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ELECTROCHEMISTRY

19965 The amount of silver deposited by passing $241.25$ coulomb of current through silver nitrate solution is

1

2.7 g

2

2.7 m g

3

0.27 g

4

0.54 g

Explanation:

(c)Given, Current $= 241.25$ columb
$1$ coulomb current will deposite $= 1.118 \times 10^{- 3} g m A g$ .
$∴$ $241.25$ current will deposite = $1.118 \times 10^{- 3} \times 241.25$
$= 0.27 g m$ silver.

ELECTROCHEMISTRY

19966 When $1 F$ of electricity is passed through acidulated water, $O_{2}$ evolved is ............. ${dm}^{3}$

1

11.2

2

5.6

3

22.4

4

1

Explanation:

(b)Reaction for electrolysis of water is
$2 H_{2} O$ $⇌$ $4 H^{+} + 2 O^{2 -}$
$2 O^{2 -} \to O_{2} + 4 e^{-}$
$4 e^{-} + 4 H^{+} \to 2 H_{2}$
$n = 4$ so $4$ Faraday charge will liberate
$1$ mole $= 22.4 d m^{3}$ oxygen
$∴$ $1$ Faraday charge will liberate $\frac{22.4}{4} = 5.6 d m^{3} O_{2}$ .

ELECTROCHEMISTRY

19967 Charge required to liberate $11.5 g$ sodium is

1

0.5 F

2

0.1 F

3

1.5 F

4

96500

coulombs

Explanation:

(a) $N a^{+} + e^{-} \to N a$
Charge (in $F$ ) $=$ moles of $e^{-}$ used $=$ moles of $N a$ deposited
$= \frac{11.5}{23} g m = 0.5 F a r a d a y$ .

ELECTROCHEMISTRY

19968 In the electrolysis of water, one Faraday of electrical energy would evolve

1 One mole of oxygen

2 One

g

atom of oxygen

3

8 g

of oxygen

4

22.4

litres of oxygen

Explanation:

(c)Hydrolysis of water : $2 H_{2} O$ $\Leftrightarrow$ $4 H^{+} + 4 e^{-} + O_{2}$
$4 F$ charge will produce $= 1$ mole $O_{2} = 32 g m O_{2}$
$1 F$ charge will produce $= \frac{32}{4} = 8 g m O_{2}$ .

ELECTROCHEMISTRY

19965 The amount of silver deposited by passing $241.25$ coulomb of current through silver nitrate solution is

1

2.7 g

2

2.7 m g

3

0.27 g

4

0.54 g

Explanation:

(c)Given, Current $= 241.25$ columb
$1$ coulomb current will deposite $= 1.118 \times 10^{- 3} g m A g$ .
$∴$ $241.25$ current will deposite = $1.118 \times 10^{- 3} \times 241.25$
$= 0.27 g m$ silver.

ELECTROCHEMISTRY

19966 When $1 F$ of electricity is passed through acidulated water, $O_{2}$ evolved is ............. ${dm}^{3}$

1

11.2

2

5.6

3

22.4

4

1

Explanation:

(b)Reaction for electrolysis of water is
$2 H_{2} O$ $⇌$ $4 H^{+} + 2 O^{2 -}$
$2 O^{2 -} \to O_{2} + 4 e^{-}$
$4 e^{-} + 4 H^{+} \to 2 H_{2}$
$n = 4$ so $4$ Faraday charge will liberate
$1$ mole $= 22.4 d m^{3}$ oxygen
$∴$ $1$ Faraday charge will liberate $\frac{22.4}{4} = 5.6 d m^{3} O_{2}$ .

ELECTROCHEMISTRY

19967 Charge required to liberate $11.5 g$ sodium is

1

0.5 F

2

0.1 F

3

1.5 F

4

96500

coulombs

Explanation:

(a) $N a^{+} + e^{-} \to N a$
Charge (in $F$ ) $=$ moles of $e^{-}$ used $=$ moles of $N a$ deposited
$= \frac{11.5}{23} g m = 0.5 F a r a d a y$ .

ELECTROCHEMISTRY

19968 In the electrolysis of water, one Faraday of electrical energy would evolve

1 One mole of oxygen

2 One

g

atom of oxygen

3

8 g

of oxygen

4

22.4

litres of oxygen

Explanation:

(c)Hydrolysis of water : $2 H_{2} O$ $\Leftrightarrow$ $4 H^{+} + 4 e^{-} + O_{2}$
$4 F$ charge will produce $= 1$ mole $O_{2} = 32 g m O_{2}$
$1 F$ charge will produce $= \frac{32}{4} = 8 g m O_{2}$ .

ELECTROCHEMISTRY

19965 The amount of silver deposited by passing $241.25$ coulomb of current through silver nitrate solution is

1

2.7 g

2

2.7 m g

3

0.27 g

4

0.54 g

Explanation:

(c)Given, Current $= 241.25$ columb
$1$ coulomb current will deposite $= 1.118 \times 10^{- 3} g m A g$ .
$∴$ $241.25$ current will deposite = $1.118 \times 10^{- 3} \times 241.25$
$= 0.27 g m$ silver.

ELECTROCHEMISTRY

19966 When $1 F$ of electricity is passed through acidulated water, $O_{2}$ evolved is ............. ${dm}^{3}$

1

11.2

2

5.6

3

22.4

4

1

Explanation:

(b)Reaction for electrolysis of water is
$2 H_{2} O$ $⇌$ $4 H^{+} + 2 O^{2 -}$
$2 O^{2 -} \to O_{2} + 4 e^{-}$
$4 e^{-} + 4 H^{+} \to 2 H_{2}$
$n = 4$ so $4$ Faraday charge will liberate
$1$ mole $= 22.4 d m^{3}$ oxygen
$∴$ $1$ Faraday charge will liberate $\frac{22.4}{4} = 5.6 d m^{3} O_{2}$ .

ELECTROCHEMISTRY

19967 Charge required to liberate $11.5 g$ sodium is

1

0.5 F

2

0.1 F

3

1.5 F

4

96500

coulombs

Explanation:

(a) $N a^{+} + e^{-} \to N a$
Charge (in $F$ ) $=$ moles of $e^{-}$ used $=$ moles of $N a$ deposited
$= \frac{11.5}{23} g m = 0.5 F a r a d a y$ .

ELECTROCHEMISTRY

19968 In the electrolysis of water, one Faraday of electrical energy would evolve

1 One mole of oxygen

2 One

g

atom of oxygen

3

8 g

of oxygen

4

22.4

litres of oxygen

Explanation:

(c)Hydrolysis of water : $2 H_{2} O$ $\Leftrightarrow$ $4 H^{+} + 4 e^{-} + O_{2}$
$4 F$ charge will produce $= 1$ mole $O_{2} = 32 g m O_{2}$
$1 F$ charge will produce $= \frac{32}{4} = 8 g m O_{2}$ .

ELECTROCHEMISTRY

19965 The amount of silver deposited by passing $241.25$ coulomb of current through silver nitrate solution is

1

2.7 g

2

2.7 m g

3

0.27 g

4

0.54 g

Explanation:

(c)Given, Current $= 241.25$ columb
$1$ coulomb current will deposite $= 1.118 \times 10^{- 3} g m A g$ .
$∴$ $241.25$ current will deposite = $1.118 \times 10^{- 3} \times 241.25$
$= 0.27 g m$ silver.

ELECTROCHEMISTRY

19966 When $1 F$ of electricity is passed through acidulated water, $O_{2}$ evolved is ............. ${dm}^{3}$

1

11.2

2

5.6

3

22.4

4

1

Explanation:

(b)Reaction for electrolysis of water is
$2 H_{2} O$ $⇌$ $4 H^{+} + 2 O^{2 -}$
$2 O^{2 -} \to O_{2} + 4 e^{-}$
$4 e^{-} + 4 H^{+} \to 2 H_{2}$
$n = 4$ so $4$ Faraday charge will liberate
$1$ mole $= 22.4 d m^{3}$ oxygen
$∴$ $1$ Faraday charge will liberate $\frac{22.4}{4} = 5.6 d m^{3} O_{2}$ .

ELECTROCHEMISTRY

19967 Charge required to liberate $11.5 g$ sodium is

1

0.5 F

2

0.1 F

3

1.5 F

4

96500

coulombs

Explanation:

(a) $N a^{+} + e^{-} \to N a$
Charge (in $F$ ) $=$ moles of $e^{-}$ used $=$ moles of $N a$ deposited
$= \frac{11.5}{23} g m = 0.5 F a r a d a y$ .

ELECTROCHEMISTRY

19968 In the electrolysis of water, one Faraday of electrical energy would evolve

1 One mole of oxygen

2 One

g

atom of oxygen

3

8 g

of oxygen

4

22.4

litres of oxygen

Explanation:

(c)Hydrolysis of water : $2 H_{2} O$ $\Leftrightarrow$ $4 H^{+} + 4 e^{-} + O_{2}$
$4 F$ charge will produce $= 1$ mole $O_{2} = 32 g m O_{2}$
$1 F$ charge will produce $= \frac{32}{4} = 8 g m O_{2}$ .