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ELECTROCHEMISTRY

19918 What is the amount of chlorine evolved when $2$ amperes of current is passed for $30$ minutes in an aqueous solution of $N a C l$ ............. $g$

1

66

2

1.32

3

33

4

99

Explanation:

(b)At Andoe $C l^{-} \to \frac{1}{2} C l_{2} + e^{-}$
$E_{C l_{2}} = \frac{35.5 \times 2}{2} = 35.5$
$W_{C l_{2}} = \frac{E_{C l_{2}} \times I \times t}{96500}$ $= \frac{35.5 \times 2 \times 30 \times 60}{96500} = 1.32 g m$ .

ELECTROCHEMISTRY

19919 On passing a current through $K C l$ solution, $19.5 g$ of potassium is deposited. If the same quantity of electricity is passed through a solution of aluminium chloride, the amount of aluminium deposited is ............ $g$

1

4.5

2

9

3

13.5

4

27

Explanation:

$K^{+} + e^{-} = K$
$A l^{3 +} + 3 e^{-} = A l$
Aluminium requires 3 times the amount of electricity to deposit one mole
Moles of $K$ deposited $= 19.5 / 39 = 0.5$
Moles of $A l$ deposited $= 0.5 / 3 = 1 / 6$
Mass of $A l$ deposited $= 27 \times 1 / 6$
$= 4.5 g$

ELECTROCHEMISTRY

19920 Electrolysis rules of Faraday’s states that mass deposited on electrode is proportional to

1

m \propto I^{2}

2

m \propto Q

3

m \propto Q^{2}

4 None of these

ELECTROCHEMISTRY

19958 In order to separate oxygen from one mole of $H_{2} O$ the required quantity of coulomb would be

1

1.93 \times 10^{5}

2

9.6 \times 10^{4}

3

1.8

4

3.2

Explanation:

we have $1 m o l e$ of $H_{2} O$
$1 m o l e$ of $H_{2} O$ given one atom $O$
$H_{2} O \to H_{2} + 1 / 2 O_{2}$
charge on $O = - 2$
electricity required on oxidation of mole of $H_{2} O$ and $O_{2} = n f$ where $n = 2$
$= 2 \times 96487 c o u l o m b$
$= 192974 c o u l o m b$
$= 1.93 \times 10^{5}$

ELECTROCHEMISTRY

19918 What is the amount of chlorine evolved when $2$ amperes of current is passed for $30$ minutes in an aqueous solution of $N a C l$ ............. $g$

1

66

2

1.32

3

33

4

99

Explanation:

(b)At Andoe $C l^{-} \to \frac{1}{2} C l_{2} + e^{-}$
$E_{C l_{2}} = \frac{35.5 \times 2}{2} = 35.5$
$W_{C l_{2}} = \frac{E_{C l_{2}} \times I \times t}{96500}$ $= \frac{35.5 \times 2 \times 30 \times 60}{96500} = 1.32 g m$ .

ELECTROCHEMISTRY

19919 On passing a current through $K C l$ solution, $19.5 g$ of potassium is deposited. If the same quantity of electricity is passed through a solution of aluminium chloride, the amount of aluminium deposited is ............ $g$

1

4.5

2

9

3

13.5

4

27

Explanation:

$K^{+} + e^{-} = K$
$A l^{3 +} + 3 e^{-} = A l$
Aluminium requires 3 times the amount of electricity to deposit one mole
Moles of $K$ deposited $= 19.5 / 39 = 0.5$
Moles of $A l$ deposited $= 0.5 / 3 = 1 / 6$
Mass of $A l$ deposited $= 27 \times 1 / 6$
$= 4.5 g$

ELECTROCHEMISTRY

19920 Electrolysis rules of Faraday’s states that mass deposited on electrode is proportional to

1

m \propto I^{2}

2

m \propto Q

3

m \propto Q^{2}

4 None of these

ELECTROCHEMISTRY

19958 In order to separate oxygen from one mole of $H_{2} O$ the required quantity of coulomb would be

1

1.93 \times 10^{5}

2

9.6 \times 10^{4}

3

1.8

4

3.2

Explanation:

we have $1 m o l e$ of $H_{2} O$
$1 m o l e$ of $H_{2} O$ given one atom $O$
$H_{2} O \to H_{2} + 1 / 2 O_{2}$
charge on $O = - 2$
electricity required on oxidation of mole of $H_{2} O$ and $O_{2} = n f$ where $n = 2$
$= 2 \times 96487 c o u l o m b$
$= 192974 c o u l o m b$
$= 1.93 \times 10^{5}$

ELECTROCHEMISTRY

19918 What is the amount of chlorine evolved when $2$ amperes of current is passed for $30$ minutes in an aqueous solution of $N a C l$ ............. $g$

1

66

2

1.32

3

33

4

99

Explanation:

(b)At Andoe $C l^{-} \to \frac{1}{2} C l_{2} + e^{-}$
$E_{C l_{2}} = \frac{35.5 \times 2}{2} = 35.5$
$W_{C l_{2}} = \frac{E_{C l_{2}} \times I \times t}{96500}$ $= \frac{35.5 \times 2 \times 30 \times 60}{96500} = 1.32 g m$ .

ELECTROCHEMISTRY

19919 On passing a current through $K C l$ solution, $19.5 g$ of potassium is deposited. If the same quantity of electricity is passed through a solution of aluminium chloride, the amount of aluminium deposited is ............ $g$

1

4.5

2

9

3

13.5

4

27

Explanation:

$K^{+} + e^{-} = K$
$A l^{3 +} + 3 e^{-} = A l$
Aluminium requires 3 times the amount of electricity to deposit one mole
Moles of $K$ deposited $= 19.5 / 39 = 0.5$
Moles of $A l$ deposited $= 0.5 / 3 = 1 / 6$
Mass of $A l$ deposited $= 27 \times 1 / 6$
$= 4.5 g$

ELECTROCHEMISTRY

19920 Electrolysis rules of Faraday’s states that mass deposited on electrode is proportional to

1

m \propto I^{2}

2

m \propto Q

3

m \propto Q^{2}

4 None of these

ELECTROCHEMISTRY

19958 In order to separate oxygen from one mole of $H_{2} O$ the required quantity of coulomb would be

1

1.93 \times 10^{5}

2

9.6 \times 10^{4}

3

1.8

4

3.2

Explanation:

we have $1 m o l e$ of $H_{2} O$
$1 m o l e$ of $H_{2} O$ given one atom $O$
$H_{2} O \to H_{2} + 1 / 2 O_{2}$
charge on $O = - 2$
electricity required on oxidation of mole of $H_{2} O$ and $O_{2} = n f$ where $n = 2$
$= 2 \times 96487 c o u l o m b$
$= 192974 c o u l o m b$
$= 1.93 \times 10^{5}$

ELECTROCHEMISTRY

19918 What is the amount of chlorine evolved when $2$ amperes of current is passed for $30$ minutes in an aqueous solution of $N a C l$ ............. $g$

1

66

2

1.32

3

33

4

99

Explanation:

(b)At Andoe $C l^{-} \to \frac{1}{2} C l_{2} + e^{-}$
$E_{C l_{2}} = \frac{35.5 \times 2}{2} = 35.5$
$W_{C l_{2}} = \frac{E_{C l_{2}} \times I \times t}{96500}$ $= \frac{35.5 \times 2 \times 30 \times 60}{96500} = 1.32 g m$ .

ELECTROCHEMISTRY

19919 On passing a current through $K C l$ solution, $19.5 g$ of potassium is deposited. If the same quantity of electricity is passed through a solution of aluminium chloride, the amount of aluminium deposited is ............ $g$

1

4.5

2

9

3

13.5

4

27

Explanation:

$K^{+} + e^{-} = K$
$A l^{3 +} + 3 e^{-} = A l$
Aluminium requires 3 times the amount of electricity to deposit one mole
Moles of $K$ deposited $= 19.5 / 39 = 0.5$
Moles of $A l$ deposited $= 0.5 / 3 = 1 / 6$
Mass of $A l$ deposited $= 27 \times 1 / 6$
$= 4.5 g$

ELECTROCHEMISTRY

19920 Electrolysis rules of Faraday’s states that mass deposited on electrode is proportional to

1

m \propto I^{2}

2

m \propto Q

3

m \propto Q^{2}

4 None of these

ELECTROCHEMISTRY

19958 In order to separate oxygen from one mole of $H_{2} O$ the required quantity of coulomb would be

1

1.93 \times 10^{5}

2

9.6 \times 10^{4}

3

1.8

4

3.2

Explanation:

we have $1 m o l e$ of $H_{2} O$
$1 m o l e$ of $H_{2} O$ given one atom $O$
$H_{2} O \to H_{2} + 1 / 2 O_{2}$
charge on $O = - 2$
electricity required on oxidation of mole of $H_{2} O$ and $O_{2} = n f$ where $n = 2$
$= 2 \times 96487 c o u l o m b$
$= 192974 c o u l o m b$
$= 1.93 \times 10^{5}$