Extra Credit 18

Q19.1.16

Predict the products of each of the following reactions. (Note: In addition to using the information in this chapter, also use the knowledge you have accumulated at this stage of your study, including information on the prediction of reaction products.)

When predicting products of a reaction, it's key to realize that there are certain patterns associated with types of reactions. If the reaction in question has N₂ in the reactant then you will form a Nitride. If the reaction has H₂, then the product will be some type of hydride. If the reactant is O₂, the product will be some oxide. In any case, the product needs to be formed so that the oxidation states balance out.

1. \( \mathrm{MnCO_3(s)}+\mathrm{HI(aq)}\rightarrow\)
   1. Identify the metal if there is one: In this case, it's Manganese (Mn).
   2. Identify the oxidation state of the metal: This is a transition metal and therefore by nature has a lot of oxidation states. So, you can't determine what its oxidation state will be just by looking at the periodic table but because it's in a compound, you can mathematically figure it out. Oxygen has an oxidation state of -2. Carbon has an oxidation state of 4+. Therefore for the overall compound to be neutral, all the O.S. have to add up to 0. Therefore: \( \mathrm{3*2^-}+\mathrm{4^+}+x =\mathrm{0}\) and \(\mathrm{x}=\mathrm{2^+}\). Another way to look at this would be to realize that Manganese is bonded to 1 mole of \( \mathrm{CO^{2-}_3}\) and therefore you'd know that Manganese should have 2⁺ oxidation state.
   3. Identify what type of reaction this is: This is actually an acid-base reaction. However, an easier way to go about predicting the products in this case would be to realize that this reaction has two reactants, where both reactants are composed of two different ions. Therefore an easier method would be double substitution, where the ions of each reactant are switched to form the final product. Take for instance these reactant. \( \mathrm{HI(aq)}+\mathrm{MnCO_3(s)}\rightarrow\) is composed of \( \mathrm{H^+}\) and \(\mathrm{I^-}\) and \( \mathrm{Mn^{2+}}\) and \(\mathrm{CO_3^{-2}}\). So, once you switch the ions to get new compounds for the products, you'd get: \( \mathrm{MnCO_3(s)}+\mathrm{HI(aq)}\rightarrow\mathrm{HCO_3(aq)}+\mathrm{MnI(s)} \).
   4. Balance the oxidation states: \( \mathrm{MnCO_3(s)}+\mathrm{HI(aq)}\rightarrow\mathrm{HCO_3(aq)}+\mathrm{MnI(s)} \) is actually incorrect because the oxidation states do not cancel out. \( \mathrm{Mn^{2+}}\) and \(\mathrm{I^-}\) should not form a compound with a 1:1 ratio. The charges aren't equal, so to balance them out so that the product is neutral, there needs to be 2 moles of Iodine as 2 is the smallest common factor between the absolute value of the two charges. Thus the new predicted reaction would be: \( \mathrm{MnCO_3(s)}+\mathrm{HI(aq)}\rightarrow\mathrm{HCO_3(aq)}+\mathrm{MnI_2(s)} \)
   5. Balance the reaction: If there are now 2 moles of Iodine in the products, there needs to be two moles of iodine in the reactants. Balance out the reaction so that each side of the reaction has the same number of moles of each element. The new predicted reaction will be: \( \mathrm{2HI(aq)}+\mathrm{MnCO_3(s)}\rightarrow\mathrm{H_2CO_3(aq)}+\mathrm{MnI_2(s)} \)
   6. Does it make sense: \(\mathrm{H_2CO_3^{-2}}\) is actually unstable at room temperature and standard pressure. This means that it won't actually stay in this state. As a rule of thumb to memorize: \( \mathrm{H_2CO_3(s)}\rightarrow\mathrm{H_2O}+\mathrm{CO_2} \). Therefore the actual final reaction will be:

\( \mathrm{MnCO_3(s)}+\mathrm{2HI(aq)}\rightarrow\mathrm{CO_2(g)}+\mathrm{MnI_2(s)}+\mathrm{H_2O(l)} \)

1. \( \mathrm{CoO(s)}+\mathrm{O_2(g)}\rightarrow\)
   1. Identify the metal if there is one. In this case, it's Cobalt (Co).
   2. Find the oxidation state of the metal. In this case, it's \( \mathrm{Co^{2+}}\) because it's bonded to an oxygen atom and the overall charge of the compound is neutral so the charges of the metal and the oxygen must cancel out.
   3. Realize what type of reaction this is: You're reacting a metal (oxide) with oxide and will therefore get a metal oxide. Generally with such a reaction, you can use the general format of \( \mathrm{M_xO}\) to predict the product. However, in this case, it's better to logically reason what the logistics of the reaction are. When you react a metal with an oxide, it's becoming even more oxidized. Since the O.S. is already 2+, when oxidized in this reaction, the o.s. will simply go up to the next integer. In this case that would result in \( \mathrm{Co^{3+}}\).
   4. Balance the oxidation states: So now you know the product's Co will have O.S. of 3⁺. This means that for the formed product to be neutral, you need to find the smallest common integer between the oxidation states of the \( \mathrm{Co^{3+}}\) and \( \mathrm{O^{2-}}\). That means that the product will be \( \mathrm{2CoO_3}\)
   5. Balance the reaction: If the product's side of the reaction has 2 moles of Cobalt, then the reactant's side should too. If there are 6 moles of oxygen in the product's side, the reactant's side should as well. The final product will thus be:

\( \mathrm{2CoO(s)}+\mathrm{2O_2(g)}\rightarrow\mathrm{2CoO_3(s)}\)

1. La(s)+O₂(g)⟶
   1. Identify the metal if there is one. In this case, it's Lanthanum (La).
   2. Find the oxidation state of the metal. In this case, it's 3+ because on the periodic table, it's lined up in period 3, meaning that it has 3 valence electrons to lose. So your metal ion is: \( \mathrm{La^{3+}}\)
   3. Realize what type of reaction this is: This chemical reaction is the reaction between La and O₂. You're combining a metal and oxide and will therefore form a metal oxide. In reactions where a metal oxide is being formed, the product has the formula of \( \mathrm{M_xO}\). Keep in mind that here, the Oxygen that the metal is reaction with is actually the ion : \( \mathrm{O^{2-}}\) .
   4. Using the above format, balance the oxidation states. \( \mathrm{La^{3+}}\) and \( \mathrm{O^{2-}}\) won't combine to a 1:1 ratio. Rather, you need to find the smallest common multiple between the two charges. In this case, \( \mathrm{2} * \mathrm{3}= \mathrm{6}\). To get the \( \mathrm{La^{3+}}\) to a total charge of 6⁺, you will need two moles. To get \( \mathrm{O^{2-}}\) to a total charge of 6^-, you need to 3 moles. Thus your product will be: \( \mathrm{La_2O_3}\).
   5. Balance your reaction: There are 2 moles of La (s) on the products side so you need to multiply the reactant La by 2 and since the products has 6 oxygen atoms but the reactant's side only has 2 atoms, multiply the reactant's oxygen by 3. So in the end, the final product is:

\( \mathrm{2La(s)}+\mathrm{3O_2(g)}\rightarrow\mathrm{La_2O_3(s)}\)

1. \( \mathrm{V(s)}+\mathrm{VCl_4(s)}\rightarrow\)
   1. Identify the metal: The metal in the reactants is Vanadium (V)
   2. Identify the oxidation states of the metal: The first Vanadium is a free element and thus has an oxidation state of 0, and the second vanadium is bonded to \( \mathrm{Cl^{4-}_4}\). Therefore to get the compound neutral, the oxidation state of the second vanadium needs to be 4⁺.
   3. Identify the reaction: In this case, there isn't a reaction that is easily recognizable, so you'd want to try to use an alternative strategy. However since there aren't two reactants that are composed of two ions each, we can't use double substitution. So the only alternative left is to simply combine the reactants all together in the most logical way. Let's look at the bonding ability of each of the Vanadium by looking at the electron configurations.

V= [Ar]4s²3d³

V⁴⁺=[Ar]4s¹3d³

Now, the preferred state of Vanadium is V²⁺ because the 2 electrons in the S orbital are easy to remove. It's most stable with an oxidation state of 2⁺. So the reaction will most likely proceed in a way to create a product where Vanadium is stable.

V²⁺=[Ar]3d³

To get the desired electron configuration, the easiest way to think about is that in VCl₄, Vanadium has the oxidation state of 4⁺, in VCl₃, Vanadium will have an oxidation state of 3⁺, and in VCl₂, vanadium will have the oxidation state of 2⁺. Therefore, you want the Vanadium in the products to be structured so that it looks like VCl₂.

Thus the predicted products and reactions would be: \( \mathrm{V(s)}+\mathrm{VCl_4(s)}\rightarrow\mathrm{VCl_2(s)}+\mathrm{VCl_2(s)}\). Now simplify.

\( \mathrm{V(s)}+\mathrm{VCl_4(s)}\rightarrow\mathrm{2VCl_2(s)}\)

1. \( \mathrm{Co(s)}+\mathrm{XeF_2(aq)}\rightarrow\)
   1. Identify the metal: The metal here is Cobalt (Co).
   2. Identify the oxidation state of the metal: Co is a free element and thus has the oxidation state of 0.
   3. Identify the reaction: In this case, there isn't a reaction that is easily recognizable (besides the fact that it is a redox reaction), so you'd want to try to use an alternative strategy. However since there are two reactants that are composed of two ions each, we can't use double substitution. So the only alternative left is to simply combine the reactants all together in the most logical way. Notice that Xenon, a noble gas is present and noble gases are usually highly unreactive. The only exception to this is with highly electronegative halogens, which fluorine is. However, cobalt is not a halogen. So it makes sense that cobalt and fluorine would bond but xenon would not. As a result, it would get released as a free element. Thus the propsed reaction is : \( \mathrm{Co(s)}+\mathrm{XeF_2(g)}\rightarrow\mathrm{CoF_2(s)}+\mathrm{Xe(g)}\).
   4. Balance Oxidation states and elements: Everything is balanced. Therefore the final reaction is:

\( \mathrm{Co(s)}+\mathrm{XeF_2(g)}\rightarrow\mathrm{CoF_2(s)}+\mathrm{Xe(g)}\)

1. \( \mathrm{CrO_3(s)}+\mathrm{CsOH(aq)}\rightarrow\)
   1. Identify the metal, if there is one: In this case the metal is Chromium (Cr)and Caesium (Cs).
   2. Identify the oxidation state of the metal: To get the oxidation state of Cr, know that the oxidation state of oxygen is 2^- and to get the compound is neutral: \( \mathrm{3*2^{2-}} + x= \mathrm{0}\). So, x=6⁺. The oxidation state of Chromium is \( \mathrm{Cr^{6+}}\). The oxidation state of Cs is \( \mathrm{Cs^+}\) because its bonded to \(\mathrm{OH^-}\). To get the compound neutral, use the equation: \( \mathrm{1^{-}} + x= \mathrm{0}\).
   3. Identify the reaction: Because there are 2 reactants and both reactants are made of 2 ions, you can thus use the double substitution method. The reactants are actually made of the ions: \(\mathrm{Cr^{6+}}\) and \(\mathrm{O^{2-}}\) and \( \mathrm{Cs^+}\) and \(\mathrm{OH^-}\). Now match the reactants together so that new compounds are formed. Remember to still match the ions so that anions are bonded to cations or vice versa. Thus your propsed products are: \( \mathrm{CrO_3(s)}+\mathrm{CsOH(aq)}\rightarrow\mathrm{CrOH(aq)}+\mathrm{CsO(s)}\).
   4. Balance the oxidation states: Notice that Cs has the oxidation state of 1⁺ on the reactants side. If the ratio of the products of CsOH is 1:1, then that would mean that Caesium's oxidation state has become 2⁺. So to balance the oxidation states, there needs to be 2 moles of Caesium. Also, the Cr's oxidation state on the reactant's side is 6⁺, so for the oxidation states of Cr to match on both sides of the reaction, there needs to be 6 moles of \(\mathrm{OH^-}\). Thus your reaction now looks like: \(\mathrm{CrO_3(s)}+\mathrm{CsOH(aq)}\rightarrow\mathrm{Cr(OH)_6(aq)}+\mathrm{Cs_2O(s)}\)
   5. Balance the elements: In \(\mathrm{CrO_3(s)}+\mathrm{CsOH(aq)}\rightarrow\mathrm{Cr(OH)_6(aq)}+\mathrm{Cs_2O(s)}\) , the moles of each element on both sides of the reacion do not equal each other. If there are 6 moles of Hydrogen on the products side, there needs to be 6 moles of Hydrogen on the reactants side, so there needs to be a coeffecient of 6 in front of \(\mathrm{CsOH(aq)}\). Now the moles of Cs are unblanced. So if there are 6 moles of Cs on the reactants side, there needs to be 6 moles on the products side as well. The final products and overall reaction will thus look like this:

\(\mathrm{CrO_3(s)}+\mathrm{6CsOH(aq)}\rightarrow\mathrm{Cr(OH)_6(aq)}+\mathrm{3Cs_2O(s)}\)

S19.3.8

For complexes of the same metal ion with no change in oxidation number, the stability increases as the number of electrons in the t_2g orbitals increases. Which complex in each of the following pairs of complexes is more stable?

It is stated that to the complex with more electrons in the t_2g Orbital will be more stable. In other words, larger the crystal field splitting energy, the more stable a complex is. This is because having more electrons in the orbital will mean that the complex is low spin and a complex is generally low spin if the crystal field splitting energy is higher than the pairing energy. It should also be noted that complexes with low spin have less energy than those with high spin. So the real question here isn't which complex is more stable, but rather which has a higher crystal field splitting energy and therefore a more stable, lower energy complex. There are four factors that affect the crystal field splitting energy.

1. Oxidation State of the Metal

2. Identity of the metal

3. Geometry of the complex

4. The identity of the ligands attached

As stated above, the oxidation states will not change and as you'll see below, neither does the identity of the metal or the geometry of the complexes between the two complexes in comparison. So the only real factor we can take into account when deciding which has a higher crystal field splitting energy and thus which one is more stable is the identity of the ligand attached. Ligands can be divided into two categories: strong field and weak field. Strong field ligands increase the size of the distance between the t_2g orbitals and e_g the orbitals. So the key here is to look at the spectrochemical series and see which of the ligands attached are a stronger field. Another way to look at this is that stability of the complex increases as basicity increases. The more basic a ligand, the higher the charge:radius ratio the ligand has. Either way, basic ligands will still be grouped together as strong field ligands, so both concepts go hand in hand.

Also keep in mind, the chelate effect. Complexes with ligands that are polydentate are more stable than complexes with monodentate ligands.

1. \( \mathrm{[Fe(H_2O)_6]^{2+}}\) or \( \mathrm{[Fe(CN)_6]^{4-}}\)

Because the metal, oxidation state of the metal, and the geometry are the same for both complexes, we must compare the ligands to determine which one has a larger crystal field splitting energy and is thus more stable. In this case, the two complexes: \( \mathrm{[Fe(H_2O)_6]^{2+}}\) and \( \mathrm{[Fe(CN)_6]^{4-}}\) have the ligands \( \mathrm{(H_2O)}\) and \( \mathrm{(CN^-)}\) respectively. Using the spectrochemical series: notice that \( \mathrm{(CN^-)}\) is ranked as a stronger field ligand than \( \mathrm{(H_2O)}\).

This means that \( \mathrm{[Fe(CN)_6]^{4-}}\) is more stable because strong field ligands have low spin complexes.

1. \( \mathrm{[Co(NH_3)_6]^{3+}}\) or \( \mathrm{[Co(F)_6]^{3-}}\)

Because the metal, oxidation state of the metal, and the geometry are the same for both complexes, we must compare the ligands to determine which one has a larger crystal field splitting energy and is thus more stable. In this case, the two complexes: \( \mathrm{[Co(NH_3)_6]^{3+}}\) and \( \mathrm{[Co(F)_6]^{3-}}\)have the ligands \( \mathrm{(NH_3)}\) and \( \mathrm{(F^-)}\) respectively. Using the spectrochemical series: notice that \( \mathrm{(NH_3)}\) is ranked as a stronger field ligand than \( \mathrm{(F^-)}\).

This means that \( \mathrm{[Co(NH_3)_6]^{3+}}\) is more stable.

1. \( \mathrm{[Mn(CN)_6]^{4-}}\) or \( \mathrm{[Mn(Cl)_6]^{4-}}\)

Because the metal, oxidation state of the metal, and the geometry are the same for both complexes, we must compare the ligands to determine which one has a larger crystal field splitting energy and is thus more stable. In this case, the two complexes:\( \mathrm{[Mn(CN)_6]^{4-}}\) or \( \mathrm{[Mn(Cl)_6]^{4-}}\) have the ligands \( \mathrm{(CN^-)}\) and \( \mathrm{(Cl^-)}\) respectively. Using the spectrochemical series: notice that \( \mathrm{(CN^-)}\) is ranked as a stronger field ligand than \( \mathrm{(Cl^-)}\).

This means that \( \mathrm{[Mn(CN)_6]^{4-}}\) is more stable.

Q12.4.8

What is the half-life for the decomposition of NOCl when the concentration of NOCl is 0.15 M? The rate constant for this second-order reaction is 8.0 × 10⁻⁸ L/mol/s.

Let's break down the information we're given: What is the half-life for the decomposition of NOCl when the concentration of NOCl is 0.15 M? The rate constant for this second-order reaction is 8.0 × 10⁻⁸ L/mol/s.

What's important to note here is that we're given:

1. k = 8.0 x 10 L/mol/s, where k is the rate constant

2. [NOCl]= 0.15. We're starting the decomposition process with the 0.15 M of NOCl

3. We're trying to find half life. There are several formulas that can be used to find half life. But which one is correct?

4. This reaction is stated to be "second order", and we are given the initial concentration (which we know is only necessary for two half live equations. The only half life formula that corresponds to a second order reaction is : \( \mathrm{t_{1/2}} = \mathrm{(\frac{1}{k+[A]_o})}\)

As you can see, we're given all the components required to solve for the half life. So just plug and chug:

1. \( \mathrm{t_{1/2}} = \mathrm{(\frac{1}{k+[A]_o})}\)

2. \( \mathrm{t_{1/2}} = \mathrm{(\frac{1}{(8.0*10^{-8})+(.15)})}\)

3. \( \mathrm{t_{1/2}} = \mathrm{8.3*10^{-8}}\) s

Q21.2.2

Write the following isotopes in nuclide notation (e.g., " \(\ce{^{14}_{6}C}\) ")

Nuclide Notation is the way isotopes of an element a written. The basic format is : \(\ce{^{Mass Number}_{Atomic Number}Element Symbol}\) , where the mass number is the sum of the number of neutrons and protons. This can sometimes be abbreviated as: \(\ce{^{A}_{Z}X}\)

To write the nuclide notation, you need three components:

Step 1. Identify the element in question and it's corresponding element symbol on the periodic table of elements. Write this symbol on the right side. This will be the largest of all three components.

Step 2. Identify the corresponding atomic number on the periodic table of elements. Even if the mass number changes, the atomic number will remain the same for isotopes, as changing the atomic number changes the number of protons and thereby the element itself. Write this number on the bottom left corner of the element symbol. This should be written in a smaller font.

Step 3. Identify the mass number. This is usually the number accompanied with the element in the original problem. Write this number on the upper left hand corner of the element symbol in the same size font used for the atomic number.

1. oxygen-14
   1. Identify the element: Oxygen-14 essentially translates into words as "An Isotope of Oxygen with mass number of 14". Oxygen is thus the element in question. The element symbol corresponding to Oxygen is "O". Thus: \(\ce{^{A}_{Z}O}\)
   2. Identify the atomic number: Oxygen has the corresponding atomic number of "8" no matter the atomic mass. Thus: \(\ce{^{A}_{8}O}\)
   3. Identify the atomic mass: The mass is given and stated to be 14. Therefore the finished product will be:

\(\ce{^{14}_{8}O}\)

1. copper-70
   1. Identify the element: copper-70 essentially translates into words as "An Isotope of copper with mass number of 70". Copper is thus the element in question. The element symbol corresponding to copper is "Cu". Thus: \(\ce{^{A}_{Z}Cu}\)
   2. Identify the atomic number: Copper has the corresponding atomic number of "29" no matter the atomic mass. Thus: \(\ce{^{A}_{29}Cu}\)
   3. Identify the atomic mass: The mass is given and stated to be 70. Therefore the finished product will be:

\(\ce{^{70}_{29}Cu}\)

1. tantalum-175
   1. Identify the element: tantalum-175 essentially translates into words as "An Isotope of tantalum with mass number of 175". Tantalum is thus the element in question. The element symbol corresponding to tantalum is "Ta". Thus: \(\ce{^{A}_{Z}Ta}\)
   2. Identify the atomic number: Copper has the corresponding atomic number of "73" no matter the atomic mass. Thus: \(\ce{^{A}_{73}Ta}\)
   3. Identify the atomic mass: The mass is given and stated to be 175. Therefore the finished product will be:

\(\ce{^{175}_{73}Ta}\)

1. francium-217
   1. Identify the element: francium-217 essentially translates into words as "An Isotope of francium with mass number of 217". Francium is thus the element in question. The element symbol corresponding to tantalum is "Fr". Thus: \(\ce{^{A}_{Z}Fr}\)
   2. Identify the atomic number: Copper has the corresponding atomic number of "87" no matter the atomic mass. Thus: \(\ce{^{A}_{87}Fr}\)
   3. Identify the atomic mass: The mass is given and stated to be 217. Therefore the finished product will be:

\(\ce{^{217}_{87}Fr}\)

Note: Because stating the element symbol already implies what it's atomic number is and is thus only gives redudant information, some professors will not require the atomic to be added in the notation.

Q21.5.6

In usual practice, both a moderator and control rods are necessary to operate a nuclear chain reaction safely for the purpose of energy production. Cite the function of each and explain why both are necessary.

In some fission reactions, there will be an amount of excess particle produced from the original element. This excess of particles, neutrons, will then trigger more fission reactions, therefore being dubbed a nuclear chain reaction. These neutrons that are released, however, move a rate too fast to cause the proceeding nuclear reaction. A moderator, in this case, will be used to control the rate of the neutrons so that a chain reaction can occur. A moderator is substance, usually consisting of either heavy or light water, that reduces the speed of the neutrons released. This allows for the neutrons to be absorbed by the fuel rods and thus allows for the reaction to continue.

A control rod has the opposite effect. It's purpose is to control the amount of reaction that occurs. Fuel rods are continuously emitting neutrons and need the emmission from other fuel rods to continue the reaction. However, if rods were placed in between the fuel rods, then the neutrons would be absorbed by the control rods instead of the fuel rods. The amount of energy produced will therefore depend on how far the control rods are lowered and how much they aligned with the fuel rods. If they align perfectly with the fuel rods, then most of the neutrons will be absorbed by the control rods, not the fuel rods, and the reaction will stop. However, if the control rods are not lowered into the moderator all the way then the fuel rods can absorb most of the emitted neutrons, allowing for more energy to be produced. Generally, the set up of a controlled nuclear reaction involves some form of a moderator with a set of fuel rods spaced apart so that control rods can be lowered in between the fuel rods.

In other words, moderators are needed to produce energy from the nuclear reaction and control rods are necessary to control the amount of energy produced.

Q20.4.4

If the components of a galvanic cell include aluminum and bromine, what is the predicted direction of electron flow? Why?

Because the electrochemical cell is stated to be a galvanic cell, you know that the redox reaction that occurs will be spontaneous. For the reaction to be spontaneous, the electrons must flow from the element that wants it least to the element that wants it most. In other words, the electrons will flow from the electrode made of the element with the lower standard reduction potential to the electrode made of the element with the higher standard reduction potential. (From the element that is less easily reduced to that which is more easily reduced.)

Looking at the table of standard reduction potential, you'll see that the reaction:

\( \mathrm{Al^{3+}(aq)}+\mathrm{3e^-}\rightarrow\mathrm{Al(s)};\textrm{-1.676 V}\)

\( \mathrm{Br_2(l)}+\mathrm{2e^-}\rightarrow\mathrm{2Br^{-}(aq)};\textrm{+1.065 V}\)

Because \( \mathrm{+1.065 V}>{-1.676 V}>\), Bromine has a higher standard reduction potential and the electrons will flow from Aluminum to Bromine.

Q20.7.2

Why does the density of the fluid in lead–acid batteries drop when the battery is discharged?

In a lead acid battery, a redox reaction of : \(\mathrm{Pb(s)}+\mathrm{PbO_2(s)}+\mathrm{2HSO^-_4(aq)}+\mathrm{2H^+}\rightarrow\mathrm{2PbSO_4(s)}+\mathrm{2H_2O(l)}\) occurs, with an \(\mathrm{E_{cell}}=2.041v\). Here, solid lead acts as the electrodes and the sulfuric acid is a fluid that acts as a medium for the transfer of electrons. This fluid is also called an electrolyte.

In this reaction, lead and lead dioxide will combine to form lead sulfate. This buildup of lead sulfate on the electrodes is what makes this reaction reversible once it's been completely discharged. It can then proceed in the reverse reaction, thereby recharging it. However, recharging the battery would be a non-spontaneous reaction. Since the question here is asking about discharging the battery, it must be referring to the opposite of recharging the battery and therefore must be referring to the forward reaction. Looking at the forward reaction, sulfuric acid is consumed and water is produced. The density of sulfuric acid is 1.28 g/ml and the density of water is 1 g/ml. So, as the electrolyte is now more water than not sulfuric acid, the density of the fluid (electrolyte) in lead acid battery has dropped.