A led acid-battery is designed that the new battery contains 62.1 grams of lead

Question

A led acid-battery is designed that the new battery contains 62.1 grams of lead for the anode and unknown grams of the lead dioxide for the cathode. If the anode material was and completely consumed, 20.2 grams of lead dioxide remained unconsumed a) How much charge in Coulombs could the battery provide? b) what percent of the cathode material was consumed? c) If current is drawn at a constant rate of 2.0 A, how long (in minutes) will the battery last? Anode Reaction Pb + SO_4^-2 rightarrow PbSO_4 + 2e^- Cathode Reaction PbO_2 + SO_4^-2 + 4H^+ + 2e^- rightarrow PbSO_4 + 2H_2O

Explanation / Answer

(a) From the anode reaction,

1 mol of Pb i.e 207.2 g. of lead can produce 2 mol (or) 2 Faraday (or) 2 * 96485 C of charge

then, 62.1 g. of Lead can produce 62.1 * 2 * 96485 / 207.2 = 57835 C of charge.

(b)

Balanced cell equation is,

Pb (s) + PbO2 (s) + 2 SO42- (aq.) + 4 H+ (aq.) ------------> 2 PbSO4 (s) 2 H2O (l)

from the above balanced equation,

1 mol of Pb requires 1 mol of PbO2

then, 62.1 / 207.2 mol of Pb needs 62.1 * 1 / 207.2 = 0.300 mol of PbO2

So, mass of PbO2 consumed = 0.300 * (207.2 + 32) = 71.8 g.

Therefore, total mass of PbO2 at cathode initially = 71.8 + 20.2 = 92 g.

Now, % of PbO2 consumed = 71.8 * 100 / 92 = 78.0 %

(c)

According to Faraday's first law of electrolysis,

W = M c t / Z F

t = W Z F / M c

t = 62.1 * 2 * 96485 / (207.2 * 2.00)

t = 28918.56 s

t = 481.96 min

t = 8.03 h

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