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April 17, 2014

Homework Help: Chemstry

Posted by Sara on Monday, April 28, 2008 at 8:32pm.

I just copied this because i have a couple more questions on this part....

Sulfuric acid/ Lead Battery:
-the electrodes for the battery are composed of Pb(s), PbO2, and/or PbSO4. What are the half reaction occuring at each electrode? include electrolysis of water:

Process Anode/Cathode

Initial Charging:
Discharging
Recharging

Responses

* Chemistry - DrBob222, Saturday, April 26, 2008 at 5:31pm

Surely you don't want me to give you all the answers. Perhaps you just need a hint or two. Tell me what you don't understand about the next step if you get stuck.
The battery in its fully charged state is Pb is one pole and PbO2 is the other. Pb is oxidized to Pb^+2 where it reacts with the H2SO4 to form PbSO4.
The other pole is PbO2 and it is reduced to Pb^+2 where it reacts with H2SO4 to form PbSO4. At its fully discharged state, both electrodes are PbSO4. This should get you started.

* Chemistry - sara, Saturday, April 26, 2008 at 6:16pm

1)I know that initially charging, PbO2 is transferring e- to Pb. So when I charge lead electrodes with Battery, am I sending proton to the lead and receiving back the e-? so Pb is the battery?

since Pb is accepting e-, it's being oxidized [anode] PbO2 is donating e-, so being reduced [cathode]

2)for discharging: Pb is transferring e- to PbO2. Pb is cathode and PbO2 is anode?

3) recharging: does this process involve sulfuric acid? why are we using sulfuric acid???

THANKS

* Chemistry - DrBob222, Saturday, April 26, 2008 at 7:10pm

For a battery that is fully charged and you are using it (discharging it),
Pb + H2SO4 ==> PbSO4 + 2e + 2H^+
PbO2 + 2H2SO4 2e ==> PbSO4 + 2H2O + SO4^=

Remember that oxidation is the loss of electrons and reduction is the gain of electrons. For discharging, Pb is the anode and negatively charged while PbO2 is the cathode and is + charged.

H2SO4 is the electrolyte .

Please check my answers:
Initially charging: Pb is an anode: it's half rxn would then be: Pb-->2e- + Pb^2+

H is the cathode: so the half rxn would be: H2O + e- ---> H2

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