Electrolysis Driving Me Crazy!

Chemistry Level 2

The electrode reactions for the charging of a lead storage battery are:

P b S O X 4 + 2 e X P b + S O X 4 X 2 \ce{PbSO4 + 2 e- -> Pb + SO_4^2- } P b S O X 4 + 2 H X 2 O P b O X 2 + S O X 4 X 2 + 4 H X + + 2 e X \ce{PbSO4 + 2 H2O -> PbO2 + SO4^2- + 4H+ + 2e- }

The electrolyte in the battery is an aqueous solution of sulphuric acid.Before charging,the specific gravity of liquid was found to be 1.11 (15.7% H X 2 S O X 4 \ce{H2 SO4} by weight).After charging for 100 hours,the sp.gravity was found to be 1.28 (36.9 % H X 2 S O X 4 \ce{H2 SO4} by wt.).

If the battery contained 2 litres of the liquid calculate the average current used for charging the battery.

2.829 A 1.729 A 2.529 A 1.629 A

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1 solution

Ayon Ghosh
Dec 11, 2017

Initially ; mass of H X 2 S O X 4 = 1.11 2000 15.7 / 100 = 348.58 g \ce{H2SO4} = 1.11*2000*15.7/100 = 348.58 g .

Finally; mass of H X 2 S O X 4 = 1.28 2000 36.9 / 100 = 944.64 g \ce{H2SO4} = 1.28*2000*36.9/100 = 944.64 g .

Now note that the eq.wt of H X 2 S O X 4 \ce{H2SO4} in this reaction is 98 g 98 g (not 49 g 49 g ) because 2 moles are reacting with 2 electrons in the second reaction.

Intially ; Equivalents of H X 2 S O X 4 = 348.58 / 98 = 3.5565 \ce{H2SO4} = 348.58 / 98 = 3.5565

Finally ; Equivalents of H X 2 S O X 4 = 944.64 / 98 = 9.6391 \ce{H2SO4} = 944.64 / 98 = 9.6391

Change = 6.0826 \text{Change} = 6.0826 equivalents \text{equivalents} ; thus the Average current I = 6.0826 96500 / 3600 100 1.63 A I = 6.0826 * 96500 / 3600*100 \approx 1.63 A .

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