QBManager

ELECTROCHEMISTRY

19897 Unit of Faraday is

1 Ampere

2 Coulomb

3 Coulomb

m o l e^{- 1}

4 Coulomb

{Sec}^{- 1}

Explanation:

In the International System of Units ( $S I$ ), the coulomb ( $C$ ) is the preferred unit of electric charge quantity. It is equivalent to one ampere-second ( $1 A . s$ ) and represents approximately $6.24 \times 10^{18}$ electric charge carriers. To convert from coulombs to faradays, multiply by $1.04 \times 10^{- 5}$ .

ELECTROCHEMISTRY

19898 On passing $0.1$ Faraday of electricity through aluminium chloride, the amount of aluminium metal deposited on cathode is $(A l = 27)$ ............ $gm$

1

0.9

2

0.3

3

0.27

4

2.7

Explanation:

(a) At cathode; $A l^{3 +} + 3 e^{-} \to A l$
$E_{A l} = \frac{27}{3} = 9$
$W_{A l} = E_{A l} \times N o . o f f a r a d a y s$ $= 9 \times 0.1 = 0.9 g m$ .

ELECTROCHEMISTRY

19899 Which of the following represents the first law of Faraday

1

E = m c^{2}

2

E = h v

3

m = e c t

4

P V = n R T

Explanation:

The mass of the substance $(m)$ deposited or liberated at any electrode is directly proportional to the quantity of electricity or charge ( $Q$ ) passed.

ELECTROCHEMISTRY

19900 $5$ amperes is passed through a solution of zinc sulphate for $40$ minutes. Find the amount of zinc deposited at the cathode ............. $gm$

1

40.65

2

4.065

3

0.4065

4

65.04

Explanation:

(b) $W = z i t$ ;
$W = \frac{32.69 \times 5 \times 60 \times 40}{96500}$
$= 4.065 g m$ .

ELECTROCHEMISTRY

19901 In an electroplating experiment $m g$ of silver is deposited, when $4$ amperes of current flows for $2$ minutes. The amount (in $g m s$ ) of silver deposited by $6$ amperes of current flowing for $40$ seconds will be

1

4 m

2

m / 2

3

m / 4

4

2 m

Explanation:

(b) $m = Z \times 4 \times 120$ ; $M = Z \times 6 \times 40$
$\frac{M}{m} = \frac{6 \times 40}{4 \times 120} = \frac{1}{2}; M = m / 2$ .

ELECTROCHEMISTRY

19897 Unit of Faraday is

1 Ampere

2 Coulomb

3 Coulomb

m o l e^{- 1}

4 Coulomb

{Sec}^{- 1}

Explanation:

In the International System of Units ( $S I$ ), the coulomb ( $C$ ) is the preferred unit of electric charge quantity. It is equivalent to one ampere-second ( $1 A . s$ ) and represents approximately $6.24 \times 10^{18}$ electric charge carriers. To convert from coulombs to faradays, multiply by $1.04 \times 10^{- 5}$ .

ELECTROCHEMISTRY

19898 On passing $0.1$ Faraday of electricity through aluminium chloride, the amount of aluminium metal deposited on cathode is $(A l = 27)$ ............ $gm$

1

0.9

2

0.3

3

0.27

4

2.7

Explanation:

(a) At cathode; $A l^{3 +} + 3 e^{-} \to A l$
$E_{A l} = \frac{27}{3} = 9$
$W_{A l} = E_{A l} \times N o . o f f a r a d a y s$ $= 9 \times 0.1 = 0.9 g m$ .

ELECTROCHEMISTRY

19899 Which of the following represents the first law of Faraday

1

E = m c^{2}

2

E = h v

3

m = e c t

4

P V = n R T

Explanation:

The mass of the substance $(m)$ deposited or liberated at any electrode is directly proportional to the quantity of electricity or charge ( $Q$ ) passed.

ELECTROCHEMISTRY

19900 $5$ amperes is passed through a solution of zinc sulphate for $40$ minutes. Find the amount of zinc deposited at the cathode ............. $gm$

1

40.65

2

4.065

3

0.4065

4

65.04

Explanation:

(b) $W = z i t$ ;
$W = \frac{32.69 \times 5 \times 60 \times 40}{96500}$
$= 4.065 g m$ .

ELECTROCHEMISTRY

19901 In an electroplating experiment $m g$ of silver is deposited, when $4$ amperes of current flows for $2$ minutes. The amount (in $g m s$ ) of silver deposited by $6$ amperes of current flowing for $40$ seconds will be

1

4 m

2

m / 2

3

m / 4

4

2 m

Explanation:

(b) $m = Z \times 4 \times 120$ ; $M = Z \times 6 \times 40$
$\frac{M}{m} = \frac{6 \times 40}{4 \times 120} = \frac{1}{2}; M = m / 2$ .

ELECTROCHEMISTRY

19897 Unit of Faraday is

1 Ampere

2 Coulomb

3 Coulomb

m o l e^{- 1}

4 Coulomb

{Sec}^{- 1}

Explanation:

In the International System of Units ( $S I$ ), the coulomb ( $C$ ) is the preferred unit of electric charge quantity. It is equivalent to one ampere-second ( $1 A . s$ ) and represents approximately $6.24 \times 10^{18}$ electric charge carriers. To convert from coulombs to faradays, multiply by $1.04 \times 10^{- 5}$ .

ELECTROCHEMISTRY

19898 On passing $0.1$ Faraday of electricity through aluminium chloride, the amount of aluminium metal deposited on cathode is $(A l = 27)$ ............ $gm$

1

0.9

2

0.3

3

0.27

4

2.7

Explanation:

(a) At cathode; $A l^{3 +} + 3 e^{-} \to A l$
$E_{A l} = \frac{27}{3} = 9$
$W_{A l} = E_{A l} \times N o . o f f a r a d a y s$ $= 9 \times 0.1 = 0.9 g m$ .

ELECTROCHEMISTRY

19899 Which of the following represents the first law of Faraday

1

E = m c^{2}

2

E = h v

3

m = e c t

4

P V = n R T

Explanation:

The mass of the substance $(m)$ deposited or liberated at any electrode is directly proportional to the quantity of electricity or charge ( $Q$ ) passed.

ELECTROCHEMISTRY

19900 $5$ amperes is passed through a solution of zinc sulphate for $40$ minutes. Find the amount of zinc deposited at the cathode ............. $gm$

1

40.65

2

4.065

3

0.4065

4

65.04

Explanation:

(b) $W = z i t$ ;
$W = \frac{32.69 \times 5 \times 60 \times 40}{96500}$
$= 4.065 g m$ .

ELECTROCHEMISTRY

19901 In an electroplating experiment $m g$ of silver is deposited, when $4$ amperes of current flows for $2$ minutes. The amount (in $g m s$ ) of silver deposited by $6$ amperes of current flowing for $40$ seconds will be

1

4 m

2

m / 2

3

m / 4

4

2 m

Explanation:

(b) $m = Z \times 4 \times 120$ ; $M = Z \times 6 \times 40$
$\frac{M}{m} = \frac{6 \times 40}{4 \times 120} = \frac{1}{2}; M = m / 2$ .

ELECTROCHEMISTRY

19897 Unit of Faraday is

1 Ampere

2 Coulomb

3 Coulomb

m o l e^{- 1}

4 Coulomb

{Sec}^{- 1}

Explanation:

In the International System of Units ( $S I$ ), the coulomb ( $C$ ) is the preferred unit of electric charge quantity. It is equivalent to one ampere-second ( $1 A . s$ ) and represents approximately $6.24 \times 10^{18}$ electric charge carriers. To convert from coulombs to faradays, multiply by $1.04 \times 10^{- 5}$ .

ELECTROCHEMISTRY

19898 On passing $0.1$ Faraday of electricity through aluminium chloride, the amount of aluminium metal deposited on cathode is $(A l = 27)$ ............ $gm$

1

0.9

2

0.3

3

0.27

4

2.7

Explanation:

(a) At cathode; $A l^{3 +} + 3 e^{-} \to A l$
$E_{A l} = \frac{27}{3} = 9$
$W_{A l} = E_{A l} \times N o . o f f a r a d a y s$ $= 9 \times 0.1 = 0.9 g m$ .

ELECTROCHEMISTRY

19899 Which of the following represents the first law of Faraday

1

E = m c^{2}

2

E = h v

3

m = e c t

4

P V = n R T

Explanation:

The mass of the substance $(m)$ deposited or liberated at any electrode is directly proportional to the quantity of electricity or charge ( $Q$ ) passed.

ELECTROCHEMISTRY

19900 $5$ amperes is passed through a solution of zinc sulphate for $40$ minutes. Find the amount of zinc deposited at the cathode ............. $gm$

1

40.65

2

4.065

3

0.4065

4

65.04

Explanation:

(b) $W = z i t$ ;
$W = \frac{32.69 \times 5 \times 60 \times 40}{96500}$
$= 4.065 g m$ .

ELECTROCHEMISTRY

19901 In an electroplating experiment $m g$ of silver is deposited, when $4$ amperes of current flows for $2$ minutes. The amount (in $g m s$ ) of silver deposited by $6$ amperes of current flowing for $40$ seconds will be

1

4 m

2

m / 2

3

m / 4

4

2 m

Explanation:

(b) $m = Z \times 4 \times 120$ ; $M = Z \times 6 \times 40$
$\frac{M}{m} = \frac{6 \times 40}{4 \times 120} = \frac{1}{2}; M = m / 2$ .

ELECTROCHEMISTRY

19897 Unit of Faraday is

1 Ampere

2 Coulomb

3 Coulomb

m o l e^{- 1}

4 Coulomb

{Sec}^{- 1}

Explanation:

In the International System of Units ( $S I$ ), the coulomb ( $C$ ) is the preferred unit of electric charge quantity. It is equivalent to one ampere-second ( $1 A . s$ ) and represents approximately $6.24 \times 10^{18}$ electric charge carriers. To convert from coulombs to faradays, multiply by $1.04 \times 10^{- 5}$ .

ELECTROCHEMISTRY

19898 On passing $0.1$ Faraday of electricity through aluminium chloride, the amount of aluminium metal deposited on cathode is $(A l = 27)$ ............ $gm$

1

0.9

2

0.3

3

0.27

4

2.7

Explanation:

(a) At cathode; $A l^{3 +} + 3 e^{-} \to A l$
$E_{A l} = \frac{27}{3} = 9$
$W_{A l} = E_{A l} \times N o . o f f a r a d a y s$ $= 9 \times 0.1 = 0.9 g m$ .

ELECTROCHEMISTRY

19899 Which of the following represents the first law of Faraday

1

E = m c^{2}

2

E = h v

3

m = e c t

4

P V = n R T

Explanation:

The mass of the substance $(m)$ deposited or liberated at any electrode is directly proportional to the quantity of electricity or charge ( $Q$ ) passed.

ELECTROCHEMISTRY

19900 $5$ amperes is passed through a solution of zinc sulphate for $40$ minutes. Find the amount of zinc deposited at the cathode ............. $gm$

1

40.65

2

4.065

3

0.4065

4

65.04

Explanation:

(b) $W = z i t$ ;
$W = \frac{32.69 \times 5 \times 60 \times 40}{96500}$
$= 4.065 g m$ .

ELECTROCHEMISTRY

19901 In an electroplating experiment $m g$ of silver is deposited, when $4$ amperes of current flows for $2$ minutes. The amount (in $g m s$ ) of silver deposited by $6$ amperes of current flowing for $40$ seconds will be

1

4 m

2

m / 2

3

m / 4

4

2 m

Explanation:

(b) $m = Z \times 4 \times 120$ ; $M = Z \times 6 \times 40$
$\frac{M}{m} = \frac{6 \times 40}{4 \times 120} = \frac{1}{2}; M = m / 2$ .

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