The electrolyte in the lead storage battery is dilute sulphuric acid. The concentration of sulphuric acid in a lead-storage battery must be between $ 4.8\text{ }M $ and $ 5.3\text{ }M $ for most efficient functioning: A $ 5\text{ }mL $ sulphuric acid sample of a particular battery requires $ 50\text{ }mL $ of $ 1.0\text{ }M\text{ }NaOH $ for complete neutralization. Which of the following statements about the functioning of the battery is the most appropriate?
(A) The acid concentration in the battery is not in the most effective range.
(B) The acid concentration in the battery is in the most effective range.
(C) The acid concentration in the battery is hardly in the most effective range.
(D) Only a good mechanic can tell whether or not the acid concentration in the battery is in the most effective range.
Answer
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Hint: We know that a lead storage battery is also known as lead acid battery. The battery makes use of the electrochemical reaction which converts chemical energy into electrical energy by losing electrolytes when dissolved in sulphuric acid.
Complete answer:
Lead acid batteries have a small energy-to-volume ratio and a very low energy-to-weight ratio; its ability to supply high contents reveals that the cell has a large power-to-weight ratio. Lead-acid batteries are categorized by secondary batteries. The chemical reactions, which occur in secondary cells, are reversible. The commercial forms of lead acid batteries consist of six (or) twelve lead storage cells linked together. Lead storage batteries are otherwise called lead-acid batteries. It is a secondary voltaic cell. The anode is made up of spongy porous lead and the cathode is made up of lead dioxide. The electrolyte used is sulfuric acid.
The active materials present on the plate of the battery (electrodes) are lead and lead dioxide. They react with sulfuric acid (electrolyte) to form lead sulfate. The formed lead sulfate is in amorphous state, finely divided and when the battery recharges it again forms lead, lead dioxide and sulfuric acid.
Volume of sample of sulphuric taken $ =5\text{ }mL $ ; Volume of $ NaOH $ required for neutralization $ =50\text{ }mL $ Normality of $ NaOH, $ required for neutralization $ =1.0\text{ }M=1.0\text{ }N $
[Since, n-factor of $ NaOH $ is 1 and $ {{H}_{2}}S{{O}_{4}} $ is two] $ \because neq={{N}_{1}}{{V}_{1}}={{N}_{2}}{{V}_{2}} $
Also here, $ {{N}_{1}}=\dfrac{1\times 50}{5}=10N. $
$ \therefore Molarity=\dfrac{normality}{n-factor}=\dfrac{10}{2}=5M. $
Thus, the acid concentration in lead storage battery is in the most effective range between $ 4.8\text{ }to\text{ }5.3\text{ }M. $
Therefore, correct answer is option B.
Note:
Remember that lead storage batteries are cost efficient, have low internal impedance and are tolerant to overcharging. They are also heavy and bulky with a cycle life of $ 300 $ to $ 500 $ cycles. There are varieties of lead acid batteries such as lead calcium battery, lead antimony battery etc.
Complete answer:
Lead acid batteries have a small energy-to-volume ratio and a very low energy-to-weight ratio; its ability to supply high contents reveals that the cell has a large power-to-weight ratio. Lead-acid batteries are categorized by secondary batteries. The chemical reactions, which occur in secondary cells, are reversible. The commercial forms of lead acid batteries consist of six (or) twelve lead storage cells linked together. Lead storage batteries are otherwise called lead-acid batteries. It is a secondary voltaic cell. The anode is made up of spongy porous lead and the cathode is made up of lead dioxide. The electrolyte used is sulfuric acid.
The active materials present on the plate of the battery (electrodes) are lead and lead dioxide. They react with sulfuric acid (electrolyte) to form lead sulfate. The formed lead sulfate is in amorphous state, finely divided and when the battery recharges it again forms lead, lead dioxide and sulfuric acid.
Volume of sample of sulphuric taken $ =5\text{ }mL $ ; Volume of $ NaOH $ required for neutralization $ =50\text{ }mL $ Normality of $ NaOH, $ required for neutralization $ =1.0\text{ }M=1.0\text{ }N $
[Since, n-factor of $ NaOH $ is 1 and $ {{H}_{2}}S{{O}_{4}} $ is two] $ \because neq={{N}_{1}}{{V}_{1}}={{N}_{2}}{{V}_{2}} $
Also here, $ {{N}_{1}}=\dfrac{1\times 50}{5}=10N. $
$ \therefore Molarity=\dfrac{normality}{n-factor}=\dfrac{10}{2}=5M. $
Thus, the acid concentration in lead storage battery is in the most effective range between $ 4.8\text{ }to\text{ }5.3\text{ }M. $
Therefore, correct answer is option B.
Note:
Remember that lead storage batteries are cost efficient, have low internal impedance and are tolerant to overcharging. They are also heavy and bulky with a cycle life of $ 300 $ to $ 500 $ cycles. There are varieties of lead acid batteries such as lead calcium battery, lead antimony battery etc.
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