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ELECTROCHEMISTRY

20057 The acid used in lead storage battery is

1

H_{2} S O_{4}

2

H_{3} P O_{4}

3

H C l

4

H N O_{3}

Explanation:

(a)Dil. $H_{2} S O_{4}$ is used in lead in lead storage battery as electrolyte.

ELECTROCHEMISTRY

19825 Two electrolytic cells containing $C u S O_{4}$ and $A g N O_{3}$ respectively are connected in series and a current is passed through them until $1 m g$ of copper is deposited in the first cell. The amount of silver deposited in the second cell during this time is approximately ......... $mg$ [Atomic weights of copper and silver are respectively $63.57$ and $107.88$ ]

1

1.7

2

3.4

3

5.1

4

6.8

Explanation:

(b) $\frac{m_{1}}{m_{2}} = \frac{E_{1}}{E_{2}}$ (By faraday law for same current and time)
Where $E_{1}$ and $E_{2}$ are the chemical equivalents and $m_{1}$ and $m_{2}$ are the masses of copper and silver respectively.
$E = \frac{A t o m i c w e i g h t}{V a l e n c y}$ .
$E_{1} = \frac{63.57}{2} = 31.79$ and $E_{2} = \frac{107.88}{1} = 107.88$
$\frac{1 m g}{m_{2}} = \frac{31.79}{107.88}$
$\Rightarrow$ $m_{2} = \frac{107.88}{31.79} m g = 3.4 m g$

ELECTROCHEMISTRY

19826 A current $I$ is passed for a time $t$ through a number of voltameters. If m is the mass of a substance deposited on an electrode and $z$ is its electrochemical equivalent, then

1

\frac{z I t}{m} = c o n s t a n t

2

\frac{z}{m I t} = c o n s t a n t

3

\frac{I}{z m t} = c o n s t a n t

4

\frac{I t}{z m} = c o n s t a n t

Explanation:

(a) $m = Z i t \Rightarrow \frac{m}{Z i t} = 1 (c o n s t a n t)$

ELECTROCHEMISTRY

19827 For electroplating a spoon, it is placed in the voltameter at

1 The position of anode

2 The position of cathode

3 Exactly in the middle of anode and the cathode

4 Anywhere in the electrolyte

ELECTROCHEMISTRY

19828 If nearly $10^{5}$ coulomb liberate $1 g m$ equivalent of aluminium, then the amount of aluminium (equivalent weight $9$ ) deposited through electrolysis in $20$ minutes by a current of $50 a m p$ will be .............. $gm$

1

0.6

2

0.09

3

5.4

4

10.8

Explanation:

(c) $m = Z i t$ or $m \propto i t$
$\frac{m_{1}}{m_{2}} = \frac{i_{1} t_{1}}{i_{2} t_{2}}$ $\Rightarrow$ $\frac{9}{m_{2}} = \frac{10^{5}}{50 \times 20 \times 60}$ $\Rightarrow$ $m_{2} = 5.4 g m$

ELECTROCHEMISTRY

20057 The acid used in lead storage battery is

1

H_{2} S O_{4}

2

H_{3} P O_{4}

3

H C l

4

H N O_{3}

Explanation:

(a)Dil. $H_{2} S O_{4}$ is used in lead in lead storage battery as electrolyte.

ELECTROCHEMISTRY

19825 Two electrolytic cells containing $C u S O_{4}$ and $A g N O_{3}$ respectively are connected in series and a current is passed through them until $1 m g$ of copper is deposited in the first cell. The amount of silver deposited in the second cell during this time is approximately ......... $mg$ [Atomic weights of copper and silver are respectively $63.57$ and $107.88$ ]

1

1.7

2

3.4

3

5.1

4

6.8

Explanation:

(b) $\frac{m_{1}}{m_{2}} = \frac{E_{1}}{E_{2}}$ (By faraday law for same current and time)
Where $E_{1}$ and $E_{2}$ are the chemical equivalents and $m_{1}$ and $m_{2}$ are the masses of copper and silver respectively.
$E = \frac{A t o m i c w e i g h t}{V a l e n c y}$ .
$E_{1} = \frac{63.57}{2} = 31.79$ and $E_{2} = \frac{107.88}{1} = 107.88$
$\frac{1 m g}{m_{2}} = \frac{31.79}{107.88}$
$\Rightarrow$ $m_{2} = \frac{107.88}{31.79} m g = 3.4 m g$

ELECTROCHEMISTRY

19826 A current $I$ is passed for a time $t$ through a number of voltameters. If m is the mass of a substance deposited on an electrode and $z$ is its electrochemical equivalent, then

1

\frac{z I t}{m} = c o n s t a n t

2

\frac{z}{m I t} = c o n s t a n t

3

\frac{I}{z m t} = c o n s t a n t

4

\frac{I t}{z m} = c o n s t a n t

Explanation:

(a) $m = Z i t \Rightarrow \frac{m}{Z i t} = 1 (c o n s t a n t)$

ELECTROCHEMISTRY

19827 For electroplating a spoon, it is placed in the voltameter at

1 The position of anode

2 The position of cathode

3 Exactly in the middle of anode and the cathode

4 Anywhere in the electrolyte

ELECTROCHEMISTRY

19828 If nearly $10^{5}$ coulomb liberate $1 g m$ equivalent of aluminium, then the amount of aluminium (equivalent weight $9$ ) deposited through electrolysis in $20$ minutes by a current of $50 a m p$ will be .............. $gm$

1

0.6

2

0.09

3

5.4

4

10.8

Explanation:

(c) $m = Z i t$ or $m \propto i t$
$\frac{m_{1}}{m_{2}} = \frac{i_{1} t_{1}}{i_{2} t_{2}}$ $\Rightarrow$ $\frac{9}{m_{2}} = \frac{10^{5}}{50 \times 20 \times 60}$ $\Rightarrow$ $m_{2} = 5.4 g m$

ELECTROCHEMISTRY

20057 The acid used in lead storage battery is

1

H_{2} S O_{4}

2

H_{3} P O_{4}

3

H C l

4

H N O_{3}

Explanation:

(a)Dil. $H_{2} S O_{4}$ is used in lead in lead storage battery as electrolyte.

ELECTROCHEMISTRY

19825 Two electrolytic cells containing $C u S O_{4}$ and $A g N O_{3}$ respectively are connected in series and a current is passed through them until $1 m g$ of copper is deposited in the first cell. The amount of silver deposited in the second cell during this time is approximately ......... $mg$ [Atomic weights of copper and silver are respectively $63.57$ and $107.88$ ]

1

1.7

2

3.4

3

5.1

4

6.8

Explanation:

(b) $\frac{m_{1}}{m_{2}} = \frac{E_{1}}{E_{2}}$ (By faraday law for same current and time)
Where $E_{1}$ and $E_{2}$ are the chemical equivalents and $m_{1}$ and $m_{2}$ are the masses of copper and silver respectively.
$E = \frac{A t o m i c w e i g h t}{V a l e n c y}$ .
$E_{1} = \frac{63.57}{2} = 31.79$ and $E_{2} = \frac{107.88}{1} = 107.88$
$\frac{1 m g}{m_{2}} = \frac{31.79}{107.88}$
$\Rightarrow$ $m_{2} = \frac{107.88}{31.79} m g = 3.4 m g$

ELECTROCHEMISTRY

19826 A current $I$ is passed for a time $t$ through a number of voltameters. If m is the mass of a substance deposited on an electrode and $z$ is its electrochemical equivalent, then

1

\frac{z I t}{m} = c o n s t a n t

2

\frac{z}{m I t} = c o n s t a n t

3

\frac{I}{z m t} = c o n s t a n t

4

\frac{I t}{z m} = c o n s t a n t

Explanation:

(a) $m = Z i t \Rightarrow \frac{m}{Z i t} = 1 (c o n s t a n t)$

ELECTROCHEMISTRY

19827 For electroplating a spoon, it is placed in the voltameter at

1 The position of anode

2 The position of cathode

3 Exactly in the middle of anode and the cathode

4 Anywhere in the electrolyte

ELECTROCHEMISTRY

19828 If nearly $10^{5}$ coulomb liberate $1 g m$ equivalent of aluminium, then the amount of aluminium (equivalent weight $9$ ) deposited through electrolysis in $20$ minutes by a current of $50 a m p$ will be .............. $gm$

1

0.6

2

0.09

3

5.4

4

10.8

Explanation:

(c) $m = Z i t$ or $m \propto i t$
$\frac{m_{1}}{m_{2}} = \frac{i_{1} t_{1}}{i_{2} t_{2}}$ $\Rightarrow$ $\frac{9}{m_{2}} = \frac{10^{5}}{50 \times 20 \times 60}$ $\Rightarrow$ $m_{2} = 5.4 g m$

ELECTROCHEMISTRY

20057 The acid used in lead storage battery is

1

H_{2} S O_{4}

2

H_{3} P O_{4}

3

H C l

4

H N O_{3}

Explanation:

(a)Dil. $H_{2} S O_{4}$ is used in lead in lead storage battery as electrolyte.

ELECTROCHEMISTRY

19825 Two electrolytic cells containing $C u S O_{4}$ and $A g N O_{3}$ respectively are connected in series and a current is passed through them until $1 m g$ of copper is deposited in the first cell. The amount of silver deposited in the second cell during this time is approximately ......... $mg$ [Atomic weights of copper and silver are respectively $63.57$ and $107.88$ ]

1

1.7

2

3.4

3

5.1

4

6.8

Explanation:

(b) $\frac{m_{1}}{m_{2}} = \frac{E_{1}}{E_{2}}$ (By faraday law for same current and time)
Where $E_{1}$ and $E_{2}$ are the chemical equivalents and $m_{1}$ and $m_{2}$ are the masses of copper and silver respectively.
$E = \frac{A t o m i c w e i g h t}{V a l e n c y}$ .
$E_{1} = \frac{63.57}{2} = 31.79$ and $E_{2} = \frac{107.88}{1} = 107.88$
$\frac{1 m g}{m_{2}} = \frac{31.79}{107.88}$
$\Rightarrow$ $m_{2} = \frac{107.88}{31.79} m g = 3.4 m g$

ELECTROCHEMISTRY

19826 A current $I$ is passed for a time $t$ through a number of voltameters. If m is the mass of a substance deposited on an electrode and $z$ is its electrochemical equivalent, then

1

\frac{z I t}{m} = c o n s t a n t

2

\frac{z}{m I t} = c o n s t a n t

3

\frac{I}{z m t} = c o n s t a n t

4

\frac{I t}{z m} = c o n s t a n t

Explanation:

(a) $m = Z i t \Rightarrow \frac{m}{Z i t} = 1 (c o n s t a n t)$

ELECTROCHEMISTRY

19827 For electroplating a spoon, it is placed in the voltameter at

1 The position of anode

2 The position of cathode

3 Exactly in the middle of anode and the cathode

4 Anywhere in the electrolyte

ELECTROCHEMISTRY

19828 If nearly $10^{5}$ coulomb liberate $1 g m$ equivalent of aluminium, then the amount of aluminium (equivalent weight $9$ ) deposited through electrolysis in $20$ minutes by a current of $50 a m p$ will be .............. $gm$

1

0.6

2

0.09

3

5.4

4

10.8

Explanation:

(c) $m = Z i t$ or $m \propto i t$
$\frac{m_{1}}{m_{2}} = \frac{i_{1} t_{1}}{i_{2} t_{2}}$ $\Rightarrow$ $\frac{9}{m_{2}} = \frac{10^{5}}{50 \times 20 \times 60}$ $\Rightarrow$ $m_{2} = 5.4 g m$

ELECTROCHEMISTRY

20057 The acid used in lead storage battery is

1

H_{2} S O_{4}

2

H_{3} P O_{4}

3

H C l

4

H N O_{3}

Explanation:

(a)Dil. $H_{2} S O_{4}$ is used in lead in lead storage battery as electrolyte.

ELECTROCHEMISTRY

19825 Two electrolytic cells containing $C u S O_{4}$ and $A g N O_{3}$ respectively are connected in series and a current is passed through them until $1 m g$ of copper is deposited in the first cell. The amount of silver deposited in the second cell during this time is approximately ......... $mg$ [Atomic weights of copper and silver are respectively $63.57$ and $107.88$ ]

1

1.7

2

3.4

3

5.1

4

6.8

Explanation:

(b) $\frac{m_{1}}{m_{2}} = \frac{E_{1}}{E_{2}}$ (By faraday law for same current and time)
Where $E_{1}$ and $E_{2}$ are the chemical equivalents and $m_{1}$ and $m_{2}$ are the masses of copper and silver respectively.
$E = \frac{A t o m i c w e i g h t}{V a l e n c y}$ .
$E_{1} = \frac{63.57}{2} = 31.79$ and $E_{2} = \frac{107.88}{1} = 107.88$
$\frac{1 m g}{m_{2}} = \frac{31.79}{107.88}$
$\Rightarrow$ $m_{2} = \frac{107.88}{31.79} m g = 3.4 m g$

ELECTROCHEMISTRY

19826 A current $I$ is passed for a time $t$ through a number of voltameters. If m is the mass of a substance deposited on an electrode and $z$ is its electrochemical equivalent, then

1

\frac{z I t}{m} = c o n s t a n t

2

\frac{z}{m I t} = c o n s t a n t

3

\frac{I}{z m t} = c o n s t a n t

4

\frac{I t}{z m} = c o n s t a n t

Explanation:

(a) $m = Z i t \Rightarrow \frac{m}{Z i t} = 1 (c o n s t a n t)$

ELECTROCHEMISTRY

19827 For electroplating a spoon, it is placed in the voltameter at

1 The position of anode

2 The position of cathode

3 Exactly in the middle of anode and the cathode

4 Anywhere in the electrolyte

ELECTROCHEMISTRY

19828 If nearly $10^{5}$ coulomb liberate $1 g m$ equivalent of aluminium, then the amount of aluminium (equivalent weight $9$ ) deposited through electrolysis in $20$ minutes by a current of $50 a m p$ will be .............. $gm$

1

0.6

2

0.09

3

5.4

4

10.8

Explanation:

(c) $m = Z i t$ or $m \propto i t$
$\frac{m_{1}}{m_{2}} = \frac{i_{1} t_{1}}{i_{2} t_{2}}$ $\Rightarrow$ $\frac{9}{m_{2}} = \frac{10^{5}}{50 \times 20 \times 60}$ $\Rightarrow$ $m_{2} = 5.4 g m$

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