ElectrochemistryHard

Question

In the lead storage battery, the anode reaction is Pb(s) + HSO4− + H2O → PbSO4(s) + H3O+ + 2e−. How many grams of Pb will be used up to deliver 1 A for 100 h? (Pb = 208)

Options

A.776 g
B.388 g
C.194 g
D.0.1 g

Solution

$n_{eq} = \frac{Q}{F} \Rightarrow \frac{w}{208} \times 2 = \frac{1 \times 100 \times 3600}{96500} \Rightarrow w \approx 388\text{ gm}$

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