In a lead storage battery, the anode reaction is, Pb+H2SO4→PbSO4+2H++2e− Weight in g of Pb would be used up for battery to use for 1 ampere hour is___________. (write the nearest integer value)
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Solution
The current passed is 100 ampere hour. The number of Faradays passed is I(A)×t(s)96500=1×360096500=3.73×10−2. 1 mole of Pb will be deposited by 2 Faradays of electricity. The molar mass of Pb is 207 g/mol. The mass of lead deposited is 3.73×10−2×2072=3.86g≈4.