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Faraday's First Law

During the di...

Question

During the discharge of a lead storage battery, density of $H_{2} S O_{4}$ fell from $1.294$ to $1.139$ \ g/mL $. S u l p h u r i c a c i d o f d e n s i t y$ 1.294 $i s$ 39H_{2}SO_{4} $b y w e i g h t a n d t h a t o f d e n s i t y$ 1.139\ g/mL $i s$ 20H_{2}SO_{4} $b y w e i g h t . T h e b a t t e r y h o l d$ 3.5\ litre $o f t h e a c i d a n d v o l u m e r e m a i n s p r a c t i c a l l y c o n s t a n t d u r i n g d i s c h a r g e . C a l c u l a t e a m p e r e - h o u r o f w h i c h t h e b a t t e r y m u s t h a v e b e e n u s e d . T h e c h a r g i n g a n d d i s c h a r g i n g r e a c t i o n s a r e :$ Pb + SO_{4}^{2-} \rightarrow PbSO_{4} + 2e^{-} $(c h a r g i n g)$ PbO_{2} + 4H^{+} + SO_{4}^{2-} + 2e^{-}\rightarrow PbSO_{4} + 2H_{2}O$ (discharging).

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Solution

Weight of solution before discharge $= 3500 \times 1.294$
$= 4529 g$
Weight of $H_{2} S O_{4}$ before discharge $= \frac{39}{100} \times 4529$
$= 1766.31 g$
Weight of solution after discharge $= 3500 \times 1.139$
$= 3986.5 g$
Weight of $H_{2} S O_{4}$ after discharge $= \frac{20}{100} \times 3986.5$
$= 797.3 g$
Loss in mass of $H_{2} S O_{4}$ during discharge
$= 1766.31 - 797.3 = 969.01 g$
Now from first law of electrolysis,
$W = \frac{Q \times E}{96500}$
$969.01 = \frac{Q \times 98}{96500}$
$Q = 954178.21 c o u l o m b$
$A m p e r e - h o u r = \frac{C o u l o m b}{3600} = \frac{954178.21}{3600}$
$= 265.04 a m p e r e - h o u r$ .

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Q. During the discharge of a lead storage battery, the density of sulphuric acid fell from 1.294 to 1.139 g

m l^{- 1}

and sulphuric acid of the density of 1.294 g

m l^{- 1}

is 39% by mass and that of the density of 1.139 g

m l^{- 1}

is 20% by mass. The battery holds 3.5 litre of acid and the volume practically remained constant during the discharge. Calculate the no. of ampere hour for which the battery must have been used. The charging and discharging reactions are:

P b + S O_{4}^{2 -} \to P b S O_{4} + 2 e ⟹

P b O_{2} + 4 H^{+} + S O_{4}^{2 -} + 2 e \to P b S O_{4} + 2 H_{2} O ⟹

Q. During the discharge of a lead storage battery, the density of sulphuric acid fell from

1.294

1.139 {g mL}^{- 1}

. Sulphuric acid of density

1.294 {g mL}^{- 1}

39 % H_{2} S O_{4}

by mass and that of density

1.139 {g mL}^{- 1}

20 % H_{2} S O_{4}

by mass. The battery holds

3.5 L

of the acid and the volume remained practically constant during the discharge. The charging and discharging reactions are

P b + S O_{4}^{2 -} \to P b S O_{4} + 2 e^{-}

P b O_{2} + 4 H^{+} + S O_{4}^{2 -} + 2 e^{-} \to P b S O_{4} + 2 H_{2} O

(discharging)
The number of ampere-hour for which the battery must have been used is

Q. During the discharge of a lead storage battery, the density of sulphuric acid fell from 1.294 to 1.139

g m L^{- 1}

H_{2} S O_{4}

1.294 g m L^{- 1}

is 39% and that of density

1.139 g m L^{- 1}

is 20% by weight.The battery holds 3.5 L of acid and the volume practically remains constant during discharge. The number of ampere hours for which the battery must have been used are........................
The discharging and charging reactions are:

P b + S O_{4}^{2 -} \to P b S O_{4} + 2 e^{-}

P b O_{2} + 4 H^{+} + S O_{4}^{2 -} + 2 e^{-} \to P b S O_{4} + 2 H_{2} O

During the discharge of a lead storage battery, the density of sulphuric acid fell from $1.294$ to $1.139$ gm $L^{- 1}$ . Sulphuric acid of density $1.294$ gm $L^{- 1}$ is $39$ % by weight and that of density 1.139 gm $L^{- 1}$ is $20$ % by weight. The battery holds $3.5$ litres of acid and the column practically remained constant during the discharge.

Calculate the number of ampere-hours for which the battery must have been used. The charging and discharging reactions are :

$P b + S O {_{4}}^{2 -} \to P b S O_{4} + 2 e^{-}$ (charging)

$P b O_{2} + 4 H^{+} + S O {_{4}}^{2 -} + 2 e^{-} \to P b S O_{4} + 2 H_{2} O$ (discharging)

Q. During the discharge of a lead storage battery, the density of sulphuric acid fell from

1.294

1.139 g m l^{- 1}

. Sulphuric acid of density

1.294 g m l^{- 1}

is 39% by weight and that of density

1.139 g m l^{- 1}

is 20% by weight. The battery holds 3.5 L of the acid and the volume remained practically constant during the discharge. Find the number of ampere-hours for which the battery must has been used.

The charging and discharging reactions are as follows:

$P b + S O_{4}^{2 -} \to P b S O_{4} + 2 e^{-}$ (charging)

$P b O_{2} + 4 H^{\oplus} + S O_{4}^{2 -} + 2 e^{-} \to P b S O_{4} + 2 H_{2} O$ (discharging)

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