# Question 9b0c6

Apr 3, 2017

$6.90310880829 \text{mole of } {H}_{2} S {O}_{4}$

#### Explanation:

First convert the units to the appropriate units

$\left(5 \cancel{\text{hour")/1 * (60cancelmin)/cancel("hour}}\right) \cdot \frac{60 \sec}{\cancel{\min}} = 18 , 000 \sec$

3.2A means 3.2coulombs/sec. Now calculate coulombs passed in 5 hours.

$\text{3.2coulombs"/cancelsec) xx 18,000cancelsec = "57,600 coulombs}$

Now what is happening at the anode

Anode is the place where the oxidation takes place. The Pb is oxidized to form $P {b}^{2 +}$ and two electrons are lost per atom of lead. At the end the lead anode will lose mass.

Use the Faraday's constant to calculate the moles of electrons lost and gained in total.

(57,600 cancel"Coulombs")xx ("1 mole of electrons") / (96500 cancel"Coulombs") = "0.59689119171 mole of electrons"

0.59689119171 mole of electrons were released from lead and was used to reduce PbO2 to Pb2+ ions

For calculating the moles of Pb reduced to Pb2+ we must do some calculation

$\text{0.59689119171"cancel" moles of electron" xx ("1 mole of" Pb^(2+))/"2 "cancel"mole of electron} =$0.29844559585 of Pb^+2

An equal amount of Pb2+ ions is formed from PbO2 giving a total of 0.59689119171mole of Pb+2

Pb and PbO2 reacts with the ${H}_{2} S {O}_{4}$ and to form 1mole PbSO4 1mole of H2SO4 is required. So for 0.59689119171mole of Pb+2 0.59689119171mole of H2SO4 is required and this amount of H2SO4 leaves the solution.

So calculate the moles of ${H}_{2} S {O}_{4}$ in 1.5L of 5M solution

1.5cancelL xx (5 "mole")/cancelL H_2SO_4 = 7.5"moles of" H_2SO_4

$7.5 \text{moles of" H_2SO_4 - 0.59689119171"mole of" H_2SO_4 = 6.90310880829"mole of } {H}_{2} S {O}_{4}$