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CHXII03:ELECTROCHEMISTRY

329927 When lead storage battery is charged

1 lead dioxide dissolves

2 sulphuric acid is regenerated

3 the lead electrode becomes coated with lead sulphate

4 the amount of sulphuric acid decreases

Explanation:

During the charging of a lead storage battery, the reactions at the anode and cathode are :
$A t a n o d e$
${PbSO}_{4} + 2 e^{-} ⟶ Pb + {SO}_{4}^{2 -}$
$A t c a t h o d e$
$P b S O_{4} + 2 H_{2} O \to P b O_{2}$
$+ 4 H^{+} + {S O}_{4}^{2 -} + 2 e_{2}^{-}$
In both the reactions, $H_{2} {SO}_{4}$ is regenerated.

CHXII03:ELECTROCHEMISTRY

329928 Lithium is generally used as an electrode in high energy density batteries. This is because

1 Lithium is the lightest element

2 Lithium has quite high negative reduction potential

3 Lithium is quite reactive

4 Lithium does not corrode easily

CHXII03:ELECTROCHEMISTRY

329929 Some of the batteries are rechargeable:
I. Fuel cell
II. Dry cell,
III. Lead-storage
IV. Nickel-Cadmium,
V. Lithium
Select the batteries which can be recharged

1 II, IV and V

2 III, IV and V

3 I, II and III

4 None of above

CHXII03:ELECTROCHEMISTRY

329931 During the discharge of a lead storage battery, the density of $40 % H_{2} {SO}_{4}$ by weight fell from 1.225 to 0.98 (which is $20 %$ by weight). What is the change in molarities of $H_{2} {SO}_{4}$ ?

1 1

2 2

3 3

4 4

Explanation:

Discharging reaction:
$Pb O_{2} + 4 H^{\oplus} + S {O_{4}}^{2 -} + 2 e^{-} ⇌ PbS O_{4} + 2 H_{2} O$
$M_{1} = \frac{% by weight \times 10 \times d}{M w_{2}} = \frac{40 \times 1.225 \times 10}{98}$
$= 5 M$
$M_{2} = \frac{% by weight \times 10 \times d}{M w_{2}} = \frac{20 \times 10 \times 0.98}{98}$
$= 2 M$
Change is molarities $= M_{1} - M_{2} = 5 - 2 = 3 M$

CHXII03:ELECTROCHEMISTRY

329927 When lead storage battery is charged

1 lead dioxide dissolves

2 sulphuric acid is regenerated

3 the lead electrode becomes coated with lead sulphate

4 the amount of sulphuric acid decreases

Explanation:

During the charging of a lead storage battery, the reactions at the anode and cathode are :
$A t a n o d e$
${PbSO}_{4} + 2 e^{-} ⟶ Pb + {SO}_{4}^{2 -}$
$A t c a t h o d e$
$P b S O_{4} + 2 H_{2} O \to P b O_{2}$
$+ 4 H^{+} + {S O}_{4}^{2 -} + 2 e_{2}^{-}$
In both the reactions, $H_{2} {SO}_{4}$ is regenerated.

CHXII03:ELECTROCHEMISTRY

329928 Lithium is generally used as an electrode in high energy density batteries. This is because

1 Lithium is the lightest element

2 Lithium has quite high negative reduction potential

3 Lithium is quite reactive

4 Lithium does not corrode easily

CHXII03:ELECTROCHEMISTRY

329929 Some of the batteries are rechargeable:
I. Fuel cell
II. Dry cell,
III. Lead-storage
IV. Nickel-Cadmium,
V. Lithium
Select the batteries which can be recharged

1 II, IV and V

2 III, IV and V

3 I, II and III

4 None of above

CHXII03:ELECTROCHEMISTRY

329931 During the discharge of a lead storage battery, the density of $40 % H_{2} {SO}_{4}$ by weight fell from 1.225 to 0.98 (which is $20 %$ by weight). What is the change in molarities of $H_{2} {SO}_{4}$ ?

1 1

2 2

3 3

4 4

Explanation:

Discharging reaction:
$Pb O_{2} + 4 H^{\oplus} + S {O_{4}}^{2 -} + 2 e^{-} ⇌ PbS O_{4} + 2 H_{2} O$
$M_{1} = \frac{% by weight \times 10 \times d}{M w_{2}} = \frac{40 \times 1.225 \times 10}{98}$
$= 5 M$
$M_{2} = \frac{% by weight \times 10 \times d}{M w_{2}} = \frac{20 \times 10 \times 0.98}{98}$
$= 2 M$
Change is molarities $= M_{1} - M_{2} = 5 - 2 = 3 M$

CHXII03:ELECTROCHEMISTRY

329927 When lead storage battery is charged

1 lead dioxide dissolves

2 sulphuric acid is regenerated

3 the lead electrode becomes coated with lead sulphate

4 the amount of sulphuric acid decreases

Explanation:

During the charging of a lead storage battery, the reactions at the anode and cathode are :
$A t a n o d e$
${PbSO}_{4} + 2 e^{-} ⟶ Pb + {SO}_{4}^{2 -}$
$A t c a t h o d e$
$P b S O_{4} + 2 H_{2} O \to P b O_{2}$
$+ 4 H^{+} + {S O}_{4}^{2 -} + 2 e_{2}^{-}$
In both the reactions, $H_{2} {SO}_{4}$ is regenerated.

CHXII03:ELECTROCHEMISTRY

329928 Lithium is generally used as an electrode in high energy density batteries. This is because

1 Lithium is the lightest element

2 Lithium has quite high negative reduction potential

3 Lithium is quite reactive

4 Lithium does not corrode easily

CHXII03:ELECTROCHEMISTRY

329929 Some of the batteries are rechargeable:
I. Fuel cell
II. Dry cell,
III. Lead-storage
IV. Nickel-Cadmium,
V. Lithium
Select the batteries which can be recharged

1 II, IV and V

2 III, IV and V

3 I, II and III

4 None of above

CHXII03:ELECTROCHEMISTRY

329931 During the discharge of a lead storage battery, the density of $40 % H_{2} {SO}_{4}$ by weight fell from 1.225 to 0.98 (which is $20 %$ by weight). What is the change in molarities of $H_{2} {SO}_{4}$ ?

1 1

2 2

3 3

4 4

Explanation:

Discharging reaction:
$Pb O_{2} + 4 H^{\oplus} + S {O_{4}}^{2 -} + 2 e^{-} ⇌ PbS O_{4} + 2 H_{2} O$
$M_{1} = \frac{% by weight \times 10 \times d}{M w_{2}} = \frac{40 \times 1.225 \times 10}{98}$
$= 5 M$
$M_{2} = \frac{% by weight \times 10 \times d}{M w_{2}} = \frac{20 \times 10 \times 0.98}{98}$
$= 2 M$
Change is molarities $= M_{1} - M_{2} = 5 - 2 = 3 M$

CHXII03:ELECTROCHEMISTRY

329927 When lead storage battery is charged

1 lead dioxide dissolves

2 sulphuric acid is regenerated

3 the lead electrode becomes coated with lead sulphate

4 the amount of sulphuric acid decreases

Explanation:

During the charging of a lead storage battery, the reactions at the anode and cathode are :
$A t a n o d e$
${PbSO}_{4} + 2 e^{-} ⟶ Pb + {SO}_{4}^{2 -}$
$A t c a t h o d e$
$P b S O_{4} + 2 H_{2} O \to P b O_{2}$
$+ 4 H^{+} + {S O}_{4}^{2 -} + 2 e_{2}^{-}$
In both the reactions, $H_{2} {SO}_{4}$ is regenerated.

CHXII03:ELECTROCHEMISTRY

329928 Lithium is generally used as an electrode in high energy density batteries. This is because

1 Lithium is the lightest element

2 Lithium has quite high negative reduction potential

3 Lithium is quite reactive

4 Lithium does not corrode easily

CHXII03:ELECTROCHEMISTRY

329929 Some of the batteries are rechargeable:
I. Fuel cell
II. Dry cell,
III. Lead-storage
IV. Nickel-Cadmium,
V. Lithium
Select the batteries which can be recharged

1 II, IV and V

2 III, IV and V

3 I, II and III

4 None of above

CHXII03:ELECTROCHEMISTRY

329931 During the discharge of a lead storage battery, the density of $40 % H_{2} {SO}_{4}$ by weight fell from 1.225 to 0.98 (which is $20 %$ by weight). What is the change in molarities of $H_{2} {SO}_{4}$ ?

1 1

2 2

3 3

4 4

Explanation:

Discharging reaction:
$Pb O_{2} + 4 H^{\oplus} + S {O_{4}}^{2 -} + 2 e^{-} ⇌ PbS O_{4} + 2 H_{2} O$
$M_{1} = \frac{% by weight \times 10 \times d}{M w_{2}} = \frac{40 \times 1.225 \times 10}{98}$
$= 5 M$
$M_{2} = \frac{% by weight \times 10 \times d}{M w_{2}} = \frac{20 \times 10 \times 0.98}{98}$
$= 2 M$
Change is molarities $= M_{1} - M_{2} = 5 - 2 = 3 M$

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