Let a fully charged lead storage battery contains 1.5 L of 5 M H 2 S O 4 . What will be the concentration of H 2 S O 4 in the battery after 2.5 ampere current is drawn from the battery for 6 hour?

Step-by-step solution

Number of moles of H 2 S O 4 before electrolysis = M V 1000 M × V m L = 5 × 1.5 = 7.5 Write cell rection ⇒ P b 0 + P b O 2 + 4 + 2 H 2 S O 4 → 2 P b S O 4 + 2 + 2 H 2 O n = 2 For 2 moles of electrons 2 moles of H 2 S O 4 are used Moles of electrons used = I i n A × t i n s e c 96500 = 2.5 × 6 × 3600 96500 = 0.56 Hence, moles of H 2 S O 4 used = 0.56 Remaining mol of H 2 S O 4 = 7.5 - 0.56 = 6.94 Final molarity = n V × 1000 = 6.94 1.5 × 1000 × 1000 = 4.626