.46 M

2

0.92 M

3

0.625 M

4

1.25 M

Explanation:

Moles of $Ni {({NO}_{3})}_{2}$ in $500 mL$ of
$2 M Ni {({NO}_{3})}_{2}$ is $\frac{2 \times 500}{1000} = 1 mol$
Charge $=$ it $= 96.5 \times 18 \times 60 = 104220 C$
${Ni}^{2 +} + 2 e^{-} \to Ni$
$∵ 2 \times 96500 C$ deposits 1 mole of $Ni {(N O_{3})}_{2}$
$∴ 104220 C$ will deposit $= \frac{104220}{2 \times 96500} = 0.54 mol$
So, moles of $Ni {({NO}_{3})}_{2}$ left
$= 1.0 - 0.54 = 0.46 mol$
Thus, molarity of $Ni {({NO}_{3})}_{2}$ , solution
$= 0.46 \times \frac{1000}{500} = 0.92 mol / L$

AIIMS - 2007

CHXII03:ELECTROCHEMISTRY

330167 In the electrolysis of acidulated water, it is desired to obtain 1.12 cc of hydrogen per second under S.T.P condition. The current to be passed is

1 9.65 A

2 19.3 A

3 0.965 A

4 1.93 A

Explanation:

No. of moles of $H_{2} = \frac{1 .12}{22400}$
$∴$ No. of equivalents of hydrogen
$= \frac{1 .12 \times 2}{22400} = 1 0^{- 4}$
No. of Faradays required $= 1 0^{- 4}$
$∴$ Current to be passed in 1 sec. $= 96500 \times 1 0^{- 4}$
= 9.65 A

KCET - 2009

CHXII03:ELECTROCHEMISTRY

330168 The number of Faradays needed to reduce $4 g$ equivalents of ${Cu}^{2 +}$ to $Cu$ metal will be

1 1

2 2

3

1 / 2

4 4

Explanation:

$\begin{matrix} C u^{2 +} & + & 2 e^{-} & \to & Cu \\ 1 mol & 2 F & 1 mol \\ \frac{1}{2} mol & 1 F & \frac{1}{2} mol \\ 1 g equiv . & 1 F & 1 g equiv . \end{matrix}$
Thus, to reduce $4 g$ equivalents of $C u^{2 +}$ into $C u$ four Faradays are required.

CHXII03:ELECTROCHEMISTRY

330164 How many Faraday of electricity is required to deposit $0.8 g$ of calcium at cathode by the electrolysis of ${CaCl}_{2}$ ?
(Molar mass of $Ca = 40 g {mol}^{- 1}$ )

1

4 F

2

0.04 F

3

2.5 F

4

2 F

Explanation:

${Ca}^{2 +} + 2 e^{-} \to Ca$
$40 g$ of $Ca$ requires $2 F$ of electricity
$∴ 0.8 g$ of $Ca$ requires $= \frac{2 F \times 0.8 g}{40 g} = 0.04 F$ of electricity.

MHTCET - 2021

CHXII03:ELECTROCHEMISTRY

330165 When during electrolysis of a solution of $A g N O_{3}, 9650$ coulombs of charge pass through the electroplating bath, the mass of silver deposited on the cathode will be:

1 108 g

2 10.8 g

3 1.08 g

4 21.6 g

Explanation:

$w = \frac{E . Q .}{96500} = \frac{108 \times 9650}{96500} = 10 .8 g$

CHXII03:ELECTROCHEMISTRY

330166 A current of $96.5 A$ is passed for $18 \min$ between nickel electrodes in $500 mL$ solution of $2 M Ni {(N O_{3})}_{2}$ . The molarity of solution after electrolysis would be

1

0.46 M

2

0.92 M

3

0.625 M

4

1.25 M

Explanation:

Moles of $Ni {({NO}_{3})}_{2}$ in $500 mL$ of
$2 M Ni {({NO}_{3})}_{2}$ is $\frac{2 \times 500}{1000} = 1 mol$
Charge $=$ it $= 96.5 \times 18 \times 60 = 104220 C$
${Ni}^{2 +} + 2 e^{-} \to Ni$
$∵ 2 \times 96500 C$ deposits 1 mole of $Ni {(N O_{3})}_{2}$
$∴ 104220 C$ will deposit $= \frac{104220}{2 \times 96500} = 0.54 mol$
So, moles of $Ni {({NO}_{3})}_{2}$ left
$= 1.0 - 0.54 = 0.46 mol$
Thus, molarity of $Ni {({NO}_{3})}_{2}$ , solution
$= 0.46 \times \frac{1000}{500} = 0.92 mol / L$

AIIMS - 2007

CHXII03:ELECTROCHEMISTRY

330167 In the electrolysis of acidulated water, it is desired to obtain 1.12 cc of hydrogen per second under S.T.P condition. The current to be passed is

1 9.65 A

2 19.3 A

3 0.965 A

4 1.93 A

Explanation:

No. of moles of $H_{2} = \frac{1 .12}{22400}$
$∴$ No. of equivalents of hydrogen
$= \frac{1 .12 \times 2}{22400} = 1 0^{- 4}$
No. of Faradays required $= 1 0^{- 4}$
$∴$ Current to be passed in 1 sec. $= 96500 \times 1 0^{- 4}$
= 9.65 A

KCET - 2009

CHXII03:ELECTROCHEMISTRY

330168 The number of Faradays needed to reduce $4 g$ equivalents of ${Cu}^{2 +}$ to $Cu$ metal will be

1 1

2 2

3

1 / 2

4 4

Explanation:

$\begin{matrix} C u^{2 +} & + & 2 e^{-} & \to & Cu \\ 1 mol & 2 F & 1 mol \\ \frac{1}{2} mol & 1 F & \frac{1}{2} mol \\ 1 g equiv . & 1 F & 1 g equiv . \end{matrix}$
Thus, to reduce $4 g$ equivalents of $C u^{2 +}$ into $C u$ four Faradays are required.

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