Question

In the lead storage battery, the anode is $Pb\left( s \right) + HS{O_4}^ - + {H_2}O \to PbS{O_4}\left( s \right) + {H_3}{O^ + } + 2{e^ - }$. How many grams of Pb will be used up to deliver 1A for 100h? (Pb=208)
A.776 g
B.388 g
C.194 g
D.0.1 g

Answer 1

Hint:We can calculate the mass of lead by multiplying the number of Faradays passed with the molar mass of lead divided by number of electrons transferred. We can calculate the number of Faradays passed by taking the product of current and time divided by Faraday's constant.
Complete step by step answer:
The given chemical equation is,
$Pb\left( s \right) + HS{O_4}^ - + {H_2}O \to PbS{O_4}\left( s \right) + {H_3}{O^ + } + 2{e^ - }$
Given data contains,
Current (in Ampere)=1 A
Time taken is 100h.
Molar mass of lead is 208g/mol.
We have to convert the time taken in hours to seconds,
$t = 100h \times \dfrac{{3600s}}{{1h}}$
$t = 360000s$
Let us now calculate the amount of Faradays passed. Faraday's pass is calculated by taking the product of current and time. The product is divided by Faraday's constant.
The value of Faraday's constant is $96500C \cdot mo{l^{ - 1}}$.
We can calculate the amount of Faradays passed as,
Amount of Faradays passed=\[\dfrac{{I\left( A \right) \times t\left( s \right)}}{{96500C \cdot mo{l^{ - 1}}}}\]
Let us now substitute the values of current and time in the equation.
Amount of Faradays passed=$\dfrac{{1\left( A \right) \times 360000\left( s \right)}}{{96500C \cdot mo{l^{ - 1}}}}$
Amount of Faradays passed=$3.7305$
One mole of lead would be deposited by two Faradays of electricity.
We know that the molar mass of lead is 208 g/mol.
We can now calculate the mass of lead deposited by taking the product of the amount of Faradays passed with molar mass. The product is divided by the number of electrons transferred.
The number of electrons transferred is two.
We can calculate the mass of lead as,
Mass of lead=$\dfrac{{3.7305 \times 208g/mol}}{{2mol}}$
Mass of lead=$388g$
The mass of lead deposited is $388g$.
Therefore, option (B) is correct.

Note: Lead acid batteries are used in large backup power supplies for telephone and compound centers, grid energy storage and as emergency lighting and to power sump pumps in power failure. Lead storage batteries are cost efficient. They 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.