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
You are correct that the Pb electrode is the anode and that it is negatively charged. The half reaction is correct also, although that doesn't show the next step to form PbSO4 from the Pb^+. Technically, that is not part of the half reaction, though. For the cathode, what you have written is not correct. The PbO2 is the cathode and it is positively charged. PbO2 goes to Pb^+2 but that isn't all of the half reaction for it.
Your answers for the initially charging scenario are correct:
Anode: Pb --> 2e- + Pb^2+
Cathode: H2O + e- --> H2
For discharging, the reactions are:
Anode: Pb + H2SO4 --> PbSO4 + 2e- + 2H+
Cathode: PbO2 + 2H2SO4 + 2e- --> PbSO4 + 2H2O + SO4^2-
And for recharging, the reactions are the reverse of the discharging reactions:
Anode: PbSO4 + 2e- + 2H+ --> Pb + H2SO4
Cathode: PbSO4 + 2H2O + SO4^2- --> PbO2 + 2H2SO4 + 2e-
Actually, in the initially charging process, the half-reactions at each electrode would be as follows:
Anode (negative electrode): Pb(s) --> Pb^2+(aq) + 2e^-
Cathode (positive electrode): PbO2(s) + 4H^+(aq) + 2e^- --> Pb^2+(aq) + 2H2O(l)
In the discharging process, the half-reactions would be reversed:
Anode (negative electrode): Pb^2+(aq) + 2e^- --> Pb(s)
Cathode (positive electrode): Pb^2+(aq) + 2H2O(l) + SO4^2-(aq) --> PbSO4(s) + 4H^+(aq)
In the recharging process, the half-reactions would be the same as in the initially charging process. The sulfuric acid acts as the electrolyte, providing a medium for the flow of ions during the electrochemical reactions. It helps facilitate the transfer of electrons and maintains the charge balance in the battery.