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# Lead storage battery

Explore the process of balancing redox reactions in acidic solutions using a lead storage battery as an example. Understand the assignment of oxidation states, the creation of oxidation and reduction half-reactions, and how to balance atoms and charge. Discover how this reaction powers car batteries and can be reversed for recharging. Created by Jay.

## Want to join the conversation?

• WHy did he add SO4 in both sides at ? Why did he do it again for the reduction at ?
(22 votes)
• He has no choice. The Pb exists as PbSO₄ on the right hand side. If he has sulfate on the right, he must have sulfate on the left.

For the OXIDATION half-reaction
Skeleton ionic equation: Pb → PbSO₄
Balance sulfate: Pb + SO₄²⁻ → PbSO₄
Balance charge: Pb + SO₄²⁻ → PbSO₄ + 2e⁻

For the REDUCTION half-reaction
Skeleton ionic equation: PbO₂ → PbSO₄
Balance sulfate: PbO₂ + SO₄²⁻ → PbSO₄
Balance O: PbO₂ + SO₄²⁻ → PbSO₄ + 2H₂O
Balance H: PbO₂ + SO₄²⁻ + 4H⁺ → PbSO₄ + 2H₂O
Balance charge: PbO₂ + SO₄²⁻ + 4H⁺ + 2e⁻ → PbSO₄ + 2H₂O
(54 votes)
• why cant we right it as H2SO4 instead of 2H+ and 2HSO4-
(14 votes)
• In a solution, the H2SO4 (sulphuric acid) is mainly present in its ionised form H+ and HSO4- as it is a strong acid, so you could write H2SO4 if you wanted, but H+ and HSO4- is more accurate in terms of what is exactly in the solution
(19 votes)
• At , he mentions voltaic cell and electrolytic cell. What is the difference between the two?
(6 votes)
• voltaic cells are also called galvanic cells. They are spontaneous reactions so recall when the spontaneous reactions give off energy and so these generate power, through the half cell reactions, as I understand this is your base set up for a battery so that's why you hear people compare it to a battery. Electrolytic cells require energy to run so a battery or some sort of power source is needed for the reaction to occur.
(11 votes)
• How did he get those numbers in the first 1.25 minutes of the video? I"ve always had trouble figuring this out.
(7 votes)
• the oxidation numbers ? molecules are of neutral charge so we simply balance charges....the sum of the oxidation multiplied by the number of atoms of that particular atom in a molecule is 0
in case of PbO2 Pb is bonded to two oxygen atoms...each oxygen atom has an oxidation state of -2 so net become -4 since the molecule is neutral the oxidation state of Pb has to be +4............
in non polar and elemental compounds like pure metal or n2 molecule or o2 molecule the oxidation state is taken to be 0.....
you could say that oxidation state refers to the number of electrons that are being shared....get me ?
(6 votes)
• how exactly is the reverse mechanism going on?
(7 votes)
• If I understand your question correctly, To reverse the reaction in a rechargeable battery, a current is applied to the battery, and it creates a non spontaneous reaction where the precipitated lead moves back so that the cell has electrical potential.
(2 votes)
• ah so we are to assume that this is in a solution of sulfuric acid? would it not be more correct to put the h2so4 under the arrow to tell the reader that this is all happening in that acid?
(5 votes)
• The sulfuric acid is part of the chemical reaction and so it is not a catalyst. Usually reagents placed under the arrow are catalysts that are either not consumed during the process or recreated during the reaction.
(3 votes)
• So, would the shorthand notation for this reaction be Pb | PbSO4 || PbO2 | PbSO4?
(6 votes)
• To expand on the first question, at , PbO2 is +4 and -4 (multiply O by 2), but on the product side, PbSO4 is +2 and -2. Is the reason that we don't multiply O on the product side by 4 is that SO4 -2 is an ion and so it stays -2, (I guess 2- because it is an oxidation state we're talking about, not a charge). So to do Redox reactions, I really need to know my ions, don't I?
(4 votes)
• In this case, you should know that Hydrogen has charge 1+, so if you take H2SO4 and split it to H2 and SO4, than SO4 has 2- charge, because it is missing 2 hydrogens. So if you know where have you got that ion from than you can derive its charge.
(2 votes)
• How did he get the 2V and 12V near the end of the video? And why did he say that it is reverse?
(4 votes)
• You can get this from a table of Standard Electrode Potentials.
Once you have 2v/cell, you can add 6 cells in series to get 12V. This is what happens in car batteries. It is made up of 6 such cells.
(3 votes)
• At , did he mean to write 2HS04 (2-)? If not, why is the charge only -1 now?
(1 vote)
• Because he combined H^+ and SO4^2-

The + charge cancels out one of the - charges, meaning it is HSO4^-
(3 votes)

## Video transcript

- [Voiceover] Here's the reaction for a lead storage battery. Solid lead plus lead dioxide gives lead sulfate. Our goal is to balance this redox reaction in acidic solution. For a lead storage battery, you're dealing with sulfuric acid. In a previous video I've already gone over how to balance a redox reaction in acidic solution. So this video is just a review of that one using the lead storage battery as an example. The first thing we need to do is assign oxidation states. And we'll start with solid lead. The oxidation state of an element is zero, so it's zero for solid lead. Move on to lead dioxide. For oxygen, the oxidation state is negative two. And we have two of them, giving us negative four for the total. The sum of the oxidation states in a neutral compound is zero, so it must be plus four for lead. We move onto lead sulfate. We know that sulfate is SO four two minus, therefore we're dealing with lead two plus. The lead at two plus cations. So the oxidation state is plus two. Next we write our oxidation half-reaction. So what is being oxidized here? Solid lead is going from an oxidation state of zero to an oxidation state of plus two. That's an increase in the oxidation state. So that's oxidation. So we write down here solid lead is going to lead two plus ion for our oxidation. We have sulfate present, so we need to have sulfate on both sides. So if we have solid lead plus SO four two minus, that goes to lead sulfate, PbSO four. Next we look at our atoms. And we have one lead on the left and one lead on the right. One sulfur on the left, one sulfur on the right. Four oxygens on the left and four oxygens on the right. All of our atoms are balanced. We move on to balancing the charge. You balance charge by adding electrons. So let's figure out how many electrons we need to add and where do we add them? On the left side we have two negative charges. On the right side our overall charge is zero. So we need to make those charges equal. And we can do that by adding two electrons to the right side. So now the charges are balanced. Remember, oxidation is loss of electrons. So this is the half-reaction that occurs at the anode of our battery. Next we want to do the reduction half-reaction. So we need to write down our reduction half-reaction. We can see that we're going from plus four to plus two. That's a decrease in the oxidation state. So for our reduction half-reaction, it would be PbO two going to Pb two plus. Once again we need to add in our sulfate, since that is present. So on the left side we would have PbO two plus SO four two minus. And then going to PbSO four on the right. Next we need to balance our atoms. So we'll start with lead. One lead on the left, one lead on the right. And once that's done and the sulfur's balanced too, one sulfur on the left, one sulfur on the right, we look at oxygen. So let's look at the left side here. So we have two oxygens here and four here, for a total of six oxygens. So six oxygens on the left. And on the right we have four oxygens. So four oxygens on the right. You balance oxygen by adding water. So we need two oxygens on the right side. So we can get that by adding two waters. So if we add two waters, now we have the proper number of oxygens. We just added two more oxygens to the right side. Next we balance hydrogens by adding H plus. So let's look at the right side. We have four hydrogens on the right. And on the left we have zero. We have zero hydrogens so we need, we need to add four protons to the left side of our half-reaction. So if we add four protons to the left side, now hydrogen balances. We have four on both sides. Next we balance charge. So what's the overall charge on the left side? We have four positive charges here and two negative charges. So we have a total of two plus on the left side. On the right side, it would be zero. So we need to add, we need to add two electrons to the left side. If we add two electrons to the left side, we get our charges balanced. So now we have our reduction half-reaction. Our reduction half-reaction occurs at the cathode of our battery. So let's add our two half-reactions together. And we can do this because we have the same number of electrons. So let me, let me box this half-reaction. So for our oxidation half-reaction we lost two electrons. And for our reduction half-reaction we gained two electrons. So those electrons cancel out. So we're gonna add our two half-reactions together to get the overall reaction for a lead storage battery. So let's add all of our reactants. Let's start with these up here. So we're going to have solid lead and sulfate. Let's start by writing that. So we have Pb plus SO four two minus. And then for our reactants down here we have four H plus plus PbO two plus SO four two minus. So we have four H plus plus PbO two plus SO four two minus. For the products we have PbSO four, so we have PbSO four, and then down here we have PbSO four and two H two O. So PbSO four plus two H two O. We've already cancelled out the electrons. So we don't need to worry about that. Let's simplify this. So how could we simplify this? On the left side we have lead, so Pb, plus PbO two. So Pb plus PbO two. Then we have four H plus, so four H plus. And then we have two sulfates. We have two sulfates here. So we'll write two SO four two minus. And then for the products we have two PbSO fours. So that would be two PbSO four plus, that's a four here, plus two H two O. Alright, so we've finally done it. This is our overall reaction for a lead storage battery. And sometimes you might see this written a little bit different in a textbook. You might see two of these protons added on to these two sulfates to give you two HSO four minus, leaving two protons left over. So two H plus, so that's just another way to write what's happening here. So this is, this is a voltaic cell. You get a voltage out of this spontaneous redox reaction. And one of these cells has about positive two volts. So the cell potential is about positive two volts. So if you take six of these cells and add them together, you get a car battery. And a car battery has a total of 12 volts. So six cells, each one with two volts. When the car battery discharges, the spontaneous redox reaction delivers a current and your car engine starts. Once your car is running, the reaction is reversed. And that allows you to recharge your lead storage battery. The reverse reaction is ready to go, because your lead sulfate here is a solid. So if we go back and looks at our half-reactions, for our half-reactions here we have lead sulfate for both of them. And it's a solid that precipitates on your electrodes. And so therefore the reverse reaction is ready to go. And you can recharge your battery. So first you could think about this as a voltaic cell, which produces a current using a spontaneous redox reaction. And then you can think about an electrolytic cell. A current is used to recharge your battery. A current is used to drive a non-spontaneous reaction. And so a lead storage battery is rechargable and that makes it very useful in car batteries. They last an extremely long time.