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ELECTROCHEMISTRY

19930 Faraday's laws of electrolysis are related to the

1 Atomic number of cation

2 Atomic number of anion

3 Equivalent weight of the electrolyte

4 Speed of the cation

Explanation:

(c) $w \propto E$ if $i$ and $t$ are constant.

ELECTROCHEMISTRY

19931 The electric charge for electrode decomposition of one gram equivalent of a substance is

1 One ampere per second

2

96500

coulombs per second

3 One ampere for one hour

4 Charge on one mole of electrons

Explanation:

One gram equivalent of a substance requires $96500 c o u l o m b s s^{- 1}$ of electricity.

ELECTROCHEMISTRY

19932 The number of electrons passing per second through a cross-section of copper wire carrying $10^{- 6}$ amperes of current per second is found to be

1

1.6 \times 10^{- 19}

2

6 \times 10^{- 35}

3

6 \times 10^{- 16}

4

6 \times 10^{12}

Explanation:

(d) Charge (Coulombs) pass per second $= 10^{- 6}$
number of electrons passed per second
$= \frac{10^{- 6}}{1.602 \times 10^{- 19}} = 6.24 \times 10^{12}$ .

ELECTROCHEMISTRY

19933 The electrolytic cells, one containing acidified ferrous chloride and another acidified ferric chloride are connected in series. The ratio of iron deposited at cathodes in the two cells when electricity is passed through the cells will be

1

1 : 1

2

3 : 1

3

2 : 1

4

3 : 2

Explanation:

(d) At cathode;
$F e^{2 +} + 2 e^{-} \to F e$ ; $F e^{3 +} + 3 e^{-} \to F e$
$(E_{F e})_{1} = \frac{A t o m i c . w e i g h t}{2}; (E_{F e})_{2} = \frac{A t o m i c . w e i g h t}{3}$
Ratio of weight of $F e$ liberated
$= \frac{A t o m i c w e i g h t}{3} : \frac{A t o m i c w e i g h t}{2} = 3 : 2$ .

ELECTROCHEMISTRY

19930 Faraday's laws of electrolysis are related to the

1 Atomic number of cation

2 Atomic number of anion

3 Equivalent weight of the electrolyte

4 Speed of the cation

Explanation:

(c) $w \propto E$ if $i$ and $t$ are constant.

ELECTROCHEMISTRY

19931 The electric charge for electrode decomposition of one gram equivalent of a substance is

1 One ampere per second

2

96500

coulombs per second

3 One ampere for one hour

4 Charge on one mole of electrons

Explanation:

One gram equivalent of a substance requires $96500 c o u l o m b s s^{- 1}$ of electricity.

ELECTROCHEMISTRY

19932 The number of electrons passing per second through a cross-section of copper wire carrying $10^{- 6}$ amperes of current per second is found to be

1

1.6 \times 10^{- 19}

2

6 \times 10^{- 35}

3

6 \times 10^{- 16}

4

6 \times 10^{12}

Explanation:

(d) Charge (Coulombs) pass per second $= 10^{- 6}$
number of electrons passed per second
$= \frac{10^{- 6}}{1.602 \times 10^{- 19}} = 6.24 \times 10^{12}$ .

ELECTROCHEMISTRY

19933 The electrolytic cells, one containing acidified ferrous chloride and another acidified ferric chloride are connected in series. The ratio of iron deposited at cathodes in the two cells when electricity is passed through the cells will be

1

1 : 1

2

3 : 1

3

2 : 1

4

3 : 2

Explanation:

(d) At cathode;
$F e^{2 +} + 2 e^{-} \to F e$ ; $F e^{3 +} + 3 e^{-} \to F e$
$(E_{F e})_{1} = \frac{A t o m i c . w e i g h t}{2}; (E_{F e})_{2} = \frac{A t o m i c . w e i g h t}{3}$
Ratio of weight of $F e$ liberated
$= \frac{A t o m i c w e i g h t}{3} : \frac{A t o m i c w e i g h t}{2} = 3 : 2$ .

ELECTROCHEMISTRY

19930 Faraday's laws of electrolysis are related to the

1 Atomic number of cation

2 Atomic number of anion

3 Equivalent weight of the electrolyte

4 Speed of the cation

Explanation:

(c) $w \propto E$ if $i$ and $t$ are constant.

ELECTROCHEMISTRY

19931 The electric charge for electrode decomposition of one gram equivalent of a substance is

1 One ampere per second

2

96500

coulombs per second

3 One ampere for one hour

4 Charge on one mole of electrons

Explanation:

One gram equivalent of a substance requires $96500 c o u l o m b s s^{- 1}$ of electricity.

ELECTROCHEMISTRY

19932 The number of electrons passing per second through a cross-section of copper wire carrying $10^{- 6}$ amperes of current per second is found to be

1

1.6 \times 10^{- 19}

2

6 \times 10^{- 35}

3

6 \times 10^{- 16}

4

6 \times 10^{12}

Explanation:

(d) Charge (Coulombs) pass per second $= 10^{- 6}$
number of electrons passed per second
$= \frac{10^{- 6}}{1.602 \times 10^{- 19}} = 6.24 \times 10^{12}$ .

ELECTROCHEMISTRY

19933 The electrolytic cells, one containing acidified ferrous chloride and another acidified ferric chloride are connected in series. The ratio of iron deposited at cathodes in the two cells when electricity is passed through the cells will be

1

1 : 1

2

3 : 1

3

2 : 1

4

3 : 2

Explanation:

(d) At cathode;
$F e^{2 +} + 2 e^{-} \to F e$ ; $F e^{3 +} + 3 e^{-} \to F e$
$(E_{F e})_{1} = \frac{A t o m i c . w e i g h t}{2}; (E_{F e})_{2} = \frac{A t o m i c . w e i g h t}{3}$
Ratio of weight of $F e$ liberated
$= \frac{A t o m i c w e i g h t}{3} : \frac{A t o m i c w e i g h t}{2} = 3 : 2$ .

ELECTROCHEMISTRY

19930 Faraday's laws of electrolysis are related to the

1 Atomic number of cation

2 Atomic number of anion

3 Equivalent weight of the electrolyte

4 Speed of the cation

Explanation:

(c) $w \propto E$ if $i$ and $t$ are constant.

ELECTROCHEMISTRY

19931 The electric charge for electrode decomposition of one gram equivalent of a substance is

1 One ampere per second

2

96500

coulombs per second

3 One ampere for one hour

4 Charge on one mole of electrons

Explanation:

One gram equivalent of a substance requires $96500 c o u l o m b s s^{- 1}$ of electricity.

ELECTROCHEMISTRY

19932 The number of electrons passing per second through a cross-section of copper wire carrying $10^{- 6}$ amperes of current per second is found to be

1

1.6 \times 10^{- 19}

2

6 \times 10^{- 35}

3

6 \times 10^{- 16}

4

6 \times 10^{12}

Explanation:

(d) Charge (Coulombs) pass per second $= 10^{- 6}$
number of electrons passed per second
$= \frac{10^{- 6}}{1.602 \times 10^{- 19}} = 6.24 \times 10^{12}$ .

ELECTROCHEMISTRY

19933 The electrolytic cells, one containing acidified ferrous chloride and another acidified ferric chloride are connected in series. The ratio of iron deposited at cathodes in the two cells when electricity is passed through the cells will be

1

1 : 1

2

3 : 1

3

2 : 1

4

3 : 2

Explanation:

(d) At cathode;
$F e^{2 +} + 2 e^{-} \to F e$ ; $F e^{3 +} + 3 e^{-} \to F e$
$(E_{F e})_{1} = \frac{A t o m i c . w e i g h t}{2}; (E_{F e})_{2} = \frac{A t o m i c . w e i g h t}{3}$
Ratio of weight of $F e$ liberated
$= \frac{A t o m i c w e i g h t}{3} : \frac{A t o m i c w e i g h t}{2} = 3 : 2$ .