When lead-acid accumulator battery is discharged- the following reaction takes place- Pbs-PbO2-2H2SO4aq- 2PbSO4s-2H2OIf 1-25A average current is drawn for a period of 1930 minutes- the mass of H2SO4 consumed is -

The correct option is B 73.5 gmThe mass of H_2SO_4 consumed is equal to the mass of H_2SO_4 #160;deposited. W=Zit=Eit96500=49× 1.25× 1930× 60 ...

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Standard XII

Chemistry

Faraday's First Law

When lead-aci...

Question

When lead-acid accumulator (battery) is discharged, the following reaction takes place.
$P b (s) + P b O_{2} + 2 H_{2} S O_{4} (a q) \to 2 P b S O_{4} (s) + 2 H_{2} O$
If $1.25 A$ average current is drawn for a period of 1930 minutes, the mass of $H_{2} S O_{4}$ consumed is :

196 gm

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73.5 gm

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98 gm

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147 gm

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Solution

The correct option is B 73.5 gm
The mass of $H_{2} S O_{4}$ consumed is equal to the mass of $H_{2} S O_{4}$ deposited.
$W = Z i t = \frac{E i t}{96500} = \frac{49 \times 1.25 \times 1930 \times 60}{96500} = 73.5 g m$ .

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Similar questions

During discharging of lead-storage acid battery following reaction takes place:

P b (s) + P b O_{2} (s) + 2 H_{2} S O_{4} \to 2 P b S O_{4} (s) + 2 H_{2} O

If 2.5 amp of current is drawn for 965 minutes, $H_{2} S O_{4}$ consumed is:

Q. In the following reaction

P b

undergoes:

P b (s)

P b O_{2} (s)

2 H_{2} S O_{4} (a q)

\to

2 P b S O_{4} (s)

2 H_{2} O (l)

Q. A lead storage cell is discharged which causes the

H_{2} S O_{4}

electrolyte from a concentration of

34.6

% by mass (density 1.261 g

m l^{- 1}

28^{o} C

) to one of

27

% by mass. The original volume of electrolyte is one litre. How many Faraday have left the anode of battery? Note the water is produced by the cell reaction and

H_{2} S O_{4}

is used up. Overall reaction is :

P b (s) + P b O_{2} + 2 H_{2} S O_{4} (l) \to 2 P b S O_{4} (s) + 2 H_{2} O

Q. Identify the substance oxidised, reduced, oxidising agent and reducing agent for each of the following reactions :

P b_{(s)} + P b O_{2 (s)} + 2 H_{2} S O_{4 (a q)} \to 2 P b S O_{4 (s)} + 2 H_{2} O_{(l)}

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)

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When lead-acid accumulator (battery) is discharged, the following reaction takes place. Pb(s)+PbO2+2H2SO4(aq)→2PbSO4(s)+2H2OIf 1.25A average current is drawn for a period of 1930 minutes, the mass of H2SO4 consumed is :

The correct option is B 73.5 gmThe mass of H2SO4 consumed is equal to the mass of H2SO4 deposited.W=Zit=Eit96500=49×1.25×1930×6096500=73.5 gm.

When lead-acid accumulator (battery) is discharged, the following reaction takes place.
$P b (s) + P b O_{2} + 2 H_{2} S O_{4} (a q) \to 2 P b S O_{4} (s) + 2 H_{2} O$
If $1.25 A$ average current is drawn for a period of 1930 minutes, the mass of $H_{2} S O_{4}$ consumed is :

The correct option is B 73.5 gm
The mass of $H_{2} S O_{4}$ consumed is equal to the mass of $H_{2} S O_{4}$ deposited.
$W = Z i t = \frac{E i t}{96500} = \frac{49 \times 1.25 \times 1930 \times 60}{96500} = 73.5 g m$ .