Question 4 In a lead-acid battery, the following reaction takes place: Pb(s) + PbO2 (s) + 2H₂SO4 (aqueous) 2PbSO4(s) + 2H₂0 (1) (a) Identify relevant information in the table below and calculate standard cell potential for the battery. Half reaction Eº, V -0.1262 -0.580 1.455 -0.537 0.247 Pb²+ + 2e → Pb PbO + H₂O +2e → Pb+20H™ PbO₂ + 4H* +2e¯ → Pb²+ + 2H₂O HPbO₂ + H₂O +2e¯¯ →Pb+30H™ PbO₂ + H₂O +2e → PbO+20H™ PbO₂ +SO¼¯ +4H* +2e¯ → PbSO4 + 2H₂O|1.6913 PbSO4 +2e¯ → Pb+SO²/¯ -0.3588 (b) Assume the activity of water is always 1 and the activities of H+ and SO4²- are equal to the molality (i.e. concentration in mole per 1000 g of H₂O). You can further assume that the concentration of H+ is always twice that of SO4²-. When the battery is in the discharged state, the cell potential drops to 1.7 V. Estimate the concentration of H+ when the battery is in the discharged state [Hint: Don't worry about the change in the amount of water during the reaction. You can assume that we always have the same amount of water].
Question 4 In a lead-acid battery, the following reaction takes place: Pb(s) + PbO2 (s) + 2H₂SO4 (aqueous) 2PbSO4(s) + 2H₂0 (1) (a) Identify relevant information in the table below and calculate standard cell potential for the battery. Half reaction Eº, V -0.1262 -0.580 1.455 -0.537 0.247 Pb²+ + 2e → Pb PbO + H₂O +2e → Pb+20H™ PbO₂ + 4H* +2e¯ → Pb²+ + 2H₂O HPbO₂ + H₂O +2e¯¯ →Pb+30H™ PbO₂ + H₂O +2e → PbO+20H™ PbO₂ +SO¼¯ +4H* +2e¯ → PbSO4 + 2H₂O|1.6913 PbSO4 +2e¯ → Pb+SO²/¯ -0.3588 (b) Assume the activity of water is always 1 and the activities of H+ and SO4²- are equal to the molality (i.e. concentration in mole per 1000 g of H₂O). You can further assume that the concentration of H+ is always twice that of SO4²-. When the battery is in the discharged state, the cell potential drops to 1.7 V. Estimate the concentration of H+ when the battery is in the discharged state [Hint: Don't worry about the change in the amount of water during the reaction. You can assume that we always have the same amount of water].
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ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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![Question 4
In a lead-acid battery, the following reaction takes place:
Pb(s) + PbO2 (s)
+ 2H₂SO4(aqueous) → 2PbSO4(s) + 2H₂0 (1)
(a) Identify relevant information in the table below and calculate standard cell potential for the
battery.
Eº, V
-0.1262
-0.580
1.455
-0.537
PbO₂ + H₂O +2e¯ →PbO+20H™
0.247
4
PbO₂ +SO²¯ +4H* +2e¯ → PbSO +2H₂O 1.6913
PbSO4 +2e → Pb+SO²¯
-0.3588
Half reaction
Pb²+ + 2e → Pb
PbO + H₂O +2e¯¯ → Pb+ 2OH¯
2
PbO₂ +4H* +2e¯ → Pb²+ + 2H₂O
HPbO₂ + H₂O +2e¯ → Pb+30H¯
(b) Assume the activity of water is always 1 and the activities of H+ and SO4²- are equal to the
molality (i.e. concentration in mole per 1000 g of H₂O). You can further assume that the
concentration of H+ is always twice that of SO4². When the battery is in the discharged
state, the cell potential drops to 1.7 V. Estimate the concentration of H+ when the battery
is in the discharged state [Hint: Don't worry about the change in the amount of water
during the reaction. You can assume that we always have the same amount of water].](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fca6b5de9-d666-4be4-bec5-372f49facd74%2Fbc110dd1-a44b-4e75-8e4f-7a47920440a8%2F9nn8z8h_processed.png&w=3840&q=75)
Transcribed Image Text:Question 4
In a lead-acid battery, the following reaction takes place:
Pb(s) + PbO2 (s)
+ 2H₂SO4(aqueous) → 2PbSO4(s) + 2H₂0 (1)
(a) Identify relevant information in the table below and calculate standard cell potential for the
battery.
Eº, V
-0.1262
-0.580
1.455
-0.537
PbO₂ + H₂O +2e¯ →PbO+20H™
0.247
4
PbO₂ +SO²¯ +4H* +2e¯ → PbSO +2H₂O 1.6913
PbSO4 +2e → Pb+SO²¯
-0.3588
Half reaction
Pb²+ + 2e → Pb
PbO + H₂O +2e¯¯ → Pb+ 2OH¯
2
PbO₂ +4H* +2e¯ → Pb²+ + 2H₂O
HPbO₂ + H₂O +2e¯ → Pb+30H¯
(b) Assume the activity of water is always 1 and the activities of H+ and SO4²- are equal to the
molality (i.e. concentration in mole per 1000 g of H₂O). You can further assume that the
concentration of H+ is always twice that of SO4². When the battery is in the discharged
state, the cell potential drops to 1.7 V. Estimate the concentration of H+ when the battery
is in the discharged state [Hint: Don't worry about the change in the amount of water
during the reaction. You can assume that we always have the same amount of water].
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