ElectrochemistryHard

Question

A lead storage battery has initially 200 g of lead and 200 g of PbO2 plus excess H2SO4. Theoretically, how long could this cell deliver a current of 10.0 A, without recharging, if it were possible to operate it so that the reaction goes to completion? (Pb = 208)

Options

A.16083.33 s
B.8041.67 s
C.44.68 s
D.18557.7 s

Solution

Cell reaction during discharge:

Pb + PbO2 + 2H2SO4 ?2PbSO4 + 2H2O

$n_{Pb}\text{ taken = }\frac{200}{208}\text{ and }n_{PbO_{2}}\text{ taken} = \frac{200}{240}$

Hence, PbO2 is L.R.

Now, $n_{eq}PbO_{2} = \frac{Q}{F} \Rightarrow \frac{200}{240} \times 2 = \frac{10 \times t}{96500}$

⇒ t = 16083.33 sec

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