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JEE Class main Answered

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Asked by swayamagarwal2114 | 12 Aug, 2022, 10:22: AM
Expert Answer
Dear Student,
 

Net cell reaction for the discharge of lead storage battery is:

Pb(s) + PbO2(s) + 2H2SO4(aq) → 2PbSO4(s) + 2H2O(l)

             y mole                          y mole

Initial mass of H2SO4,

w1 =       1000 × 1.26 × 40/100 = 504 gm

 

Final mass of H2SO4,

w2 = (1260 – 98y + 18y) × 28/100  = (352.8 – 22.4y) g

From reaction, 504 – 98y

= 352.8 – 22.4y

y = 2 moles

Thus, final mass w2 = 308 is decreased from initial mass.

No. of moles of H2SO4  reacted,

y = 2 moles

As, 2 moles of electrons have been released in the anode reaction.

Therefore, charge released is 2F

i.e. 2 × 96500 = 1.93 × 105 Coulomb

 

So, all the given options are correct.

  

Answered by | 13 Aug, 2022, 10:31: AM
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