Question

20/20 MC

± Acid Concentration in a Lead Battery A lead-acid battery uses a redox reaction in which lead(0) and lead(IV) are both converted to lead(II). This reaction is facilitated by the presence of sulfuric acid, H2SO4, as shown by the reaction Pb+PbO2+2H2SO4→2PbSO4+2H2O Part A Suppose that a fully charged lead-acid battery contains 1.50 L of 5.00 M H2SO4. What will be the concentration of H2SO4 in the battery after 2.60 A of current is drawn from the battery for 9.50 hours ? Express your answer with the appropriate units. Hints SubmitMy AnswersGive Up

Answer 1

First, calculate total acid:

mol of acid = Macid*Vacid = 5*1.5 = 7.5 mol of H2SO4 present

find concentraion when:

2.6 A have worked for 9.5 h = 34,560 seconds

calcualte total charge:

C = I*t = 2.6*34,560 = 89,856 C

recall that:

1 mol of e- = 96500 C (faraday constant)

x mol of e- = 89,856 C

x = 89,856/96500 = 0.9311 mol of e- has been transferred

therefore...

initially we have:

Pb(s) + PbO 2(s) + 2H2SO 4(aq) → 2PbSO 4(s) + 2H 2O(l)

moles being transferred = 2 mol of e-

then...

2 mol of e- = 2 mol of H2SO4

0.9311 mol of e- = 0.9311 mol of H2SO4

now...

this implies that 0.9311 mol of H2SO5 have reacted

mol of H2SO4 left = 7.5-0.9311 = 6.5689 mol of H2SO4

Final concentration

[H2SO4]= mol of aicd / Vtotal = 6.5689/1.5 = 4.37926 M