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ELECTROCHEMISTRY

19817 The amount of charge required to liberate $9 g m$ of aluminium (atomic weight = $27$ and valency = $3$ ) in the process of electrolysis is ......... $coulombs$ (Faraday's number $= 96500 c o u l o m b s / g m$ equivalent)

1

321660

2

69500

3

289500

4

96500

Explanation:

(d) Equivalent weight of aluminium $= \frac{27}{3} = 9$
So $1$ faraday = $96500 C$ are required to liberate $9 g m$ of $A l$ .

ELECTROCHEMISTRY

19818 In an electroplating experiment, $m g m$ of silver is deposited when $4 a m p e r e$ of current flows for $2$ minute. The amount (in $g m$ ) of silver deposited by $6 a m p e r e$ of current for $40$ second will be

1

4 m

2

m / 2

3

m / 4

4

2 m

Explanation:

(b) By Faraday's law, $m \propto i t .$
$\frac{m_{1}}{m_{2}} = \frac{i_{1} t_{1}}{i_{2} t_{2}}$ $\Rightarrow$ $\frac{m}{m_{2}} = \frac{4 \times 120}{6 \times 40}$ $\Rightarrow$ $m_{2} = \frac{m}{2}$

ELECTROCHEMISTRY

19819 In electrolysis, if the duration of the passage of current is doubled, the mass liberated is

1 Doubled

2 Halved

3 Increased four times

4 Remains the same

Explanation:

(a) $m \propto i t$

ELECTROCHEMISTRY

19820 A current of $16 a m p e r e$ flows through molten $N a C l$ for $10$ minute. The amount of metallic sodium that appears at the negative electrode would be ............ $gm$

1

0.23

2

1.15

3

2.3

4

11.5

Explanation:

(c) Amount of metallic sodium appears $m = Z i t = (\frac{A}{V F}) i t$
$= (\frac{23}{1 \times 96500}) \times 16 \times 10 \times 60$ = $2.3 g m$

ELECTROCHEMISTRY

19817 The amount of charge required to liberate $9 g m$ of aluminium (atomic weight = $27$ and valency = $3$ ) in the process of electrolysis is ......... $coulombs$ (Faraday's number $= 96500 c o u l o m b s / g m$ equivalent)

1

321660

2

69500

3

289500

4

96500

Explanation:

(d) Equivalent weight of aluminium $= \frac{27}{3} = 9$
So $1$ faraday = $96500 C$ are required to liberate $9 g m$ of $A l$ .

ELECTROCHEMISTRY

19818 In an electroplating experiment, $m g m$ of silver is deposited when $4 a m p e r e$ of current flows for $2$ minute. The amount (in $g m$ ) of silver deposited by $6 a m p e r e$ of current for $40$ second will be

1

4 m

2

m / 2

3

m / 4

4

2 m

Explanation:

(b) By Faraday's law, $m \propto i t .$
$\frac{m_{1}}{m_{2}} = \frac{i_{1} t_{1}}{i_{2} t_{2}}$ $\Rightarrow$ $\frac{m}{m_{2}} = \frac{4 \times 120}{6 \times 40}$ $\Rightarrow$ $m_{2} = \frac{m}{2}$

ELECTROCHEMISTRY

19819 In electrolysis, if the duration of the passage of current is doubled, the mass liberated is

1 Doubled

2 Halved

3 Increased four times

4 Remains the same

Explanation:

(a) $m \propto i t$

ELECTROCHEMISTRY

19820 A current of $16 a m p e r e$ flows through molten $N a C l$ for $10$ minute. The amount of metallic sodium that appears at the negative electrode would be ............ $gm$

1

0.23

2

1.15

3

2.3

4

11.5

Explanation:

(c) Amount of metallic sodium appears $m = Z i t = (\frac{A}{V F}) i t$
$= (\frac{23}{1 \times 96500}) \times 16 \times 10 \times 60$ = $2.3 g m$

ELECTROCHEMISTRY

19817 The amount of charge required to liberate $9 g m$ of aluminium (atomic weight = $27$ and valency = $3$ ) in the process of electrolysis is ......... $coulombs$ (Faraday's number $= 96500 c o u l o m b s / g m$ equivalent)

1

321660

2

69500

3

289500

4

96500

Explanation:

(d) Equivalent weight of aluminium $= \frac{27}{3} = 9$
So $1$ faraday = $96500 C$ are required to liberate $9 g m$ of $A l$ .

ELECTROCHEMISTRY

19818 In an electroplating experiment, $m g m$ of silver is deposited when $4 a m p e r e$ of current flows for $2$ minute. The amount (in $g m$ ) of silver deposited by $6 a m p e r e$ of current for $40$ second will be

1

4 m

2

m / 2

3

m / 4

4

2 m

Explanation:

(b) By Faraday's law, $m \propto i t .$
$\frac{m_{1}}{m_{2}} = \frac{i_{1} t_{1}}{i_{2} t_{2}}$ $\Rightarrow$ $\frac{m}{m_{2}} = \frac{4 \times 120}{6 \times 40}$ $\Rightarrow$ $m_{2} = \frac{m}{2}$

ELECTROCHEMISTRY

19819 In electrolysis, if the duration of the passage of current is doubled, the mass liberated is

1 Doubled

2 Halved

3 Increased four times

4 Remains the same

Explanation:

(a) $m \propto i t$

ELECTROCHEMISTRY

19820 A current of $16 a m p e r e$ flows through molten $N a C l$ for $10$ minute. The amount of metallic sodium that appears at the negative electrode would be ............ $gm$

1

0.23

2

1.15

3

2.3

4

11.5

Explanation:

(c) Amount of metallic sodium appears $m = Z i t = (\frac{A}{V F}) i t$
$= (\frac{23}{1 \times 96500}) \times 16 \times 10 \times 60$ = $2.3 g m$

ELECTROCHEMISTRY

19817 The amount of charge required to liberate $9 g m$ of aluminium (atomic weight = $27$ and valency = $3$ ) in the process of electrolysis is ......... $coulombs$ (Faraday's number $= 96500 c o u l o m b s / g m$ equivalent)

1

321660

2

69500

3

289500

4

96500

Explanation:

(d) Equivalent weight of aluminium $= \frac{27}{3} = 9$
So $1$ faraday = $96500 C$ are required to liberate $9 g m$ of $A l$ .

ELECTROCHEMISTRY

19818 In an electroplating experiment, $m g m$ of silver is deposited when $4 a m p e r e$ of current flows for $2$ minute. The amount (in $g m$ ) of silver deposited by $6 a m p e r e$ of current for $40$ second will be

1

4 m

2

m / 2

3

m / 4

4

2 m

Explanation:

(b) By Faraday's law, $m \propto i t .$
$\frac{m_{1}}{m_{2}} = \frac{i_{1} t_{1}}{i_{2} t_{2}}$ $\Rightarrow$ $\frac{m}{m_{2}} = \frac{4 \times 120}{6 \times 40}$ $\Rightarrow$ $m_{2} = \frac{m}{2}$

ELECTROCHEMISTRY

19819 In electrolysis, if the duration of the passage of current is doubled, the mass liberated is

1 Doubled

2 Halved

3 Increased four times

4 Remains the same

Explanation:

(a) $m \propto i t$

ELECTROCHEMISTRY

19820 A current of $16 a m p e r e$ flows through molten $N a C l$ for $10$ minute. The amount of metallic sodium that appears at the negative electrode would be ............ $gm$

1

0.23

2

1.15

3

2.3

4

11.5

Explanation:

(c) Amount of metallic sodium appears $m = Z i t = (\frac{A}{V F}) i t$
$= (\frac{23}{1 \times 96500}) \times 16 \times 10 \times 60$ = $2.3 g m$